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SELECTED QUESTIONS AND PROBLEMS IN PHYSICS
P. A . fjjaflKOBa, H . II. KyTMJiOBCKaa
C6opHHK 3a«aH n BonpocoB no
«BwcniaH uiKOJia» MocKBa
R.GLADKOVAand N.KUTYLOVSKAYA
Selected
Questions and Problems
in Physics
Mir Publishers Moscow
Translated from Russian by N atalia W adhwa
First published 1989 Revised from the 1986 Russian edition
Ha aHBAUUCKOM Jl3blKe
P r i n t e d in th e U n io n o f S o v i e t S o c i a l i s t R e p u b l i c s
ISBN 5-03-000908-6
© H3AaTejibCTBO «Bucuiafl njKOjia», 1986 © E nglish translation, Mir Publishers, 1989
Contents
Preface
7
General M ethodical Instructions
8
Chapter I. Fundamentals of Molecular PhysicsandThermodynamics §
1. M olecular K in etic Theory. M otion of M olecules, Their Size and Mass § 2. V elo cities of M olecules. B asic E q u ation in the K in etic Theory of Gases § 3. E quation of State for an Ideal Gas. Isotherm al, Isochoric, and Isobaric Processes § 4. The Change in the Internal E nergy D uring H eat Trans fer and Due to M echanical Work § 5. Properties of Vapours § 6. Properties of L iquids § 7. Properties of Solids. D eform ations § 8. Therm al E xpansion of B odies
10 19 24 37 50 62 68 78
Chapter II. Fundamentals of Electrodynamics § 9. E lectric F ield § 10. E lectric Current in M etals. O hm ’s Law. E lectric R e sistance § 11. W ork, Power, and the Therm al Effect of Current § 1 2 . E lectric Current in E lectrolytes § 13. E lectric Current in Gases and in Vacuum § 14. E lectric Current in Sem iconductors § 15. E lectrom agnetism § 16. E lectrom agnetic Induction
87 106 132 141 148 153 155 169
Chapter III. Oscillations .and Waves § 17. M echanical V ib ration s and W aves. Sound and U ltrasound § 18. A ltern atin g Current § 19. E lectrom agnetic O scillation s and W aves
178 195 211
Chapter IV . Optics. Fundamentals of the Special Theory of Relativity § 20. G eom etrical O ptics § 21. P h otom etry
217 243
6
Contents
§ 22. Phenom ena E xp lain ed by the W ave Properties of R adia tion . Interference. D iffraction § 23. R adiation and Spectra § 24. Phenom ena E xp lain ed by the Q uantum Properties of R adiation . P h otoelectric E ffect § 25. F undam entals of the Sp ecial T heory of R e la tiv ity
251 260 263 269
Chapter V. Physics of A tom ic N ucleus § 26. Structure of A tom ic N ucleus. A tom ic E nergy and Its A pp lication
276
Chapter VI. General Remarks on A stronom y § 27. Fundam entals of A stronom y
288
A ppendices
297
Answers
330
Preface
This collection of questions and problems in physics is in tended for the students of correspondence courses and evening classes in interm ediate colleges and is in accord w ith the existing curriculum . The purpose of this book is to teach the students how to solve problem s in physics. This should stim ulate corre spondence course students to work independently, encourage them to accum ulate an adequate theoretical background, and develop in them the requisite ap titu d e for practical a c tiv ity in various branches of the economy. Each section begins w ith a brief description of the basic theoretical concepts, laws, and form ulas. T his provides the m axim um possible help to correspondence course students in solving problem s. A large num ber of problem s are supplied w ith detailed solutions and an analysis of the results, while in some cases different approaches are used to solve the same problem so th a t the student can discover the m ost ratio n al form of independent study. The theoretical m aterial is presented in a lucid form, and m ost problem s are of m edium com plexity. However, each section contains tougher problem s as well. T heir solution requires a broader range of theoretical d ata, and w ill fa c ilita te a deeper understanding of the physics course. In keeping w ith the existing curriculum , problem s in astronom y have also been included in the collection. T heir solution requires the use of a star chart, which is printed on the fly-leaf. The authors are grateful to A. L. Kosorukov, a researcher a t the In s titu te of A pplied M athem atics of the USSR Acad emy of Sciences, for the help in com piling the problems. R . Gladkova N . Kutylovskaya
General Methodical Instructions
Before startin g a new chapter in this course, go through the contents of the textbook for this chapter, and w rite down the topics in a separate notebook. H aving carefully read the list of topics, w rite down the num bers of the relevant sections in the m argin. W hile studying the m aterial in a section, it is necessary to read through the entire section w ithout stopping at the difficult parts. D uring the second reading, think about the m eaning of each sentence, and w rite down the derivation of the form ulas, the definitions of the physical q u an tities and th eir units, as well as the form ulation of the relevant laws of physics. Use the textbook whenever necessary. Your stu d y of the section should be com pleted by revising the m aterial, repeating the examples and explanations, and m aking notes of the derivations and diagram s. If you can reproduce the entire m aterial w ithout looking a t either the textbook or your notes, it is safe to assume th a t you have grasped the subject. If difficulties are encountered while reading the th eo re ti cal m aterial, and if it is not possible to overcome these d if ficulties independently, you m ust ask your teacher and get oral or w ritten consultation before revising the m ain and supplem entary m aterial. I t is not possible to get a firm grip over theoretical m a te ria l w ithout solving problems. Moreover, the solution of problem s helps when m em orizing laws and form ulas. Hence a tte n tio n m ust be paid to the solution of problem s. Before solving a problem , read the conditions and w rite them in your notebook w ithout abbreviations after under standing the essence of the problem. Using the standard
General M ethodical Instructions
9
notation for physical q u antities, w rite down the given quan tities in SI u n its in the notebook. A fter this, w rite down the q u an tities to be determ ined. Use the tables in the A ppen dices for the extra q u an tities required to solve the problem . H aving understood the physical phenomenon and using the physical laws applicable to a current problem , w rite down the necessary form ulas to express the required q u a n tity . In other words, solve the problem in the general form ana lytically. N um erical values should be su b stitu ted in the final expression together w ith the u n its of the quantities. The u n its of q u an tities should be sim plified first, followed by the num erical values. Obey the rules for operations w ith num bers. W henever possible, use m athem atical tables. All calculations should be made w ith a pocket calculator or a slide rule. The cor rectness of the solution can be verified by com paring the result w ith the actual values of the q u antities. E xplanatory notes and figures should accom pany each solution. The an swers m ust be based on the laws studied before solving the problem , and should reveal the essence of the phenomenon involved. Solutions of typical problems and detailed expla nations are included to help students tackle the problem s. W hile studying the course on physics, students m ust pass tests to determ ine the level of their understanding of a to p ic, and also do practical work to strengthen th eir th eo reti cal background as well as to acquire a practical knowledge of the way equipm ent is handled. A t the end of the course, correspondence course students have to pass exam inations, where they m ust dem onstrate a sound knowledge of theory, the form ulation of laws, form u las, and u n its of physical q u antities. They m ust also learn how to identify problems in everyday practice on the basis of th eir knowledge of physical laws, and also how to solve physical problem s.
Chapter I
Fundamentals of Molecular Physics and Thermodynamics § 1. MOLECULAR KINETIC THEORY. MOTION OF MOLECULES, THEIR SIZE AND MASS
Basic Concepts and Form ulas The m olecular kinetic theory explains the structure and prop erties of bodies through the m otion and interaction of the sm all particles con stitu tin g them . The theory is based on three postulates: 1. All substances consist of very sm all particles, viz. m olecules.1 M olecules, in turn, consist of still sm aller p a r ticles, viz. atom s, which them selves are m ade up of pro tons, electrons, and neutrons. 2. Molecules are in a state of perpetual random m otion known as therm al m otion. As a result, every molecule pos sesses a kinetic energy. 3. Molecules of a substance in teract w ith one another. The forces of interaction (attra c tiv e and repulsive) are electrical in origin. They depend both on the nature of the m olecules and on their separation. Consequently, m olecules possess a potential energy. The internal energy of a body is defined in m olecular phys ics as the sum of the kinetic energies of all the m olecules and the p o tential energies of their interaction. The num ber of m olecules in any body is extrem ely large, w hile th eir size and mass are very sm all. The SI u n it of the am ount of substance, viz. the mole, is a base u n it. The mole is the am ount of substance containing the same num ber of stru ctu ral elem ents as the num ber of 12C atom s in a mass of 0.012 kg. A mole of any substance contains the same num ber of m olecules. This num ber is known as the Avogadro constant 1 A m olecule is the sm allest particle of a substance that retain s it s chem ical properties.
§ 1. M olecular K inetic Theory
11
iVA. The mass of a molecule is ?n0 = M I N A, where M is the m olar mass. The num ber of m olecules in 1 m 3 of a substance is deter m ined from the form ula n = N Ap!M, where p is the density of the substance. The num ber of molecules in a given mass of a substance is N = m N JM , where m is the mass of the substance and m /M = v is the num ber of moles. A gas whose m olecules do not a ttra c t one another is called an ideal gas. Under norm al conditions (T Q = 273 K and p 0 = 101325 P a), the m olar volumes of all ideal gases are the same: Vm = 22.4 X 10“3 m 3/m ol. The L oschm idt num ber iVL is the num ber of gas molecules in 1 m 3 of a substance under norm al conditions: = N J V m, where Vm is the m olar volume defined as Vm = M l p0. The m ean distance I between molecules is defined as i= y v jN
,
where V IN is the elem entary volume per molecule. The m ean free path X is the mean distance covered by a m olecule between two successive collisions: X = v fz, where v is the arithm etic mean velocity of the molecules and z is the average num ber of collisions per second: z = y 2 Tid\uvn0. H ere d ett is the effective diam eter of a m olecule and n0 is the num ber of molecules in 1 m 3 of a substance.
12
Ch. I. Fundam entals of M olecular Physics
Worked Problems Problem 1. D eterm ine the am ount of substance (in moles) contained in (a) 1 kg of mercury, and (b) 5.6 dm 3 of oxygen under norm al conditions. Given: m x = 1 kg is the mass of m ercury, V0 = 5.6 dm 3 = 5.6 X 10“3 m3 is the volume of oxygen under norm al condi tions. From tables, we find the m olar mass of m ercury, M x = 200.6 X 10~3 kg/mol, the m olar mass of oxygen, M 2 = 32 X 10"3 kg/mol, and the density of oxygen under normal conditions, p0 = 1.43 kg/m 3. F ind : the am ount of substance v x in 1 kg of m ercury and the am ount of substance, v2, in 5.6 dm 3 of oxygen under normal conditions. Solution. The am ount of substance in 1 kg of m ercury can be determ ined as follows: v41 = m Ji Ni I i% ’
i
= 200.6 onn ax 10-3 kg/m ,—;— s- = 4.98 mol. ol
In order to determ ine the am ount of substance in 5.6 dm a of oxygen, we calculate the mass m 2 of oxygen: m 2 = p0JV This gives v
=
2
PqFq
M2
’
2
1.43 k g/m 3 X 5.6 X 10~3 m 3 _ Q ^ 32 X 10~3 kg/m ol
^
Answer. One kg of the m ercury contains about 5 mol, and 5.6 dm 3 of the oxygen contain about 0.25 mol. Problem 2. Calculate the num ber of molecules contained in 0.5 kg of oxygen and in 5.0 cm3 of carbon dioxide under norm al conditions. Given: m x = 0.5 kg is the mass of oxygen, V0 = 5.0 cm3 = 5.0 X 10“6 m 3 is the volume of carbon dioxide under norm al conditions. From tables, we find the m olar mass of oxygen, M x — 32 x 10“3 kg/mol, and the m olar mass of carbon dioxide, M 2 = 44 X 10~3 kg/mol (it is equal to the sum of the m olar masses of carbon and oxygen), the Avogadro con stan t N a = 6.022 x 1023 m ol"1, the m olar volume V m = 22.4 x 10~3 m 3/m ol, and the Loschm idt num ber N l = 2.68 x 1025 m "3. F in d : the num ber N x of molecules in 0.5 kg of oxygen and the num ber N 2 of molecules in 5.0 cm3 of carbon dioxide under normal conditions.
13
§ 1. Mblecular K inetic Theory
Solution. The am ount of substance in moles is v = m J M x. Knowing th a t one mole contains N A molecules, we deter mine the num ber of molecules in 0.5 kg of oxygen: N i = vNA= N 1i =
QO0 -^ 53 k»g ;
1
32 X 10-3 k g /m o l
X 6.022 < 1023 mol"* ~ 9.4 x 10“ .
In order to find the num ber of carbon dioxide molecules, N 2, contained in 5.0 cm3 under normal conditions, we use the relation N 2 = N AV0/V m, where N J V m is the num ber of molecules in 1 m3. This gives Nn — 906-?22yn13° 2a 3mol7 t x 5 x 10"6 in3,-- 1.34 22.4 X 10 3 in3 - m o l-1
x 102°.
Rem ark. The second part of the problem can be solved quite easily if we consider th a t 1 m3 of a gas contains A L molecules under norm al conditions. Then N 2 = N h V0 = 2.68 >C 1025 m~3 x 5.0 x 10~6 m3 = 1.34 x 1020. Answer. The num ber of molecules in 0.5 kg of oxygen is approxim ately 9.4 x 1024, while 5.0 cm3 of carbon dioxide contain 1.34 X 1020 molecules. Problem 3. The m olar mass of oxygen is 32 X 10~3 kg/mol. D eterm ine the m olar mass of air if the densities of oxygen and air under norm al conditions are 1.43 and 1.29 kg/m 3 re spectively. Given: the m olar mass of oxygen, M x = 32 X 10“3 kg/mol, the density of oxygen, p01 = 1.43 kg/m3, and the density of air, p02 = 1.29 kg/m 3. F ind: the m olar mass l\I2 of air. Solution. The m olar mass M can be expressed in term s of the mass m 0 of a m olecule and the Avogadro constant N A: M = m 0N A. The density p0 of a gas under normal conditions is d eter m ined from the relation p0 = m0/VL, where A L is the Loschm idt num ber. Let us divide term wise the expressions for m olar mass and density: M l p0 = N J N h . B ut the ratio A a /A l of the two constants is a constant. Consequently, the m olar masses of
14
Ch. I. Fund&pientals of M olecular P hysics
two gases are directly proportional to their densities under norm al conditions: Mi/poi = ^2^Po2 = • • • = N J N ^ . This gives Mo == p02^l/p01’ M 2=
|~43
i f f
x 32 x
10-3 kg/mol ^ 29
x
10-3 kg/m ol.
Answer. The m olar mass of air is approxim ately 29 X 10"3 kg/mol. Problem 4. Determ ine the mass of an acetylene molecule C2H 2 and the density of acetylene under norm al conditions. Given: the chemical form ula of acetylene is C2H 2. From tables, we find th a t the m olar masses of carbon and hydrogen are M c t = 24 X 10~3 kg/mol and M h 2 = 2 X 10"3 kg/m ol respectively. The Loschm idt num ber is N h = 2.68 X 1025 m~3 and the Avogadro constant is N A = 6.022 X 1023 m ol"1. Find: the mass m 0 of an acetylene molecule and the density p0 of acetylene under norm al conditions. Solution. The mass of an acetylene molecule can be d e te r m ined from the relation m0 = M /N a . The acetylene molecule C2H 2 consists of carbon and hydro gen. Therefore, the m olar mass of acetylene can be d eter m ined from the m olar masses of its components: M = M c 2+ M h 2 = 26 X 10"3 kg/mol. R em ark. The m olar mass of any chemical compound, for exam ple, am m onia N H 3, can be determ ined in this way: M = M n + 3M h = 17 x 10"3 kg/mol. For an acetylene molecule, we have m — 0—
26 x 10~3 k g/m ol 6.022 x 10*s m ol"1
_ 4 32 y — X
jQ"26 kff g‘
Know ing the mass m 0 of a molecule and the num ber of m olecules in 1 m3 (equal to /VL), we can determ ine the den sity p0 of acetylene under norm al conditions: p0 = m 0N h , p0 = 4.32 x 10"26 kg x 2.68 x 1026 m "3 ^ 1.16 kg/m 3.
15
§ 1. M olecular K inetic Theory
Answer. The mass of an acetylene molecule is approxim ate ly 4.3 X 10“26 kg. The density of acetylene under norm al conditions is 1.16 kg/m 3. Problem 5. Calculate the approxim ate size of a water molecule assum ing th a t molecules are spherical and touch each other. Given: the chemical form ula of w ater is H 20 . From tables, we find the den sity of w ater, p = 1 X 103 kg/m 3, at room tem perature and the m olar mass of w ater, M = 18 x 10-3 kg/mol (see Problem 4). The Avogadro constant is N a = 6.022 x 1023 m ol"1. F ind: the volume V and the diam e ter d of a w ater molecule. Solution. Knowing the density and the m olar mass of w ater, we can determ ine its m olar volume V m: V m = M /p,
y
m
1 8 x l 0 ~ 3 k g/m ol 1 X 103 k g /m 3
1.8 X 10”5 m 3/mol.
One mole of any substance is known to contain N A m ole cules. Consequently, the volume of a w ater molecule can be determ ined by dividing the m olar volume by the Avogadro constant: V = Vm Na
y ’
1 . 8 x 1 0 5 m 3/m o l _ ^
6.022 x l O 23 m o l" 1
q 3 x jQ -2 8
m 3
A
I t was stated in the problem th a t w ater m olecules are spherical and closely packed (Fig. 1), i.e. the gaps between the molecules are negligibly sm all. Hence we can assume th a t the elem entary volume V taken from the m olar volume V m w ill contain one molecule whose diam eter is approxi m ately equal to the edge of the cube: < 2=3/ F , d = V 0.3 x lO"28 m3 ~ 3 x 10"10 m. In the general form, the solution is d = y v = y v j N ~ A -.= j w W a Answer. The volume of a w ater molecule is approxim ately 3 X 10“29 m 3, and the diam eter of a molecule is about 3 X 10“10 m.
16
Ch. I. Fundam entals of Molecular Physics
Problem 6. C alculate the mean distance between the cen tres of molecules of an ideal gas under normal conditions. Solution. The m olar volume under norm al conditions is V m = 22.4 X 10~3 m3/m ol. One mole of any substance con tain s N a = 6.022 x 1023 molecules. If the molecules are assumed to be d istributed uniform ly over the entire v o l ume, the volum e per molecule is V = VmIN A. This volume can be regarded as a cube whose edge is equal to the mean distance d between the molecules (see Fig. 1): j
3 / t T — T at—
d —
^
n i^ A ,
j
X 1 0 " 3 m 3/m ol 6 0 2 2 x 1023 m o l-1 ~
t ,^ 2 2 .4
d — y
./-VQ
OQ
X
m*
Answer. The mean distance between gas molecules under norm al conditions is approxim ately 3.3 X 10“9 m. Problem 7. The arithm etic mean velocity of nitrogen m ol ecules under norm al conditions is 453 m /s. Find the mean free path, the mean free tim e, and the mean momentum of a molecule if it undergoes 7.55 X 109 collisions per second. Given: the arithm etic mean velocity v = 453 m/s of a nitrogen molecule under normal conditions, the average num ber of collisions of the molecule per second, z = 7.55 X 109 s'"1. From tables, we find the m olar mass of nitrogen, M = 28 X 10~3 kg/mol, and the Avogadro constant, N A = 6.022 X 1023 m ol"1. F in d : the mean free path A, the mean free tim e t , and the m ean m om entum p of a molecule. Solution. As a result of the random m otion, the values of A, t , and p for molecules do not rem ain constant. Therefore, we m ust determ ine the mean values of these quantities: * v X==T ’
i - £v ,
t
453 m/s n a i _a ~ 7"55 x 10* s-1 = X m,
t = g-P453 * 10~8 , m = 1.3 x 10~10 s m /s
In order to determ ine the momentum of a nitrogen m ole cule, we m ust find its mass: m 0 = M / N A.
§ 1. M olecular K inetic Theory
The mean m omentum lated thus /» m0v = ( M i N A) v, "
6 ^ l ° < d . Tg1 S F
17
p of the molecule can be calcu
X 45*
« f c * 2 .1 1
x
1 0 - k e ., „ / S.
Answer. The mean free path and the mean free tim e are 6.0 X 10“8 m and 1.3 X 10"10 s respectively. The mean momentum of a nitrogen molecule is 2.11 X IQ-23 kg-m /s. Questions and Problems 1.1. W hat experim ental facts clearly confirm the random nature of m olecule m otion and the relationship between the intensity of this m otion and the tem perature? 1.2. W hy is diffusion in liquids much slower than it is in gases? 1.3. W hat physical process occurs when the surface of a solid is painted? 1.4. Common sa lt placed in w ater is uniform ly d is trib u t ed over the entire volume a certain tim e after being added. Kxplain this phenomenon. 1.5. W hy do gauge blocks (Johansson blocks) stick together when their end faces are brought in contact (Fig. 2)? 1.6. W hat phenomenon is reF ig . 2 sponsible for gluing solids? 1.7. One of the largest nuggets of gold, w ith a mass of 62.3 kg, was found at the m outh of the Amazon river. How much substance is contained in it? 1.8. Determ ine the mass of 1 kmol of carbon, nitrogen, and helium . 1.9. W hat is the mass of 50 m ol of oxygen? 1.10. How m any molecules are contained in 32 kg of oxygen and in 2 g of hydrogen? 1.11. W hat is the volume occupied by 7 X 1028 molecules of carbon dioxide under norm al conditions? 1.12. D eterm ine the am ount of substance contained in 6 g of carbon dioxide. How m any molecules constitute th is m ass? 1.13. W hat is the volume occupied by 0.6 x 1023 atom s of graphite? The density of graphite is known. 2-0530
18
Ch. I. Fundam entals of Molecular Physics
1.14. The density of brass is 8500 kg/m 3. W hat does this mean? 1.15. The density of alum inium is 2.7 X 103 kg/m 3. How m uch substance is contained in 1 m3 of alum inium ? 1.16. Determ ine the mass of a molecule and of an atom of oxygen, nitrogen, and helium . 1.17. C alculate the m olar mass of m ethane and the mass of a m ethane molecule C H 4. 1.18. A l-fim layer of silver is deposited on the surface of a m etal m irror. How m any silver atom s would be in a sur face layer w ith area 25 cm 2? 1.19. A grain of common salt having a mass of 3 X 10~3 g is dissolved in 10 1 of w ater and uniform ly distributed over the entire volume. How m any molecules of salt are contained in 5 cm 3 of the solution? 1.20. W hat is the ra tio between the masses of a 10~8-g dust particle and an air molecule? The m olar mass M of a ir is 29 X 10“3 kg/m ol. / 1.21. Com paring the densities of air and hydrogen, find the ratio between the masses of th eir molecules. 1.22. D eterm ine the mass of a propane molecule C3H 8 and the density of propane under normal conditions. 1.23. Calculate the mass of a butane molecule C4H 10 if its density under norm al conditions is 2.67 kg/m 3. 1.24. Given the density of hydrogen under norm al condi tions, determ ine its m olar mass. 1.25. A drop of m ineral oil w ith a mass of 0.023 mg is poured on the surface of w ater and forms a film 60 cm 2 in area. Assuming th a t the molecules in the film are arranged in one row, determ ine the size of a molecule. 1.26. D eterm ine the approxim ate mass and size of a m ol ecule of carbon disulfide CS2 assuming th a t the molecules are closely packed and spherical. 1.27. W hat w ill be the length of a chain of the molecules contained in 1 mg of w ater and closely arranged in a row? How m any tim es can such a chain be wound around the globe along the equator if the equator is 4 X 107-m long and the diam eter of a molecule is 2.69 X 10~10 m? 1.28. The densities of hydrogen and m ethane under nor m al conditions are 0.09 and 0.72 kg/m 3 respectively. Find the m olar mass of m ethane if the m olar mass of hydrogen is 2 X 10-3 kg/mol.
§ 2. V elocities of M olecules
19
1.29. D eterm ine the fraction of volume occupied by the gas molecules in a vessel containing a gas under norm al conditions. The diam eter of a molecule should be assumed l.o be 10~10 m. 1.30. The m ean velocity of a carbon dioxide molecule is ,162 m/s, and the average num ber of collisions of a m olecule per second is 9 X 109. C alculate the m ean free path. 1.31. The m ean free path of m olecules in a high vacuum Is about 5000 km. W hat is the average num ber of collisions of gas molecules per second if the m ean m olecular velocity is 5(10 m/s? 1.32. The m axim um altitu d e of the o rb it of the Vostok spaceship is about 327 km above the ground. The m ean free path of gas m olecules at such an a ltitu d e is about 5 km. W hat would be the velocity of molecules if the num ber of co lli sions per second were approxim ately 0.11 s " 1? 1.33. The m ean velocity of an oxygen molecule under normal conditions is 425.1 m /s. D eterm ine the m ean free path of the molecule if it undergoes on the average 6.57 X 10° collisions per second. 1.34. D eterm ine the mean free p ath of air m olecules u n der norm al conditions. The effective diam eter of the m ole cules should be taken to be 3 X 10~10 m. 1.35. D eterm ine the m ean free path of helium atom s under the conditions such th a t the num ber of atom s per unit volume (number density) is 3.2 X 1024 m -3, and the effective diam eter of helium atom s is 1.9 X 10~10 m. 1.36. A nitrogen m olecule moves under norm al condi tions a t a mean velocity of 454 m/s. Determ ine the m ean m o mentum of the molecule. 1.37. A carbon dioxide molecule having a m om entum of 2.7 x 10~23 kg-m /s undergoes 9.5 X 109 collisions per sec ond. D eterm ine the mean distance covered by the m ole cule between collisions. § 2. VELOCITIES OF MOLECULES. BASIC EQUATION IN THE KINETIC THEORY OF GASES
Basic Concepts and Formulas In th eir therm al m otion, the molecules of m onatom ic gases undergo translation. Molecules consisting of several atom s are in tran slato ry as well as in rotary m otion.
20
Ch. I. Fundam entals of M olecular P hysics
Brownian m ovem ent and m olecular collisions point to a perm anent change both in m agnitude and in direction of m olecular velocities. For this reason, the properties of gases are studied by using the sta tistic a l approach, which makes it possible to calculate the mean values of velocities of molecules, their energy, and other param eters. The arithm etic m ean velocity of all the m olecules is I 'r r y r r ; . :„~L.^>T. N
'U
v = 1 / SRT . ~ j.6 1/ — V Mx ~ 1,0 V M ’
where M is the m olar mass of the gas, R = 8.31 J/(m o l-K ) is the m olar gas constant, and N is the num ber of the m ole cules. The root-inean-square velocity of molecules is ,/ ^rins— [ /
SRT
M
t ijry
/~ R T
— 1.73 J /
M .
The m ost probable velocity of molecules is 1 / 2R T A /4 l f ' W vp = V s r — 141 V n r The basic equation in the kinetic theory of gases establishes a relatio n between the pressure of gas molecules and the kinetic energy of their translatory m otion: P
2 „ IT 2 _ '4 n0 ^ k — o 0
9
1
where n0 is the num ber of molecules per m 3 (number densi ty) and Ex is the m ean kinetic energy of a molecule. In an isochoric process, the pressure of a gas is proportion al to its therm odynam ic tem perature: p j p 2 = T XI T 2, and
This gives p = w0&7\ where k = R / N A = 1.38 X 10~23 J/K is the B oltzm ann constant. The m ean values of the kinetic energy of translatory m o tion of molecules of different gases at the same tem perature
§ 2. V elocities of M olecules
21
urn the same: mol"?msl
m«zvlms2
( ’.nusequently, Lrmsi __ 1 / mo2 VTmS2 V '""I Worked Problems Problem 8. Determ ine the mean value of the kinetic ener gy and the root-m ean-square velocity of helium molecules under norm al conditions. Given: the pressure and tem perature of helium under nor mal conditions are p 0 = 1.013 X 106 Pa and T 0 = 273 K respectively. From tables, we find the m olar mass of h e li um, M = 4 X 10”3 kg/mol, the Boltzm ann constant k = 1.38 x 10~23 J /K , the Avogadro constant N A = 6.02 X I023 m ol”1, and the density of helium under norm al condi tions, p0 = 0.18 kg/in3. Find: the m ean kinetic energy of a helium m olecule, E k, and the root-m ean-square velocity of helium molecules, v Tm6. Solution. We express the mean kinetic energy of a helium molecule in term s of the tem perature: 2?k = (3/2) k T 0. For monatomic gases like helium , this w ill be the to ta l kinetic energy of molecules: £ = |x
1.38 x 10~23 J/K x 273 K ~ 5.65 x 10“2‘ J.
Since E - /re0i>?ms/2, we have vrm s= ^ 2 E / m 0, where m 0 is the mass of a helium molecule, which can be expressed as the ratio of the m olar mass and the Avogadro constant: m0 = M / N A. Finally, we obtain
,-y.
2E N A M ’
- / ,x™ x^
x, £ S lgg I -
m/8'
Answer. The mean kinetic energy of a molecule is 5.65 X 10-21 J, the root-m ean-square velocity is about 1300 m /s.
22
Ch. I. Fundam entals of Molecular P h ysics
Remark. The root-m ean-square velocity can also be found from the form ula vrms = l^SpJpQ. Problem 9. C alculate the num ber of air molecules in a room of size 6 x 4 x 2.5 m3 a t a tem perature of 27°C and a pressure of 99.8 kPa. Given: V = 60 m3 is the volume of air in the room, T = 300 K is the air tem perature, p = 99.8 X 103 Pa is the air pressure. We know the B oltzm ann constant k = 1.38 x 10”23 J/K from tables. F ind: the num ber N of air molecules. Solution. The num ber N of air molecules in the room can be determ ined from their num ber density (their num ber in 1 m3) and the volume of the air: N = n0V. In order to find n 0, we shall use the basic equation p = n0k T in the kinetic theory of gases, whence n 0 = p /k T . This gives N = ^k ~T
'
S ubstitu tin g the num erical values, we get 1V —
. ,, q 27 ^ A ’
99-8 X 103 Pa X 60 m 3 1.38 X 1 0 -23 J/K X 300 K
Answer. The room contains 1.45
X
1027 air molecules.
Questions and Problems 2 .1 . D eterm ine the arithm etic m ean and the root-m eansquare velocity of air and oxygen molecules a t a tem perature of 300 K. 2 .2 . H elium and neon have the same tem perature. Mole cules of which of the gases have a higher m ean kinetic ener gy? 2 .3 . A t w hat tem perature is the root-m ean-square veloci ty of oxygen molecules equal to 500 m /s? 2 .4 . W hat is the ratio between the root-m ean-square ve locities of helium and neon molecules a t the same tem pera ture? 2 .5 . The critical tem perature of hydrogen is 32 K, w hile the hydrogen in the S un’s atm osphere is at about 6000 K.
§ 2. V elocities of M olecules
23
Determine the root-m ean-square velocities of hydrogen mnlocules at these two tem peratures. 2.(5. C alculate the m ean kinetic energy of hydrogen molem los at the tem peratures m entioned in Problem 2.5. 2.7. W hat is the tem perature of a m onatom ic gas if the mean kinetic energy of its m olecules is 0.8 X 10“19 J? 2.8. D eterm ine the m ean k inetic energy of translatory motion of all the neon m olecules in 1 mol and in 1 kg at 1000 K. 2.9. W hat m ust be the tem perature of hydrogen for its molecules to have the same root-m ean-square velocities as Hint of helium at 580 K? 2.10. One cubic m etre of a gas contains 2.4 X 1010 m ole cules at 27°C. Determ ine the gas pressure. W hat is the term npplied to such a degree of rarefaction? 2.11. How m any molecules are there in 0.5 m 3 of a gas a t ;J00 K and a pressure of 120 kPa? 2.12. D eterm ine the mean kinetic energy of m onatom ic gas molecules at 310 K. How m any m olecules are there in I m3 of such a gas at 0.4 M Pa? 2.13. D eterm ine the pressure of nitrogen in an am poule if I lie num ber density of the molecules in it a t 0°C is 3.5 X 1014. 2.14. Find the root-m ean-square velocity and the m ean kinetic energy of helium m olecules at 20°C. 2.15. The pressure of the rarefied air between the w alls of a Dewar flask is 1.33 X 10~2 Pa at 0°C. How m any m ole cules does a cubic centim etre contain? 2.16. A vessel whose volume is 2 m3 contains 2.4 kg of a gas. W hat is the gas pressure if the root-m ean-square veloc ity of its molecules is 500 m/s? 2.17. How m any air m olecules are contained in a 4 X 5 X 3 m3 room at 20°C under a pressure of 90 kPa? 2.18. W hat is the increm ent in the mean kinetic energy of translation of molecules in a m onatom ic gas heated from 0°C to 373 K? 2.19. A gas has leaked from a 5-1 cylinder because of a faulty valve. As a result, the gas pressure dropped by 2.9 kPa. The tem perature in the cylinder rem ained unchanged at 17°C. How m any molecules escaped from the cylinder? 2.20. After an electric heater had been switched on, the air tem perature in the room was raised from 17 to 22°C a t a
24
Ch. I. Fundam entals of M olecular P hysics
constant pressure. By w hat am ount (in percent) was the num ber of air molecules in the room reduced? 2.21. Hydrogen leaked from a 1-m3 cylinder because of a damaged valve. In itia lly , when the hydrogen was under a pressure of 5 MPa, its tem perature was 280 K. After some tim e, the tem perature rose to 290 K at the same pressure. How m any hydrogen molecules escaped from the cylinder? W hat was the decrease in hydrogen mass? 2.22. The root-m ean-square velocity of therm al m otion of gas molecules was increased by a factor of two. Determ ine the change in the therm odynam ic tem perature of the gas and in the kinetic energy of therm al m otion of molecules if the gas was m onatom ic. § 3. EQUATION OF STATE FOR AN IDEAL GAS. ISOTHERMAL, ISOCIIORIC, AND ISORARIC PROCESSES
Basic Concepts and Form ulas The physical q uantities characterizing the state of a body are known as param eters. In order to characterize the state of a gas, three param eters are needed: pressure p, volume F, and tem perature T . The equation relating the three param eters is known as the equation of state for an ideal gas. For a fixed mass of a gas, the equation of state has the form - ^ - = co n st,
or
Pl^
=
•
If the gas was first under norm al conditions (p0, F 0, T 0) and then changed to a state for which the param eters became p , F, and 7\ the equation of state has the form PoVo/To = p V /T . I t should be noted th a t p 0 and T 0 are known and equal to 1.013 X 105 Pa and 273 K. Therefore, in order to determ ine F 0 it is sufficient to know the values of p, F, and T . The equation of sta te of a gas is applicable to isotherm al, isobaric, and isochoric processes. Indeed, by cancelling out the param eter which is constant for a given process, we ob tain :
§ 3. Equation of State for an Ideal Gas
25
for an isotherm al process p 1V1 = p 2V 2 for T = const, m = const, for an isobaric process V J T X = V 2/ T 2 or V J V 2 = T J T 2 for p = const, m = const, and for an isochoric process Pi _
P2
or
= —i-
for V = const, m = const.
In the general case when the mass m is known or is to be determ ined, the equation of state of an ideal gas is
where M is the m olar mass of the gas and R = 8.314 J / (m ol-K ) is the m olar gas constant. Using th is equation, we can determ ine the density of a gas as a function of its tem perature and pressure. D ividing both sides of the equation by volume V and su b stitu tin g density p for m ! V , we obtain
I'or a fixed m ass of a gas, M IR is constant. Consequently, the gas density is directly proportional to pressure and in versely proportional to therm odynam ic tem perature. D a lto n 's law . If a vessel contains a m ixture of several gases which do not react chem ically, the pressure of the gas m ixture is equal to the sum of the partial pressures of each gas taken separately, i.e. P — Pi + P 2 + Pb + • • •» where p Xi p 2, p 3, . . . are the partial pressures, i.e. the pressures exerted by each gas separately as if it alone were occupying the entire vessel. Worked Problems Problem 10. A cylinder contains a gas at a tem perature of 17°C and a pressure of 1.0 M Pa. W hat w ill be the change in the pressure if the tem perature is lowered to —23°C?
26
Ch. I. Fundam entals of Molecular P hysics
Given: t x = 17°C and p x = 1.0 MPa are respectively the tem perature and pressure of the gas in the first state, and t2 = —23°C is the gas tem perature in the second s ta te . F in d : the change in pressure, Ap, in the cylinder as a re su lt of the transition of the gas from the first to the second state. Solution. We use the equation of state for the gas, pyV J T j = p 2V 2/ T 2. Since the change in gas pressure occurs w ith decreasing tem perature but a t a constant volume (an isochoric process), we have V1 = V2 and p j p 2 = T J T 2. H ence p 2 = p 1T 2/ T 1 and Ap = Pl — p 2. We shall w rite the param eters for the first and second states of the gas separately. The param eters of the first state are p x = 1.0 X 106 Pa, T x = 290 K, and the param eters of the second sta te are T 2 = 250 K, p 2 = ? Let us determ ine the gas pressure p 2 after the tem perature has been lowered and calculate the pressure difference Ap: ft = W
x i0 x W
Ap = (1.0 — 0.86)
Pa = 0.86
x 106 Pa,
x 10* Pa = 0.14 x 10« Pa.
Answer. The gas pressure in the cylinder dropped by 0.14 M Pa. Problem 11. A 20-1 vessel is filled w ith air at a pressure of 0.4 MPa and connected to another vessel from which all the a ir has been pum ped out. The pressure in the two vessels equalizes at 1.0 X 106 P a. Determ ine the volume of the second vessel, assum ing th a t the process is isotherm al. Given: p x = 0.4 MPa = 0.4 X 106 Pa and Vx = 20 1 = 0.02 m 3 are respectively the pressure and the volume of the gas in the first state and p 2 = 1.0 X 105 Pa is the gas pres sure after the second vessel has been connected to the first one. Find: volume V of the second vessel. Solution. An isotherm al process occurs a t a constant tem p etature (T 1 = T 2) and obeys B oyle’s law, which is form a lized by the equation p V = const, or p xVr = p 2V 2. After the second vessel is connected to the first, the gas occupies the volume V2 = V1 + V. Consequently, p 1V l = p 2 (V1 +
§ 3. Equation of State for an Ideal Gas
27
V), or Pi
whence v = -Er r - y "
v = 0'i v ' l < Z x V l? = 0.06 m3.
m’ - ° « 2 m‘
Answer. The volume of the second vessel is 60 1. Problem 12. W hat was the tem perature of a gas if as a result of isobaric heating through 1 K its volume increased by 0.0035 of the in itial value? Given: A T = 1 K is the change in the gas tem perature and A V = 0.0035 V1 is the increm ent in the gas volume. F ind: T x, the in itia l tem perature of the gas. Solution. As a result of heating, the gas tem perature has increased by 1 K, hence r 2 = r i + A7\ The gas volume lias increased thereby from Vx to V2, i.e. V2 = Vx + AV. Since the process occurs a t a constant pressure (px = p 2), the equation of state can be transform ed as follows: TT =
or
TT = T 7 T ^ F (C harles’ ,aw)-
Let us solve th is equation for T x: VXT X + Vx A T = VyTy^ + A V T X, whence rp
1i~
V i h. T
AF
rri
’
1 1~
^ l X l
K
O Q fi
If
0.0035F! “
Answer. The gas tem perature before heating was 286 K . Problem 13. Two cylinders having volumes 3 1 and 7 1 are filled respectively w ith oxygen at a pressure of 200 kPa and nitrogen a t a pressure of 300 kPa a t the same tem pera ture. The cylinders are connected and after a certain tim e each w ill contain a gas m ixture at the same tem perature. Determ ine the pressure of the gas m ixture in the cylinders. Given: F 10x = 3 1 and p 10x = 200 kPa are the volume and the pressure of the oxygen prior to the connection of the cylinders, Fim t = 7 1 and Pm it = 300 kP a are the volume and the pressure of the nitrogen prior to the con nection of the cylinders. Find: the pressure p of the gas m ixture.
28
Ch. I. Fundam entals of M olecular P h ysics
Solution. According to D alton’s law, the pressure of a gas m ixture is equal to the sum of the p artial pressures of the gases: p = p 2ox P 2nit> where /?2ox and P 2n\t are the p a rtial pressures of the oxygen and the nitrogen. After the cylinders have been connected, each gas occupies a volume equal to the sum of the volumes of the cylinders: ^2ox =
^ 2 n it =
F io x
F ln it *
Let us w rite the param eters of the two states for the oxy gen and nitrogen in SI units: for oxygen state
state
1
2
F 10x - 3 X 10-3 m3,
Fsox = 3 X lO"3 m3
Piox = 2 X 105 Pa,
+ 7 x 10“3 m3 = 1 x 10-2 m3* p'lox = ? for nitrogen
state
state
1
F lnlt = 7 X 10-3 m3, Plnlt = 3 X10s Pa,
2
FJnit = 1 X 10-2 m 3, P2nit = ?
The process is isotherm al, hence T = const. In order to find />2ox and /?2nit, we use B oyle’s law Plox^lox = P 2ox^2oxi P in it^ in lt = P 2nitP 2nit for each gas separately: P io x F io x P2ox=
' ^ 2ox P in itF m it
P2m==
V'tnit
2 X IV
P a X 3 X 10“ 3 m 3
n
v 105 P t
= ---------- 1 X 10-* m*----------= U . b X l U 3 x 1 0 * P a x 7 x 1 0 - 3 m3
0 t ^
= --------- 1 x 1 0 ^ m*------------ 2.1 X 1 0
Pa, Do
Pa.
Finally, we obtain p = 0.6 X 105 P a + 2.1 X 105 Pa = 2.7 X 106 Pa. Answer. The pressure of the gas m ixture in the cylinders is 270 kPa. Problem 14. The air in a balloon at a tem perature of 20°C and a pressure of 99.75 kPa has a volume of 2.5 1. W hen the balloon is immersed in w ater at a tem perature of 5°C, the air pressure in it increases to 2 X 106 Pa. W hat is the change in the volume of the air in the balloon?
29
§ 3. Equation of State for an Ideal Gas
Given: t x = 20°C, p 1 = 99.75 kPa, and Vx = 2.5 1 are the tem perature, pressure, and volume of the air before the balloon is immersed in w ater, t2 = 5°C and p 2 = 2 X 105 Pa are the tem perature and pressure of the air after the im m er sion of the balloon in water. Find: the change A V in the volume of the air in the bal loon. Solution. Before the balloon is immersed in w ater, the state of the air in it is characterized by the param eters />!, V1, and 7V After im m ersion, the param eters are p 2, K2, and 7V Let us w rite the param eters of the gas in states 1 and 2 separately, expressing them in SI units: state 1
state 2
p x = 9.975 X 104 Pa, Vx = 2.5 X 10-3 m3, Tx = 293 K,
p 2 = 2 X 105 Pa, V2 = ?
T 2 = 278 K.
As the air changes from state 1 to state 2, all three param eters change. Consequently, we m ust use the equation of state to determ ine the final volume V 2: P\V\ _ T1
P2V2 T2
w hen ce
Pi Vi T2
T iP2 ’ 9.975 X 104 Pa x 2.5 X 10"3 m 3 X 278 K 293 K X 2 X 105 Pa
AV = 2.5 X 10-3 m3 -
= 1.2
X
10 3 m3.
1.2 x 1 0 '3 m3 = 1.3 X l O ”3 m 3.
Answer. The change in the volume of the a ir is 1.3 X 10-3 m 3 = 1.3 1. Problem 15. A cylinder having a volume of 0.6 m3 con tains oxygen at a tem perature of 27°C. A pressure gauge on
Ch. I. Fundam entals of M olecular P h ysics
30
the cylinder indicates 11.7 MPa of excess pressure.2 Reduce the volum e of the oxygen to normal conditions and d eter mine its mass. Given: V± = 0.6 m3 is the volume and T x = 27°C is the tem perature of the oxygen in the cylinder, p g = 11.7 MPa is the reading of the pressure gauge, T 0 = 273 K, p 0 = 1.013 X 106 Pa are the tem perature and pressure under nor m al conditions. From tables, we take the density p0 = I.43 kg/m 3 of oxygen under norm al conditions, the m olar mass of oxygen, M = 32 X 10~3 kg/mol, and the m olar gas constant R = 8.31 J/(m o l-K ). Find: the volume V0 of the oxygen under norm al condi tions and the mass m of the oxygen in the cylinder. Solution. From the reading of the pressure gauge, we de term ine the gas pressure in the cylinder: p x = p g + Po — I I .7 X 106 Pa + 0.1013 X 106 Pa - 11.8 x 106 Pa. R e ducing the volume to norm al conditions means determ ining the volum e the gas would occupy at a tem perature of 273 K and a pressure of 1.013 X 105 Pa. We shall w rite the param eters of the oxygen in SI units for the tw o statas: state 1 p x = 11.8 x 106 Pa, V t = 0 .6 m 3,
T x = 300 K,
state 2 Po = 1.013 X 105 Pa, T0 = 273 K, Vo = ?
We solve the problem by using the equation of state for an ideal gas: PiV i Ti
PqVq T0
whence T/’ °“
P \V \T
T iPo
o ’
T/ 11.8 X 106 Pa X 0.6 m 3 X 273 K ao a 3 Vo= 300 K X 1.013 X 105-Pa-------= 6 3 ‘6 m
2 The excess pressure in d icated on a pressure gauge is the difference betw een the pressure of the gas in the cy lin d er and atm ospheric pres sure.
§ 3. Equation of State for an Ideal Gas
31
From V0 and p0, we can determ ine the mass of the oxygen: m = p0F 0, m = 1.43 kg/m3 X 63.6 m3 = 91 kg. Answer. The volume of the oxygen under norm al condi tions is 63.6 m3, the mass is approxim ately 91 kg. Remark. The problem can be solved by using the equation PjF, = ^ R T X, from which we first determ ine the mass ni = ^H 1 ! We can find the gas volume from the density of the oxygen under normal conditions: Vq =
m /p 0 .
Problem 16. A cylinder contains acetylene a t 27°C under a pressure of 4.05 M Pa. W hat w ill the pressure in the c y lin der be after half the mass of the gas has been used up if the tem perature has fallen thereby to 12°C? Given: T x = 300 K and p x = 4.05 X 106 Pa are the in i tial tem perature and pressure of the gas in the cylinder, T2 = 285 K is the tem perature of the rem aining gas, m 2 = 0.5 m x is the mass of the consumed gas. Find: the pressure p 2 of the gas rem aining in the cylinder. Solution. In this problem, the tem perature, pressure, and mass of the gas change. Therefore, it is best to use the equa tions of state for the two cases in the form p .V ^ ^ -R T ,
and
R T 2.
Dividing the first equation by the second term wise and c a n celling out m 1, M , R , and the unknown volum e Vj, we obtain
Pi _ p2
7*1 o.5r 2 ’
whence Pi Pi
0-5 P l T 2 Tx
’
0.5 X 4.05 X 106 1’a X 285 K 300 K
192.4 x 104 Pa
cz. 1.92 x 106 Pa. Answer. The pressure of the acetylene rem aining in the cylinder is approxim ately 1.92 MPa.
Ch. I. Fundam entals of M olecular P hysics
32
Problem 17. D eterm ine the density of hydrogen a t a tem perature of 17°C and a pressure of 204 kPa. Given: T = 290 K and p = 2.04 x 105 Pa are the tem perature and pressure for which the density of hydrogen has to be determ ined. From tables, we obtain the m olar mass of hydrogen, M = 2 X 10~3 kg/mol, and the m olar gas con sta n t R = 8 .3 1 J/(m ol-K ). F in d : the hydrogen density p. Solution. We w rite the equation of state p F = - ^ - R T . D iv id in g both sides by V , we obtain p = RT
m /V is the density. Then p = p M PM
P = RT
YX\f
Jt^. ^ 1
, where
whence
_ 2.04 X 105 Pa x 2 X 10~3 kg/m ol 8.31 J/ (m ol-K ) X 290 K
= 0.17 kg/m 3.
Answer. The density of hydrogen is 0.17 kg/m 3. Problem 18. P lo t the graphs for an isotherm al, isobaric, and isochoric process in p - V coordinates. Solution. An isotherm al process occurs a t a constant tem perature T = const and obeys B oyle’s law. According to P,MPq Is o b a r
0.2
0.1
<5 -
___________ i____________
0.7
Fig. 3
OA
Vm*
F ig. 4
this law, the pressure of a given mass of a gas varies in in verse proportion to its volum e. Such a dependence is repre sented by a hyperbola, which is known in physics as an isotherm . In order to plot the graph, it is sufficient to keep the product p V constant, say, p V = 1.2. Ascribing a rb it rary values 2, 4, 6, . . . to the volume, we calculate the corresponding pressures: 0.6, 0.3, 0.2, . . . . H aving chosen a scale, we plot the graph (Fig. 3)
§ 3. Equation of State for an Ideal Gas
33
The volum e of a gas in an isobaric process varies w ith tem perature a t constant pressure p = const. The process is graphically represented by a straig h t line parallel to the K-axis. For an isochoric process, the pressure varies at a constant volum e V = const. This de pendence is represented by a straig h t line parallel to the p-axis (Fig. 4). Problem 19. W hich of the two iso therm s (Fig. 5) plotted for the same mass of a gas corresponds to a higher tem perature? Solution. We construct the isochor for a certain volume Vx. The isochor intersects the isotherm corresponding to the tem perature T x a t a pressure p u and the isochor intersects the isotherm for T 2 at a pressure p 2. Since the same volume corresponds to states 1 and 2 , we can w rite p 1/ T 1 = p J T 2 from GayLussac’s law. Since p 2 > p XJ T 2 > T x. Consequently, the upper isotherm corresponds to the higher tem perature. (Questions and Problems 3.1. A vessel contains 10.2 1 of a gas under norm al condi tions. W hat volume w ill the gas occupy a t a tem perature of 40°C and a pressure of 1 M Pa? 3.2 . A gas occupies a volum e of 4 1 a t a tem perature of r>0°C and a pressure of 196 kP a. At w hat pressure w ill this gas occupy a volume of 16 1 if heated to 20°C? 3.3. At w hat tem perature do 4.0 m 3 of a gas produce a pressure of 150 kPa if the same mass of the gas under norm al conditions has a volume of 5 m 3? 3.4. A 45-1 cylinder contains oxygen a t a tem perature of M 'C and a pressure of 1.52 M Pa. W hat volume would the gas occupy under norm al conditions? 3.5. Compressed oxygen for w elding is stored in 20-1 cy l inders a t a pressure of 9.8 M Pa and a tem perature of 290 K . Mcduce the volum e of oxygen to norm al conditions. 3.6. A 6-1 cylinder contains 0.1 kg of a gas a t a tem per ature of 300 K and a pressure of 9.44 X 105 Pa. D eterm ine th e m olar mass and identify the gas. i o'kio
34
Ch. I. Fundam entals of Molecular P hysics
3 .7 . W hat am ount of gas (in moles) is contained in a 10-1 cylinder at a pressure of 0.29 MPa and a tem perature of 17°C? 3 .8. Determ ine the mass of carbon dioxide stored in a cylinder having a volume of 40 1 at a tem perature of 13°C and a pressure of 2.7 M Pa. 3.9 . Determ ine the am ount of gas (in moles) occupying a volum e of 25 1 a t a pressure of 1.4 X 105 Pa and a te m p e ra ture of 300 K. 3.10. A m ountain-clim ber takes in 5 g of air w ith each breath under norm al conditions. W hat volume of air m ust he inhale in the m ountains where the atm ospheric pressure is lower than th a t at sealevel and is 79.8 kPa a t a tem perature of —13°C? 3.11. Under norm al conditions neon occupies a volume of 12.4 1. How m any tim es w ill the pressure of the gas increase if it is placed in a vessel having a volume of 5.6 1 a t a tem perature of 318 K? 3.12. An ideal gas occupies a volume of 2 1 at a pressure of 1.33 kPa and a tem perature of 15°C. W hat w ill be the pressure if the tem perature is doubled, w hile the volume de creases by 0.25 of the in itia l value? 3 .1 3 . A 40-1 cylinder contains 64 g of oxygen under a pressure of 213 kPa. D eterm ine the tem perature of the gas. 3.14. 42 g of acetylene are kept in a 20-1 cylinder a t a tem perature of 17°C. D eterm ine the am ount of substance in the gas and its pressure. 3.15. A 40-1 cylinder contains 1.98 kg of carbon dioxide a t 0°G. W hen the tem perature is increased through 48 K, the cylinder explodes. At w hat pressure does the explosion occur? 3.16. C alculate the m olar mass of butane if 2 1 of the gas have a mass of 4.2 g a t a tem perature of 15°C and a pressure pf 87 kPa. C alculate the num ber of gas molecules in 1 m 3. 3.17. At the beginning of the compression stroke in a d ie sel engine, the tem perature of the air was 40°C and the pres sure was 78.4 kPa. As a result of compression, the volume was reduced by a factor of 15, and the pressure increased to 3.5 M Pa. D eterm ine the tem perature of the air a t the end of the compression stroke. 3.18. The volume of 265 g of a gas a t a tem perature of 273 K and a pressure of 5 MPa is 60 1. W hat gas is it?
§ 3. Equation of State for an Ideal Gas
35
3.19. D eterm ine the mass of air in a room of 6 X 5 X .'I in3 at a tem perature of 293 K and a pressure of 1.04 X K)ft Pa. The m olar mass M of air is 29 X 10“3 kg/mol. 3.20. W hat w ill the final tem perature of a gas m ixture in mi internal com bustion engine be if it occupies a volume of 40 dm 3 in the cylinder at a tem perature of 50°C under nor mal atm ospheric pressure and is compressed by the piston to n volume of 5 dm 3 and a pressure of 15.2 X 105 Pa? 3.21. A cylinder contains a gas at a tem perature of 7°C mid a pressure of 91.2 M Pa. W hat w ill be the pressure after 0,25 of the mass of the gas has flown out of the cylinder and the tem perature has risen to 27°C? 3.22. A closed vessel of 2-m3 volume contains 1 kg of nitrogen and 1.5 kg of oxygen. Determ ine the pressure of flm m ixture in the vessel if the tem perature of the m ixture In 17°C. 3.23. D eterm ine the density of oxygen a t a tem perature of 47°C and a pressure of 105 Pa. 3.24. The pressure of air 10 km above the surface of the Knrth is about 30.6 kPa and the tem perature is 230 K. D e term ine the density of the air, the num ber density of the molecules, and their root-m ean-square velocity a t this a lt i tude. 3.25. 7 g of a gas contained in a cylinder a t 27°C produce n pressure of 4.9 X 104 P a. 4 g of hydrogen a t 60°C produce a pressure of 43.5 X 104 Pa in the same volum e. D eterm ine the molar mass of the unknown gas and identify it. 3.26. W hat w ill be the increase in the mass of air in a room if the atm ospheric pressure changes from 9.84 X 104 to 10.1 X 104 Pa and the air tem perature rem ains unchanged mid equal to 273 K? The size of the room is 4 X 5 X 2.5 m 3. 3.27. A vessel containing 10 1 of air under a pressure of 1 MPa is connected to a 4-1 em pty vessel. D eterm ine the final air pressure in the vessels, assuming th a t the process is Isotherm al. 3.28. A vessel contains a gas under a pressure of 5 X 106 Pa. W hat w ill the gas pressure be if 3/5 of the mass of the gas has flown out, the tem perature being m aintained constant? 3.29. P lo t the graphs of an isotherm al process in V -T mid p -T coordinates. 3.30. The bladder of a football having a volume of 2.5 1 ;i*
36
Ch. I. Fundam entals of Molecular P hysics
m ust be inflated to a pressure of 300 kP a. The pump takes in 0.14 1 of air under norm al atm ospheric pressure during one stroke. How m any strokes are required if the bladder was in itia lly em pty? 3.31. Given two equations of isotherm al processes for the same mass of a gas: p l V1 = 6 and p 2V 2 = 9. P lo t the graphs in p-V coordinates and answer the following ques tions: (1) in which units is the product p V expressed? (2) w hich of the isotherm s w ill be further from the coordinate axes and why? 3.32. D ry atm ospheric air consists of oxygen, nitrogen, and argon. D isregarding other components whose concen tratio n s are very sm all, determ ine the masses of these gases
T
F ig. 6
F ig. 7
in 1 m3 of atm ospheric air under normal conditions if their p artial pressures are 2.1 X 104 Pa for oxygen, 7.8 X 104 Pa for nitrogen, and 103 Pa for argon. 3.33. R epresent an isobaric process graphically in V - T , p-V, and p - T coordinates. 3.34. Figure 6 shows two isobars plotted for the same mass of a gas. W hich of the two isobaric processes occurs at a higher pressure and why? 3.35. A gas is heated isobarically from T x to T 2 (Fig. 7). W hat changes occur in the gas? 3.36. W hat volume w ill a gas occupy a t 348 K if its volume a t 35°C is 7.5 1? The process is isobaric. 3.37. A gas occupies a volume of 10 1 a t 27°C. To w hat tem perature should it be cooled isobarically to reduce its volume by 0.25 of the in itia l value? 3.38. Through how m any degrees K elvin should the tem perature of a gas be increased in an isobaric process to in-
§ 4. The Change in the Internal Energy
37
crease its volume by a factor of 1.3 in com parison w ith the volume occupied by it at 0°C? 3.39. A certain mass of a gas is taken through a closed cycle 1-2-3-1 in which the state of the gas changes isobaricnlly-isochorically-isotherm ally. Represent these processes in V T and p -V coordinates. 3.40. A balloon of volume 4500 m 3 is filled w ith helium at 290 K . The mass of the envelope is 677 kg. D eterm ine the lifting force of the balloon at a tem perature of 27°C, assuming th a t the atm ospheric pressure rem ains unchanged iiiid is 102 kP a.
§ 4. THE CHANGE IN THE INTERNAL ENERGY DURING HEAT TRANSFER AND DUE TO MECHANICAL WORK
llasic Concepts and Formulas When several bodies having different tem peratures come into contact, heat transfer takes place, as a result of which the tem perature of the bodies equalizes, and therm al equilib rium sets in. The in tern al energy of the bodies being heated increases a t the expense of the energy given away by the bodies w ith higher tem peratures. I t has been established that the change in the internal energy of a body is propor tional to its mass and to the change in its tem perature: AU = Q = cm A T , where Q is the am ount of heat (a m ea sure of the change in the in tern al energy) expressed in joules (J), and c is the proportionality factor known as the specific heat of a substance. For exam ple, the specific heat of a lu m inium is 920 J/(k g -K ), which means th a t the in te rn al energy of 1 kg of alum inium increases by 920 J when its tem perature is raised through 1 K . In order to determ ine a specific heat, the heat balance equation is employed. This equation is constructed by estab lishing the processes in which the energy is released (during cooling or as a result of fuel combustion) and the processes in which energy is absorbed (as a result of heating). For exam ple, a body of mass m x and specific heat cx has a tem perature T x and is exchanging heat w ith a body of mass m2, which has a specific heat c2 and a tem perature T 2, It is known th a t T x > T 2. Then the first body gives away
38
Ch. I. Fundam entals of Molecular P hysics
the following am ount of heat: (?giv — c1m 1 (T x
0 ).
The second body receives as a result of the heat transfer the following am ount of heat: Qrec = C2m 2 (0 — T 2), where 0 is the final tem perature of the two bodies. Since energy is conserved, we can w rite @glv + C*rec = 0. This gives c1m 1 (7 \ — 0 ) = c2m 2 (0 — T 2). This is the heat balance equation for the system of two bodies. A body can be heated a t the expense of the energy released during fuel com bustion. The am ount of heat liberated is pro portional to the mass of the burnt fuel and depends on which fuel it is: Q = qmf, where q is the specific heat of combustion of the fuel in joules per kilogram (J/kg) and m t is the mass of the burnt fuel. N ot all the heat liberated during the com bustion of a fuel is usefully spent on heating. A fraction of the heat dissi pates. This is taken into account by introducing the efficiency of a heater: 7j — ^useful $spent (?urefui = cm
and Qspent = qmt, we have cm A T
n = -----------
The internal energy of a body or a system of bodies m ay change when a m echanical work is done. The m echanical energy of a body or a system of bodies can be transform ed com pletely into internal energy, i.e. be spent on heating: Q — AZ?k + A2?p, where AE k and Ai?p are the changes in the kinetic and po te n tia l energies of the body. The first law of therm odynam ics (the law of energy conser vation). The am ount of heat transferred to a closed system
§ 4. The Change in the Internal Energy
39
is spent on increasing its internal energy and on the m echan ical work done against external forces: Q = AC/ + A, where Q is the am ount of heat transferred to the system , \ U is the change in its internal energy, and A is the work done by the system. For isochoric processes, V = const, and hence A = p AF = 0, Q = A *7, i.e. the heat transferred to a gas is com pletely used to in crease its internal energy. In isobaric processes, p = const, and hence Q = A t/ + 4 , i.e. the heat transferred to a gas is spent on increasing its internal energy and on the work done to expand the gas. For isotherm al processes, T = const, and hence the in ternal energy does not change. Therefore, Q = A . The heat supplied to a gas during an isotherm al process is com pletely spent on the m echanical work done by the gas. A diabatic processes occur in the absence of any heat exchange w ith the surroundings, i.e. Q = 0. Therefore, the first law of therm odynam ics in this case will have the form Q = A U + A , or A = - A U, i.e. an adiabatically expanding gas does work at the expense of the change in its internal energy, being cooled thereby. Worked Problems Problem 20. In order to determ ine the specific heat of copper, a copper cylinder of mass 0.5 kg is heated to 100°C and then placed in an alum inium calorim eter of mass 40 g, containing 300 g of w ater at a tem perature of 15°C. As a result of heat exchange, the calorim eter stabilizes at tem perature 26°C. W hat is the specific heat of copper in this experim ent? Compare the obtained result w ith the ta b u la t ed value and determ ine the absolute and the relative error. Given: m c = 0.5 kg is the mass of the copper cylinder, T = 373 K is the in itia l tem perature of the cylinder, ma = 0.04 kg is the mass of the calorim eter, m w = 0.3 kg is the
40____________Ch. I. F u n d a m e n tals of M olecular P h y sics
m ass of the w ater, 71, = 288 K is the in it ia l tem perature of th e w ater and ca lorim eter, and 0 = 299 K is the final te m perature of th e w ater, ca lorim eter, and c y lin d er. From ta b le s, w e tak e th e specific h ea t of alu m in iu m , ca — 880 J/(k g - K ), and the specific h e a t of w ater, cw = 4187 J /(k g -K ). T h e tab u la ted specific h ea t of copper is ctab — 380 J /(k g -K ). F ind: the specific heat of copper cc, the absolute error Ac
and the relative error Ac/ctab
the measurem ent.
S o lu tio n . A s a r e su lt of h ea t e xch ange, th e tem perature of a ll the b odies in the calo rim eter eq u a liz e s. T he heated c y l inder g iv e s off heat @g i v and is cooled from T to 0 :
Cgiv = ccm c (T — 0 ). T he ca lo rim eter and w ater receive h eat, and their t e m perature rises from T l to 0 :
T l),
or Qrec = (c*ma + cwm w) (0 — Tj). P roceed in g from th e energy co n servation law , we equate th e h ea t g iv e n off by the copper cy lin d e r and the heat re ceiv e d by the ca lorim eter and w ater: (?p iv = (?rcc- C onse q u en tly , ccm c (T — 0 ) = (cama + c wm w) ( 0 — T,). W e o b ta in ed the hea t b alance e q uation from w hich we deter m in e cc: mc ( r - 0 )
S u b stitu tin g the num erical v a lu es of the know n q u a n ti tie s, w e ca lc u la te th e specific h eat for copper: c
(880 J/)k g - K) X 0.0 4k g -1-4187 J / (k g -K ) X 0.3 kg) X (290 K - 288K) 0.5 kg (373 K — 299 K)
~ 3 8 4 J /( k g - K ) . The ab so lu te error is
Ac = cc — c lab, Ac = 384 J /(k g -K ) — 380 J /(k g -K ) ~ T he re la tiv e error is
Ac
4 J/(kg-K)
4 J /(k g -K ).
§ 4. T he C hange in the In te rn a l E n erg y
41
A nsw er. T he specific hea t of copper is 384 J /(k g -K ), the absolu te error is a p p ro x im a tely 4 J /(k g -K ), and the r ela tiv e error is ab out 1.1% . Problem 2 1 . 1 5 0 g of w ater is co n tained in a brass c a lo r im eter of m ass 0 .2 kg at a tem perature of 12°C. Find the final tem p eratu re in the ca lo rim eter after an iron w eigh t h avin g a m ass of 0 .5 kg and h eated to 100°C is im m ersed in w ater. P lo t the tem perature versus the a m o u nt of heat graph for th is h eat transfer process. Given: m w = 150 g = 0 .1 5 kg is the m ass of w ater in the ca lorim eter, m ca| = 0 .2 kg is the m ass of the calorim eter, /'w = 285 K is the in itia l tem perature of the w ater and calorim eter, m ( = 0 .5 kg is th e m ass of the w eig h t, and /', = 373 K is the in itia l tem p eratu re of the w eig h t. From tables, w e take the specific h ea t of iron, — 46 0 J /(k g -K ), the specific h ea t of w a ter, 4187 J /(k g -K ), and the specific beat of brass, c h — 380 J /(k g -K ). Find: the final tem p erature 8 in th e c a lorim eter. S o lu tio n . A s a resu lt of h ea t ex chan ge in the calorim eter, the in tern a l energy of the w e ig h t decreases, w h ile the in te r nal energy of the calorim eter and w ater increases. The am ount of h ea t is a m easure of the change in the internal energy. The a m o un t of h ea t Q t = c ^ ij {T \ — 8 ) liberated d ur ing the c o o lin g of the w e ig h t is absorbed by b e a tin g the w ater (Q w = c wm w ( 8 — ?'„.)) and the calo rim eter (Qcn, =
'■cThe TOca,h(e - 7'w)). eat balance
equ a tio n is
r,m | (T , — 8 ) = c wm w ( 8 — 7’,v) + r ca,mca, ( 8 — 7 \v). W e sh a ll so lv e th is eq u a tio n for the unknow n tem perature 8 in th e calorim eter a t therm al e q u ilib riu m . For th is pur pose, w e open the parentheses: c,/n j 7’ | — CimiQ = c wm w8 — c wm w T w
r crn./?tra|8 —cca;mr.nT’wA ll the term s co n ta in in g 8 can be transferred to the righthand sid e of the equation: c Jm 17’1 -j- c wm w f w 4 cca,m.ra|7’w = c wm w8 4- c tm i8
T Ccalmcal^*
42
Ch. I. Fundam entals of Molecular P h ysics
We can now w rite the expression for 0 in the general form: 0
c \ m \ T i 4~
~t~ c c a l /yic a l ) ^ w
c i m i ~r c w m w ~r c c a l m c a l
S u bstituting the num erical values, we obtain 0 _
4 6 0 x 0 . 5 x 3 7 3 + ( 4 1 9 0 x 0 .1 5 + 3 8 0 X 0 .2 ) 285 4190 x 0 .1 5 + 4 6 0 X 0.5 + 380 x 0.2
x -+J-kg^-K^-kg r l ;K: I;kr K
^ 30 7
k,
0
= 34°c.
The graph of t = / (Q) is represented in Fig. 8. The straight line A B shows the change in the tem perature of the weight (the tem perature decreases from 100 to 34°C since the pro cess is accom panied by the lib e r ation of heat). S tra ig h t line BC shows the change in the tem per ature of the calorim eter and w a ter (the tem perature increases from 12 to 34°C, the process in volves the absorption of heat). S tra ig h t lines CE and CD show the change in the tem perature of the w ater and the calorim eter respectively. Even though the tem perature difference is the same in both cases (22 K), the lines CD and CE have differ ent slopes. This is because the water absorbs more heat than the calorim eter. Answer. The equilibrium tem perature in the calorim eter is about 307 K. Problem 22. D eterm ine the efficiency of a m elting furnace in which 70 kg of A -l grade coal was burnt to heat 0.5 t of alum inium from 282 K to the m elting p oint. Given: m a = 0.5 t = 500 kg is the mass of the alum ini um, T x = 282 K is the in itia l tem perature of the alu m in ium, m coai = 70 kg is the mass of the coal. From tables, we o b tain the m elting point of alum inium T m = 932 K , the specific heat of alum inium ca = 880 J/(k g -K ), and the specific heat of com bustion of coal q = 2.05 X 107 J/kg. F ind: the efficiency rj of the furnace. Solution. The am ount of heat (?usefUi required to heat the alum inium from T x to its m elting point can be deter
§ 4. The Change in the Internal Energy
43
mined from the formula C^useful
(^m
^l)*
The am ount of heat (?spent liberated as »uinbustion of the coal is given by ^ spent =
a result of the
QM coal*
The efficiency of the furnace is defined as the ratio of the amount of heat spent in heating the alum inium to the amount of heat obtained as a result of the fuel com bustion: =
^useful x 100%? 0l. vspent
=
cama (Tm - T 1) ^ 10Q%_ Qm coa\
S u b s t i t u t i n g th e n u m e r ic a l v a lu e s , w e o b ta in , .^ 0/ ^ 20%
880 J/ (kg-K) x 500 kg X 650 K 1
2.05 X 107 J /k g
x 70
kg
^
“
Answer. The efficiency of the furnace is about 20% . Problem 23. The engine of a scooter develops a power of 3.31 kW at a speed of 58 km /h. How far w ill the scooter go l>v consuming 3.2 1 of petrol if the efliciency of the engine is 20% ? Given: P = 3.31 kW = 3310 W is the power developed by the engine, v = 58 km /h = 16.1 m/s is the speed of the scooter, V = 3.2 1 = 3.2 X 10~3 m3 is the volume of con sumed petrol, and i] = 20% is the efficiency of the engine. From tables, we obtain the specific heat of com bustion of petrol, q = 4.6 X 107 J/kg, and the density of petrol, (> = 700 kg/m 3. Find: the distance s covered by the scooter. Solution. The energy Q = qm liberated by burning the petrol is spent on the work done to move the scooter and the driver. Only 20% of the energy received from the fuel w ill lie spent in doing m echanical work. Using the law of energy conservation, we can w rite r\Q = A , or r\qm = P t . We e x press the tim e required for the m otion in terms of the distance covered and the velocity: t = s/v. Then v\qm = Ps/vy whence ii qmv
5= J V-In order to determ ine the mass of the petrol, we express it in term s of density and volume (m = pV^) and substitute
44
Ch. I. Fundam entals of M olecular P hysics
it into the expression for the distance: x\qpVv
S ~~
P
*
S u b stituting the num erical values, we obtain rt__ 0.2 X 4.6 X 107 J/kg X 700 k g/n i3 X 3.2 X 10~3 m3 x l 6 . i m /s ,S' — 3.31 x 103 W = 1 0 5 in = 100 k m .
Answer. There is enough petrol to cover 100 km. Problem 24. A freely falling steel ball h its the ground w ith a velocity of 41 m/s. After the im pact it bounces to a height of 1.6 m. D eterm ine the change in the tem perature of the ball during the im pact, assuming th a t only the internal ener gy of the ball changed as a result of the im pact against the ground. Given: v = 41 m/s is the velocity of the ball when it h its the ground, and h = 1.6 m is the height to which the ball rises after the im pact. From tables, we obtain the specific heat of steel, c = 460 J/(k g -K ), and the free fall acceler ation g = 9.8 m /s2. F ind : the change in the b a ll’s tem perature, AT, as a re su lt of im pact. Solution. The kinetic energy acquired by the ball by the m om ent of im pact is E k = rnv2l2. A fraction Q of this ener gy is spent on increasing the internal energy of the ball, i.e. on heating it, while the rem aining energy is spent on the ascent, i.e. converted into the potential energy of the ball E v = mgh. According to the energy conservation law, we can w rite E k = Q + E p. Since Q = c m AT, we can w rite -- cm AT -f- mgh, or £
mgh = cm AT. We elim inate
—2----- gh
m from
both sides:
= c \ T , and solve this equation for AT: ^ —2g/& 2c
AT
(41 m / s ) 2 — 2 x 9.8 m /s 2 x 1.6 m A T = 2 x 4 6 0 J/ (kg-K) -
, v 1’8 K ‘
Answer. The tem perature of the ball has increased through about 1.8 K.
§ 4 The Change in the Internal Energy
45
Problem 25. The volume of 80 g of oxygen at an in itia l tem perature of 300 K increases by a factor of 1.5 as a result of isobaric expansion. D eterm ine the am ount of heat spent heating the oxygen, the work done during its expansion, and the change in its internal energy. Given: m = 80 g = 0.08 kg is the mass of the oxygen, Tx = 300 K is its in itia l tem perature, V 2 = 1.5V { is the final volum e of the gas. From tables, we find the m olar mass of oxygen, M = 32 X 10~3 kg/mol, the m olar gas con stant R = 8.31 J/(m o l-K ), and the specific heat of oxy gen a t constant pressure cp = 0.92 X 103 J/(k g -K ). Find: the am ount of spent heat Q, the work A done during the isobaric expansion, and the change in the in ternal ener gy, A t/, of the gas. Solution. The am ount of heat required to heat the gas is Q = cpm A 7\ where A T — T 2 — T x. The tem perature T 2 can be found from Charles’ law, which governs the isobaric process: V2/ V x = T 2/ T 1. Since V 2 = I.5 V 1 by hypothesis, 7\> = 1.5 7\ = 1.5 X 300 K = 450 K. After su b stitu tin g the num erical values, we obtain Q = 0.92 x 103 J/(k g -K ) x 0.08 kg x 150 K = 11 040 J = 11.04 k J . In an isobaric process, A = p AV. A pplying the equation of state, we can w rite A = p AF =
M
R A7*,
■-1 — -3 2 x ° y , , / m . l * 8 -31 > /< " > ° '-K )X l5 0 K = 3116 J = 3.12 kJ. The first law of therm odynam ics for isobaric processes has the form Q = A U + A , and hence A U = Q - A , AU = 11.04 kJ - 3.12 kJ = 7.92 k J . Answer. The am ount of heat spent heating oxygen is II.0 4 k J , 3.12 kJ is spent on the expansion work, and 7.92 kJ on increasing the internal energy.
46
Ch. I. Fundam entals of Molecular P hysics
Questions and Problems 4.1. How m uch energy is required to raise the tem pera ture of 1 kg of tin through I K ? 4.2. The tem perature of a 1-kg copper w eight has decreased from 293 K to 19°C. W hat is the decrease in its in tern al energy due to cooling? 4 .3 . Two m etal bars, one alum inium and one nickel, w ith the same mass, have their tem peratures lowered through 1 K . W hich bar liberated a larger am ount of heat? W hat is the ratio of the am ounts of liberated heat? 4 .4 . 1 kg of w ater and 1 kg of steel are heated through 1 K . W h a t is the change in their internal energy? 4.5 . Three cylinders having the same volume are made of lead, copper, and alum inium . W hich of them has the largest h e a t capacity? 4 .6 . A copper w eight and an iron w eight of the same m ass were dropped from the same height to the ground. W hich of the w eights had the higher tem perature after the jm pact? Does the answer depend on the mass of the w eights? 4.7. Gases have larger specific heats a t constant pressure than a t co n stan t volum e. How can th is be explained? 4 .8 . How much energy w ill be liberated as a result of the com plete com bustion of 1 kg of Donetsk coal? 4.9. How much fuel oil has to be burnt to o b tain 4 X 107 J of heat? 4.10. How much heat w ill be liberated as a result of the com plete com bustion of 5 m 3 of n atu ral gas? 4.11. How much heat m ust be spent in heating a copper p late of mass 180 g through 15°C? 4.12. 200 1 of w ater are draw n a t 283 K for a b ath . Hqw much boiling w ater has to be added to elevate the w ater tem perature to 37°C? 4.13. To fill an aquarium , 20 1 of w ater a t 15°C are m ixed w ith 2 1 of w ater a t 70°C. D eterm ine the w ater tem p eratu re in the aquarium . 4.14. W h at w ill be the change in the in tern al energy of 1 1 of m ercury as a result of heating it from 283 K to 50°C? The change in the m ercury’s density w ith tem perature should be neglected. 4.15. 8 1 of w ater a t 20°C were poured into an alum inium
§ 4. The Change, in the Internal Energy
47
kettle whose mass is 1.5 kg. How much heat is required to bring the w ater to boil? 4.16. 240 g of w ater a t 288 K were poured into an a lu minium calorim eter of mass 40 g. A fter a lead bar having a mass of 100 g and heated to 100°C had been placed into the calorim eter w ith w ater, a tem perature of 289 K was established. W rite the heat balance equation and determ ine the specific heat of lead. 4.17. In order to cool a copper com ponent having a te m perature of 373 K , it was placed into 420 g of w ater at a tem perature of 15°C. D eterm ine the mass of the com ponent if the w ater gets heated to 18°C as a result of the h e a t tra n s fer. 4.18. C alculate the specific heat of brass if 334.4 J of beat are required to heat a brass w eight having a m ass of 200 g from 285 K to 289.4 K. 4.19. The am ount of heat required to raise the tem p e ra ture of 200 g of m ercury through 58.8°C is the same as th a t required to raise the tem perature of 50 g of w ater through 7°C. Using these d ata, determ ine the specific h e a t of m e r cury. 4.20. D eterm ine the tem perature of a furnace in w hich 0.5 t of steel was hardened if 175 MJ of heat were spent to heat the steel from 20°C to the hardening tem perature. 4.21. A steel com ponent having a mass of 0.3 kg was heated to a high tem perature and then hardened in 3 kg of m achine oil having a tem perature of 283 K. D eterm ine the in itia l tem perature of the com ponent if its final te m perature a t therm al equilibrium was 303 K. 4.22. A m etal cylinder having a mass of 146 g and heated to 100°C is immersed in a brass calorim eter having a mass of 128 g and containing 240 g of w ater a t 8.5°C. The tem p e ra ture established as a result of heat transfer was 283 K . D e term ine the specific heat of the m etal of which the cy lin d er is made and identify the m etal. 4.23. To heat 3 1 of w ater from 20 to 100°C on a gas burner, 0.06 m 3 of n atu ral gas was burnt. D eterm ine the efficiency of the burner. 4.24. D eterm ine the am ount of coke required to heat 1.5 t of scrap iron from 20°C to its m elting point, if the efficiency of the cupola is 60% . 4.25. A bullet having a mass of 9 g and shot from a gun
Ch. I. Fundam entals of Molecular P hysics
acquires a velocity of 800 m/s. D eterm ine the mass of the powder charge if the efficiency of the shot is 24% . 4.26. W hat is the change in the tem perature of 2.0 m 3 of w ater in a boiler if 25 kg of E kibastuz coal have been bu rn t in the furnace whose efficiency is 50% ? 4.27. A tum bler contains 250 g of w ater a t 80°C. W hat w ill the decrease in the w ater tem perature be if a silver spoon having a mass of 50 g and a tem perature of 293 K is immersed in it? 4.28. A blast furnace consumes 2200 m 3 of air per m in ute. The air is heated in Cowper stoves by burning the b la s t furnace gas. Determ ine the volume of the gas burnt daily to heat air from 273 K to 1200°C if the energy losses am ount to 30 %. 4.29. A tra in having a mass of 2 X 10° kg and moving a t a velocity of 54 km /h comes to a halt. How much heat is liberated in the brakes? 4.30. A lead bullet flies a t a velocity of 300 m/s. W hat w ill the change in its tem perature be when it is stopped a b ru p tly ? Assume th a t 50% of its energy is spent heating the b u llet. 4.31. To w hat height should it be possible to lift a load whose mass is 0.5 t if the entire energy given away by a brass w eight of 10 kg upon cooling by 100 K were spent for lifting? 4.32. W hat will be the change in the tem perature of wa ter falling from 120 m if 60% of its potential energy is spent to heat it. 4.33. D eterm ine the change in the tem perature of w ater falling from 96 m onto the blades of the turbine in the B ratsk hydroelectric plan t, assum ing th a t 50% of the energy of the falling w ater is spent on increasing the w ater’s internal energy. 4.34. The striker of a ham m er whose mass is 104 kg freely falls from a height of 2.5 m onto a forged iron piece whose mass is 200 kg. How m any blows does the ham m er strike if the forged piece is heated through 20 K? Assume th a t 30% of the energy of the ham m er is spent on heating. 4.35. A worker pins together wooden stru ts by bu ttin g a 500-g iron nail in to them w ith a 3-kg ham m er m oving at 12 m/s before the im pact. Assuming th a t the entire energy
§ 4. The Change in the Internal Energy
49
of the ham m er is spent on heating the nail, determ ine the change in its tem perature after 20 blows. 4.36. Two identical copper balls have received the same am ount of energy, as a result of which the first b all, rem a in ing a t rest, is heated through 40 K, and the second is set in motion w ith o u t heating. Determ ine the velocity of the second ball. 4.37. A lead bullet passes through a wooden w all so th a t its velocity is 400 m /s at the m om ent of im pact against the wall and 100 m/s when it emerges from it. D eterm ine the change in the tem perature of the bullet assum ing th a t 40% of the m echanical energy spent by the bullet piercing the wall is converted into heat. 4.38. D eterm ine the efficiency of a tractor engine which consumes 292 g of diesel fuel per kilow att per hour. 4.39. The engine of a crawler tractor develops a power of 73.6 kW by consuming 285 g of diesel fuel per kilo w att per hour. D eterm ine the efficiency of the engine. 4.40. A locom otive driven by an in tern al com bustion e n gine has an efficiency of 25% a t a power of 3 MW. D e te r mine the consum ption of the diesel fuel per hour during operation a t m axim um power. 4.41. Modern cars consume on the average 330 g of petrol per hour per k ilow att of developed power. D eterm ine the ef ficiency of a car engine. 4.42. Determ ine the power developed by the engine of a “Zaporozhets” car if it consumes 74 g of petrol per kilom etre a t a velocity of 60 km /h. The efficiency of the engine is 30% . 4.43. The capacity of the fuel tank of a car is 60 1. How far can the car go on the fuel contained in a full tank a t a constant speed if the mass of the loaded car is 1800 kg and the efficiency of the engine is 20% ? The drag coefficient is 0.04. 4.44. A volume of a gas has increased by 0.02 m 3 as a re su lt of heating, w hile its internal energy has increased by 1280 J. How much heat was supplied to the gas if the process occurred a t a constant pressure of 1.5 X 105 Pa? 4.45. 3 m3 of air is stored under a pressure of 2 X 106 Pa at 0°C. D eterm ine the work done by the air during isobaric heating through 12 K. 4.46. How m uch work is done to heat 160 g of oxygen through 20 K a t constant pressure? '*- 0 5 3 0
50
Ch. I. Fundam entals of M olecular P hysics
4.47. 580 g of air is heated isobarically through 10 K . How much heat is supplied to the air? W hat work is done in the process? 4.48. 2 kg of air are contained in a cylinder w ith a piston a t 289 K . W hat work w ill be done by it during isobaric heating to 373 K? 4.4 9 . 3.47 kg of a gas does 144 kJ of work when isobari cally heated through 159 K. Determ ine the m olar mass of the gas and identify it. 4.50. 22 g of carbon dioxide are contained in a cylinder under a heavy piston. W hat work is done by the gas as a result of heating it from 17 to 117°C? 4.51. W hat work is done by an ideal gas filling a rubber balloon when heated from 10 to 70°C? The in itia l volume of the balloon is 5 1 and the atm ospheric pressure is 105 P a. The elasticity of the balloon should be neglected. 4.52. A gas occupying a volume of 30 1 at a pressure of 1.2 X 105 Pa is isobarically heated from 300 to 450 K . D e term ine the work done by the gas. 4.53. W hat work is done by 1 mol of an ideal gas when heated isobarically through I K ? Does this work depend on the pressure and in itia l tem perature? 4.54. Oxygen is contained in a cylinder under a piston. D eterm ine the mass of the oxygen if the work done to h eat it from 273 to 473 K is 16 k J. Friction should be neglected. 4.55. Can the tem perature of a gas be changed w ithout any heat exchange w ith the environm ent? 4.56. The volume of a gas can be reduced either by isother m al or by adiabatic compression. In which case w ill the change in pressure be higher? § 5. PROPERTIES OF VAPOURS
Basic Concepts and Form ulas The conversion of a liquid to vapour is known as evapora tion, and the reverse process is called condensation. V apour is formed as the result of either the evaporation or boiling of a liquid. During evaporation, vapour is only formed a t the free surface of the liquid, which is possible at any tem per ature. D uring boiling, vapour is formed in the bulk of the liquid (vaporization).
§ 5. Properties of Vapours
51
Every liquid boils at a certain tem perature known as its boiling point. E vapbration is always accom panied by an absorption of energy. V aporization during boiling is accom panied by absorption of energy from outside. E vaporation occurs due to a decrease in the internal energy of the evapo rating liquid (which is cooled thereby). The specific laten t heat of vaporization r is the am ount of boat required for the vaporization of 1 kg of a liquid a t a constant tem perature (boiling point): Q = rm, r = Q/m. The specific latent heat of vaporization is m easured in joules per kilogram (J/kg). The specific laten t heat of vaporization depends on th e pressure and tem perature. I t decreases w ith increasing pres sure. The problem s on vaporization are solved by using the heat balance equation. Since vaporization during boiling occurs at a certain tem perature, w hile w riting the heat balance equation one should take into account not only Q = r m , but also Qx = cm ( T b — T x), where T b is the boiling point of a liquid. One should distinguish between the processes of evapora tion and condensation: h e a t is absorbed in the form er case and liberated in the la tte r case. Special atten tio n should be paid to the p h y sic a l m eaning of the specific heat of vaporization. For exam ple, the specific latent heat of vaporization of am m onia is 1.37 X 106 J/k g . This means th a t the conversion of 1 kg of am m onia into v a pour a t the boiling point (—33.4°C) requires 1.37 X 106 J of energy (the same am ount of energy is liberated during con densation). If the tem perature of am m onia is below the boil ing point, an ad d itio n al am ount of heat is required to heat the am m onia to th is tem perature. Vapour is called saturated if its pressure and density are the m axim um possible a t a given tem perature. U nsaturated vapours have properties close to those of gases and hence obey all the laws for ideal gases. An unsaturated vapour can be converted into a saturated vapour either by reducing its volume or by decreasing the tem perature.
52
Ch. I. Fundam entals of M olecular P h ysics
The h um idity of air is characterized by the presence of vapour in the atm osphere. The absolute hum idity pa is a q u a n tity equal to the density of the w ater vapour in air, or to its pressure p a. The absolute hum idity is measured in kilogram s per cubic m etre (kg/m3). The relativ e hum idity B is equal to the ratio of the abso lute h u m id ity pa (or the pressure p a of the w ater vapour in air) to the density ps (or pressure p s) of the saturated vapour a t a given tem perature: 5 = -^-1 0 0 % , or Ps
=
Ps
100%.
The values of density ps or pressure p s of saturated vapours are given in Table 6. W hen the tem perature of air drops to the dew point, the relativ e hum idity becomes 100%. Worked Problems Problem 26. How much heat is required to raise the tem perature of 200 g of w ater from 10°C to the boiling point and to vaporize 10% of it? Assume th a t there are no energy los ses. Given: m w = 0.2 kg is the mass of the cold w ater, t = 10°C, or T = 283 K , is the tem perature of the cold w ater, ttisteam = 0.1 m w is the mass of steam . From tables, we find the specific heat of w ater, cw = 4187 J/(k g -K ) ~ 4190 J/(k g -K ), the boiling point of w ater, t b = 100°C, or T h = 373 K , and the specific late n t heat of vaporization of w ater, r = 2.26 x 106 J/k g . F in d: the am ount of heat Q (spent energy). Solution. Since the w ater tem perature is below the boiling point, it m ust be heated from T to T b, for which the am ount Qx = cwm w ( T h — T) of heat is required. The am ount of heat required to vaporize the w ater is Q2 = w steamr * or Q2 = 0.1 /rcwr. The process is known to occur w ithout energy losses. Consequently, the to tal am ount of heat th a t m ust be spent is Q = Qi + (?2> Qx = 4190 J/(kg-K ) x 0.2 kg x 90 K = 75420 J = 75.42 kJ,
53
§ 5. Properties of Vapours
Q2 = 0.1 x 0.2 kg x 2.26 x 106 J/kg = 4.52 = 45.2 k J, Q = 75.42 kJ + 45.2 kJ = 120.62 kJ.
x 104 J
This process can be represented graphically (Fig. 9). We shall plot the am ount of spent heat along the abscissa axis. For the sake of sim plicity, we round off the values of Qx and Q2 to integers: 75 and 45 k J. The 700' tem perature in degrees Celsius (this is more convenient for cho osing a scale) is plotted along the ordinate axis. When analyzing the graph, we pay atten tio n to the fact th a t vaporization occurs 10 at a constant tem perature (in th is o I T k T case, at 100°C). Consequently, the F ig . 9 9 water m ust be heated to this ternFigperature. The intercepts OA and A B on the abscissa correspond to the values Qx and Q2. Answer. About 121 kJ of energy is required to heat the w ater and vaporize part of it. Problem 27. How much charcoal m ust be burnt in order to heat 6 t of w ater taken a t 283 K to boiling and vaporize 1 t of it? The efficiency of the boiler is 70% . Given: raw = 6 X 103 kg is the mass of w ater in the boiler, T = 283 K is the in itia l tem perature of the w ater, T h = 373 K is the boiling point of w ater, m sieam = 103 kg is the mass of the steam , and r| = 70% = 0.7 is the efficiency of the boiler. From tables, we take the specific heat of w a ter, c = 4187 J/(k g -K ) ~ 4190 J/(k g -K ), the specific late n t heat of vaporization of w ater, r = 2.26 X 106 J/kg, and specific heat of com bustion of charcoal, q — 3.1 X 107 J/k g . Find: the mass m c of charcoal. Solution. The energy Q = qmc liberated during the com bustion of the charcoal is spent on heating the w ater to b oil ing point, Qx = cmw ( T h — T), and on vaporizing 1 t of the w ater, Q2 = rm steamFrom the energy conservation law and given th a t only 70% of energy liberated by burning the charcoal is spent on heating and vaporizing the w ater, we w rite the heat balance equation v\Q = Qx + Q2, or v\qmQ = cmw ( T h — T) +
54
Ch. I. Fundam entals of M olecular P h ysics
Whence
( r b - r ) + m steam _
S ubstitu tin g the num erical values, we obtain 4190 J/ (kg-K ) X 6 X 103 k g x 9 0 K + 2.26 x 106 J/kg X 103 kg 0.7 X 3.1 X 1 0 7 J/kg “
= 208. Answer. About 208 kg of charcoal are spent. Problem 28* 200 kg of steam at a tem perature of 373 K are passed through 4 t of w ater at a tem perature of 293 K . To w hat tem perature w ill the w ater be heated? Energy losses should be neglected. Graph the function t = / (Q). Given: raw = 4 X 103 kg is the mass of the w ater, T w = 293 K is the tem perature of the w ater, msteam = 200 kg is the mass of the steam , r 6team = 373 K is the tem perature of the steam . From tables, we take the specific heat of w ater, cw = 4187 J/(k g -K ) ~ 4190 J/(kg*K ), and the specific la ten t heat of vaporization of w ater, r = 2.26 X 106 J/k g . Find: the tem perature 0 a t the end of the process. Solution. W e have here a heat exchange: steam a t its boil ing point (which equals its condensation tem perature) gives to w ater the am ount of heat Qt = rrasteam and is converted into w ater a t the same tem perature (during condensation, as in boiling, the tem perature rem ains constant). The w ater obtained from the steam is cooled from T h to 0 , liberating the am ount of heat Q2 = Cw^steam (^steam — ©)• Since the final tem perature 0 is the same for all com ponents, the in ternal energy of the cold w ater increases by Q = cwm w X
(e - rw).
According to the energy conservation law, we can w rite
@1 4 “
Q2
=
Qi o r
f^ s t e a m 4 “ ^w ^steam (^ s te a m
= cwm w (0 — T w). We transform the heat balance equation rem oving the pa rentheses and gathering the term s containing the unknown tem perature on the right-hand side: rtfls te a m
C w ^ s te a m ^ 6 te a m 4" Cw fflw T w — CwW w0 4~ c w ^ s te a m © =
( ^ w ^ w 4" ^ w ^ s te a r n )
whence q
__
r/yi8team ~f~ cw (msteam^steam~Hmw^tw )
cw (mw“f 'msteam)
§ 5. Properties of Vapours
55
S ubstituting the num erical values, we determ ine 0: 0 =
+
2.2 6 x 106 J/kg x 200 kg 4190 J/(kg*K ) (4 x 103 k g + 2 0 0 kg) 4190 J/(k g -K ) (200 kg x 373 K + 4 x 103 kg x 293 K) 4190 J/(kg-K ) (4 x 103 k g + 200 kg)
_ ~
o o o if ^ A
or 0 = 49°C. Figure 10 shows the tem perature versus the am ount of heat, t = f (iQ). The tem perature rem ains unchanged over segment A B (the evolution of heat during condensation occurs due to a decrease in the poten tia l energy of interaction between the m olecules). The tem perature of condensate falls from ^steam to © 1 over segment BC (cooling the con densate liberates heat). The tem per- A ature of the cold w ater increases from T to 0 over segment CD (the < process involves the absorption of heat). Answer. The final tem perature 0 Fi£- 10 is approxim ately 49°C. Problem 29. 1.5 1 of w ater at 20°C are poured into an a lu m inium pot whose mass is 600 g and put on an electric hot plate whose efficiency is 75% . In 35 m in the w ater boils and 20% of it is converted into steam . W hat is the power of the hot plate? Given: m a = 0.6 kg is the mass of the alum inium pot, F = 1.5 1 = 1.5 x 10“3 m3 is the volume of the w ater, t0 = 20°C is the in itia l tem perature of the w ater and the pot, t] = 75% = 0.75 is the efficiency of the hot plate, t = 35 X 60 s is the duration of the process, and m stea m = 0.2m w is the mass of the steam formed. From tables, we take the boiling point of w ater, = 100°C, the specific heat of alum inium , ca = 880 J/(k g -K ), the density of w ater, p = 103 kg/m 3, the specific heat of w ater, cw ~ 4190 J/(kg-K ), and the specific late n t heat of vaporization for water, r = 2.26 X 106 J/kg. Find: the power P of the hot plate. Solution. The am ount of heat required to heat the w ater in the pot and to convert part of it into steam (which w ill be
5a
Ch. I. Fundam entals of Molecular P hysics
regarded as the useful heat @usefui) *s ^ u sefu l == c a m ai (^b
*o) 4 ” c w m w (^b
^o) 4 “ ^ s t e a m r •
The am ount of heat liberated by the hot plate w ill be re garded as the spent heat @Spent* Proceeding from the energy conservation law and taking into account the efficiency of the hot plate, we can w rite 'H^spent = ^useful* H e n c e ^ s p e n t = ^useful/*]•
The required power of the hot plate can be determ ined by dividing the spent energy by the tim e: p __ (ga^ a + g w % ) (*b-~*o)4-0-2m w r
The unknown mass of the w ater can be determ ined from the form ula for density: p = m w/ V : m w = p F = 103 kg/m 3 X 1.5 X 10“3 m3 = 1.5 kg. S u b stitu tin g the num erical values, we obtain p _
[880 J/(kg*K) X 0 .6 kg + 4190 J/(k g-K ) X 1 .5 kg] x 80 K 0 .7 5 X 35 x 60 s I
0 .2 x 1 .5 kg x 2 .2 6 x 106 J/k g 0 .7 5 X 35 x 60 s “
7 QA w /OU
Answer. The power of the hot plate is approxim ately 780 W. Problem 30. The relative hum idity of air in a 5 X 4 X 3 m 3 room a t 20°C is 70% . Determ ine the dew point. How much m oisture w ill be liberated from the air in the form of dew if its tem perature is reduced to 11°C? W hat w ill be the rela tiv e hum idity of the air? Given: V = 60 m 3 is the volume of the air in the room, t x = 20°C is the in itia l tem perature of the air, B x = 70% = 0.7 is the relative hum idity of the air, and t2 = 11°C is the final tem perature of the air. From tables, we take the satu rated vapour density psl = 17.3 X 10~3 kg/m 3 for f1?and the saturated vapour density pF2 = 10 X 10“3kg/m 3 for t2. Find: the dew point td, the mass Am of vapour th a t has condensed, and the relative hum idity of the air, B 2, after the dew has precipitated. Solution. The tem perature a t which the w ater vapour in the air becomes saturated is known as the dew point. In or der "o determ ine this tem perature, we m ust know the abso
§ 5. Properties of Vapours
57
lute hum idity of the air, pa . Using the form ula B 1 = pa/p Si for relative hum idity, we obtain !>a = BiPsi = 0.7 X 17.3 X 10“3 kg/m 3 = 12.1 X 10"3 kg/m 3. Using Table 6, we find th a t a vapour having a density of 12.1 X 10"3 kg/m 3 is saturated a t 14°C. Consequently, the dew point is 14°C. In order to answer the second question, we m ust use the same table. The saturated vapour density at 11°C, i.e. the m aximum density, is p S2 = 10 X 10-3 kg/m 3. Before the dew precipitates, the mass of the air in the room is = (>aF, and after th at, m 2 = pS2 ^* Consequently, the mass of the vapour precipitated in the form of dew is Am = m x — m2, Am = paF — p S2V = V (pa — p s2), Am = 60 m3 (12.1 X 10“3 kg/m 3 — 10 X 10“3 kg/m 3) = 126 x 10-3 kg. Since the absolute h um idity at 11°C is the m axim um ad m issible density a t this tem perature, the relative h u m id ity is B 2 = 100%. Answer. The dew point is 14°C, the mass of the condensed vapour is 126 g, and the relative hum id ity is 100% . Problem 31. The relative hum idity of air a t a tem perature of 16°C is 54% . W hat is the reading of the w et-bulb therm o m eter of a psychrom eter? W hat is the absolute h u m id ity of the air? Given: B = 54% = 0.54 is the relative hum idity of the air and t = 16°C is the air tem perature. From tables, we take psl6 = 13.6 X 10~3 kg/m 3. Find: the reading £w of the w et-bulb therm om eter and the absolute hum idity of the air, pa. Solution. We shall use Table 20. In the first colum n, we find 16°C (the reading of the dry-bulb therm om eter). From the same row, we take a relative hum idity of 54% . I t lies in the column for which the tem perature difference between the readings of the dry-bulb and the wet-bulb therm om eters is 5°C. Consequently, t — t w = 5°C, whence fw = f - 5°C, t w = 16°C - 5°C = 11°C. In order to find the absolute hum idity, we w rite B = Pa/Psie from which Pa =
^ P s i 6*
pa = 0.54 x 13.6 x 10~3 kg/m 3 = 7.3 X 10~3 kg/m 3.
58
Ch. I. Fundam entals of M olecular P h ysics
Answer. The reading of the w et-bulb therm om eter is 11°C and the absolute hum idity is approxim ately 7.3 X 10 “ 3 kg/m 3. Problem 32. At a tem perature of 30°C, the p a rtial pressure of w ater vapour in air is 4.1 kPa. D eterm ine the absolute h u m id ity of the air. Given: T = 303 K is the air tem perature and p = 4.1 k P a = 4.1 x 103 Pa is the partial pressure of the w ater v a pour. From tables, we take the m olar mass of w ater vapour, M = 18 X 10 " 3 kg/m ol, and the m olar gas constant R = 8.31 J/(m ol*K ). Find: the absolute hum idity of the air, pa . Solution. Since unsaturated vapours obey the gas laws, we can use the equation of state p V = ^ R T to determ ine the density of the w ater vapour in air. From this equation, we express the ratio of the vapour mass to its volume, i.e. the absolute hum idity of the air: m
T
^
pM
~ Pa — ~RT~ '
S u b stitu tin g the num erical values, we obtain o _
4.1 X 103 Pa X 18 X 10~3 k g/m ol _ q q o q k{T/ir3 8.31 J /(m o l• K) 303 K g '
Answer. 1 m 3 of air contains about 29 g of w ater vapour. Questions and Problems 5 .1 . Can w ater be m ade boil w ithout heating it? 5.2. Equal masses of w ater and steam are a t 100°C. Are th eir internal energies equal? 5.3. In the south, drinking w ater in sum m er is stored in vessels m ade of porous clay. W hy? 5.4. At a pressure below 0.1 MPa (lower th an atm ospher ic pressure), w ater boils at a tem perature lower th an 100°C. E xplain why the steam tem perature in a boiler, where a pressure gauge indicates a pressure of 0.07 M Pa, is higher than 100°C. 5.5. A herm etically sealed vessel contains w ater and w a te r vapour. How will the num ber density of the w ater vapour molecules change as a result of heating?
§ 5. Properties of Vapours
5.6 . W hy does a m ist appear in lowlands after a hot sum mer day? 5 .7 . W hen petroleum is distilled , first petrol is liberated, followed by nap h th a, kerosene, and other com ponents. How can this be explained? 5 .8. W hich puts out a fire more quickly, boiling or cold water? 5 .9 . Is the tem perature of the boiling w ater in a deep ves sel the same at the surface and at the bottom ? 5.10. W hen a gas is released from a cylinder, dew or even frost is formed a t the valve. W hat is the reason behind this phenomenon? 5.11. I t is easier to tolerate hot w eather in C entral Asia than a t m edium latitu d es (at the same air tem perature). W hy? 5.12. W ater and ether have different specific late n t heats of vaporization. Is this alw ays true? 5.13. How much heat is liberated during the condensation of 1 g of petrol vapour having a tem perature of 150°C? 5.14. In which case is more energy liberated: during the condensation of 1 kg of w ater vapour or 1 kg of m ercury vapour? W hat is the ratio of the am ounts of heat? The v a pours are both a t th eir boiling points. 5.15. How much heat is required to vaporize 5 kg of w a ter a t 373 K? How much extra heat is needed if the tem perature of the w ater is 0°C? 5.16. How m uch heat is required to vaporize 10 g of turpentine having a tem perature of 100°C? 5.17. 100 g of steam at 100°C condense into w ater a t 20°C. How much heat is liberated in the process? 5.18. 9.9 X 105 J of heat are spent to heat 2.24 1 of w a ter taken a t 19°C. The w ater is heated to 100°C and some of it is converted into steam . D eterm ine the mass of the steam . 5.19. W hat tem perature becomes established in a b ath containing 80 1 of w ater a t 20°C after 2.2 kg of steam at 373 K is added? The energy taken to heat the bath itself should be neglected. 5.20. S tripped steam a t 100°C is mixed w ith 2 t of w ater whose tem perature is 293 K . How much steam is required to raise the tem perature of the w ater to 309 K? 5.21. 0.3 kg of steam a t 100°C are m ixed w ith 6 kg of water. A fter the steam condenses, the tem perature of the
GO
Ch. I. Fundam entals of M olecular P hysics
w ater rises to 40°C. Determ ine the in itia l tem perature of the w ater. 5.22. 150 g of steam at 373 K are m ixed w ith 1.65 1 of w ater whose tem perature is 20°C. Assuming th a t heat ex change occurs w ithout energy dissipation, determ ine the final tem perature. 5.23. A copper body at 993 K is immersed in 1.75 kg of w ater a t 291 K. As a result, the w ater is heated to 100°C, and 75 g of it are vaporized. D eterm ine the mass of the body, neglecting energy losses. P lot the graph of T = / (()). 5.24. The tem perature in a refrigerator is m aintained by vaporizing a refrigerant freon-12. W hat w ill be the change in the in tern al energy of the air in the refrigerator cham ber if 50 g of freon-12 are vaporized? The freon is at its boiling point. 5 .2 5 . 21 g of dry w ater vapour a t 100°C is let into a cop per calorim eter having a mass of 200 g and containing 400 g of w ater a t 283 K. As a result, the tem perature of the w ater in the calorim eter rises to 40°C. D eterm ine the specific la ten t heat of vaporization of the w ater and compare the result w ith the tab u lated value. Calculate the absolute and re la tiv e errors of m easurem ent. 5.26. A 5-kg iron w eight a t 500°C is placed in a boiler containing 101 of w ater a t 20°C. Some of the w ater is vapor ized, and the final tem perature in the boiler is 25°C. The heat spent during the heat exchange to heat the boiler is 21.85 k J . D eterm ine the mass of the steam formed. 5.27. A 3-kg steel bar a t 450°C is immersed in a vessel containing 3 1 of w ater a t 20°C. As a result, 70 g of the w ater are vaporized, and the tem perature in the vessel becomes 50°C. Determ ine the heat losses during the heat exchange. 5.28. 2 1 of w ater are heated in a 600-g alum inium teapot from 20 to 100°C. 50 g of the w ater are converted into steam . How much n atu ral gas was burnt if the burner efficiency is 60%? 5.29. 1.2 1 of w ater are heated from 283 to 373 K on an electric hot plate. 3% of the w ater is converted into steam . How long does the heating last if the power of the hot p late is 800 W and its efficiency is 65% ? 5.30. 48 1 of w ater a t 277 K were poured in a still. D eter m ine the am ount of wood burnt in the furnace to o b tain 2 0 1 of distilled water. The efficiency of the s till is 15% .
§ 5. Properties of Vapours
HI
5.31. W ater boils at 470.4 K under a pressure of 1.47 MPa (see Table 8 ). How much Donetsk coal m ust be b urnt in the furnace of a boiler to obtain 50 kg of steam under these conditions? The in itial tem perature of the w ater is 10°C, and the efficiency of the boiler is 80% . 5.32. A 1.2-kg alum inium pot containing 2 1 of w ater at 15°C is heated on a gas burner. The w ater in the pot is heated to 373 K ,a n d 200 g of it are vaporized. W hat is the efficiency of the burner if 0 . 1 m 3 of n atu ral gas was burnt? 5.33. 1 m 3 of air at 15°C contains 10 g of w ater vapour. Determ ine the absolute and relative hum idities of the a ir. 5.34. The tem perature of air is 30°C, the relative hum id ity is 64% . Determ ine the absolute hum idity and the dew point. 5.35. The relativ e hum idity of air a t 25°C is 75% . How much w ater vapour is contained in a cubic m etre of the air? 5.36. The absolute hum idity of a ir a t 5°C is 5.2 X 10 “ 3 kg/m 3. To w hat tem perature should th is a ir be cooled for dew to s ta rt to precipitate? 5.37. The relative hum idity of air at 18°C is 50% . A t w hat tem perature w ill dew precipitate? 5.38. How much w ater vapour is contained in 1 1 of air at 17°C if the dew point is 10°C? D eterm ine the relative hum idity of the air. 5.39. The relative h um idity of air in a room at 17°C is 70%. To w hat tem perature has the window glass been cooled if it is covered w ith m oisture? 5.40. The tem perature of a ir is 22°C and its dew point is 10°C. D eterm ine the absolute and relative h um idities of the air. 5.41. In which case w ill m oisture be felt more: if 1 m 3 of air contains 10 g of w ater vapour at 25°C or 3.8 g a t 4°C? 5.42. Fog was observed in the m orning, when the air tem perature was 12°C. W hat was the tem perature during the night if the relative hu m id ity of air did not change and was 70% ? 5.43. The dry-bulb therm om eter of a psychrom eter shows 21°C, w hile the reading of the w et-bulb therm om eter is 16°C. W hat is the relative h um idity of air and how much w ater vapour is contained in 1 m 3 of the air? 5.44. The air tem perature in a room is 23°C and its re la tive hum idity is 55% . W hat is the reading of the wet bull)
62
Ch. I. Fundam entals of Molecular P hysics
therm om eter of a psychrom eter? W hat is the dew point? 5.45. The relative h um idity of air in the halls of a m u seum is 65% . The difference between the readings of the w et-bulb and dry-bulb therm om eters is 4°C. W hat is the tem perature in the halls? How w ill the relative h u m idity change if the psychrom etric difference decreases? 5.46. A cylinder contains air a t 15°C. The relative h u m id ity of the air is 63% . After the air was dried by calci um chloride, the mass of the cylinder decreased by 3.243 g. Determ ine the volume of the cylinder. 5.47. The relative hum idity of air at 12°G is 78% . W hat w ill the change in the relativ e hum idity be if the tem pera ture increases to 18°C? 5.48. The tem perature in a room of size 8 X 5 X 2.5 m 3 is 20°C and the relative h u m idity is 70% . How much w ater vapour is contained in the room? How much vapour is con densed as a result of a tem perature drop to 10°C? 5.49. The relative h um idity of air in a room of volume 30 m 3 is 60% a t 20°C. D eterm ine the saturated vapour pres sure at th is tem perature if the mass of w ater evaporated in the room is 310 g. 5.50. The tem perature of air in a room is 27°C and the p a rtial pressure of w ater vapour in it is 1.7 kPa. D eterm ine the absolute and rela tiv e hum idities of the air. § 6. PROPERTIES OF LIQUIDS
Basic Concepts and Form ulas Substances in the liquid sta te have constant volum es, are fluid, and hence acquire the shape of the vessel in which they are contained. Since the molecules of a liquid are packed more densely than gas molecules, the interm olecular in teraction forces are significant. The range of these forces does not exceed 1 0 nm. The m olecules in the surface layer of a liquid experience the action of forces whose resu lta n t is directed inside the liquid. In order to bring a molecule from the bulk to the surface of a liquid, work m ust be done. Therefore, the molecules of the surface layer have excess potential energy,
§ 6. Properties of Liquids
63
which is responsible for the stretched sta te of the sur face layer. Surface tension a is a q u a n tity defined as the ratio of the work required for increasing the surface area to the incre ment of th is area: A
° = AS • The u n it of surface tension is the joule per m etre squared (J/m 2). Surface tension can also be defined as the ra tio of the force F acting in the surface (force of surface tension) to the length I of the boundary of the liquid surface: a - F/L In th is case, a is expressed in newtons per m etre (N/m). A liquid is term ed w etting if the forces of interm olecular interaction between a solid and the liquid are stronger th an the forces acting between the liquid molecules. The m eniscus (curved surface of the liquid) is concave for w etting liquids, and the w etting angle 0 (i.e. the angle between the m eniscus and the surface of the solid) is acute. For nonw etting liquids, the meniscus is convex, and the w etting angle 0 is obtuse. The curved surface of a liquid produces an additional pressure (called Laplacian pressure) 2a P ad
j
where R is the radius of the spherical surface. The additional pressure in capillaries causes w etting liquids to rise and nonw etting liquids to fall by height ,
2a n cos 0 . pgR
For com plete w etting, we have 0 = 0, cos 0 = 1, and 7
2a
Worked Problem* Problem 33. A fram e in the form of equilateral triangle w ith sides 4 cm long is carefully placed on the surface of water. W hat force keeps the fram e on the surface? W hat
Ch. I. Fundam entals of Molecular P h ysics
force m ust be applied to separate the fram e from the w ater surface? The mass of the fram e is 2 g. Given: r = 4 x 10 " 2 m is the length of a side of the fram e, and m = 2 X 10 “ 3 kg is the mass of the frame. From tables, we take the surface tension of w ater, a = 0.072 N /m , and the free fall acceleration g — 9.81 m /s2. F in d : the force keeping the fram e on the surface and the force F 2 required to separate the fram e from the w ater surface. Solution: The fram e is kept on the w ater surface by the force of surface tension, F u which can be determ ined from the form ula a = F J l : F, = a /, where I is the length of the inner and outer boundaries of the liq u id surface, the length being equal to twice the per im eter of the triangle: 1 = 2 X 3 r . This gives Fx =
6 or,
F 1 = 6 x 0.072 N/m x 4 x 10 ~2 m = 1.73 x 10 ~2 N.
In order to separate the frame from the w ater surface, we m ust overcome the force of g ravity acting on the frame in addition to the force of surface tension, i.e. F 2 = F\ + F 2 = 1.73 x 10 - 2 N + 2 -1 0 - 3 kg x 9.81 m /s 2 - 3.7 x 10 - 2 N. Answer. The force keeping the frame on the surface of wa ter is 1.73 X 10 -2 N. The force required to separate the fram e from the surface is 3.7 X 10 -2 N. Problem 34. The surface tension of w ater was determ ined in a lab o rato ry by using the drop weight m ethod. 1 0 0 drops wrere released from a burette the inner diam eter of whose opening is 1.8 mm. The mass of the droplets was 3.78 g. Using these results, determ ine the surface tension of the water and, com paring it w ith the tab u lated value, calculate the relativ e error of the m easurem ents. Given: d 0 = 1.8 X 10 “ 3 m is the inner diam eter of the burette opening, n = 1 0 0 is the num ber of drops, and
§ 6. Properties o f Liquids
65
m = 3.78 X 10 -3 kg is the mass of 100 drops. From tables, we take the free fall acceleration g = 9.81 m /s 2 and the tabulated value of the surface tension for w ater,
= 0.0728 N/m = 72.8 x 10' 3 N/m. The absolute error Aa is equal to the difference a — a t : Aa - 72.8 x 10 - 3 N/m - 72.0 x 10 ~3 N/m = 0.8 x lO - 3 N/m . The rela tiv e error is equal to the ratio of the absolute error to the tab u lated value of the surface tension: 6n = - ^ 100%, 6„ = Answer. The surface tension of w ater is 72.8 x 10“3 N/m , and the relative error of m easurem ent is about 1 . 1 %. Problem 35. To w hat height will w ater rise in a capillary whose inner diam eter is 3.0 mm? W hat will be the h eig h t qf m ercury colum n in the same capillary? The capillaries are made of glass. Given: d = 3.0 x 10 ' 3 m is the inner diam eter of the capil lary. From tables, we take the surface tension of w ater, cTj = 0.072 N/m , the surface tension of m ercury, o 2 = 0.47 N/m , the density of w ater, p 1 = 103 kg/m 3, the density 5-0530
66
Ch. I. Fundam entals of M olecular P h ysics
of m ercury, p 2 = 1.36 x 104 kg/m 3, and th e free fall ac celeration g = 9.81 m /s2. Find: the heights h x and h 2 of the w ater and m ercury col um ns in the capillaries. Solution. W ater is a liquid th a t wets glass, and hence it has a concave meniscus. For w etting liquids, th e L aplacian pressure is directed upw ards, and equals />l = 2 o /r for com plete w etting. The L aplacian pressure causes w ater to rise in the capillary u n til the hydrostatic and L aplacian pressures equalize: p L = p ht p h = pghy 2
Pi£r
i, _ 2 x 0 .0 7 2 N/m _ n q w *n-a 1 _ 103 kg/m3x 9.81 m/s* x 1 .5x 1 0 -3 m m.
M ercury is a liquid which does not wet glass. Therefore, the Laplacian pressure is directed downwards, into the bulk of the liquid. Hence the m ercury is forced down the c a p il lary. The height to which a nonw etting liquid is forced down is determ ined by using the same form ula 2g 2 P2 gr h
2
= ________________2 x 0 . 4 7 N /m ________________ ^ 1 .3 6 X 1 0 4 k g /m 3 x 9.81 m /s2 X 1 .5 x 1 0 - 3 m —
4 7
X
10~3 m
11K
Answer. The w ater rises in the capillary by about 9.8 m m, w hile the m ercury is lowered by 4.7 mm. Questions and Problems 6.1. Touch the surface of w ater between two floating m atches w ith a piece of soap. Repeat the experim ent w ith a piece of sugar. W hy do the m atches move ap art in the first case and approach each other in the second case? How do sugar and soap affect the surface tension of w ater? 6.2. Sm all pellets are made by pouring m olten lead into a vessel containing w ater. W hy do the pellets become spherical? 6.3. A cap illary tube is immersed in a vessel of h ot w ater. W ill the level of the w ater in the cap illary be affected by a fall in the tem perature of the w ater?
§ 6. Properties of Liquids
67
6.4 . W hy can oily spots not be washed away by w ater? 6 .5 . W hy can tin solder copper and not alum inium ? 6 . 6 . W hy is an area to be soldered thoroughly cleaned from grease, d irt, and oxides? 6.7. W hat liquid can be poured into a tu m b ler to above the brim ? 6 . 8 . W hy are the foundations of buildings covered w ith tar paper? 6 .9. W hy does caked soil during a drought dry more than plowed soil? 6.10. I t is expedient to break up the soil between rows of crops regularly. W hy is th is type of cu ltiv atio n sometimes called “dry irrigation” ? 6.11. A hollow m etal cube w ith a capacity of 1 dm 3 is filled w ith kerosene. D eterm ine the force of the surface tension. 6.12. A cylindrical tum bler 9 cm in height contains 250 cm 3 of m ilk. W hat is the force of surface tension? 6.13. W ater and petrol are poured into a test tube w ith a diam eter of 1.5 cm so th a t the height of the petrol colum n is 10 cm. D eterm ine the force of surface 5 cm tension and the force of the pressure exerted by the petrol on the w ater. The curvature of the surface of the petrol should be neglected. 6.14. A m atch 4 cm long floats on the surface of w ater whose tem perature is 20°C. If castor oil is poured to one side of the m atch, it sta rts m ov ing. D eterm ine the force acting on the m atch and the direction of its mo^ig- 12 tion. 6.15. A wire ring having a radius of 6.0 cm is brought in contact w ith the surface of blue v itriol. The mass of the ring is 5 g. W hat force should be applied to separate the ring from the surface of the solution? 6.16. D eterm ine the poten tial energy of the surface layer of w ater having an area of 2 0 cm2. 6.17. A soap film is formed on a wire fram e w ith a m ov able side A B = 10 cm (Fig. 12). W hat work m ust be done to stretch the film by m oving A B by 5 cm? The friction be tween A B and the fram e should be neglected. 5*
68
Ch. I. Fundam entals of Molecular P hysics
6.18. 532 drops of castor oil were dripped from a pipette w ith a tip diam eter of 1.2 mm into a m easuring glass. The volum e of castor oil was found to be 7 cm3. D eterm ine the surface tension of castor oil. 6.19. C alculate the surface tension of a liquid and id en ti fy it if a force of 0.035 N is required to separate a square fram e w ith a side of 8.75 cm from the liquid surface. The mass of the fram e is 2 g. 6.20. How high can w ater rise in a capillary tube w ith an inner diam eter of 1 0 ~ 3 m? 6.21. D eterm ine the mass of alcohol rising in a capillary tube immersed in a vessel containing alcohol. The inner di am eter of the capillary is 0.4 mm. The surface tension of e th y l alcohol should be taken to be 0.02 N/m. 6.22. D eterm ine the mass of m ercury forced down a capil lary having an inner diam eter of 0 .1 mm and immersed in m ercury. 6.23. 100 drops of pure w ater and the same num ber of drops of w ater m ixed w ith ether are poured into two test tubes. In which test tube w ill the liquid level be higher? 6.24. A liquid rises 4.25 cm up a capillary w ith an inner diam eter of 0.6 mm. D eterm ine the density of the liquid if its surface tension is 0.071 N/m. 6.25. D eterm ine the difference between the levels of m ercury in two com m unicating capillaries having inner di am eters of 1 and 2 mm. § 7. PROPERTIES OF SOLIDS. DEFORMATIONS
Basic Concepts and Form ulas All solids are elastic 3 and reta in their volum e and shape. Solids are characterized by the long-range order in the arrangem ent of the particles of which they are composed. These particles are atom s, m olecules, or ions. The cry stallin e stru ctu re of a solid is a resu lt of the or dered arrangem ent of its particles. The physical properties of a crystal depend on th e o rientation of the sym m etry axis in the crystal (anisotropy). U nder the action of external forces, solids are deformed. The deform ations disappearing after the forces stop acting on 8 Am orphous bodies are treated as supercooled liq u id s.
§ 7. Properties of Solids. D eform ations
69
a body are known as elastic. There are a large num ber of different deform ations, but extension (compression) and shear are im portant. Extension is characterized by an absolute deform ation AI: M = I — l0 and a relativ e deform ation (strain)
M echanical stress a is defined as the ratio of the internal force emerging in a body as a result of deform ation to the cross-sectional area of the body: a — F/S. The u n it of m echanical stress is the pascal (Pa). Hooke’s law establishes a relatio n between elastic deform a tions and the in ternal forces. The m echanical stress a is directly proportional to the stra in e: Z o = k7e , or o = E17 - A r—, where E is Y oung’s m odulus. Y oung’s m odulus is m easured in the same un its as stress, viz. in pascals. The elastic lim it is the m axim um stress em erging in a m aterial for which Hooke’s law rem ains in force. The safety factor is defined as the ratio of the m axim um (ultim ate) stress a u of a construction to the adm issible stress a ad:
For an elastic deform ation, the potential energy E p of a body is equal to the work done during the deform ation (extension or compression) of a body: £ P= - i r - = - l r ( A 0 2W hen a solid changes its sta te of aggregation (when i t m elts), the separation between the particles in the crystal lattic e increases, and the la ttic e is destroyed. The potential energy of the in teractio n between molecules (particles) increases. To m elt 1 kg of a solid a t its m elting point, a certain am ount of heat X is required, which is known as the specific
70
Ch. I. Fundam entals of Molecular P hysics
la te n t heat of fusion: k = Qlm. The specific late n t heat of fusion is measured in joules per kilogram (J/kg). In order to m elt a crystalline substance, heat m ust be spent to heat it to its m elting point and to convert it into a liquid: Q = cm (T m — T) -f- km. W orked Problems Problem 36. D eterm ine the elongation of a copper rod hav ing a length of 6 m and a cross-sectional area of 0.4 cm 2 under the action of a force of 2 kN. Given: I = 6 m is the length of the rod, S = 0.4 x 10 ” 4 m 2 is its cross-sectional area, and F = 2 X 103 N is the applied force. From tables, we take Y oung’s m odulus for copper E = 130 x 109 Pa. Find: the elongation A/ of the rod. Solution. We shall solve the problem using H ooke’s law -y - = a, or - y - = , whence ai
FI ES
XJ
’
2 x 103 N X 6 m 130 X 109 Pa X 0 .4 X 10”4 m 2 “
9 Qv X
ia . 0
m*
Answer. The copper rod elongates by approxim ately 0.23 cm. Problem 37. A chandelier of mass 250 kg is suspended on an alum inium rod w ith an u ltim ate stress of 0.11 GPa. W hat m ust the cross-sectional area of the rod be for the safety factor to be 4? W hat is the strain in the rod? Given: m = 250 kg is the mass of the chandelier, n = 4 is the safety factor, and cru = 1.1 X 108 Pa is the ultim ate stress. From tables, we take the free fall acceleration g = 9.81 m /s 2 and Y oung’s m odulus for alum inium , E — 7 x 1 0 10 Pa. Find: the cross-sectional area S of the rod and the stra in e. Solution. The deform ation of the rod is caused by the force of g rav ity G = mg. The cross-sectional area of the rod is chosen depending on the m echanical stress a emerging in the rod: a = G/S = m g I S , whence S = mg/a.
§ 7. Properties of Solids. Deform ations
71
Given th e safety factor n and the u ltim a te stress a u, we can determ ine the adm issible stress: n = a u/cr, a = o j n . F inally, S= o _
mg
”,
<*u 250 kg X 9.81 m /s2 x 4 ^ 1.1X 10® Pa —
8.9 X 10" 6 m2.
The strain in the rod is au nE
_
e==
1.1 X 10® Pa 4 x 7 X 1010 Pa
10~ '\
Answer. The cross-sectional area of the rod is 0.89 cm 2 and the strain is about 4 X 10~4. Problem 38. How much heat is required to convert 0.8 kg of ice a t —10°C into steam at 100°C? P lot the graph of t = f (Q). Given: m = 0.8 kg is the mass of the ice, tx = —10°C is the in itia l tem perature of the ice and ^2 = 1 0 0 °C is th e tem perature of the steam . From tables, we take the m elting point for ice, t 0 = 0°C, the boiling point for w ater, t h = 100°C, the specific heat for ice, = 2090 J/(kg*K ), the specific heat for w ater, cw = 4187 J/(k g -K ), the specific la te n t heat of fusion for ice, A,=3.35 x 105 J/kg, and the spe cific late n t heat of vaporization of w ater, r = 2.26 x 106 J/kg. Find: the am ount of heat Q required to convert the ice into steam . Solution. The required am ount of heat Q is determ ined by sum m ing the am ounts of heat Q = Qx + @2 + (?3 + (?4 (Fig. 13), where Qx is the heat required to heat the ice to the m elting point, Q, =
(t 0 — f,),
Qi = 2090 J/(kg- K) x 0.8 kg x 10 K = 16720 J - 16.72 k J,
72
Ch. I. Fundam entals of M olecular P h ysics
Q 2 is the am ount of heat required to m elt the ice, Q2 = km, Q 2 = 3.35 x 105 J/k g x 0.8 kg - 2.6 x 105 J = 260 k J, Qs is the am ount of heat required to heat the w ater obtained from the ice to the boiling point, Q 3 = CwJ7l (t 2 £q), Q 3 = 4187 J/(kg* K) x 0.8 kg x 100 K =
334960 J
~ 335 k J, and Q 4 is the am ount of heat required to evaporate the w ater Q 4 = r m , Qt = 2.26 x 106 J/k g x 0.8 kg = 1.808
x 106 J ~ 1810 k J. The to ta l am ount of heat is Q = 16.72 kJ + 260 kJ + 335 kJ + 1810 kJ ~ 2420 k J. Answer. In order to convert 0.8 kg of ice into steam , an energy of 2420 kJ = 2.42 MJ is required. Problem 39. A certain am ount of steam a t 100°C is in tro duced into a vessel containing 0.5 kg of w ater and 20 g of ice at 0°C. As a result, all the ice is m elted, and the w ater is heated to 19°C. Find the mass of the steam . The heat capac ity of the vessel should be neglected. Plot the graph of t = f {QY Given: m w = 0.5 kg is the mass of the w ater, m, == 0.02 kg is the mass of the ice, t 0 = 0°C is the in itia l tem perature of the w ater and ice, 0 = 19°C is the tem perature established in the vessel, and £stoam = 100°C is the tem per ature of the steam . From tables, we take the specific late n t h eat of vaporization (condensation) for w ater, r = 2.26 X 106 J/k g , the specific heat for w ater, c = 4187 J/(kg*K ), and the specific la te n t heat of fusion for ice, % = 3.35 x 105 J/kg. Find: the mass w steam of the steam . Solution. To solve the problem , we m ust w rite the heat balance equation. The steam delivered to the vessel con tain in g w ater and ice condenses, lib erating heat Ql = rmsteam- The w ater obtained from the steam is cooled from Jsteam = 100°C to 0, liberating heat Q 2 = m s1eamX (*steam — 0)- Thus, the heat given a w a y i s ^ giv = Qx + Q2.
§ 7. Properties of Solids. D eform ations
73
The received heat (?rec is the sum of the heat spent to m elt the ice, and the heat spent to heat the cold w ater and the w ater obtained from m elting ice: ()4 = c (m{ + m w) x (0 — *<,)• Consequently, Qrec = Q 3 + QA. According to the energy conservation law, Qglv = Qrec. Let us w rite the heat balance equation and determ ine the mass of the steam : r^ s te a m + <™steam (*steam - 6 ) = X lflj + C (m{ + m w ) (0 -
t0) ,
Xm\ + c (m\ + mw) (0 —J0) ste a m -
m steam
—
r + c (*steam “ 9)
3 .3 5 x 1 0 s J/kg x 0 .0 2 k g + 4187 J/(k g-K ) X 0 .5 2 k g X l9 K 2 .26 X 106 J/kg + 4187 J/(k g-K ) X 81 K
= 1.85 X 10" 2 kg. The graph of t = f (Q) is presented in Fig. 14. Over re gion 7, heat Qx is liberated during the condensation of the steam a t constant tem perature (the potential energy of interaction between the particles decreases). In region 2 , heat Q2 is liberated because the w ater obtained from the steam cools (the kinetic energy of the particles decreases). In region 5, heat Q 3 is absorbed to m elt the ice (the potential energy of the in teraction between the particles in creases). In region 4, heat QA is ab sorbed by the cold w ater and the ice w ater (the kinetic energy of the particles increases), their tem perature increasing to 0 . Answer. About 19 g of steam are required to m elt the ice and heat the w ater. Problem 40. How much gray cast iron taken at 20°C can be m elted in a furnace w ith an efficiency of 2 0 % by burning 1.94 t of A -l grade coal? Given: rj = 20% = 0.2 is the efficiency of the furnace, m c = 1.94 x 10 3 kg is the mass of coal spent, and tx = 20°C is the in itia l tem perature of the cast iron. From tables, we take the specific heat of com bustion of coal, q = 2.05 x 107 J/kg, the specific heat of cast iron, c = 550 J/(k g -K ),
74
Ch. I. Fundam entals of Molecular P hysics
the m elting point for cast iron, t 2 = 1150°C, and the specific la te n t heat of fusion for cast iron, X = 9.7 X 104 J/kg. Find: the mass of the m olten cast iron. Solution. The am ount of heat (^useful required to heat the cast iron and m elt it can be determ ined from the form ula ^useful = cmx(t 2 — *1) + On the other hand, the useful heat is ju st 2 0 % of the spent heat, i.e. 2 0 % of the heat liberated by burning the coal: rj = ^ - ful , where vspent
C?spent — qmc. Hence $ usefUi = W e w rite the heat balance equation and determ ine the mass of the cast iron:
m i1 =
!i /k l eg 550 J /(k g -K ) x 1130 K + 9 .7 xX|! 104J
= 11070 kg& ~
1 1 .1
t.
Answer. About 11.1 t of cast iron can be m elted in the furnace. Problem 41. W hat m ust be the m inim um velocity of an iron m eteorite for it to be com pletely evaporated in the E a rth ’s atm osphere? Assume th a t the in itia l tem perature of the m eteorite before it enters the atm osphere is 3 K and th a t 50% of its kinetic energy is converted into internal energy. Given: T - 3 K is the in itia l tem perature of the m eteorite, r] = 50 % = 0.5. From tables, we take the specific heat of iron in the solid state, c = 460 J/(k g -K ), the m elting point of iron, r m = 1803 K , the specific late n t heat of fusion of iron, X = 2.7 x 105 J/kg, the specific heat of iron in the liquid state, c{ = 830 J/(k g -K ), the boiling point of iron, T h = 3323 K , and the specific late n t heat of vaporization for iron, r = 5.8 x 104 J/kg. Find: the m inim um velocity v of the m eteorite at which it enters the E a rth ’s atm osphere. Solution. The m eteorite is m ade of iron, so all extra data needed are taken to be the tab u lated values for iron. W hen the m eteorite enters the atm osphere, its kinetic en ergy is spent on heating, fusing, and evaporating it. The problem states th a t these processes consume 50% of the ki-
§ 7. Properties of Solids. Deform ations
75
netic energy of the m eteorite. Therefore, we can w rite 0.5
= cm (Tm - T ) + km + c{m (Tb - T J + rm.
E lim inating the mass of the m eteorite from the equation (we take it out of the parentheses on the right-hand side of the equation and cancel it out), we obtain i;2 = 4 [c ( T m — T) + X + cx ( T b -
T m) + rj.
We w rite the solution in the general form and then carry out the calculations: v = 2 V c ( T m — T )-\-k + c{ (Tb — T m) + r, v = 2 V (460 x 1800 + 2.7 x 105 + 830 x 1520 + 5.8 x 104)m 2/s 2 ~ 3100 m/s. Answer. The m inim um velocity of the m eteorite m ust be approxim ately 3.1 km/s. Questions and Problems 7.1. W hy should a tool being sharpended be periodically cooled? 7.2. Two cubes, one cut from a single crystal and one from glass, are placed in a vessel containing hot w ater. W ill they retain their shape? 7.3. The hardness of glass and tool steel is the same. W hy are cu ttin g tools not m ade of glass? 7.4. W hy is lead wire used for fuses rath er th an some other m etal? 7.5. Under w hat conditions can lead be m elted in w ater? 7.6. W hy does precipitation in sum m er have the form of rain or hail and not snow? 7.7. W hy is sa lt sprinkled over pavem ents covered w ith ice in w inter? 7.8. W hat is the diam eter of a rod if the m echanical stress caused in it by a force of 2 x 103 N is 160 M Pa? 7.9. Tensile forces of 150 N each act on the two ends of an iron wire of 2-m length and 1-mm 2 cross-sectional area. De term ine the absolute elongation and strain of the wire. 7.10. D eterm ine the elongation of a 4-m long steel wire
76
Ch. I. Fundam entals of Molecular P hysics
having a diam eter of 2 mm and loaded w ith 70 kg. D eter m ine the energy of elastic deform ation of the wire. The mass of the wire should be neglected. 7 .1 1 . W hat w ill be the difference in the elongation of a wire if it is replaced by the same length of wire made of the same m aterial and w ith the same load but w ith tw ice the diam eter? 7 .1 2 . A wire having a cross-sectional area of 1 mm- and a length of 1 m is elongated by 1 mm under a load of 200 N. W hat w ill the elongation be for a wire made from the sam e m aterial, 3 m long, w ith a cross-sectional area of 0.5 m m 2, and loaded by a force of 300 N? 7 .1 3 . A residual deform ation appears in a copper wire having a cross-sectional area of 1.5 mm 2 when loaded by a tensile force of 45 N. W hat is the elastic lim it of the wire m aterial? 7.14. A brass wire has a diam eter of 1 mm and a length of 3.6 m. C alculate Y oung’s m odulus for brass if a 19-kg load elongates the wire by 8 mm. 7.15. W hat m ust the m inim um diam eter of a steel rope be for it to be able to hold a load of 2.45 x 104 N? The u ltim a te stress for steel is 1.25 GPa. 7 .1 6 . W hat is the m axim um load th a t can be borne by a steel rope w ith a cross-sectional area of 12 m m 2? The u lti m ate stress should be taken as 785 MPa. 7 .1 7 . How many piles having a diam eter of 10 cm each are required to support a platform having a mass of 2 X 105 kg if the adm issible compressive stress is 10 7 Pa? 7.18. D eterm ine the cross-sectional area of the steel rope of a crane uniform ly liftin g a 6 -t load if the u ltim a te stress of a m aterial is 780 M Pa and the safety factor is 10. 7.19. W hat m ust the m inim um length of a steel rope sus pended a t one end be for it to ru p tu re a t the point of sus pension under the force of gravity? The u ltim a te stress of the rope is 320 M Pa. 7 .2 0 . Under a 120-N force, a spring is elongated by 5 cm. W hat is the potential energy of elastic deform ation of the spring? 7 .2 1 . W hat m ust the elongation be of an iron rod having a length of 1.8 m, a cross-sectional area of 7.8 m m 2, and a po te n tia l energy of elastic deform ation of 3.9 X 10 -2 J? 7 .2 2 . A flask containing ice a t 0°C is placed in a vessel
§ 7. Properties of Solids. Deformations
77
containing w ater at the same tem perature. W ill the ice m elt? 7.23. W hy are alcohol and not m ercury therm om eters used to m easure tem peratures in the north? 7.24. How much heat is required to m elt 2.6 kg of lead at 300 K ? 7.25. How much heat m ust be spent to convert 125 g of ice at 268 K into steam ? 7.26. 0.15 kg of m olten lead at its m elting point are poured into 300 g of w ater at 285 K. D eterm ine the tem pera ture at therm al equilibrium , neglecting heat losses. 7.27. E qual volumes of lead and tin are taken at their m elting points. W hat is the ratio between the am ounts of heat required to liquefy them ? The tem perature dependence of their densities should be neglected. 7.28. To m ake shot, 50 kg of lead at its m elting point are sprayed in jets into water. D eterm ine the final tem perature in 19 1 of w ater taken at 283 K. P lot the graph t = f (Q). 7.29. 58 g of m olten lead at 357°C were poured into an indentation in a piece of ice at 273 K. How m uch ice is m elted if the lead is cooled to 273 K ? P lo t the graph t = /(
78
Ch. L Fundam entals of Molecular P h ysics
271 K from 200 g of w ater taken a t 15°C. D eterm ine the power of the refrigerator. 7.36* C alculate the efficiency of a sm elting furnace in which 1.1 t of Donetsk coal are burnt to obtain 7.2 t of w hite cast iron. The in itia l tem perature of the w hite cast iron is 0°C. 7.37. In frictional welding, alum inium rods at 17°C are pressed tig h tly together, and then one is rotated by a spindle. As a resu lt, the tem perature of the alum inium where the rods are in contact rises. W hat mass of the alum inium is fused in 10 s if the force of the pressure is 1260 N at a ro ta tional speed of the spindle of 90 m /m in? 7.38. A skidding car moves over a snow field. The skid ding consume 5 kW of power. How much snow w ill m elt under the wheels in 1 m in if the tem perature of snow is 0°C? 7.39. A lead b u llet having a velocity of 450 m /s and a tem perature of 117°C strikes a steel beam. W hat fraction of the bullet fuses if 40% of its kinetic energy before the im pact is spent on heating the bullet? 7.40. W hile m aking ice in a frige, the tem perature of w ater is reduced from 278 to 273 K in 5 m in and it takes 1 h 20 m in then for the w ater to be converted into ice. Using these d ata, determ ine the specific la te n t heat of fusion for ice assum ing th a t the specific heat of w ater is 4190 J/(kg - K). 7.41. A 300-g lum p of wet snow is placed into 1.2 1 of water #t 21°C. A fter all the snow has m elted, the tem perature of the w ater drops to 6 °C. How much w ater was contained in the lum p of snow? The specific heat of w ater should be taken as 4190 J/(kg- K).
§ 8. THERM AL EXPANSIO N OF BODIES
Basic Concepts and Formulas Solids, liq u id s and gases all expand upon heating. Solids expand m uch less th an liq u id s and gases which expand the m ost. Solid bodies w ith one dom inant dim ension are character ized by th e coefficient of linear expansion a . If the length of a solid body at t 0 = 0°C is l 0 and be comes It as a result of it being heated to a tem perature t ,
§ 8. Therm al E xpansion of Bodies
79
the coefficient of linear expansion is defined as a =
to l$to
where AI = l t — l 0 and At = t — t 0. The coefficient of linear expansion is measured in reciprocal kelvins (K -1). If the in itia l tem perature is 0°C, At = t. The length of a body at different tem peratures can be de term ined from the form ula h = h (1
a 0-
Since the value of a is very sm all, we can use the form ula h = h (1 4“ where At = t 2 — tx, and l 2 and lx are the lengths of the body at these tem peratures. Solid three-dim ensional bodies (like a cube or a sphere) and liquids are characterized by a coefficient of volum e expansion a AF The coefficient of volum e expansion is also m easured in reciprocal kelvins (K _1). By analogy w ith linear expansion, the volum e of a body at any tem perature can be expressed as follows: Vt = V 0 (1 + p*). For two nonzero tem peratures, we can w rite V 2 = Vj (1 + pAf), where At = t 2 — tx. The following relation can be established between the coefficient of lin ear expansion and the coefficient of volum e expansion: p 3a. The densities of solids and liquids also change as a resu lt of heating. As the volum e increases due to heating, the density decreases: n —
P° 1 + PA* ‘
80
Ch. I. Fundam entals of Molecular P h ysics
At constant pressure, the coefficients of volum e expansion for all gases are the same and equal to P = 0.00 361 K " 1 = ___K " 1 273.15 ^
'
Worked Problems Problem 42. A g uitar is tuned in a room at 293 K , the length of its steel string being 0.7 m. W hat will be the change in the length of the string outdoors where the tem perature is 263 K ? W hat w ill the additional m echanical stress be? W hat are the elastic force and the po ten tial energy of the elastically deformed string? The cross-sectional area of the string is 0.85 mm2. Given: lY = 0.7 m is the in itia l length of the strin g , T x = 293 K is the in itia l tem perature, T 2 = 263K tem perature outdoors, and S = 8.5 X 10 -7 m 2 is the crosssectional area of the string. From tables, we take the coeffici ent of linear expansion for steel a = 1.2 x 10 " 5 K _1, and Y oung’s m odulus for steel E = 2.2 x 1011 Pa. Find: the absolute decrease AI in the length of the strin g , the additional m echanical stress a , the elastic force F , and the p o tential energy E p of elastically deformed string. Solution. As the tem perature decreases, the length of the strin g decreases by A I. The value of AI can be determ ined from the form ula AI = ZjCtA 7\ where A T = T 2 — 7 \, AI = 0.7 m x 1.2 x 10 - 5 K " 1 ( - 3 0 K) = — 2.52 x 10 ~4 m ~
— 0.252 mm.
The m inus sign in front of AZ in this case indicates th a t the string has become shorter. The measure of the stressed state of a deformed body is the m echanical stress a , which can be determ ined using Hooke’s law: a^ l£ L lx
p= ’
2 2 X 1011 p«>< 2.52 x l0 - « m = 7 92 x 1 0 ? p a. 0 .7 in
The elastic force F emerging during the compression of the string is proportional to the absolute deform ation:
§ 8. Thermal E xpansion of Bodies
F=
/x
AZ. Since
81
E-^ 1 = a, we have
F = 7 .9 2 x 1 0 7 P a x 8 . 5 x l 0 - 7m 2 ~ 67
N.
Know ing the elastic force and the decrease in the string length, we can determ ine the energy of the elastic deform a tion of the string: 67 N x 2 . 5 2 X 1 0 - 4 m
2
~ 8.4 X 10- 3 J.
Answer. As a resu lt of the decrease tintem p eratu re, the string contracts by about 0.25 mm. The ad d itio n al stress emerging thereby is 7.92 x 107 Pa. The elastic force is approxim ately 67 N, and the potential energy of the elastic deform ation is about 8.4 m J. Problem 43. A tank car contains 20 t of petrol at 298 K. How much sm aller will the volum e of the petrol at the ter m inal be if the am bient tem perature is 248 K? The change in the capacity of the tank car as a result of the change in tem perature should be neglected. Given: T l = 298 K is the in itia l tem perature of the p et rol, m = 20 t = 2 x 104 kg is the mass of the petrol, T 2 = 248 K is the final tem perature of the petrol. From tables, we take the coefficient of volum e expansion of petrol, p — 10 ~3 K " 1, and the density of petrol at 273 K , p 0 7 x 1 0 2 kg/m 3. Find: the change A F in the volum e of the petrol. Solution. As the tem perature falls, the volum e of the pet rol decreases by AF = V 2 — F lt where Vx is the volum e of the petrol at T ly and V 2 is its volum e at T 2. The volumes Vx and F 2 can be found from the form ulas Vl = V 0 [1 + P (T, -
T 0)],
F 2 = F 0 [1 + p (T 2 -
T 0) l
The unknown volume F 0 at T 0 — 273 K can be calculated from the form ula F 0 = m lp0. Then AF = - ^ l l + P (T 2 - r 0) - l - p ( 7 \ - J o ) ] , Po
or Po
6-0530
82
Ch. I. Fundam entals of Molecular P h ysics
The m inus sign indicates th a t the volum e has decreased as a resu lt of the fall in the tem perature. Answer. The volum e of the petrol is reduced by 1.43 m 3 due to the tem perature drop. Problem 44. A steel ingot has a volum e of 2.8 dm 3 at 0°C. D eterm ine its volum e at 525°C. Find the density of steel at this tem perature. How m uch heat is spent to heat it? Given: V 0 = 2.8 x 10 ~3 m 3 is the volum e of the ingot at 0°C, t 0 = 0°C is the in itia l tem perature of the ingot, t = 525°C is the tem perature of the hot ingot. From tables, we take the coefficient of linear expansion for steel, a = 1 . 2 x 10 ~ 5 K ”1, the density of steel at 0°C, p 0 = 7.8 X 10 3 kg/m 3, and the specific heat of steel, c = 460 J/(k g -K ). F ind : the volum e V of the hot ingot, the density p of steel at 525°C, and the am ount of heat Q spent to heat the ingot. Solution. The volum e of the ingot heated to 525°C can be determ ined from the form ula V = V 0 (1 + pAT), where AT1 = t — t 0. Since the coefficient of volum e expansion for solids is P — 3a (w ith a sm all error), we can find the volume: V = 2 . 8 x 1 0 - 3 m 3 (1 + 3 = 2.85 x lO " 3 m3.
x 1.2 x
10 "5
K _1 x 525°C)
The density of steel decreases w ith increasing tem pera ture. I t can be determ ined from the form ula P = Po/(l + PA7T)» o— 7 .8 x 103 kg/m 3 _ 7 gg x j o 3 kg/m 3 P _ 1 + 3 .6 x 10*"5 K"1 x 525 K X g/ ’ Pay a tte n tio n to the fact th a t the tem perature difference can be expressed either in degrees Celsius or in kelvins. The heat spent to heat the ingot is Q = cm (t — t 0). Since mass can be expressed in term s of density, m = p0Po = pF , we have Q = Pov 0c (t — t0), Q = 7.8 x lO 3 kg/m 3 x 2.8 xlO ' 3 m3x 460 J/(kg*K ) x 5 2 5 K = 5.3x10® J. Answer. The volum e of the hot ingot is 2.85 X 10 -3 m3, the density of steel at 525°C is 7.65 x 103 kg/m 3, and the am ount of heat spent to heat it is 5.3 M J.
§ 8. Thermal E xpansion of Bodies
83
Remark. The volume of the hot ingot can be found in a different way: first we can determ ine A V -- F 0pA 7\ and then V = V 0 + AV. Questions and Problems 8.1. W hy is Invar used in clockwork m echanism s? 8.2. W hy are the holes in the cover plates connecting rail sections together oval? 8.3. The filam ents fused into the glass parts of vacuum tubes or incandescent lam ps m ust have the same therm al expansion coefficients as th a t of glass. W hy? Can copper wire be used? 8.4 . A copper ring is heated in a bunsen flame. W hat will be the change in the inner diam eter of the ring? 8.5 . How can a steel rod be extracted from a tightfitting brass sleeve? 8 . 6 . Prove th a t the coefficients of volume expansion for solids are three tim es as large as the coefficients of linear expansion. Is it an exact or approxim ate relation? 8.7. Cold w ater can be poured into a red-hot flask made of quartz glass w ithout breaking it. How can this be ex plained? 8 . 8 . W hy should tanks for petrol, kerosene or petroleum not be filled to the brim in sum m er tim e? 8 .9 . D entists do not recom mend th a t we eat very hot food. W hy? 8.10. W hat w ill happen to a b im etallic strip 4 when it is heated? W here are such plates used? 8.11. A steel m easuring tape 100 m long is used for geo desic surveying. How much w ill it be elongated if the tem perature increases by 1 0 K? 8.12. W hat is the m axim um increase in tem perature of a 1 -km long alum inium wire for it to be elongated by only 230 mm? 8.13. As a result of heating through 100 K , each m etre of a wire elongates by 0.4 mm. W hat is the m aterial from which the wire is made? 4 A b im e ta llic strip is m ade from two different m etals rivetted together. 0*
84
Ch. I. Fundam entals of Molecular P hysics
8.14. The O stankino television tower in Moscow is made of reinforced concrete and is 533 m high at 273 K. How high will it be at 20°C and at — 20°C? Assume th a t a = 1.2 X 10 " 5 K " 1 for reinforced concrete. 8.15. An iron rod 60 cm long at 273 K is placed into a furnace. As a result, it elongates by 6.5 mm. D eterm ine approxim ately the tem perature in the furnace. 8.16. The diam eter of an iron tyre at 0°C is 6 mm less than the diam eter of the wheel (equal to 1 . 2 m) on which it has to be set. By how m any kelvins should the tem perature of the ty re be raised to set it onto the wheel? 8.17. The inner diam eter of a copper ring at 273 K is 5 cm. To w hat tem perature should it be heated for it to accom m odate a ball 5.01 cm in diam eter? 8.18. The transm ission line between the Volzhskaya hydroelectric statio n and Moscow is 1000 km long. The line is made of steel-and-alum inium wire. W hat will be the differ ence in the length of the steel and alum inium com ponents if the tem perature increases by 30 K? 8.19. The rod antenna of the Luna-20 probe, which is oper ating on the surface of the Moon, is 2 m long at noon when the tem perature at the M oon’s surface is 393 K. How long will the antenna be at m idnight when the tem perature on the Moon drops to 123 K? The antenna is m ade of bras>s. 8.20. A steel beam is fastened on two supports which pre vent it from expanding. The cross-sectional area of the beam is 140 cm2. W hat w ill be the force of the pressure exerted by the beam on the supports if the tem perature increases by 20 K? 8.21. An alum inium rod whose cross-sectional area is 4 cm 2 is tig h tly clam ped between two jam s. W hat m ust be the increase in tem perature for the force of pressure exerted by the rod on the jam s to be 9.7 kN? 8.22. W hat forces should be applied to the ends of a steel rod of a cross-sectional area 1 0 cm 2 to prevent it from elon gating when heated from 273 to 303 K? 8.23. Two rulers, one m ade from alum inium and one from steel, have the same length of 1 m at 0°C. A t w hat tem perature w ill the difference in th eir lengths be 5.5 mm? 8.24. How much heat is required to m ake a copper rod whose cross-sectional area is 2 cm 2 elongate by 0 . 1 mm when heated?
8 8. Thermal E xpansion of Bodies
*5
8.25. A steel ty re is set on a wheel at 300 K. W hat force w ill emerge when the tyre is cooled to 293 K if its crosssectional area is 2 0 cm2? 8.26. A cylinder is cut from quartz so th a t its axis is parallel to the sym m etry axis of the quartz crystal. At 18°C, the radius of the cylinder base is 1 0 mm and its height is 50 mm. D eterm ine the volum e of the cylinder at 573 K. The coefficient of linear expansion is 7.2 x 10 ~ 5 K _1 for the longitudinal axis and 1.32 x 1 0 ~ 5 K _1 for the transverse axis. 8.27. The coefficient of surface therm al expansion is twice the coefficient of linear expansion. W hat m ust be stipulated in this case? 8.28. A copper sheet having a size of 0.6 X 0.5 in 2 is heat ed from 293 K to 600°C. D eterm ine the area of the hot sheet. 8.29. W hat is the increase in the tem perature of an alu m inium sheet if its area has increased by 3200 mm 2 by heating? The area of the sheet at 0°C is 1.5 m2. 8.30. A vessel whose volum e is 5 1 is filled to the brim w ith kerosene at 0°C, brought into a room where the tem per ature is 18°C, and placed on a tray. How much kerosene (in litres) will overflow if the therm al expansion of the vessel is not taken into account? How much will overflow if it is taken into account, the coefficient of volum e expansion for the vessel m aterial being p = 3.6 X 10 -5 K _1? 8.31. A 2-1 alum inium k e ttle is filled w ith w ater at 4°C. How much w ater (in litres) w ill overflow from the k ettle if it is heated to 353 K? 8.32. W hat is the density of tungsten at its m elting point? 8.33. A t w hat tem perature is the density of concrete 2.19 X 103 kg/m 3? Assume th a t the coefficient of linear expansion for concrete is 1.2 X 10 “ 5 K " 1. 8.34. The volum e of a brass cylinder at 325 K is 425 cm3D eterm ine the mass of the cylinder. 8.35. The mass of a copper bar is 10 kg. At w hat tem pera ture will the bar have a volum e of 1.125 dm 3? 8.36. Alcohol at 273 K is poured into a 10-1 glass bottle. N eglecting the therm al expansion of glass, calculate the vol ume and mass of alcohol th a t can be poured into it so th a t it does not overflow at 50°C.
86
Ch. I. Fundam entals of Molecular P hysics
8.37. Solve Problem 8.36 taking into account the expan sion of the glass bottle. 8.38. How much petrol can be poured at 273 K into a 60m 3 iron tank so th a t it does not overflow when heated to 40°C? D eterm ine the mass of the petrol. 8.39. W hat volume is occupied by w ater in a steam boiler at 100°C if the mass of the w ater is 2000 kg? 8.40. W hat is the m inim um capacity of the cooling sys tem of a transform er if the tem perature of the oil in it does not exceed 363 K during operation of the transform er? The mass of the oil is 3000 kg. 8.41. A 500-cm3 flask m ade of quartz glass is filled w ith m ercury. Determ ine the coefficient of volum e expansion for m ercury if 8.91 cm 3 of the m ercury overflows as a result of heating the flask from 273 to 373 K. 8.42. How much heat is spent if the volum e of a mass of m ercury increases when heated by 4.5 cm3? H eat losses should be neglected.
Chapter II
Fundamentals of Electrodynamics § 9. ELECTRIC FIELD
Basic Concepts and Form ulas U nder norm al conditions, any physical body is electrically n eu tral, i.e. contains equal num bers of elem entary electric charges of opposite signs. The u n it of electric charge is the coulomb (C). W hen a body is electro statically charged, the electric charges are red istrib u ted so th a t the body has an excess charge of one sign; it becomes electrically charged. For exam ple, if a body has an excess num ber of electrons, it is negatively charged. The algebraic sum of the charges in an isolated system rem ains constant (the law of charge conservation). Two bodies bearing like charges repel each other, while oppositely charged bodies a ttra c t each other. The force F w ith which two point charges Qx and Q 2 in te rac t is given by Coulom b’s law ^
1
\Q i\ \Qt \
4ae0
er2
’
where r is the separation between the charges, e is the per m ittiv ity of the m edium , and e 0 is the electric constant. It is good to rem em ber th a t 1 /( 4 j i e 0) = 9 X 1 0 9 m /F. E lectric charges in te rac t through electric fields, which are characterized by a vector q u a n tity known as the electric field strength E:
where Qt is a test charge in the field. The electric field strength produced by a point charge is p * “
f
Q
4jte0 er2 ’
where Q is the electric charge producing the field.
88
Ch. II. Fundam entals of Electrodynam ics
W hen a charge moves in an electric field, work is done. Therefore, the energy characteristic of electric field is the electric potential
The u n it of potential is the volt (V). The electric potential of the field produced by a point charge or electrically charged sphere is ^sph
_
1
Q
4^
gr
The work A done in an electric field to move a charge depends on the potential difference forthe points between which the charge Q moves: A = Q (
(Pl ~ d
2
U ~
d '
where d is the separation between points 1 and 2 measured along the field line. This form ula gives the dimensions of the electric field strength, viz. volt per m etre (V/m). W hen an isolated conductor is electrostatically charged, its potential increases in proportion to the charge Q im p art ed to it: Q = C
The u n it of capacitance is the farad (F). The electric capacitance of an isolated sphere is C sPh = 4jr.e0 er. E lectric energy can be stored in capacitors. The capacitance of a parallel-plate capacitor is calculated by the form ula ynr
£q£iS
c = ~~T~ ’
§ 9. E lectric Field
89
while th e energy accum ulated in it is given by
W hen capacitors are connected in parallel to form a bank, the to ta l capacitance is equal to the sum of the capacitances of in d iv id u al capacitors: C = Ci + C 2 + C 3 + • • • • The poten tial difference (voltage) across all the capacitors is then the same. W hen a bank is formed by series-connected capacitors, the to ta l capacitance is determ ined from the form ula 1 1
1 1 cT + 'cT Jp • • • •
W orked Problems Problem 45. A conducting sphere bearing a charge of 1.8 x 10 -8 C is brought in contact w ith two sim ilar spheres, one of which has a charge of — 0.3 X 10 “ 8 C and the other is neu tra l. How w ill the charge be d istrib u ted among the spheres? W hat w ill be the force of in teractio n between two such spheres in vacuum at a distance of 5 cm from each other? Given: ^ = 1.8 x 10 ~8 C, Q 2 = — 0.3 x 10 -8 C, and @3 = 0 are the charges of the spheres before they are brought in contact, and r = 5 x 10 ~2 m is the distance at which two spheres interact. From tables, we take the elec tric constant e 0 = 8.85 X 10 " 12 F/m and the p e rm ittiv ity of vacuum , e = 1 . Find: the electric charges @', and @' of the spheres after they have been brought in contact and the force F of e lectrostatic in teraction between two spheres. Solution. W hen the spheres are brought in contact, some of the charges neutralize each other since they have opposite signs. The rem aining charge w ill be equally d istributed between the three spheres: ^
/v
/v
VI —V2 — V3 —
3
’ Yl —
1.8xl0“8 C—0.3x10“ 8C
= 0.5
3
X
10“ 8 C.
90
Ch. II. Fundam entals of Electrodynam ics
The force of electrostatic interaction between two identical charges Q[ and Q ' in vacuum can be determ ined from Coulom b’s law: F = _ L _ (QitL 4jie0
F= 9
X
t*
*—= 9 x 10® m/F = 9 x 10® ^ 1 ,
4;ie0
O
10* N • m 2/C 2
=
9 X 1 0 -5
N-
Answer. The charge on each sphere after they have been brought in contact is 0.5 X 10 ~8 C and the force of th eir interaction is 9 X 10 -5 N. Problem 46. Two charged m etal (a ) (b) spheres negligibly sm all in size are placed in transform er oil where 1 they interact w ith a force of 2.5 x 10- 4 N. D eterm ine the separation between the spheres if their charges are 6 and 60 nC. Given: Qx = 6 x 10 ~9 C and Q 2 — 6 x 10 -8 C are the electric charges on the spheres and F = FiP* hs 2.5 < 10 ~4 N is the force of their electrostatic interaction. From ta bles, we take the p e rm ittiv ity of transform er oil, e — 2.5. Find: the distance r between the spheres. Solution. We have point charges and hence Coulom b’s law is applicable. Then the distance between the charges is QiQ2 ■ = - Y :4jie 0e F *
S u b stitu tin g the num erical values, we obtain r _ i / ~ 9 x l 0 9 m / F X 6 X 10~9x 6 x l 0 ~ 8 C2 _ ? 0 X 2.5x2.5x10-4 N —
10. 2 in.
Answer. The separation between the spheres is 7.2 cm. Problem 47. A 2-g sphere bearing a charge of 3 x 10 ~8 C is suspended in air on a silk thread. D eterm ine the tension in the thread when another sphere bearing a like charge of 2.4 x 10 ~ 7 C is placed 10 cm beneath the first (Fig. 15a). Given: m — 2 X 10 “ 3 kg is the mass of the sphere, Ql = 3 x 10 -8 C is the electric charge of the sphere, r = 10-1m is
§ 9. E lectr ic Field
91
the distance between the spheres, and Q 2 = 2.4 x 10 -7 C is the electric charge of the second sphere. From tables, we take the p e rm ittiv ity of air, e = 1 , and the free fall accel eration g = 9.81 m /s2. jFind: the tension F t of the thread. Solution. The sphere on the thread experiences three forces: the force of g ravity G = m g , the Coulomb force of repulsion F=
1
QiQz
4JIC.I
er2
’
and the tension F t in the thread (Fig. 156). Since the spheres 10
tb]
*5
,K
L/ *t\
C
Fig. 16
are at rest, the equation for the forces acting on the first sphere has the form F t + F — G = 0. Hence Ft = G - P ,
.
S u b stitu tin g the num erical values, we obtain F t = 2 x 10~3 kg x 9.81 m /s 2 - 9 x 109m/F X
3 x 1 0 -8 C X 2 .4 X 1 0 -7 C
10-2 m 2
^ 1.31 x 10“2N.
Answer. The tension in the thread is approxim ately 1.3 x lO " 2 N. Problem 48. E lectric charges Qx = 4.8 x 10 " 7 C, Q 2 = Q 3 = 1.6 x lO " 7 C, and
92
Ch. II. Fundam entals of E lectrodynam ics
Given: Qx = 4.8 x 10-7 C, Q2 = Q 3 = 1.6 x 10~7 C, and QA = —1.6 x lO ~ 7C are the charges producing the field, r == 2 x 10"2 m is the radius of the circle. From tables, we take the p e rm ittiv ity of vacuum , e =- 1. F ind: the electric field strength E and the electric poten tia l (p at point O. Solution. Since the m edium is not specified in the problem , we solve it for vacuum . Each of the charges produces at point O a field of strength E x, E 2, E 3, and E 4 respectively. The electric field vector E at O is equal to the geom etric sum of the field strengths due to the individual charges: E = Ej -j- E 2 ~r E 3 -f- E 4. The electric field strength E Y due to charge Qx can be calcu late d by the form ula
J?i = 9 x 10®m
/
F
= 10.8 x 10® V/m.
Since the charges Q2, Q3, and QA are equal in m agnitude and are at the same distance from point 0 , we can w rite E 2t = E da = E 4k = y4jie0 ^ - -er2 ^ ,’ £ 8 = £ 3= £ 4 = 9 X 109in/F
= 3' 6 X 106 V /m '
In order to determ ine the resu lta n t field stren gth, we first add the vectors directed along the same stra ig h t line (Fig. 16b) E 1 = E i — E 2 = 10.8 x 10® V/m - 3.6 x 10® V/m = 7.2 X 10® V /m , £•„ = E 2 + E t = 3.6 x 10® V/m + 3.6 x 10® V/m = 7.2 x 10® V/m . The required re su lta n t vector E can be found using the parallelogram rule (Fig. 16c). In the case under considera tion, we can use th e P ythagorean theorem since we have a right-angled triangle: E = V 2(7.2 x 10® V/m)? = 10-2 x 10® V/m,
93
§ 9. E lectric Field
The electric potential is a scalar q u an tity . Therefore, the potential of the resultant field produced by the charges Q2» Q3, and QA is equal to the algebraic (and not geom etric as for the field strength) sum of the potentials of the fields pro duced by all the charges:
9 X 108m /F
= 7 -2 * 104 V,
tp4= - 7 . 2 x 104V. The potential at point O is =
+
^3 ~
94*
Since cp3 and (p4 are equal in m agnitude, we have
cp — tpi -f- cp2?
~ 21.6 x 104 V -f- 7.2 X 104 \
= 2.88 x 105 V. Answer. The field strength at the centre of the circle is ap proxim ately 107 V/m , and the field potential is 2.9 X 105 V. Problem 49. Two point charges of 2.64 x 10~8 and 3.3 X 10”9 C are in vacuum at a distance of 0.6 m from each other. W hat work m ust be done to bring the charges closer to 25 cm? Given: Qx = 2.64 x 10~8 C, Q 2 = 3.3 X 10‘ 9 C are the electric charges, rj = 0.6 m is the in itia l distance between the charges and r 2 = 0.25 m is the separation between the charges after they are brought closer. From tables, we take the electric constant e0 = 8.85 X 10“12 F/m and the perm it tiv ity of vacuum , e = 1. F in d : the work A required to bring the charges closer to gether. Solution. We assume th a t the charge Qx produces an elec tric field and the charge Q 2 is moved in the field. Then the work of external forces done to bring the second charge closer to the first is A = Qs (
94
Ch. II. Fundam entals of Electrodynam ics
where (px and (p2 are the electric potentials of the points be tween which charge Q 2 is moved: y w _ ffi — _____ 2.64 x l 0~8 C ” 411606^ 4ji8.85x10-12 K/mx0.6 m ’ (p = —_____ 2.64xlQ~8 C 95Q y 4jie0er2 4^8.85 X10"12 F/mx 0.25 m # We can now find the work: A = 3.3
x 10-,J C (396 V - 950 V) ~ - 1.83 x 10"0 J.
The m inus sign indicates th a t the work is done against the field forces. Answer. The work done to bring the charges closer is 1.83 x 10-6 J. Problem 50. A dust particle w ith mass 10"7 g is suspended between the plates of a p arallel-plate air capacitor to which a voltage of 500 V is applied (Fig. 17). ” The separation between the plates is 5 cm. D eterm ine the electric charge 011 the dust I particle. Given: m — 10-10 kg is the mass of ~ the dust particle, U = 500 V is the Fig. 17 voltage across the capacitor plates, and d = 5 x 10"2 m is the distance between the plates. From tables, we take the free fall acceleration g = 9.8 m /s2. Find: the charge Q of the dust particle. Solution.The dust particle in the uniform field of the capac ito r is acted upon by the force of g ravity G = mg directed downwards and by the force F = QE exerted by the electric field in the upw ard direction. The particle is in equilibrium provided th a t these forces are equal: G = F, or mg = QE. Hence Q = mg/E. Using the well-known relatio n E = U/d between electric field strength and voltage, we obtain mgd U 5
ly
10”10k g x 9 .8 m /s2X 5 X 10- 2 m 500V
the
ja -is r
Answer. The charge on the dust particle is approxim ate 1 0 - 13 C.
§ 9. Electric Field
95
Problem 51. A voltage of 90 V is applied to a parallelplate air capacitor w ith a plate area of 60 cm2. The charge on the capacitor becomes 1 0 '9 C. D eterm ine the capacitance of the capacitor, the energy stored in it, and the separation between the plates. Given: S = 6 x 10’3 m 2 is the area of a plate, U = 90 V is the voltage across the plates, and Q = 10~9 C is the charge on the capacitor. From tables, we take the electric constant e 0 = 8.85 X 10-12 F/m , and the p e rm ittiv ity of air e = 1. Find: the capacitance C of the capacitor, the energy W stored in the capacitor, and the separation d of the plates. Solution. Using the form ula C = Q/U, we determ ine the capacitance of the capacitor: ^
10' ’ C 90 V
~ 1 .1
X
10-“ F ~ 11 pF.
In order to determ ine the energy stored in the capacitor, we can use any of the following three form ulas:
T his gives w = 10-9Cx90V = 4
-
x
10_8 J
The separation between the capacitor plates can be deter mined from the form ula for the capacitance of a parallel— ee0 S/d. Since e = 1, we obtain plate capacitor, C ■ C — E0 S /d . Hence eoS a -~ C ~ '
r l _ 8 . 8 5 x l 0 - 12F / m x 6 x l 0 - 3 m 2 ^ (l1.1 X 10-11F
r
1 0 -3 m m-
Answer. The capacitance of the capacitor is approxim ately 11 pF, its energy is 4.5 X 10“8 J, and the separation be tween the plates is about 5 mm. Problem 52. D eterm ine the capacitance of the capacitor bank shown in Fig. 18a if CY = 1.5 jjlF, C 2 = 2 jiF, C 3 = 3 |iF, C 4 = 4 fiF, and C5 = 2 p,F. W hat energy is stored in the bank if the voltage applied to it is 500 V? Given: Cx = 1.5 pF, C 2 = 2 pF, C 3 = 3 pF, C 4 = 4 pF, and C 5 = 2 pF are the capacitances of the capacitors, and U = 500 V is the voltage across the bank.
96
Ch. II. Fundam entals of Electrodynam ics
F ind: the capacitance C of the bank and the energy W of the electric field of the capacitor bank. Solution. In the problem , it is convenient to express the capacitances of the capacitors in m icrofarads. The capacitors m
Cl
lb) C, +o-
C2
+° — l h
(C)
Ct
+ o-
-oF ig. 18
C2, C 3, and C 4 are connected in series. They can be replaced by a single capacitor having an equivalent capacitance (Fig. 186). For series-connected capacitors, we have c t ^ Ca ^ Ct ’
Cl
Cf = T li7 ;^ " 3 iIf,+ 4 7 F ’ C1 = °-9 2 M'F The capacitors C5 and Cl are connected in parallel. There fore, th eir equivalent capacitance is Cn =
c5 +
Cn C „ = 2 pF + 0.92 pF = 2.92 pF ~ 3 pF.
As a resu lt, we obtain the capacitors Ci and ClY in series (Fig. 18c). The capacitance of the bank can be found from the form ula _L_ J L _ L c = -e r& T 7 ' The energy stored in the bank can be determ ined from the form ula W.
cm 2
, w
l x i c r 6 F x 2 5 x l0 4 V2
1.3 x 10"3 J.
Answer. The capacitance of the bank is 1 pF and the ener gy is approxim ately 1.3 X 10"3 J. Problem 53. An electron flies from point A to point fi, the p otential difference between these points being 100 V.
§ 9. Electric Field
97
W hat is the velocity acquired by the electron a t point B if the velocity a t point A is zero? The charge-to-m ass ratio for electrons is 1.76 x 1011 C/kg. Given: U = 100 V is the poten tial difference between points A and B , e / m e = 1.76 X 1011 C/kg is the charge-tomass ratio for the electron. Find: the velocity v of the electron at point B. Solution. The work done by the field in displacing the electron is A = e U . This work is converted in to the kinetic energy of the electron 2?k = m ev2l2. From the energy con servation law , we have e U = m ev2/ 2 , whence
v = 1 /2 X 100 V x 1.76 x 1011 C/kg = 5.9 x 106 m/s. Answer. The velocity of the electron a t point B is ap proxim ately 5900 km/s. Questions and Problems Conservation of electric charge. Coulomb's law 9.1. A drop of oil has an electric charge of —3.2 X 10”19 C. D eterm ine the num ber of excess electrons on th e drop. 9.2. Three electrons are m issing from an electrostatically charged body. D eterm ine the m agnitude and sign of the charge on the body. 9.3. Can an electric charge of one sign be obtained as a result of electrostatic charging by friction? 9.4 . Under w hat conditions can a brass rod be charged? 9 .5 . An electric charge on a conducting sphere has to be divided into three equal parts. How can th is be done? 9.6. W hy is a m etal chain th a t reaches the ground fixed to a lorry for transp o rtin g petrol? 9 .7 . Can a positive charge be obtained on an electroscope using a negatively charged ebonite rod? 9 .8. W hat will happen to th e surface charge density on a m etal sheet rolled in to a cylinder? 7—0530
98
Ch. II. Fundam entals of Electrodynam ics
9 .9. An elder ball is tied to a silk thread. W hat w ill hap pen when an electro statically charged rod is brought close to it? 9.10. W ill the force of interaction between two point charges change if the m agnitude of each charge and the separation between charges are halved? 9.11. Two identical conducting spheres bearing elec tric charges of 3.2 X 10-19 and —3.2 X 10-19 C are brought in contact. W hat are the new values of charge on the spheres? How m any electrons have passed from one sphere to another? 9.12. W hat is the force of interaction between two point charges of 12 nC each separated in vacuum by 3 cm? By w hat factor w ill the force of interaction be reduced if the charges are placed in w ater? 9.13. Two identical point charges in glycerol 9.0 cm ap art in teract w ith a force of 1.3 X 10“5 N. D eterm ine the m ag nitudes of the charges. 9.14. W h at is the force of in teraction between two 1-C charges 1 m apart (a) in vacuum and (b) in kerosene? 9.15. At w hat distance would two 1-C charges in teract in vacuum w ith a force of 1 N? 9.16. Using Table 14, answer the following questions: (1) W hat is the ratio of the forces of in teractio n between electric charges in vacuum and in m ica? (2) In w hat m edia is the force of interaction between charges equal to half th e force of th eir in teraction in vacuum ? (3) To w hat m edium should charges be transferred from vacuum so th a t th eir interaction is reduced by a factor of 81? 9.17. W ill the electrostatic interaction force between two charges change if they are set in ice instead of vacuum ? 9.18. Two m ercury drops on glass of mass 20 g each and 4.0 cm ap art receive charges of —6.0 x 10~8 and 2.0 x 10‘ 7 C. In w hat direction and w ith w hat acceleration w ill they s ta rt to move? W ill the m otion be uniform ly accelerated? G rav itatio n al forces should be neglected. 9.19. W ith w hat force is an electron m oving in an orbit of radius 5.0 x 10"11 m a ttrac ted to a helium nucleus? The charge of the nucleus is 3.2 X 10“19 C. The orbit should be assumed to be circular. 9.20. Two little spheres having a mass of 1 g each and suspended in vacuum from the same point on silk threads
§ 9. Electric Field
99
have equal negative charges. They repel each other and move ap art to 12 cm, form ing an angle of 22°. D eterm ine the num ber of electrons on each ball and the tension in the threads. Show the forces acting on the balls on a diagram . W ill the tension in the threads change if the charges in te rac t in zero-gravity? 9.21. Two point charges of 5.0 x 10~9 and 1.5 x 10-8 C are 4.0 cm ap art in vacuum . D eterm ine the force w ith which these charges will act on a third charge of 1.0 X 10”9C, located at the m idpoint of the line connecting the charges. 9.22. A charge of 1.57 x 1 0 '8 C is transferred to a m etal sphere of radius 5 cm. W hat is the surface charge density on the sphere? 9.23. The surface charge density of a conducting sphere is 5 x 10"5 C/m 2. D eterm ine the m agnitude of the charge on the sphere if its radius is 8 cm. 9 .2 4 . Two electric charges, one of which is tw ice the o th er, in te rac t at a distance of 0.60 m in vacuum w ith a force of 2.0 m N. C alculate the m agnitudes of charges. A t w hat distance in kerosene will the in teractio n between the charges be the same? 9.25. W hat is the ra tio of the electrostatic force of rep u l sion between two electrons to th e ir g rav itatio n a l attrac tio n ? 9.26. D eterm ine the p e rm ittiv ity of kerosene if two equal like charges in te rac t in vacuum w ith th e same force at a dis tance of 0.283 m as in kerosene a t 0.20 m. Assum ing th a t the force of in teractio n in kerosene is 3.0 X 10”2 N, deter mine the m agnitudes of the charges. Electric field strength 9.27. Do the electric field vector and the vector of the force exerted on a charge by an electric field always have the same directions? 9.28. W hy is a m etal cap som etim es put on a vacuum tube? 9.29. Can electric charges be separated on (a) a conductor, (b) a dielectric? 9.30. A cylindrical conductor is attached to a conical conductor w ith the same base area. W hat can be said about th e electric field strength near various points on the surface of the resu lta n t conductor? 7
100
Ch. II. Fundam entals of Electrodynam ics
9.31. W hy is a ball usually fixed to the rod of an electro scope? 9.32. Two electric charges of the same m agnitude and sign produce an electric field. W hat is the field strength at the m idpoint of the straig h t line connecting the conductors? 9.33. The electric field strength at a given point is 300 V/m . W hat does this mean? 9.34. D eterm ine the electric field strength at a point where a force of 5 mN acts on a charge of 0.7 X 10~6 C. 9.35. W hat is the force exerted by an electric field on a charge of 3.2 X 10-8 C at a point where the field strength is 500 V/m ? 9.36. The electric field strength near the surface of the E a rth before a stroke of lig h tning is 2 X 105 V/m . W hat is the force exerted by the field on an electron? 9.37. An electric field is produced in vacuum by a point charge of 7.5 X 10‘ 8 C. D eterm ine the field stren g th 15 cm away from the charge. 9.38. A point charge of 2.2 nC produces a field w ith a stren g th of 2.5 kV/m at 6.0 cm from the charge. D eterm ine the p e rm ittiv ity of the m edium. 9.39. A vessel of kerosene contains a conducting sphere w ith a negligibly sm all size and a charge of 8.0 x 10~8 C. D eterm ine the electric field strength 5.0 cm from the charge. W hat w ill be the change in the field stren g th at th is point if the kerosene is let out of the vessel? 9.40. The electric field strength 5.0 cm from a charge is 1.5 x 105 V/m . W hat is the electric field strength 10.0 cm from the charge? D eterm ine the m agnitude of the charge. 9.41. How m any excess electrons are contained on a dust particle acted upon by a force of 2.4 X 10"10 N in an electric field of strength 1.5 x 105 V/m ? 9.42. An electric field is produced by a point charge. W hat are the loci of the points where the m agnitude of the electric field strength is the same? 9.43. W ill an electric field rem ain uniform if a conducting sphere is placed in it? 9.44. Two identical point charges of 1.0 X 10“8 and 2.0 x 10‘ 8 C are in vacuum 20 cm ap a rt and produce an electric field. D eterm ine the field strength a t the m idpoint of the line connecting the charges. W hat w ill the electric field strength be if the charges are unlike?
§ 9. Electric Field
101
9.45. Two conducting spheres having diam eters of 10.0 and 4.0 cm are 120.0 cm ap a rt and bear charges of 3.0 x 10“® and 2.0 X 10"6 C respectively. D eterm ine the electric held strength at the m idpoint O of the stra ig h t line con necting the spheres (Fig. 19). 9.46. Two like electric charges of 7.0 X 10~8 C each are at points A and B (Fig. 20). D eterm ine the electric field
60
60 &
0 F ig. 19
stren g th at point O which is the apex of the rig h t angle AOjB; A O = BO = 5.0 cm. 9.47. D eterm ine the electric field strength of the field at a point three radii from the surface of a charged conducting sphere. The surface charge density on the sphere is 1.6 X 10“7 C/m2. 9.48. A conducting sphere of radius R is electro statically charged to a surface density a. W hat will be the field strength (a) at the centre of the sphere, (b) at h alf the radius of the sphere, (c) on the surface of the sphere? 9.49. A drop bearing a charge of 2 X 10~8 C is in equi librium in a uniform electric field of strength 49 V /m . D eter m ine the mass of the drop. 9.50. W ith w hat acceleration w ill a 10-g ball w ith charge 1.0 X 10“5 C fall in the electric field of the E a rth ? The elec tric field strength near the surface of the E a rth is 130 V /m . 9.51. A th in conducting ring of radius R has an electric charge Q. D eterm ine the electric field strength at the cen tre O of the ring and at point A (Fig. 21). 9.52. W hy is the uniform ity of the electric field of a charged p arallel-plate capacitor violated if an uncharged m etal sphere is placed between its plates?
102
Ch. II. Fundam entals of Electrodynam ics
Electric potential. Potential difference. Work done in an electric field 9.53. Considering the E a rth to be a sphere of radius 6400 km, determ ine its electric charge and poten tial if the electric field strength produced by the E a rth near the surface is 130 V/m. 9.54. The work done in displacing a 2.0 x 10~8-C charge from infinity to a point in a field is 1.13 x 10~4 J. W hat is the electric potential at this point? 9.55. D eterm ine the potential difference between two points in a field if a work of 3.0 x 10~5 J has to be done in m oving an 8.0 X 10"7-C charge between these points. 9.56. An electric field moves a positive charge of 3.0 x 10"7C between two points w ith potentials 200 V and 1200 V. W hat is the work done by the field in this case? 9.57. An electric field is produced by a point charge of 4 X 10~8 C. W hat is the electric potential 6 cm away from this charge? W hat work m ust be done against the field in order to bring a positive charge of 1 C from infinity to th is point? 9.58. Two point electric charges of 1.0 X 10"5 and 6.0 X 10"6 C are 20 cm ap art in air. D eterm ine the electric poten tia l at the m idpoint of the straig h t line connecting the charges. 9.59. A 1.6 X 10“7-C charge is moved 3.0 cm along a field line in a uniform electric field having a strength of 5.0 X 103V/m. D eterm ine the work done and the p o tential differ ence for two points between which the charge is moved. 9.60. As an electron passes between two points in an electric field, its velocity increases from 2.0 X 106 to 3.0 X 107 m/s. W hat is the potential difference between these points? W hat is the increase in the kinetic energy of the electron? 9.61. Two 8-nC charges are located at two apexes of an equilateral triangle having a side of 6.0 cm. D eterm ine the field strength and the poten tial at the th ird apex. 9.62. Two conducting spheres of radii 2.0 and 3.0 cm are charged to 30 and 40 V respectively. W hat w ill be the elec tric p otential of the spheres after they have been connected by a wire? Assume th a t the separation between the spheres is large compared to their radii.
403
§ 9. Electric Field
9.63. An electron having a velocity of 6.0 X 407 m /s flies into a p arallel-plate air capacitor m idw ay between the plates. W hat m ust the voltage across the plates be for the deviation of the electron to be m axim um ? The plates are 40.0 cm long and 3.0 cm apart. 9.64. An electron w ith a velocity of 1.6 x 106 m /s flies into a uniform electric field of strength 90 V /m and moves along a field line u n til it comes to a h alt. How long and how far does it fly in the field? Assume th a t the electron mass is 9.0 X 40~31 kg.
Capacitance. capacitor
Capacitors.
The
electrical
energy
in a
9.65. Can the potential of a charged conductor be changed w ithout changing its charge? 9.66. A p arallel-plate air capacitor of capacitance C is immersed in a m edium w ith a p e rm ittiv ity of 2. W hat will the capacitance of the capacitor be? 9.67. The capacitance of a spherical conductor is propor tional to its radius. W hat m ust the radius of a conducting sphere be for its capacitance in vacuum to be 1 F? W hat is the ratio between the radius of such a sphere and the E a rth ’s radius? 9.68. A p arallel-plate air capacitor is connected to a constant-voltage source. W hat will be the change in the capaci tance and the energy of the capacitor, the voltage across it, and the charge on the capacitor plates if the separation between the plates is reduced? 9.69. D eterm ine the capacitance of an isolated conducting sphere of radius 5.0 cm imm ersed in kerosene. Express the answer in farads, m icrofarads, and picofarads. 9.70. Assum ing the E a rth to be a sphere of radius 6400 km , determ ine its capacitance. 9.71. An isolated conducting sphere whose capacitance is 5.0 pF is electro statically charged to a potential of 4570 V. D eterm ine the radius of the sphere and the surface charge density on it. 9.72. W hy do electrolytic capacitors have a large capa citance? 9.73. Two conducting spheres having a charge of 4.0 X
104
_________ Ch. II. Fundam entals of Electrodynam ics
10“8 C each are arranged a long way apart. The capacitance of the larger sphere is 2.2 X 10-11 F, and th a t of the sm aller sphere is 5.6 X 10"12 F. W hat are their potentials? W hat will happen when they are connected by a conductor? 9.74. C alculate the capacitance of a parallel-plate capaci tor m ade from tin foil plates 15 cm2 in area w ith a mica dielectric layer 0.8 mm thick. The p e rm ittiv ity of mica is e = 6. 9>75. A parallel-plate air capacitor consists of two plates 100 cm* each in area. W hen a charge of 6.0 X 10~9 C is tra n s ferred to one of the plates, the capacitor is charged to a vol tage of 120 V. D eterm ine the separation between the plates. 9.76. The plates of a p arallel-plate air capacitor have an area of 62.3 cm2 each and their separation is 5 mm. D eterm ine the charge of the capacitor if the potential difference across its plates is 60 V. 9.77. The separation between the plates of a parallelplate air capacitor is 0.1 cm and the area of each plate is 200 cm2. The potential difference across the plates is 600 V. W hat charge is stored in the capacitor? W hat will be the change in the voltage if the space between the plates is filled w ith m ica, whose p e rm ittiv ity is 6? 9.78. A parallel-plate air capacitor w ith a plate separa tion of 1.5 mm is charged to a voltage of 150 V. How far apart should the plates be moved in order to increase the voltage to 600 V? 9.79. D eterm ine the capacitance of a capacitor in which nine mica plates having a thickness of 0.12 mm and an area of 12.56 cm2 are interleaved between tin foil plates. The p e rm ittiv ity e of mica is 6. 9.80. D eterm ine the energy of a parallel-p late paper ca pacitor w ith a plate area of 600 cm2. The charge on the capacitor is 2 X 10-7 C and the dielectric is wax paper 2.0 mm thick. 9.81. Three capacitors whose capacitances are 4, 2, and 6 fxF are connected to form a bank and to a constant-voltage source of 200 V. D eterm ine the capacitance and the energy of the bank when the capacitors are connected (1) in series and (2) in parallel. 9.82. D eterm ine the capacitance of a capacitor bank connected as shown in Fig. 22 if Cx = 1.2 pF and C 2 = C3 - 0.6. pF.
§ 9. Electric Field
105
9.83. D eterm ine the capacitance of a bank of capacitors connected as shown in Fig. 23. The capacitance of all the capacitors is the same and equal to 0.6 pF. D eterm ine the electric charge stored in the bank if a voltage of 100 V is applied to it. 9.84. C apacitors are connected as shown in Fig. 24. Given: Cx = C 2 = 2 pF, C 3 = C4 = C5 = 6 pF. Determ ine
F ig . 22
the potential difference between points A and B if the energy stored in the bank is 1.35 X 1 0 '1 J. 9.85. An uncharged 100-pF capacitor is connected in parallel to a 50-pF capacitor charged to a voltage of 300 V. Cf <-/
r—
<-2
II-------------IH
HI — II— II— G G G F ig . 24
— ii— 3
C2
c,
HI— IH o i!> F ig . 25
W hat is the voltage across the capacitors? W hat is the charge distrib u tio n between them ? 9.86. A fter a capacitor having an unknown capacitance and charged to a voltage of 600 V is connected in parallel to an uncharged 5-pF capacitor, the voltage across the bank drops to 100 V. W hat is the capacitance of the first capacitor? 9.87. Two capacitors having different capacitances are connected in parallel and to a constant-voltage source. W hich of the capacitors will accum ulate more energy? 9.88. A pulsed flash for photography is obtained by discharging a 800-pF capacitor for 2.8 ms at a voltage of
100
Ch. II. Fundam entals of Electrodynam ics
300 V across the capacitor. Determ ine the energy of the flash and the mean power. 9.89. Three capacitors are connected to make a bank as shown in Fig. 25 and then to a 200-V constant-voltage source. D eterm ine the capacitance of capacitor Cx and its charge if the capacitance of the bank is C = 6 p,F and the capacitances of the rem aining capacitors are C 2 = 2 fiF and C 3 = 3 fiF. P.90. The force w ith which the plates of a p arallel-plate capacitor a ttra c t each other is determ ined from the form ula F = Q2/(2e 0eS). Is the energy required to move the plates apart higher (1) when the capacitor rem ains connected to the source or (2) when it has been charged and then before the plates are moved apart disconnected from the source? 9.91. A spherical capacitor consists of two conducting concentric hollow spheres. If the difference between the radii of the spheres is sm all, the capacitance of the capacitor can be calculated using the form ula for a p arallel-plate capaci tor. W rite this form ula in a general form and determ ine the capacitance of an air capacitor formed by spheres of radii 5.00 and 4.95 cm. 9.92. D eterm ine the volume energy density of the uniform electric field in a mica capacitor charged to a voltage of 90 V. The separation between the plates is 1.0 mm, and the p e rm ittiv ity of mica should be taken to be 6. § 1C. ELECTRIC CURRENT IN METALS. OHM’S LAW. ELECTRIC RESISTANCE
Basic Concepfs and Form ulas An electric current is an ordered (directional) m otion of charged particles (in m etals, these particles are free electrons). The current / in a conductor is the am ount of electricity Q flowing per second through a cross section of the conductor: / = Q/t, or / = envS, where n is the num ber density of the charge carriers (e), v is the mean velocity of the charges, and S the cross-sectional area of the conductor. C urrent is a base q u a n tity in SI, its u nit being the ampere (A). The direction of current is assumed to be opposite to the direction of m otion of electrons.
§ 10. Electric Current in Metals. Ohm’s Law
107
In order to create a direct current in m etals, an electric field m ust act on the free electrons. The field m ust ensure a constant potential difference across the ends of the conduc tor (circuit). In a current source, the extraneous forces pro duce an excess of electrons at the negative term inal and an electron shortage at the positive term inal. In other words, a p otential difference is created. Each current source is characterized by an electrom otive force (einf) % which is equal to the work done by extraneous forces in m oving a positive charge of 1 C along the circuit: 8 = A tJQ . The emf u n it is the volt (V). Ohm ’s law for a conductor establishes a relation between the current flowing in it and the voltage across its ends:
where the p ro portionality factor 1/7? is called the electric conductance and R the electric resistance of the conductor. The u n it of resistance is the ohm (Q). The resistance of a conductor depends on its size, m aterial, and tem perature: R = p ± , R t = R 0 (l + aAT), where p is the resistiv ity in £2-m, and a is the tem perature resistance coefficient in K ”1: a
Mi
R0*T •
For m etal conductors, a is positive. At tem peratures close to absolute zero, superconductivity is observed for some conductors. This is a sta te in which the resistance ab ru p tly drops to zero. Ohm ’s law for a circu it establishes a relatio n between the current, electrom otive force, and the to ta l resistance of the circuit: / =
R+ r ’
where R and r are the resistances of the external part of th circuit and of the current source.
108
Ch. II. Fundam entals of Electrodynam ics
The voltage U across the term inals of a current source w ith a closed circuit is sm aller than the emf by the voltage drop w ithin the source itself: U = g - Ir. Short circuiting is observed when the resistance of the external p art of the circuit is negligibly sm all, while the current a tta in s its m axim um value. Using Ohm ’s law for a closed circuit, we can determ ine the current during short circuiting: /s h .c
=
S /r .
In d iv id u al parts of a circuit (resistors) can be connected in series or in parallel. For a connection in series, the resistors are connected one after another so th a t the current in all the subcircuits is the same, while the to ta l, or equivalent, resistance of the circuit is given by R*cr — R\ ~T # 2 “ • • • + H u ll jRjl = J?2 = . . . = R n - R ,
we have
-^ser ~ Rn» The voltage drop for a series connection is proportional to the resistances: Ul = B± U2
R2 *
For a connection in parallel, the voltage across all the parallel branches is the same, while the current in a branch is determ ined by its resistance: J l = R2_ h *1 # The to ta l, or equivalent, resistance is then given by the form ula 1 1 , 1 . , 1
§ 10. Electric Current ip M etals. Ohm's Law
109
To m easure the current, ail am m eter is connected in series to the circuit. The resistance of the instrum ent should be very low. In order to change the value of a scale division on the am m eter, a bypass (shunt) resistor R sh is connected in parallel w ith the instrum ent. The resistance of the shunt is lower than the resistance of the am m eter by a factor of n — l :
where n is the num ber by which the scale of the instrum ent is m ultiplied (Fig. 26). To measure voltage, a voltm eter is connected in parallel to the region of the circuit across which the voltage is being
-i|r *S
F ig. 20
W
F ig. 27
measured. The resistance of the voltm eter m ust be very high. If the voltage to be m easured exceeds the range of a voltm e ter, a m u ltiplying coil (series resistor) R s is connected in series w ith the instrum ent (Fig. 27): R s = R y (n — 1), where n is the num ber by which the range of the v oltm eter is increased, and R v is the resistance of the voltm eter. Sources of electric power (current sources) can be connect ed to form batteries. For a series connection, the positive term inal of the first source is connected to the negative ter m inal of the second source, and so on. The current in this case is fl + n r ’
where n is the num ber of identical current sources in the battery. For a parallel connection, all the positive term inals are connected in one junction and the negative term inals form
110___________ Ch. II. Fundam entals of Electrodynam ics
the other junction. The current in such a battery is R-{-r/n
W hile solving problem s in which Ohm’s law is applied to branched circuits, it is necessary (1) to choose a rb itra rily the directions of the currents and the direction of circum vention of subcircuits and indi cate them on a circuit diagram ; (2) to w rite equations for the currents at the junctions, the num ber of equations being one less than the num ber of junctions, the algebraic sum of currents at a junction is alw ays equal to zero if the currents flowing to a junction and from it are taken w ith opposite signs; and (3) to w rite the equations for all the closed subcircuits considering th a t the algebraic sum of e m f s in any closed subcircuit is equal to the algebraic sum of the voltage drops; if the potential increases in the direction of circum vention (we move from “m inus” to “plus”), the emf is assumed to be positive, otherwise, the emf is assumed to be negative. The voltage drop is assumed to be positive if the direction of the current coincides w ith the chosen direction of circum vention; otherwise, it is taken w ith the m inus sign. The to tal num ber of equations m ust be equal to the num ber of unknowns. W orked Problems Problem 54. A voltage of 3.6 V is applied to the ends of a steel conductor 20 m long. D eterm ine the m ean velocity of ordered m otion of the charge carriers in the conductor if their num ber density is 4.0 X 1028 m~3. Given: U = 3.6 V is the voltage drop across the conductor, n = 4.0 x 1028 m “3 is the num ber density of m obile charge carriers, and I = 20 m is the length of the conductor. From tables, we find the electron charge e = 1.6 X 10~19 C and the resistiv ity of steel, p = 1.2 X 10'7 Q -m . Find: the mean velocity v of the charge carriers. Solution. The m obile charge carriers in m etals are free, or conduction, electrons. Consequently, we m ust find the mean velocity v of the ordered m otion of free electrons.
111
§ 10. Electric Current in M etals. Ohm's Law
In the electron theory of conductivity, Ohm ’s law is for m ulated as follows: / = envS, whence
V
cnS
The current can be determ ined from the form ula I = U /R, where R should be expressed in term s of the length and the cross-sectional area: R = pUS. S u b stitu tin g the expression for current into the form ula for velocity, we obtain vs
p le n S
V —
’
»
p le n ’
3.6 V 1 .2 x l0 “7 Q - m x 2 0 m X l.6 x l0 " 19 C x 4 .0 x l 0 28m~3
2.343 x 10“4 m/s. Answer. The mean velocity of ordered m otion of the free electrons is approxim ately 0.23 m m /s. Problem 55. A telegram is sent from Leningrad to Moscow (the distance between the cities is 650 km) via a steel tele graph line carrying a current of 1.7 mA at a voltage of 150 V. The cross- Q ^ sectional area of the wire is 5.0 mm2. ~7 D eterm ine the voltage drop in the wires and the voltage at the receiving end (Fig. 28). 0--------Given: I = 6.5 X 105 m is the dis tance between Leningrad and Moscow, 28 / = 1.7 x lO " 3 A is the current in the wires, Ux = 150 V is the voltage at the beginning of the transm ission line, and S = 5 X 10-6 m2 is the cross-sec tional area of the wire. From tables, we find the resistiv ity of steel, p = 1.2 x 10“7 £2-m. Find: the voltage drop U w in the wires and the voltage U 2 at the end of the line. Solution. The length of the wire form ing the electric cir c u it (which m ust be closed) is tw ice the distance between the cities.
5
112____________Ch. II. Fundam entals of Electrodynam ics
The voltage drop in the wires is f /w = / / ? w, where Rw = p Then £/w W p f , TT U
*
=
1.7x 10-3 Ax 1.2xlO_7Q -mx2x6.5xlOs m ,7 -------------------- 5xl0~6 m2--------------------- - 5 3 V -
The voltage at the end of the transm ission line is U2 = Ux -
£/w,
U 2 = 150 V - 53 V = 97 V.
Answer. The voltage drop in the wires is approxim ately 53 V, and the voltage at the end of the line is 97 V. Problem 56. W hat is the change in the resistance of a tele graph line due to the change from w inter tem peratures to sum m er tem peratures if the line is m ade from a steel wire having a cross-sectional area of 5.0 mm 2? The tem perature changes from —30 to +30°C. The length of the wire at 0°C is 200 km. The linear expansion of the wire should be neg lected. Given: S = 5.0 X 10~6 m 2 is the cross-sectional area of the wire, tx = —30°C is the w inter tem perature, t 2 = +30°C is the sum m er tem perature, and I = 2.0 X 105 m is the length of the wire. From tables, we find the resis tiv ity of steel at 0°C, p0 = 1.2 x 10“7 Q*m, and the tem perature resistance coefficient for steel, a = 0.004 K _1. Find: the change AR in the resistance of the telegraph line due to the seasonal change in tem perature. Solution. The resistance of the line at 0°C is R 0 = p0Z/ while the resistances at tx and t 2 are R x = p0 ^ (1 + octx) and R 2 -- p0 j (1 + oct2). The change in the resistance is given by AR = R 2 — R i, \ R = p0-^-(i + a t 2 — l — a t i) ==p0
—
Since the absolute values of the tem peratures ( |*i | = \ t 2 |), we can w rite
are equal
§ 10. Electric Current in M etals. Ohm’s Law
113
A R = p0^ - 2a
x 2 x 0.004 K‘ i x 30 K
~ 1.2 k£2. Answer. The resistance of the wire has increased by about 1.2 k£2 as a result of the change from w inter to sum m er tem peratures. Problem 57. C alculate the resistance of the circuit in Fig. 29a. Given: the resistances R x = 6£2, R 2 = 5£2, R 3 = 4 Q , i?4 = 12 £2, and R 5 = 8 £2 of the resistors. Find: the total (equivalent) resistance R of the circuit. Solution. W hil^ solving problem s where loads are con nected in series and in parallel, it is expedient to replace
a branched circuit by equivalent circuits as shown in Figs. 29b-e. Since resistors R 3 and R A are connected in parallel, their equivalent resistance is D
-^3-^4
l ~' Rs + Ri ’
D
4 Q x l2 Q
q
4Q +12Q
(Fig. 296). R esistors R x and R 2 are connected in series, and their equivalent resistance is i? IX = R 2 + R j, R u = 5 £2 + 3 £2 = 8 £2 (Fig. 29c). 8 —053 0
114___________ Ch. II. Fundam entals of Electrodynam ics
R esistors and R 5 are connected in p arallel, and hence th e ir equivalent resistance is D
n =
Rn + Rb >
8 Q X 8Q 8 Q+ 8 Q =
/
(Fig. 29 d). R esistance R ul can be determ ined in a different way: R u i = R u /2 = 4 £2. The required resistance /? is R =
-(- i?m ,
i? = 6 £ 2 + 4 £ 2 = 10 £2
(Fig. 29*). Answer. The to ta l resistance of the circuit is 10 {2. Problem 58. W hen a current source w ith an emf of 4.2 V is connected to a nickeline wire 10 m long and having a d iam eter of 1.0 mm, the current in the circuit is 0.6 A. D eter m ine th e internal resistance of the current source. Given: the emf % = 4.2 V of the current source, I = 10 m is the length of the nickeline wire, d = 1.0 X 10”3 m is the diam eter of the wire cross section, and I = 0.6 A is the current in the circuit. From tables, we find the re sistiv ity of nickeline, p = 4.2 X 10~7 Q»m. F ind: the internal resistance r of the current source. Solution. We shall solve the problem hy using Ohm ’s law for a closed circuit: R-
whence I The external resistance R can be determ ined from the for m ula R — (yl/S. Considering th a t S = ndVk, we obtain n
4pZ nd*~"
4 x 4 . 2 x l0 " 7 Q -m x l0 m _ r / n 3 .1 4 x l . 0 x l 0 - 6m2
Then -5.4 £2 = 1.6 £2. Answer. The in ternal resistance of the current source is 1.6 £2. Problem 59. A b a tte ry of cells w ith an emf of 3 V and an internal resistance of 0.25 £2 supplies power to a circuit
§ 10. E lectric Current in M etals. Ohm's Law
115
consisting of four resistors R t = i.O £2, R 2 = 3.0 £2, R 3 = 1.5 £2, and i?4 = 0.75 £2 connected as shown in Fig. 30a. D eterm ine the current in the unbranched su b circu it and the voltage drop in the battery. Given: g = 3 V is the emf of the b a tte ry , r = 0.25 £2 is the interna] resistance of the battery, and / f t = 1.0 £2, R 2 = 3.0 £2,/?3 = 1.5 £2, and i?4 = 0.75 £2 are th e resist ances of the resistors. Find: th e current / in the circu it and the voltage drop U int in the b attery . Solution. W e shall solve the problem w ith th e help of Ohm’s law for a closed circuit: / = g /(fl + r). Therefore,
we m ust determ ine th e external resistance R. For sim p licity , we represent the circuit as shown in Fig. 305. I t is clear from the circu it diagram th a t th e resistors R x and R 2 are connected in parallel. T heir equivalent resistance R t can be found from th e form ula n -ft 1^2 p 1.0 Q X 3 .0 Q a T tn 1 ^ T o q +3^o q ^ The subcircuit w ith resistance R x and resistor i?4 are con nected in series. T heir equivalent resistance R lx can be de term ined as follows: R n = f?j “I- R^i R n == 0.75 £2 “j- 0.75 £2 — 1.5 £2. R esistances i? „ and R s can be replaced by an equivalent resistance R which can be determ ined from the rules gov erning parallel connection of loads. Since these resistances are equal, we have R = R u/2 = 0.75 £2. 8*
116___________ Cfe. II. Fundam entals of Electrodynam ics
The current in the unbranched circuit is 3.0 V ^.75 Q + 0.25 Q
= 3 A.
The voltage drop in the current source can be determ ined from O hm ’s law for a conductor: Ujnt - / r , Uini = 3 A X 0.25 Q = 0.75 V. Answer. The current in the unbranched circuit is 3 A and the voltage drop in the current source is 0.75 V. Problem 60. A galvanom eter w ith an internal resistance of 19.8 Q can be used to m easure currents up to 10 mA. W hat m ust be done to extend its range to 1.0 A? How can it be used to m easure voltages up to 10 V? Given: R r = 19.8 Q is the resistance of the galvanom eter, I g = 0.01 A is the current through the galvanom eter, / = 1.0 A is the current in the circuit, and U = 10 V is the voltage to be measured. F ind: the resistance of the shunt /?sh and the resistance R s of the series resistor. Solution. If a galvanom eter is used as an am m eter (in this case it is connected in series to a circuit), a sh u n t (resistor) /?sh is connected in parallel to it as shown in Fig. 26. The resistance of the shunt can be calculated by using the rules of parallel connection: r j j Rg ^sh 7?g I — Ig ^ a J*b flsh “ /g ’ *sh Ig where n ~ = ^ - , *sh = ^
,
* sh = - ^ = 0 . 2 Q .
If a galvanom eter is used as a voltm eter (in th is case it is connected in parallel to the subcircuit in which the voltage drop U has to be m easured), a resistor R s is connected in series to it as shown in Fig. 27. The voltage drop U is d istrib u ted in proportion to the resistances R g and R s: U — Uv
Re
U
a
Rs.
§ 10. Electric Current in Metals. Ohm’s Law
1° v n
~
_
117
cn
0.01 AX 19.8
B s = 49 x 19.8Q ~ 970Q. Answer. In order to measure current, the galvanom eter m ust be shunted by a resistor w ith R sh = 0.2 £2. To m easure voltage, a series resistor w ith re sistance R s 970 £2 should be connected to the galvanom eter. Problem 61. How should two g a l vanic cells having an emf of 1.45 V each and an in tern al resistance of 0.4 £2 each be connected in order to obtain the m axim um current when the circu it is closed w ith an ex ter nal resistance of 0.65 £2? Given: n = 2 is the num ber of cells in the b attery , % = 1.45 V is the emf of a cell, r = 0.4 £2 is the internal resistance of a cell, and R = 0.65 £2 is the resist ance of the external circuit. F ind: /ser, the current for the series connection and the current / par for the parallel connection of the cells. Solution. In order to find out which is the best way to connect the circuit, we determ ine the currents for the cells in series and in parallel and com pare them : , y ser
/
par
n% T _ R + nr » ser
»_
R + r/n ’ J Par
2x1.45 V 0.65 Q + 2 x ll.4 Q
/
145 V
0.65 Q + 0.4 Q/2 “
OA ’
1 7A
Answer. I t is b etter to connect the cells in series. Problem 62. The circu it shown in Fig. 31 contains three re sistors /?! = 100 £2, i?2 = 50 £2, and R 3 = 20 £2 and. g al vanic cells w ith e m f’sjfi = 2 V and £ 2. The am m eter in d i cates a current of 50 mA. D eterm ine the currents in the re sistors and the emf of the second cell. The in tern al resist ance of the am m eter and of the cells should be neglected. Given: R x = 100 £2, R 2 = 50 £2, and R 3 = 20 £2 are the resistances of the resistors, = 2 V is the emf of the first cell, / A = 0.05 A is the reading of the am m eter.
118
Ch. II. Fundam entals of Electrodynam ics
Find: the currents I x, / 2, and I 3 in the resistors and the emf g 2 of the second cell. Solution. We choose the direction of the current a rb itra rily and indicate it by arrows on the circuit diagram . Since the algebraic sum of the currents flowing to and from a junction is zero (KirchhofTs first rule), we can w r i t e / x — / 2 — / 3 = 0, whence 13 = I\ / 2. (1) The current indicated by the am m eter is I A = / 3. Let us go around the circuit clockwise and indicate this direction on the circuit diagram . Since the algebraic sum of voltages in a closed circuit is equal to the algebraic sum of e m f s (K irchhofl’s second rule), we can w rite for the circu it A BCD FA — I \ R \ — 1 2^2 ~
or I
“I" ^2^2 =
(2)
for the circuit A F M N A I XR X + I 3R 3 =
r2.
(3)
We determ ine I x from E q .(l) and su b stitu te it into Eq.(2): A = ^ 3 + ^ 2 » (^3 + ^ 2) ^1 + ^ 2^2 = £lThis gives r _
r _
’
2“
2 V —0.n5Axl(MQ 1OOQ+ 50Q
_
a
no a
The m inus sign indicates th a t the current / 2 flows in the op posite direction to th a t indicated on the diagram . The current is I 1 = I 3 + I 2 = 0.05 A -
0.02 A = 0.03 A.
S u b stitu tin g I x into Eq. (3), we obtain g 2: g 2 = 0.03 A x 100 £2 + 0.05 A x 20 £2 = 4 V. Answer. The currents in resistors R x, /?2, and R 3 are 0.03, —0.02, and 0.05 A respectively, the emf of the second cell is 4 V.
§ 10. Electric Current in Metals. Ohm*s Law
119
Questions and Problems Current. Resistance. Ohm's law for a conductor 10.1. The anode current in a vacuum tube is 12 mA. How m any electrons arrive at the anode per second? 10.2. How m any electrons pass through the cross section of the contact wire of a tram in 2 s if the current is 500 A? 10.3. D eterm ine the current in the contact copper wire of a trolleybus circuit if the num ber density of conduction electrons in copper is 3 X 1023 cm "3, and the mean velocity of their ordered m otion is 0.25 m m/s. The cross-sectional area of th e wire is 85 mm 2. 10.4. D eterm ine the num ber density of conduction elec trons in copper if the current in a copper wire w ith a crosssectional area of 105 mm2 is 500 A, and the mean velocity of the ordered m otion of conduction electrons is 0.1 m m/s. 10.5. A voltage of 18 V is applied across the ends of a cop per wire 200 m long. D eterm ine the mean velocity of the ordered m otion of electrons in th e conductor if the num ber d ensity of conduction electrons in it is 3.0 X 1023 cm ’ 3. 10.6. W hat is the electric field strength in an alum inium wire of cross-sectional area 6 m m 2 for a current of 9 A if the p o ten tial difference between the ends of the wire is 21 V, the m obility of conduction electrons in alum inium is 7 X 10‘ 3 m2/(V -s), and the electron num ber density in it is 3.7 x 1022 cm "3? How long is the conductor? 10.7. The coil of a hot p late carries a current of 2.7 A w ith an electric field strength of 3.1 V /m . D eterm ine th e m obility of th e conduction electrons if th eir num ber density in th e coil is 1023 cm ’3, and the cross-sectional area of th e w ire is 0.10 m m 2. 10.8. D eterm ine th e curren t d en sity in a conductor if 0.15 C of electricity pass per second through a cross-sectiQnal area of 6 mm 2. Is the adm issible current density im p o rtan t when choosing the wire for a residential building? 10.9. A constant cu rren t of 3.6 mA is m aintained when gold p lating an article 1.2 x 15 cm2 in area. D eterm ine the current density. 10.10. W hat m ust the diam eter of a copper wire be if it is to carry a current of 1000 A and the adm issible current density is 2.5 A/m m 2?
120___________ Ch. II. Fundam entals of E lectrodynam ics
10.11. Determ ine the current density in the excitation w inding of the traction m otor of a locom otive if the crosssectional area of the w inding is 4.7 x 25 mm2 and the nom inal current is 725 A. 10.12. W hat is the resistance of an alum inium wire 2 m long and having a cross-sectional area of 1 m m 2? W hat w ill the resistance of the same length of wire w ith twice the cross-sectional area be? 10.13. Two wires, one of nickeline and the other of nichrome, have the same length and diam eter. W hich of them has a higher resistance? W hat is the ratio between th eir resistances? 10.14. An iron wire and a tungsten wire of the same length and diam eter are connected in turn to an accum ulator. In which wire w ill the current be stronger? W hat is the ratio of th e currents? 10.15. A long conductor was cut into two halves which were then tw isted along the whole length. How does the resistance of the wire change? 10.16. Determ ine the capacitance of a capacitor the v o lt age across whose plates is 1 V as a result of charging for 12 s w ith a current of 10‘ 6 A. 10.17. An electric circu it whose conductance is 2.4 x 10-2 S is connected to a d.c. source of 50 V. D eterm ine the current in the circuit and its resistance, 10.18. D eterm ine the length of a m anganin wire required to m ake a potentiom eter rated at a m axim um resistance of 1500 Q if the diam eter of the wire is 0.3 mm. 10.19. A potentiom eter is m ade of 2.25 m of constantan wire having a diam eter of 0.15 mm. D eterm ine the resistance of the potentiom eter. 10.20. A bundle of copper wire having a mass of 3.6 kg has a resistance of 22.5 Q. D eterm ine the length of the wire in the bundle. 10.21. A coil made of 75 m of constantan wire whose d i am eter is 0.1 mm is connected to a current source supplying a voltage of 12 V. D eterm ine the current in the coil. 10.22. The adm issible current density of the w iring in a room is 6 A/m m 2. W hat is the voltage drop in the copper wire if its length is 30 m? 10.23. D eterm ine the voltage drop and the current density in the w inding of the m agnetic pole of a locom otive m otor.
§ 10. Electric Current in Metals. Ohm’s Law
121
The w inding is made of copper strip having a cross-sectional area of 28 X 4.7 mm2, the average length of a tu rn being 1.5 m, and the num ber of tu rn s is 40. The current in the w inding is 352 A. 10.24. An electric appliance is 0.5 km from a current source and is connected to it through a wire haying a crosssectional area of 5 mm2. The appliance is moved 1 km fu rth e r away from the source. W hat should the cross-sectional area of the wire be for the voltage drop to rem ain unchanged? 10.25. How much copper is required to m anufacture the contact wire of a tram circuit having a resistance of 0.2 Q if the cross-sectional area of the wire is 85 mm 2? 10.26. D eterm ine the resistance of 1 km of tra m steel rails if the m ass of the rails is 55 kg/m . 10.27. Two wires, one of copper and one of nichrom e, of the same length and diam eter are heated through 5 K. The resistance of which wire changes more? W hat is the ratio of the resistances? 10.28. The series resistance connected to the ignition coil of a “Moskvich” m otor car is 1.35 Q at 293 K and 2 Q at 373 K . D eterm ine the m aterial of which it is made. 10.29. The s ta rte r rheostat of an electric tra in is m ade of a cast iron plate and a ferro-alum inium high-resistance rib bon alloy. Its resistance is 4 Q at 298 K. W hat w ill the change in its resista n c e be when the tem perature is 723 K ? 10.30. W hat m ust the tem perature resistance coefficient of the m aterial used for potentiom eters be? 10.31. The tungsten filam ent of an incandescent lam p has a resistance of 20 £2 when cold. W hat will its operating re sistance be, i.e. when the filam ent is heated to 2100°G? 10.32. W hat is the increase in the tem perature of the cop per w inding of a d.c. m otor arm ature as a result of pro longed operation during which the resistance has increased from 2 to 2.2 £2? 10.33. An electric arc rated for a voltage of 45 V and a current of 10 A is connected to a circuit at a voltage of 110 V. D eterm ine the required resistance of the series re sistor if the resistance of the leads is 0.5 £2. 10.34. W hat is the increase in the tem perature of a poten tiom eter m ade of iron wire if its resistance has doubled? 10.35. The voltage at an electric substation is 600 V (Fig. 32). The two copper cables A B and CD connecting the
122
Ch. II. Fundam entals of Electrodynam ics
su b statio n w ith a tro lley b u s’ contacts are each 440 m long and 400 mm2 in cross-sectional area. The contact wires are m ade of copper and have a cross-sectional area of 85 mm 2. The current in the trolleybus m otor is 200 A. D eterm ine the voltage across the current collectors of the trolley. 10.36. The experim ental d.c. transm ission line between K ashira and Moscow operates a t a voltage of 200 kV and UOm Substation
“ Contacts
I-------- 1 I I
\
£
$ F ig. 32
is m ade of a single-core alum inium cable 112 km long. De term ine the cross-sectional area of the cable if the voltage drop in it is 3.1 % of the nom inal value a t a current of 150 A. 10.37. A d.c. transm ission line under a voltage of 1500 kV is 2500 km long. W hat w ill the voltage a t the end of the line be if it is m ade of an a lu m inium wire w ith a crosssectional area of 600 m m 2? The adm issible current load in the wire is 1070 A. 10.38. W hich of the ammeFig- 33 ters in Fig. 33 w ill show a higher reading? W hat w ill the change in the resistance of the circuit be when bulb B x fuses? W ill the reading of am m eters A x and A 2 change in th is case? 10.39. The sta rte r rheostat of a railw ay engine consists of 80 plates having a resistance of 0.05 Q each and connected in series. D eterm ine the resistance of the sta rte r rheostat. 10.40. The arm ature w inding of a tractio n m otor consists of 132 sections connected in series in two parallel branches. The resistance of one section is 0.001 Q. D eterm ine the resistance of the whole w inding.
§ 10. Electric Current in Metals. Ohm's Law
123
10.41. Determ ine the to ta l resistance, the current, and the voltage d istribution in a circuit (Fig. 34), if the applied voltage is 220 V. 10.42. W hat series resistor should be connected to the heating elem ent of an electric iron having a resistance R = 24 £2 and rated a t a voltage of 120 V in order to use it at a voltage of 220 V? 10.43. A circuit is constructed as shown in Fig. 34 and connected to a d.c. source of 120 V. I t is known th a t R t =
Fig. 34
Fig. 35
6 £2, R 2 = 1 5 £2, and / = 5 A. Determ ine the to ta l re sistance of the circuit and the resistance of resistor i?3. 10.44. Figure 35 shows the excitation circuit for the tra c tion m otor of a trolleybus. The resistances of the sections of the rheostatic controller are R l = 1040 £2, R 2 = 200 £2, R 3 = 90 £2, and /?4 = 10 £2. The resistance of the excitation winding is 7?ex = 160 £2. Determ ine the excitation current w ith one of the switches 7, 2, or 3 closed and w ithout the switches. 10.45. The m agnetic field in the m otor of an electric loco m otive is produced by four series-connected w indings. De term ine the resistance of the windings if the num ber of turns in one of them is 39, the average length of a tu rn is 1.5 m, and the current in the w indings is 352 A, the adm is sible current density being 2.7 A/m m 2. 10.46. How m any equal pieces should a conductor of re sistance 4 £2 be cut into and how should these pieces be con nected to obtain a resistance of 1 £2? 10.47. W hat resistor should be connected in parallel to a resistor of 100 £2 for their to ta l resistance to be 20 £2? 10.48. D eterm ine the to ta l resistance of a lam p resistor, the current in the circuit and through each lam p if the re
124___________ Ch. II. Fundam entals of Electrodynam ics
sistor contains 10 parallel-connected lam ps having a re sistance of 440 £2 each, and the voltage applied is 220 V. 10.49. Three resistors of 1, 2, and 16 £2 are connected in parallel to a d.c. source. The current in the unbranched c ir c u it is 5 A. Determ ine the to ta l resistance of the circuit and the current through each resistor. 10.50. W hen two conductors are connected in parallel, th eir equivalent resistance is 0.72 £2. The equivalent re sistance for the same conduc Af-7Si tors connected in series is 3 £2. Determ ine the resistance of # each conductor. * 10.51. Three incandescent lam ps having a resistance of 400 £2 each are connected to a circuit at a voltage of 220 V. Fig. 36 Determ ine the current passing through each lam p if they are connected in parallel and in series. D eterm ine the to ta l current in each case. 10.52. Five resistors are connected as shown in Fig. 36. They are connected to a 12-V d.c. source. D eterm ine the to ta l resistance of the circuit and the current in resistors R lJ and /?4. 10.53. A circuit is formed as shown in Fig. 36. All the resistors have the same resistance of 1.2 £2. The current through the first resistor is 10 A. D eterm ine the to ta l re sistance, the voltage across the term inals of the circuit, and the currents and voltages in each resistor. 10.54. The filament of an incandescent lam p is heated to about 2200°C and fuses more often when it is being switched on th an when it is being sw itched off. W hy? W hat conclu sions can be drawn from a com parison of the currents through the lam p in the two cases? The m aterial of the filam ent is tungsten. 10.55. Two resistors R t = 12 £2 and R 2 are connected in parallel to a current source. Determ ine the resistance R 2 and the current through it if the to ta l current is 4 A and the voltage supplied to the resistors is 12 V. 10.56. Determ ine the voltage across the current col lectors of trolleybuses if the contact wire is m ade of copper and has a cross-sectional area of 85 mm 2. The lengths of the
§ 10. Electric Current in Metals. Ohm's Law
125
wires and the currents are indicated in Fig. 37. The generator voltage UBC is 575 V. 10.57. D eterm ine the voltage drop across a railw ay track if an electric locom otive is 800 m from the cable connecting
500A 1Q00A
SOOA 800A
F ig . 38
the rails w ith the substation (Fig. 38). The cross-sectional area of the cable is 72 cm 2. 10.58. Figure 39 shows the circuit diagram of a sta rte r rheostat consisting of three sections. Determ ine the re-
srx r x
r -r R,=1.SSi
f t'T J ?
KS=2Q
in
t_ r
tLj
t j
2~ Fig. 39
Fig. 40
sistance of the rheostat for the following three positions: (a) switch 1 is closed, (b) sw itch 2 is'closed, (c) switches 1 and 4 are closed, (d) switches 2 and 4 are closed, (e) switches 7, 2, and 4 are closed, and (f) switches 2, 3, and 4 are closed. 10.59. D eterm ine the resistance of two sections of a starter rheostat if the resistance of a cast iron plate is 0.06 Q. The first section contains 10 such plates and the second 12 plates. The plates are connected as shown in Fig. 40.
126___________ Ch. II. Fundam entals of Electrodynam ics
10.60. Determ ine the resistance of a sta rte r rheostat con sisting of two series-connected sections. In the first section, 22 coils are connected in two parallel groups, each of which contains 11 series-connected coils. In the second section* 45 coils are connected into three parallel groups of 15 se ries-connected coils each. The coils are made of ferro-alum inium high-resistance ^alloy wire 1.6 m long and 3 mm in d i am eter. Fig. 41 10.61. The lighting circuit of a tram car consists of two parallel groups each containing five series-connected bulbs. D eterm ine the to tal current in the circuit and in the groups if the resistance of each bulb is 220 Q and the voltage in the circu it is 550 V. 10.62. Incandescent lam ps, w ith a resistance of 440 Q each, are connected to a circuit at a voltage of 220 V as
r<8h
-C=D—
Fig. 42
Fig. 43
shown in Fig. 41. D eterm ine the to tal resistance of the lam ps and the current and voltage in each lam p. 10.63. W hat w ill the change in the voltage and current in the lam ps (see Problem 10.62) be if one of the lam ps fuses? 10.64. The arm ature w inding in a locom otive m otor con sists of 924 copper rods 1 m long each. The rods are d istrib uted equally among four parallel branches (Fig. 42). De term ine the resistance of the arm ature w inding for a current of 352 A, assum ing th a t the current density is 5 A/m m 2. 10.65. Figure 43 presents a circuit diagram . The voltage UAB is 120 V. D eterm ine the resistance of the circuit, the unbranched current, and the current in each resistor.
§ 10. Electric Current in Metals. Ohm’s Law
127
10.66. Figure 44 shows a circuit diagram . The voltage UAB is 220 V. D eterm ine the to ta l current and the current in parallel branches. 10.67. A current up to 50 A is to be m easured w ith an am m eter whose scale is rated to 10 A. D eterm ine the re-
F ig. 45
sistance of the shunt if the resistance of the am m eter is 0.01 Q (see Fig. 26). 10.68. An 0.18-Q am m eter indicates a current of 6 A. The am m eter has a shunt whose resistance is 0.02 Q. D eterm ine the c u rre n t in the m ains (see Fig. 26). 10.69. A voltm eter w ith a scale from 0 to 150 V has to be used for m easuring voltage from 0 to 250 V. C alculate the resistance of a series resistor if the resistance of the v o lt m eter is 600 Q (Fig. 45). 10.70. A voltm eter rated for the m axim um voltage of 20 V a t a current of 8 mA has to be used for m easuring a v o lt age of 100 V. W hat series resistance is required? 10.71. A voltm eter having an in ternal resistance of 1000 Q is connected to a circu it w ith a constant voltage supply in series w ith a series resistor R s and indicates 180 V. If another identical series resistor is connected to the circuit, the v o lt m eter indicates U 2 = 150 V. D eterm ine the resistance of the series resistor R s and the voltage U in the circuit. 10.72. The m axim um current th a t can be m easured w ith a m illiam m eter having an in terval resistance R = 150 Q is I = 10 mA. How long m ust a m anganin wire of diam eter d = 0.1 mm be if it is used as the series resistor to convert the instrum ent into a voltm eter w ith a scale from 0 to 10 V? 10.73. The internal resistance of a voltm eter is 300 Q. A series resistor of 1200 Q is connected to it in a m easuring circuit. W hat is the ratio of the scale factors of the v o lt m eter w ith and w ithout the series resistor?
128
Ohm's
Ch. II. Fundam entals of Electrodynam ics
law for
a closed circuit. Connection of batteries
10.74. A voltm eter is connected to the term inals of a cur ren t source. The external circuit is closed. W hat does the voltm eter measure in th is case? 10.75. A cell whose internal resistance is 0.4 £2 is closed through an external circuit having a resistance of 2.1 £2. The current in the circu it is 0.6 A. Deter—v ) — mine the emf of the cell and the voltage drop across it. j ( 10.76. Figure 46 shows a circuit diagram . * W hen the resistance of the external circuit is 1.5 £2, the voltm eter indicates 1.65 V, while for an external resistance of 3.6 £2 the F ig. 46 reading of the voltm eter is 1.8 V. D eterm ine the emf and the internal resistance of the cell. 10.77. A galvanic cell w ith an emf of 1.45 V and an in te r nal resistance of 1.5 £2 is closed through an external re sistance of 3.5 £2. D eterm ine the current in the circuit, the voltage across the cell term inals, and the efficiency of the cell in th is circuit. 10.78. Determ ine the emf of a generator and the voltage across its term inals if its resistance is 0.05 £2 and the exter nal resistance is 11.45 £2. The current in the circuit is 20 A. 10.79. D eterm ine the current in a circuit if the emf of the generator is 230 V, its resistance is 0.1 £2, and the resistance of the external circuit is 22.9 £2. 10.80. A resistor of 2.2 £2 is connected to a generator w ith an emf of 230 V. W hat is the resistance of the generator if the voltage across its term inals is 220 V? 10.81. A galvanic b a tte ry w ith an emf of 15 V and an in tern al resistance of 5 £2 is closed through a conductor whose resistance is 10 £2. A 1-pF capacitor is connected to the b attery term inals. D eterm ine the electric charge on the capacitor. 10.82. A 2-(xF capacitor and a 3-£2 resistor are connected in parallel to a current source w ith an emf of 10 V and an in tern al resistance of 1 £2. W hat is the electric charge on the capacitor? 10.83. If a current source is closed through a 13-£2 re sistor, the current in the c irc u it is 0.8 A. If the source is closed through an 8-£2 resistor, the current in the circu it is
j 10. Electric Current in Metals. Ohm’s Law
129
1.2 A. Determ ine the emf and the internal resistance of the current source. 10.84. A d.c. generator w ith an emf of 150 V and an in te r nal resistance of 0.3 £2 supplies voltage to 20 incandescent lam ps having a resistance of 240 £2 each and connected in parallel. The resistance of the leads is 2.7 Q. D eterm ine the voltage across the generator term inals and across the lam ps. 10.85. The emergency lam ps of a tram car are fed by an accum ulator battery having an emf of 48 V and an internal resistance of 0.2 £2. Ten lam ps having a resistance of 39.5 £2 each are connected as shown in Fig. 47. Determ ine the current in each lam p and in the leads. 10.86. A circuit contains 20 parallelF ig. 47 connected bulbs. The current through a bulb is 1 A. The resistance of the wires connecting the load w ith a generator is 0.2 £2. W hat m ust the emf of the generator be for the voltage across the bulbs to be 220 V? The internal resistance of the generator is 0.05 £2. 10.87. Three electric m otors and ten parallel-connected incandescent lam ps are connected to a generator w ith emf + o
240 V and resistance 0.025 £2. A current of 50 A passes through each m otor while the current through each lam p is 1 A. The resistance of the leads is 0.1 £2. D eterm ine the voltage across the generator term inals and across the loads. 10.88. A workshop receives electric power from a collective farm ’s electric power sta tio n . There are two electric m otors (M) and four bulbs in the workshop connected as shown in Fig. 48. The current through each m otor is 10 A, and through 9-0530
130
Ch. II. Fundam entals of Electrodynam ics
each bulb 0.5 A. The distance I between the sta tio n and the workshop is 0.5 km. The internal resistance of the generator is 0.1 £2 and the voltage across its term inals is 220 V. De term ine the emf of the generator and the cross-sectional area of the copper leads if the adm issible voltage drop in them is 8% . 10.89. Determ ine the counter emf of a tractio n m otor if the resistance of its windings is 0.1 £2 and the voltage in the circu it is 550 V a t a current of 150 A. 10.90. Determ ine the emf of a generator w ith an in tern al resistance of 0.05 £2 and the counter emf of a m otor if the current in the circuit is 100 A, the voltage across the gen erator term inals is 225 V, and the resistances of the m otor w inding and the leads are 0.2 and 0.1 £2 respec tively. 10.91. The circuit diagram of a d.c. m otor is shown in Fig. 49. The voltage in the circuit is U = 550 V, and the current is / = 102 A. The resistance of the arm ature circuit is R a = 0.1 £2, th a t of the parallel excitation w inding R ex = 150 £2, and th a t of the rheostatic controller R r = 125 £2. D eterm ine the current in the parallel excitation w inding when the rheostatic controller is com pletely on, and the counter emf if the electric m otor is started w ithout a s ta rte r rheostat. 10.92. Four loads having a resistance of 10 £2 each are con nected to an accum ulator battery having an emf of 48 V and an in ternal resistance of 0.25 £2. Determ ine the current through the b attery if the loads are connected (a) in series, (b) in parallel, and (c) in two parallel branches containing two series-connected loads each. 10.93. One of two cells has an emf of 1.45 V and an in ternal resistance of 0.5 £2 and supplies voltage to a circu it w ith an efficiency of 90% , w hile the other cell has an emf of 2 V and an in tern al resistance of 0.5 £2 and operates w ith an effic iency of 80% in an identical circuit. D eterm ine the current in the two circuits. 10.94. An accum ulator w ith an emf of 1.45 V produces a current of 0.5 A in a conductor whose resistance is 2.5 £2. D eterm ine the short-circuit current. 10.95. D uring a short circuit, the current from a source of 1.8 V is 6 A. W hat m ust the external resistance be for the current to be 2 A?
§ 10. Electric Current in Metals. Ohm's Law
131
10.96. W hy should lead accum ulators not be short c ir cuited? 10.97. A galvanic cell is first connected to an external resistance of 1.9 £2 and the current in the circu it is 0.6 A. Then it is connected to an external resistance of 2.4 £2 and the current becomes 0.5 A. W hat is the short-circuit current for this cell? 10.98. Three galvanic cells, each having an emf of 1.5 V and an in tern al resistance of 0.6 £2, are connected in series to a conductor having a resistance of 1.8 £2. D eterm ine the current in the circuit. W hat w ill the current in the same conductor be if the cells are connected in parallel? 10.99. Using the d ata in Problem 10.98, determ ine the connection for which the short-circuit current is the largest. 10.100. How should six galvanic cells, each having an emf of 1.5 V and an internal resistance of 0.5 £2, be connected in parallel groups to obtain the m axim um current when con nected to a 1.0-Q resistor? 10.101. How should 12 cells, each having an emf of 1.2 V and an in tern al resistance of 0.6 £2, be connected to produce a current of 1.6 A in an external circu it whose resistance is 2.2
£2 ?
10.102. A “K rona” b a tte ry is the power source for tra n sistor radios. I t contains seven series-connected galvanic cells. W h at is the emf of a cell if the emf of the b a tte ry is 9 V? W hat w ill the change in the emf of the b a tte ry be if the co n stitu en t cells are connected in parallel? 10.103. Two alkaline accum ulators, each having an emf of 1.5 V and an in tern al resistance of 0.3 £2, are connected first in series and then in parallel and used to supply voltage to a circu it w ith a resistance of 6 £2. W hich connection produces the higher current in the external circuit, i.e. which m ethod is more advantageous? 10.104. Solve Problem 10.103 for the case when the ex ter nal resistance of the c irc u it is 0.9 £2. Using the results of the two problem s, find the ratio between the external and in te r nal resistances for which connecting the current sources in series is more advantageous. 10.105. A passenger carriage is lighted using a b attery containing 26 series-connected lead accum ulators. The emf of an accum ulator is 2 V and the in tern al resistance is 9*
132___________ Ch. II. Fundam entals of Electrodynam ics
0.004 £2. D eterm ine the emf and the voltage across the te r m inals of the b a tte ry if the current in the circu it is 20 A. 10.106. W hat is the reading of a voltm eter connected to the term inals of a battery consisting of three series-connected alk alin e accum ulators, each w ith an emf of 1.2 V and an in te rn al resistance of 0.3 £2? The external circu it consists
Fig. 50
F ig. 51
of a bulb w ith a resistance of 16 £2 and 2-m long alum inium leads w ith a cross-sectional area of 0.2 m m 2. D eterm ine the voltage on the bulb. 10.107. An accum ulator b attery , having an emf of 22 V and an in tern al resistance of 0.1 £2, is charged from a re c ti fier the voltage across whose term inals is 24 V. D eterm ine the resistance introduced into the circuit by a potentiom eter connected in series w ith the b attery if the charging current is 4 A. 10.108. Two accum ulators, having em f’s of 1.3 and 1.8 V and in tern al resistances of 0.1 and 0.15 £2 respectively, are connected in parallel. D eterm ine the current in the c irc u it and the voltage across the term inals (Fig. 50). 10.109. D eterm ine the current through a resistor R = 2 £2 connected in a circuit as shown in Fig. 51. Given: = 2 V, rj = 0.50 £2, g 2 = 4.0 V, and r 2 = 0.70 £2. § 11. WORK, POWER, AND THE THERMAL EFFECT OF CURRENT
Basic Concepts and Form ulas E lectric energy is transform ed into other kinds of energy so th a t energy is conserved. The m easure of the transform ation is the work done by electric current: A = Q U = JUt,
§ 11. Work, Power, and the Thermal Effect of Current
133
If the current or voltage are substituted using O hm ’s law for a conductor we obtain A = IUt ---- I 2R t = ^ - t . Power is the ratio of the work done by an electric cu rren t during tim e t to the tim e t: p = ^t - f p *=i u = i - r = !£-. n The u n it of work is the joule (J). In electrical engineering and everyday life, electrical work is measured in kilow atthours (1 kW h = 3.6 M J). The u n it of power is the w a tt (1 W - 1 J/s). Jo u le ’s law: the am ount of heat liberated in a conductor carrying a current is proportional to the square of the cu rren t, the tim e of its passage, and the resistance of the conductor: Q = P R t. This form ula is for series-connected loads. For a parallel connection, the form ula is
Worked Problems Problem 63. An electric m otor operating for 5 h is driven at the m ains voltage of 380 V and a current of 35 A. The resistance of the m otor w inding is 0.5 £2. D eterm ine the am ount of energy consumed, the am ount of heat liberated in the w inding during the operation, and the m echanical work done by the m otor. Given: U = 380 V is the voltage at the m otor term inals, / = 35 A is the current, R = 0.5 £2 is the resistance of the m otor w inding, and t = 5 h = 5 X 3600 s is the operation tim e. F ind: the energy A consumed by the m otor, the am ount of heat Q liberated in the w inding, and the m echanical work A mechSolution. The energy consumed or the to ta l work done by the curren t can be determ ined from the form ula A = IU t,
A = 35 A x 380 V X 5 x 3600 s ~ 2.4 x 108 J .
134
Ch. II. Fundam entals of Electrodynam ics
The am ount of heat liberated in the w inding of the m otor can be determ ined from Jo u le’s law: Q = P R t, Q = (35 A)2 x 0.5 Q X 5 X 3600 s = 1.1 X 107 J. The m echanical work done by the m otor can be d eter mined by subtracting the energy spent in heating the w ind ing from the whole energy spent: ^1 mech — A Q, ^ mech = 2.4 x 108 J - 1.1 x 107 J = 2.29 X 108 J . Answer. The energy spent by the m otor is approxim ately 2.4 X 108 J , the am ount of heat liberated in the w inding is 11 M J, and the m echanical work is 2.3 X 108 J . Problem 64. A tower crane, whose efficiency is 70% , hoists a load of 49 kN at a constant velocity of 0.55 m /s. D eterm ine the current in its electric m otors if driven from a supply voltage of 380 V. Given: t] = 70% is the efficiency of the crane, G = 4.9 X 104 N is the force of g ravity acting on the load, v = 0.55 m /s is the velocity of hoisting, and U = 380 V is the voltage in the circuit. Find: the current / . Solution . We w rite the form ula for the efficiency as the ra tio of the useful power (Gv) and the power spent (IU): *1 = ~jjj ioo% . Hence we obtain the current: / = $ -1 0 0 % , T_
4 . 9 x l ° 4 N x°.55 m/sXl00% _
7
ACXA a
— 1U1A.
Answer. The current is approxim ately 101 A. Problem 65. Two resistors of 40 and 80 Q are connected in parallel to a source of constant voltage. The am ount of h e a t liberated in the first resistor is 3.0 X 10B J . How m uch heat is liberated in the second resistor over the same tim e and in both resistors if they are connected in series? Given: R x = 40 Q is the resistance of the first resistor, R 2 = 80 Q is the resistance of the second resistor, and Qx =
j 11. W ork, Power, and the Therm al Effect o f Current
135
3.0 X 10s J is the amount of heat liberated in the first re sistor. Find : the amount of heat Q2 liberated in the second re sistor, and the amount of heat liberated in the series-con nected resistors. Solution, We use Joule’s law
D ividing these equations termwise, we obtain QxlQt =
R t/Rlt whence we get Qt : & =
4nQx^°aXinU
~ 1 -5 x 10*J.
If the resistors are connected in series, we have
Expressing UH from (1) and substituting it into Eq. (2), we obtain c — 3 ' ° X j20 QX * ° ^ = 1 -0 X 10* J .
Answer. The amount of heat liberated by the second re sistor is 1.5 X 10* J, and the heat liberated by the seriesconnected resistors is 1.0 X 10* J. Problem 66. The air in a room loses 293 MJ of heat per day. Determine the diameter of a nichrome wire 10.2 m long from which the coil of an electric heater is made such that the temperature in the room is kept constant if supplied with a voltage of 220 V. Given: Q = 2.93 X 10* J is the amount of heat lost, t = 24 x 3600 s is the time during which the heat is lost, I = 10.2 m is the length of the nichrome wire, U — 220 V is the supply voltage. From tables, we find the resistivity of nichrome, p = 1.05 x 10"* Q*m. Find: the diameter d of the wire. Solution. Let us determine the amount of heat lost by the air in the room per second, i.e. the power of heat losses Px = Q/t. The power of the electric heater can be determined from the formula P t = UVR.
136
Ch. II. Fundam entals of Electrodynam ics
In order to m aintain the tem perature in the room, the power of the heater m ust be the same as the power of h e a t losses: P 1 = P 2, Q/t = U 2/R . Hence we can find the re sistance of the heater coil: R = UHIQ. I t is well known th a t R = pUS, where S = juf2/4. Then R = Apl/nd2. E quating the expressions for /?, we obtain Hence U 'tT i
4 X 2.9 3 X 108 J X 1 .0 5 x 10“®Q-m x 10.2 m 220 V X 220 V x 24 X 3600 s x 3.14
= 9.5 x 10“'* m.
Answer. The diam eter of the wire is approxim ately 1 m m. Problem 67. How long w ill it take to boil 1 1 of w ater from 15°C using an electric imm ersion heater whose resistance is 25 Q and efficiency is 85% ? The applied voltage is 120 V. Given: V = 10“3 m3 is the volume of the w ater, tx = 15°C is the in itia l tem perature of the w ater, t 2 = 100°C is the boiling point of w ater, R = 25 Q is the resistance of the coil, U = 120 V is the applied voltage, and r\ = 0.85 is the ef ficiency of the heater. From tables, we find the density of w ater, p = 103 kg/m 3, and the specific heat of w ater, c = 4187 J/(k g -K ). F ind: the tim e t required for boiling w ater. Solution. Problem s involving a given efficiency should be solved starting w ith the formula for efficiency r) = q T~
i where QTtc = cm (t 2 — t x), and Qgw = - ^ ' t. T his
gives t) =
R- Hence we can find the tim e: t = Considering th a t the mass of water can be
expressed in term s of density and volume, m = pF, we obtain cp V ( t 2 — *i) B
1
T\U*
’
4i87 J/(k g-K ) X 103 Kg/m3 X 10”3 m3 X 85 K x 25 Q _ 70Q 0 .8 5 x 1 2 0 V x 120 V
= 12 min. Answer. I t takes 12 m in to boil the water.
§ 1 1 . Work, Power, and the Thermal Effect of Current
137
Questions and Problems 11.1. How much electricity w ill pass through the cross section of a conductor in 30 s if its resistance is 20 £2 and the voltage across it is 12 V? D eterm ine the work done by the electric current. 11.2. W h at energy is supplied by a generator to an e x te r nal circu it over 8 h if the readings of the am m eter and v o lt m eter rem ain unchanged a t 50 A and 220 V respectively? Express the answer in joules and kilow att-hours. 11.3. The reading of an electric m eter is 0981 kW h. The load consists of three bulbs of 100 W , two bulbs of 60 W , and four bulbs of 15 W . W hat w ill the m eter reading be in 30 days if the bulbs are switched on for 10 h every day? W hat w ill the electricity bill be a t a rate of 4 kopecks per kilow att-hour? 11.4. An electric m otor driven by a current of 15 A a t a constant voltage of 220 V develops a power of 3 kW . D eter mine the efficiency of the m otor and the cost of 8 h of elec tric energy, if it is charged at a rate of 4 kopecks per kW h. 11.5. A d.c. electric transm ission line between the Volzhskaya hydroelectric plan t and the Donbas is rated for a v o lt age of 800 kV w ith a nom inal current of 1000 A. D eterm ine the transm itted power and the energy transferred per year under these nominal* conditions. 11.6. D eterm ine the electric energy expenditure over 8 h of operation of two parallel-connected electric m otors if the current in one branch is 50 A and the supply voltage is 220 V. 11.7. How much energy is consumed by the m otor of a tram in one hour of continuous operation if the voltage across the collector plates of the m otor is 500 V and the current in the w inding of the m otor is 130 A? Express the answer in joules and kilow att-hours. 11.8. D eterm ine the power of the current in the circu it shown in Fig. 52. 11.9. Five resistors are connected as shown in Fig. 53 and to a voltage supply UAb = 24 V. D eterm ine the to ta l resistance of the circuit, the potential difference between points C and Z), the current in the fifth resistor, and the power of the current in the circuit.
138___________ Ch. II. Fundam entals of Electrodynam ics
11.10. An electric loader is driven by an accum ulator b a t tery consisting of 22 series-connected alkaline accum ulators w ith an emf of 1.1 V each. The power developed by the m otor during loading is 1.2 kW . D eterm ine the current in the c ir c u it if the voltage drop across the b attery and in the wires is 2.2 V. 11.11. The battery of an electrom obile consists of 42 se ries-connected accum ulators having an emf of 2.0 V each. fi,=2 S? c
Fig. 53
D eterm ine the to ta l power of the b attery if the current in each of two parallel-connected electric m otors is 50 A. 11.12. D eterm ine the power developed by an underground train consisting of six coaches if the m otors in each coach are connected in two parallel branches consisting of two se ries-connected m otors. The supply voltage is 750 V and the current in each branch is 150 A. 11.13. Determ ine the power developed by the m otors of an electric locom otive if they are connected into three parallel branches of two series-connected m otors, the supply voltage is 3 kV, and the current in each branch is 300 A. 11.14. D eterm ine the energy consum ption in 8 hours if four incandescent lam ps are connected to a circuit a t a v o lt age of 120 V (a) in parallel, (b) in series, and (c) in two parallel groups containing two series-connected lam ps. The resistance of a lam p is 120 £2. 11.15. D eterm ine the power consumed and the useful power for the m otor from Problem 10.89 if its efficiency is 90% . 11.16. Using the d ata of Problem 10.56, calculate the power consumed by the m otors of trolleybus 1 and the power losses in the circuit.
§ 1 1 . Work, Power, and the Thermal Effect of Current
139
11.17. A generator w ith an in tern al resistance of 0.01 £2 is connected to electric m otors as shown in Fig. 54. The voltm eter indicates 220 V and the am m eter 500 A. The resistance of the leads is 0.05 £2. D eterm ine the to ta l power, the power consumed by the RrQ.OSQ electric m otors, and the effic iency of the set-up. 11.18. Using the data of r=0.07A Problem 10.61, determ ine the M z > -r power of the current in the bulbs, and the energy expendi ture in ten hours of operation. 11.19. W hat m ust the crossFig- 54 sectional area of a copper wire be to tran sm it a power of 1.0 kW a t a voltage of 100 V over a distance of 50 m so th a t voltage losses are less than 6 V? 11.20. Two identical resistors are connected to a circuit a t a constant voltage first in parallel and then in series. Is the electric power the same in both cases? 11.21. The resistance of a glowing incandescent lam p is 360 £2. The supply voltage is 220 V. How m any lam ps are connected in parallel in the circuit if the power consumed by a ll the lam ps is 2.15 kW ? The resistance of the leads should be neglected. 11.22. An electric loader lifts a load of 500 kg to a height of 2 m a t a constant velocity. The m otor is driven by an accum ulator b a tte ry w ith an emf of 24 V a t a curren t of 41 A. The efficiency of the loader is 80% . D eterm ine the velocity and du ratio n of one operation, neglecting the in ternal resistance of the b a tte ry . 11.23. The m otors of a tram m oving uniform ly along horizontal rails develop a tra c tiv e force of 2 kN . The voltage across the collector plates of the m otors is 550 V and the c urrent in the circ u it is 80 A. The efficiency is 80% . D eter m ine the velocity of the tram . 11.24. A tram sta rts to move a t a constant acceleration of 1 m /s2 for 8 s w ith a tra c tiv e force of 28 kN and an effic iency of 88 %. W hat is the curren t a t the end of the eight seconds if the supply voltage is 550 V? 11.25. W hy is lead wire used in fuses? 11.26. The 8-A and 16-A fuses for the VAZ m otor car have the same length. W h at is the difference between the fuses?
140
Ch. II. Fundam entals of Electrodynam ics
11.27. Given two hot plates one of which is rated for a voltage of 127 V and the other for 220 V. W hich coil is thicker if the length and m aterial of the coils are the same? 11.28. A carriage is heated w ith eight electric heaters having a resistance of 275 Q each and connected in parallel. The voltage across the term inals of the heaters is 550 V. D eterm ine the am ount of heat given away by the heaters during 18 h of operation, neglecting heat losses. 11.29. Two bulbs rated for 110 V are connected in series and to a voltage supply of 220 V. D eterm ine the voltage across each bulb, the power, and the am ount of heat lib e r ated by a bulb during 1 h if the powers of the bulbs are (a) 60 W each, (b) 60 and 40 W, and (c) 60 and 100 W . 11.30. An electric soldering iron operates at a voltage of 220 V and a cu rren t of 0.22 A. How much tin a t 293 K can be m elted by it per m inute? 11.31. An electric sam ovar having a power of 600 W boils 1.5 1 of w ater from 283 K in 20 m in. D eterm ine the efficiency of the sam ovar and the cost of the energy if elec tric ity is charged a t a rate of 4 kopecks per kW h. 11.32. Two resistors, having resistances of 1 and 4 Q, are connected in tu rn to a current source and are found to con sume the same power. D eterm ine the in tern al resistance of the current source. 11.33. D eterm ine the emf and the internal resistance of a current source if the power of an external circu it is 230 W at a current of 10 A and 337.5 W a t a current of 15 A. 11.34. An electric m otor is connected to a voltage supply of 220 V. The resistance of the m otor w inding is 1.8 Q and the cu rren t in the m otor is 12 A. D eterm ine the power con sumed and the efficiency of the m otor. 11.35. W ater boils in an electric boiler 12 m in after it .has been switched on. The heating elem ent consists of 4.5 m of wire wound in a coil. W h at m ust be done to obtain boiling w ater in 8 m in? The energy losses should be neglected. 11.36. The heating elem ent of an electric boiler has two sections. If the two sections are connected in series, w ater boils in the boiler in 27 m in, while if connected in parallel the w ater boils in 6 m in. The resistance of one section is 40 Q. W h at is the resistance of the other section? Energy losses should be neglected. 11.37. The distance between a generator w ith an emf of
§ 12. Electric Current in E lectrolytes
141
240 V and an internal resistance of 0.1 £2 and a consumer is 50 in. How much copper is required for the cabling if a power of 22 kW is rated for a voltage of 220 V? § 12. ELECTRIC CURRENT IN ELECTROLYTES
Basic Concepts and Formulas L iquid conductors m ainly include solutions of salts, alkalis, and acids. The current carriers in liquid conductors aie the ions formed by electrolytic dissociation. This is the decom position of n eutral molecules of salts, alkalis, and acids into positive and negative ions when they dissolve in w ater or other solvent. The large p e rm ittiv ity of w ater (e = 8 1 ) and therm al m otion lead to the decom position of m olecules. The passage of a current through a liquid conductor (electrolyte) is accom panied by a chem ical transform ation of the substance and its deposition on the electrodes (this is known as electrolysis). E lectrolysis is governed by F arad ay ’s two laws. Faraday's first law of electrolysis. The mass of a substance liberated in electrolysis is proportional to the am ount of electricity passing through the electrolyte: m = kQ, where k is the electrochem ical equivalent and is the am ount of substance liberated in the electrolysis by the passage of 1 C of electricity through the electrolyte. Faraday's second law of electrolysis. Electrochem ical equivalents are proportional to the ratio of the m olar mass to the valency of the substance: h\
Mi
_
k2
rii
* n2
M% ’
where M 1 and M 2 are the m olar masses, and n x and n 2 are the valencies. I t is im p o rtan t to rem em ber th a t the charge passing through an electrolyte to liberate 1 mol of a substance is F = N Ae = 9.648456 x 104 C/mol, where F is the same num ber for all electrolytes and is known as the Faraday con stan t.
142
Ch. II. Fundam entals of Electrodynam ics
Faraday’s generalized law is 1 M n F
n
The passage of a current through an electrolyte may cause the electrodes to polarize, which induces a counter emf, i.e. decreases the current through the circuit. In such a case, problem s can be solved w ith the help of O hm ’s law for a su b circu it containing an emf: j
U Spol R
Worked Problems Problem 68. A m etal article is electrolytically plated w ith a silver layer 20 pm thick. How long did the electrolysis require for a current density of 2.5 X 10”3 A/cm 2? Given: h = 20 pm = 2 X 10~5 m is the thickness of the silver layer, and / = 2.5 X 10~3 A/cm 2 = 25 A/m 2 is the current density. From tables, we find the electrochem ical equivalent of silver, k = 1.118 X 10'° kg/C, the density of silver, p = 10.5 X 10s kg/m 3, the valency of silver, n = 1, the m olar mass of silver, M = 108 x 10~3 kg/m ol, and the Faraday constant F = 9.65 X 104 C/mol. F in d : the tim e t of the electrolysis. Solution. 1st method. We solve the problem using F a ra d a y ’s first law m = k i t . This gives t = m l k l . The mass and the current can be determ ined from the form ulas m = pSh and / = j S . S u b stitu tin g these qu an tities into the form ula for tim e, we obtain .
_
ph kj
9
l ~
10.5 X 103 k g /m 3 X 20 X 10~fl m _ 1.118 X 10”6 kg/C X 25 A /m 2
„
2nd method. If the electrochem ical equivalent is unknow n, the problem can be solved using F arad ay ’s generalized law 1
1^1
mFn
m = — -rrr v — ft I t , whence t = in / • The mass and the current can be determ ined as before: m = p F = pSh and / = jS . This gives , _
phFn
~Mj~ ’ ._ “
1 0 .5 X 1 0 3 k g/m 3 X 20 x l 0 “6 m x 9 .6 5 X l 0 4 C /m ol _ y c n n g 108 X 10“3 k g/m ol X 25 A /m 2
§ 12. Electric Current in E lectrolytes
143
Answer. The tim e required for silver plating is about 2.1 h. Problem 69. D uring the electrolysis of silver n itra te solu tion, 12 g of silver took an hour to deposit on the cathode. The voltage across the term inals of the electrolytic bath was 5.2 V, the resistance of the solution was 1.5 Q, and the po larization emf was 0.7 V. D eterm ine the valency of silver and the num ber of silver atom s liberated at the cathode. Given: t = 3600 s is the duration of the electrolysis, m = 1.2 X 10-2 kg is the mass of liberated silver, U = 5.2 V is the voltage across the term inals of the b ath , R = 1.5 £2 is the resistance of the electrolyte solution, and % = 0.7 V is the polarization emf. From tables, we find the m olar mass of silver, M = 108 X 10-3 kg/m ol, the F araday constant F = 9.65 x 104 C/mol, and the electron charge e = 1.6 X lO"19 C. Find: the valency n of silver and the num ber N of silver atom s liberated at the cathode. Solution. The valency of silver can be determ ined from F arad ay ’s generalized law m = " j r ~ I t , whence n * Since the polarization emf emerges in the circuit, the current can be determ ined from Ohm ’s law for a subcircuit w ith an emf: I = (U — %)tR. Then the expression for the valency becomes 108 X l 0 “3 k g/m ol (5 .2 V - 0 . 7 V) 3600 s 1 .2 X 10“2 kg X 9 .6 5 X 104 C /m ol X 1 .5 Q
M (U — %) t
n ~
’
mFR
n
, '
Since the valency of silver is u n ity , we can determ ine the num ber of silver atom s by dividing the charge passing through the electrolyte by the elem entary charge: N = Qle. Since Q = I t , we obtain N _(U-%)t i y ~
~
R
e
N
’ yV“
(5 .2 V - 0 . 7 V )3600 s _
? 5 ^ 1Q22
1 . 5 Q x l . 6 x l O - “ "C“ “ b- / : ) X 1 U
*
Answer. Silver is m onovalent, and 6.75 X 1022 silver atom s were liberated at the cathode. Problem 70. The current through an electrolytic bath containing the solution of zinc sulphate (Z nS 04) increases linearly I = (2 + 0.02*). How much zinc w ill be liberated at the cathode 5 m in after the current sta rts changing?
144
Ch. II. Fundam entals of Electrodynam ics
Given: / = (2 + 0.02t) is the linear increase in current, and t = 300 s is the duration of the electrolysis. From tables, we find the electrochem ical equivalent of zinc, k = 3.4 x 10-7 kg/C. F ind: the mass m of the deposited zinc. Solution. We use F a ra d a y ’s first law of electrolysis m = kQ , where Q is the am ount of electricity passing through the electrolyte. In order to determ ine Q, we plot the curFig* 55 rent versus tim e graph. The current can be determ ined from the given equation. At tim e t 0 0, we have / 0 — 2A, while at t = 300 s, we have / = (2 + 0.02 X 300) A = 8 A. The tim e v ariatio n of current is shown in Fig. 55. From the graph, we see th a t the am ount of electricity passing through the electrolyte is equal to the area of the hatched region (trapezium ): £ = /< = A + £ t,
Q = (2 A + 8A > 300 s = 1500 C.
The mass of the zinc is m = 3.4 x 10-7 kg/C x
1500 C = 51.0 x 10~5 kg.
Answer. 510 mg of zinc will be deposited on the cathode. Problem 71. D eterm ine the power consumed during the electrolysis of sulphuric acid solution if 150 mg of hydrogen are liberated over 25 m in, and the resistance of the electro lyte is 0.4 Q. The losses should be neglected. Given: t. —- 1500 s is the duration of the electrolysis, m — 0.15 x 10“3 kg is the mass of the liberated hydrogen, R —0.4 Q is the resistance of the electrolyte. From tables, we find the electrochem ical equivalent of hydrogen, k — 1.044 x lO-8 kg/C. Find: the power P consumed in the electrolysis. Solution. In order to determ ine the power of the current, we use the form ula P = P R . The current can be determ ined from F arad ay ’s first law of electrolysis: I = mlkt. S ubsti
§ 12. Electric Current in ,E lectro ly tes
145
tu tin g this q u a n tity into the form ula for power, we obtain p
p (ft/)2 ’
(0.15 X 10~3 kg )2 0.4 Q ^ (1 -044 X 10-® kg/C X 1500 s)* ~
07
w
Answer. The power consumed in the lib eratio n of the hydrogen is approxim ately equal to 37 W. Questions and Problems 12.1. A solution of the salt NaCl is electrically neutral. Can we sta te th a t there are no ions in the solution? W hy? 12.2. E lectrolytic capacitors are connected to a circuit, w ith their polarity observed. Can the polarity be ignored? W hy? 12.3. C urrent reversal is used to polish the surface of a m etal electrolytically. W hat effect does it produce? Can this m ethod be used to sharpen cu ttin g tools? 12.4. How much alum inium , silver, and copper w ill be deposited on the cathode as a result of the passage of 1 C of electricity through appropriate electrolytes? How m any electrons pass through each electrolyte thereby? 12.5. The electrolysis of copper sulphate solution yields 1 g of copper. How much alum inium can be obtained electro ly tically by passing the same am ount of electricity through an appropriate electrolyte? 12.6. Two electrolytic baths are connected in series and to a current source. The first bath contains a solution of nickel sulphate N iS 0 4 while the second contains a solution of chrom ium (II) chloride CrCl2. How much chrom ium w ill be liberated in the second b ath if 300 g of nickel are libe rated in the first bath? 12.7. Two series-connected electrolytic baths contain copper sulphate C u S 0 4 and copper chloride CuCl solutions respectively. How much copper will be liberated in each bath when 1 C of electricity is passed through them ? 12.8. A student first calculated the electrochem ical equiv alent of copper using the form ula k = M IF n and then experim entally determ ined it. W h at m ust the increase in the mass of the cathode be after 15 m in if copper sulphate solu tion is electrolyzed at a cu rren t of 1 A? The valency of copper is 2. 10-0530
146
Ch. 11. Fundam entals of Electrodynam ics
12.9. To determ ine the electrochem ical equivalent of copper, a student electrolyzed a copper sulphate solution for 20 m in a t a current of 1.5 A. The cathode increased in mass by 600 mg. W hat is the resu ltan t electrochem ical equiv alent? W hat are the absolute and relativ e errors of m easure m ent in com parison w ith the tab u lated value? 12.10. A solution was electrolyzed for 20 m in a t a current of 1.5 A. 594 mg were liberated at the cathode. W hat was liberated? 12.11. D uring an electrolysis, 503 mg of a m etal were liberated at the cathode. The process lasted 5 m in at a cur ren t of 1.5 A. W hat was the m etal and w hat is its valency? 12.12. W hat are the charges of mono-, bi-, and triv a le n t ions? 12.13. The valencies of silver and gold are 1 and 3 re spectively. D eterm ine th eir electrochem ical equivalents. 12.14. How m any silver atom s are liberated at the cathode when silver n itra te solution is electrolyzed for 1 h at a cur rent of 1 A? 12.15. Two series-connected electrolytic baths contain solutions of copper sulphate and auric chloride respectively. As a result of the electrolysis, 2 g of copper was liberated at the cathode. How much triv a le n t gold was liberated in the other bath? How m any copper and gold atom s were deposited on the cathodes? 12.16. Copper is refined at a voltage of 0.3 V across the bath term inals. How much copper is deposited on the cathode over 1 h if the resistance of the electrolyte is 3 x 10-5 Q? Copper is bivalent. 12.17. To obtain alum inium , a current of 50 A is passed through m olten cryolite w ith alum ina at a voltage of 6 V. D eterm ine the energy consum ption per ton of alum inium and the resistance of the electrolyte. 12.18. In a copper refinery the cathodes are replaced after 10 days of continuous operation of electrolytic baths. The am ount of copper accum ulated on each electrode over this period is 71 kg. D eterm ine the current density if the area of a cathode is 0.9 m2. 12.19. D uring an electrolysis, 5 X 104 C of electricity were passed through a ferric chloride (FeCl3) solution. How much iron and chlorine are liberated in the process? At which electrode is the chlorine liberated? W hy?
§ 12. Electric Current in E lectrolytes
147
12.20. How long does it take to gold-plate a w atch case electro ly tically w ith a layer thickness of 12 pm a t a current density of 0.1 A /dm 2? 12.21. The reflectivity of an autom obile’s headlight is increased by an electrolytic coating of a 10-pm th ick silver layer. How long m ust the electrolysis of a silver n itra te solution w ith a current density of 0.3 A /dm 2 be to obtain the required thickness of the layer? 12.22. A m elt of alum inium salts is electrolyzed at a voltage of 6.5 V. How much electricity is spent to o btain 1 t of alum inium and how much does it cost at a rate of 2 kop ecks per kW h if the efficiency of the electrolyzer is 75% ? 12.23. How m uch alum inium is obtained by electrolysis if 100 kW h of electricity has been spent? The electrolysis is at a voltage of 6 V w ith an efficiency of 80% . 12.24. How long does it take for a copper anode 50 X 10 X 1 mm 3 in size to com pletely dissolve if a copper sul phate solution is electrolyzed a t a current of 0.3 A? 12.25. E lectro ly tic nickel p latin g of an article is carried out at a current density of 0.8 A /dm 2. D eterm ine the rate of growth of the nickel layer. The valency of nickel in the com pound is 2. 12.26. D uring the electrolysis of weakly acidified w ater, 0.5 1 of hydrogen are obtained a t a pressure of 0.13 M Pa over 50 m in. D eterm ine the tem perature of the hydrogen if the current in the circu it is 1.6 A. 12.27. A silver n itra te solution was electrolyzed for 5 m in, and the am ount of silver deposited on the cathode was 336 mg. An am m eter in the circu it indicated 0.9 A. W as this reading correct or has a correction to be m ade? 12.28. Given the electrochem ical equivalent of oxygen, determ ine the electrochem ical equivalent of hydrogen, sodium , and m agnesium . 12.29. How much electric charge passes through a silver n itra te solution in 20 s if th e current increases from 1 to 4 A during the process? How m uch silver is deposited on the cathode? 12.30. D eterm ine the electric power spent if a w eakly acidified w ater is electrolyzed for 25 m in lib eratin g 0.5 g of oxygen. The resistance of the electrolyte is 1.8 £2 and does not change w ith tim e. to
148
Ch. II. Fundam entals of Electrodynam ics
12.31. D uring an electrolysis of a silver n itra te solution, the current in the bath varied thus: I = 0.2 + 6 X 10_3J. How m uch silver was deposited on the cathode 300 s after the current started to vary? 12.32. A t w hat current density does the thickness of the silver layer in a silver n itra te solution grow a t a rate of 3 X 10~3 pm /s? 12.33. How m any atom s of a m onovalent m etal are depos ited per square m etre of cathode surface if an electrolysis is carried out for 10 min at a current density of 5 A /m 2? 12.34. D uring the electrolysis of a nickel sulphate solution, 2.19 g of nickel are deposited after 40 m in on the surface of the cathode. D eterm ine the polarization emf if the voltage across the bath term inals is 5 V and the resistance of the solution is 1.4 Q. § 13. ELECTRIC CURRENT IN GASES AND IN VACUUM
Basic Concepts and Formulas Under norm al conditions, all gases are nonconducting. They either do not contain charge carriers or the num ber of carriers in them is sm all. H eating or irra d ia tin g a gas ionizes it: its atom s lose a valency electron and become positive ions. Some of the electrons th a t escape the atom s recom bine w ith n eutral m ol ecules to form negative ions. Thus, positive and negative ions and electrons become the m obile charge carriers in the gas. In order to detach an electron from an atom , i.e. to ionize it, the following work m ust be done: A{= where
§ 1 3 . Electric Current in Gases and in Vacuum
149
An electron m ay acquire this energy at the expense of the work done by electric field forces: mv 2l 2 = eEX, when it trav els a distance X between two collisions. Positive ions also take part in ionization. The passage of an electric current through a gas is known as an electric discharge. There are several types of gas dis charge. Conduction in vacuum is determ ined by the presence of a source of charged particles. For exam ple, in a vacuum tube, electrons em itted from the surface of a heated cathode acquire a. kinetic energy from the field, i.e. mv 2/ 2 = Ue, where U is the voltage between the electrodes of the tube and is known as the accelerating voltage. The num ber of electrons em itted by the cathode depends on its tem perature. The sa tu ra tio n current density is given by j = env, where n and v are the num ber density and velocity of the electrons respectively. Worked Problems Problem 72. U nder norm al conditions, a spark discharge in air occurs at a field strength of 3 x 106 V/m . D eterm ine the energy required for an electron to ionize air m olecules if the mean free p ath of an electron is 5 pm. W hat m ust the m inim um velocity of an electron capable of ionizing air molecules be? Given: E = 3 X 106 V/m is the electric field stren g th and X = 5 x 10“6 m is the mean free path of an electron. From tables, we find the electron charge e = 1.6 X 10“19 C and the electron mass m = 9.11 x 10”31 kg. Find: the ionization energy W { and the m inim um ve locity v required for the ionization. Solution. The energy W x required to ionize air molecules is received by an electron at the expense of the work done by electric field forces: A = e (q^ —
150
Ch. II. Fundam entals of Electrodynam ics
strength: (px —
1 0 -19
C x
3 x
106 V/m x 5x 10"6 m
10-18J.
The electron moving in the electric field acquires ak i netic energy mv 2/2. The molecules m ay ionize when the kinetic energy of the electron is higher th an or equal to the ionization energy. To determ ine the m inim um velocity, we can use the eq u ality mi?2/2 = W x, from which u can be de term ined: 2.3 x 106 m/s. Answer. The ionization energy is 2.4 X 10‘ 18 J , and the velocity of the electrons is 2.3 X 106 m/s. Problem 73. The separation between the electrodes in an ionization cham ber is 6.2 cm, and the area of each electrode is 100 cm 2. An ionizer produces 109 pairs of ions per second in 1 cm 3 of the cham ber. The ion pairs have a m obility of 3.29 X 10~4 m2/(V*s). Assum ing th a t the ions are mono valent, determ ine the satu ratio n current and the field strength in the cham ber. Given: S = 10‘ 2 m 2 is the surface area of an electrode, I = 6.2 X 10~2 m is the separation between the electrodes, n = 2 X 1015 m -3 is the num ber density of the ions, b = 3.29 X 10-4 m 2/(V -s) is the m obility of the ions, and t = I s is the tim e of ionization. From tables, we find the electron charge e = 1.6 x 10~19 C. Find: the sa tu ra tio n current I and the electric field stren g th E in the cham ber. Solution. The charge carriers in gases are ions and elec trons. A t the sa tu ra tio n current, all the charge carriers reach the electrodes. The sa tu ra tio n current can be d eter mined from the form ula I = envS, where v = lit. This gives J — en -j- S.
8 13. Electric Current in Gases and in Vacuum
151
Using the form ula b = v!E for the m obility of charges, we determ ine the field strength: V— v — 1 b bt ' S u b stitu tin g in the num erical values, we o btain
c
/ = 1 . 6 X 10-‘® x 2 X 10-1® m-3 .6.;2-><10-2 m Is E-=
6 .2 X 10"2 m 3 .2 9 X 1 0 -4 m 2/(V*s) X 1 s
1 0 -2 m 2 =
0.2 uA, r
= 188 V/m.
Answer. The satu ratio n current is 0.2 pA, and the electric field stren g th is 188 V/m. Questions and Problems 13.1. U nder the action of an ionizer, a gas has become a conductor. A charged electroscope placed nearby sta rts to discharge rapidly. W hy does the discharge cease after the ionizer is rem oved? 13.2. Figure 56 shows a dependence of the current through a gas on the applied voltage. W hat processes correspond to
Va F ig. 56
F ig. 57
different sections of the graph? W hich section describes a Self-sustained discharge? 13.3. W hat is the difference in the conduction of gases and solutions? 13.4. W hat should be done to increase the sa tu ra tio n cu r rent? 13.5. Is im pact ionization possible at a low voltage if a gas is at atm ospheric pressure?
152
Ch. H. Fundam entals of Electrodynam ics
13.6. How can you explain th a t spark discharges occur only in te rm itte n tly ? 13.7. Give exam ples when a corona discharge is (a) haim ful and (b) beneficial. 13.8. W hat type of discharge is observed in day-light lam ps? W hat particles are the charge c a rrie rs in th is dis charge? 13.9. W hy does im pact ionization in rarefied gases become more intense w ith decreasing voltage? 13.10. W hy is the surface of the cathode in a vacuum tube coated w ith a th in layer of m etal (say, barium or stro n tiu m )? 13.11. Figure 57 shows the anode current versus anode voltage for different tem peratures of the filam ent of a diode. How can the presence of the horizontal lines in the graph be explained? W hat is the reason behind the increase in th e s a tu ra tio n current at higher tem peratures? 13.12. How can an electron beam be controlled? How is it controlled in the CRT tube of a television set? 13.13. A gas and a plasm a are both electrically neutral as a whole. W hat is the difference between them ? 13.14. D eterm ine the ionization potential for silver atom s if an energy of 6.9 X 10~19 J is required to ionize them . 13.15. How will the velocity of the electrons in a CRT change if the energy of an electron changes from 700 to 1000 eV as a result of a change in the voltage between the anode and the cathode? W hat is the electron velocity in the two cases? 13.16. For obtaining cathode rays, a voltage of 30 kV is applied to the electrodes of a gas discharge tube. D eterm ine th e m axim um velocity of the electrons in the cathode beam. 13.17. A potential difference of 300 V is applied between the cathode and the anode of a diode. D eterm ine the velocity of the electrons as they move in the tube if the separation between the cathode and the anode is 10 mm. How long do they move? 13.18. W hat m ust the m inim um velocity of electrons be to cause the im pact ionization of cesium atom s, whose work function is 1.8 eV? 13.19. The ionization energy for a hydrogen atom is 13.5 eV. W hat m ust the m inim um velocity of an electron be to cause the im pact ionization of the hydrogen atom ? 13.20. On average, five ion pairs are formed per second
§ 14. Electric Current in Sem iconductors
153
per cm3 of atm ospheric air near the E a rth ’s surface due to rad io a ctiv ity of soil and cosmic rad iatio n . The separation between two electrodes is 8 cm. D eterm ine the sa tu ra tio n current density between the electrodes for singly charged ions. 13.21. The work function of barium oxide electrons is 1.0 eV. W hat m ust the m ean free p ath of the electrons in a p arallel-plate capacitor be for the electrons to ionize the barium atom s? The field strength between the plates is 3 X 105 V/m . The field should be treated as uniform . 13.22. The satu ratio n current in an air-filled tu b e is 2 X 10‘10 A at an electric field strength of 30 V/m between the electrodes. The overall m obility of m onovalent ions is 3.29 x 10~4 m2/(V -s), and the area of plane electrodes is 100 cm2. D eterm ine the num ber density of the ions. 13.23. D eterm ine the overall m obility of hydrogen ions if the satu ratio n current density is 2.8 x 10"10 A/m 2 at a field strength of 1.2 kV/m and a num ber density of ions of 104 cm -3. § 14. ELECTRIC CURRENT IN SEMICONDUCTORS
Basic Concepts and Formulas Sem iconductors are m aterials having resistiv ity between th a t of conductors and insulators; their resistiv ity decreases w ith rising tem perature and the presence of im purities. T ypical sem iconductors are elem ents of group IV of the Periodic T able, whose atom s have four valency electrons in the outer shell (like germ anium Ge and silicon Si). At low tem peratures, crystals of these elem ents do not contain free electrons and are good insulators. The covalent bonds in such crystals are ruptured w ith increasing tem perature, illum inance, or due to strong electric fields. Free electrons appear, and intrinsic electron conduction emerges in (n-type) semiconductors. In trin sic hole conduction of (p-type) sem i conductors is due to the displacem ent of holes. Im p u rity conduction in sem iconductors is due to the pres ence of “alien” group V elem ents (like arsenic or antim ony) or group I II elem ents (like boron and alum inium ). In the former case, im p u rity electron conduction is created, and in the la tte r, im p u rity hole conduction sets in. Thus, the
154 __________Ch. II. Fundam entals of Electrodynam ics
introduction of im purities into pure sem iconductors can destroy the equilibrium between the p-type and w-type con ductions. W hen p- and rc-type sem iconductors are brought in con tac t, a barrier layer emerges as a result of diffusion through the p-n junction. The application of an external field to a p-n junction m ay change its conduction and create the con ditions for u n ilateral conduction. A sem iconductor diode is a sem iconductor w ith a p-n junction. Its advantages over vacuum tube diodes include sm all size, re lia b ility , and efficiency. Questions and Problems 14.1. Figure 58 shows the tem perature dependences of resistance for conductors and sem iconductors. W hich of them corresponds to sem iconduc tors? 14.2. W hat are the m obile charge carriers in a pure sem iconductor? 14.3. W hat is the ratio between the num ber of holes and the num ber of free electrons in a pure sem i conductor? Is this ratio preserved for im p u rity conduction of sem icon ductors? 14.4. W hat w ill the type of con duction in germ anium w ith trace im purities of phosphorus or alum inium be? 14.5. How do the conductivities of germ anium and silicon vary w ith lowering tem perature? 14.6. By w hat factor will the current density in a sem i conductor change if the velocity of the electrons increases from 0.5 to 0.75 m/s as a result of a tem perature increase from 0 to 175°C, while the electron num ber density increases thereby from 1.3 x 1014 to 2.1 X 1018 m~3? 14.7. The velocity of directional m otion of free electrons in a sem iconductor is 0.25 m /s for a given tem perature. D eterm ine the m obility of the charges and their num ber den sity if the current density is 4 X 10“2 A /m 2 for a field strength of 100 V/m.
§ 15. E lectrom agnetism
155
14.8. W hat is a therm istor? W hy are therm istors called nonlinear resistances? 14.9. W hat is the difference between a therm istor and a photoresistor? 14.10. W hat is a transistor? W hich regions does the crystal of a tran sisto r contain? 14.11. The thickness of the base in a tran sisto r is very sm all (1-25 pm). W hy? 14.12. W hat is the ratio between the e m itte r, base, and collector currents? 14.13. W hat is the advantage of sem iconductor devices over vacuum tubes in radio engineering? § 15. ELECTROMAGNETISM
Basic Concepts and Formulas A m agnetic field is a special case of electrom agnetic field, being characterized by the action on a m oving charged particle of a force proportional to the charge of the particle and its velocity. The shape of a m agnetic field de pends on the shape of the currentcarrying conductor producing it. For exam ple, the m agnetic field formed around a stra ig h t currentcarrying conductor is graphically represented by m agnetic field lines in the form of concentric rings in the plane perpendicular to the d i rection of the current (Fig. 59). The direction of the m agnetic field in this case is given by Am pere’s (rightFig. 59 hand screw) rule: ro tatio n of the head of a screw indicates the direc tion of the m agnetic field lines if the m otion of the screw body coincides w ith the direction of the current. The m agnetic field of a current-carrying coil (solenoid) is sim ilar to the m agnetic field of a perm anent bar m agnet. The interaction between m agnetic fields formed by par allel current-carrying conductors is determ ined by the
156
Ch. II. Fundam entals of Electrodynam ics
form ula F= a
rm
/l/a * 2:la
'
where I x and I 2 are the currents in the conductors, I is the length of wire over which the force is acting, a is the sepa ration between the conductors, and |xm is the absolute per m eability of the m edium characterizing the dependence of the force of in teractio n of current-carrying conductors on the properties of the medium: M'm = HoHHere p 0 = 4jt x 10"7 H /m is the m agnetic constant, and p is the rela tiv e perm eability of the m edium (see T able 21). A m agnetic field exerts a force F \ (Ampere’s law) on a current-carrying conductor: FA = B I l sin a . If the conductor is perpendicular to m agnetic field lines (a = 90°), we have Fa = B IL The p ro p o rtionality factor B is called the m agnetic induction and is the force characteristic of the m agnetic field. M agnetic induction is a vector q u a n tity . The SI /* unit' m agnetic induction is the tesla (T). ( X \ ^ any P °in t *n a uniform m agnetic field, the v 5 ] m agnetic induction has the same m agnitude and direction. Therefore, such a field is graphically represented by parallel straig h t lines w ith uniform density. Vig. 60 m agnetic flux is equal to the num ber of m agnetic field lines piercing a surface of area S if the m agnetic induction vector coincides w ith the norm al to this surface: O = BS. The SI u n it of m agnetic flux is the weber (Wb). The m agnetic properties of a current loop are characterized by the m agnetic moment P mag (Fig. 60): Pmag = IS -
§ 15. Electromagnetism
157
A current-carrying loop in a uniform magnetic field of induction B experiences a rotational magnetic moment (torque) M: M = I S B sin a, where a is the angle between vectors B and Pmag. For 90°, the torque will have the maximum value M max = I S B = P masB. A closed current-carrying loop of length I is displaced by a force F \ through a distance b. Consequently, the following work is done: A = F\b = Bllb. But lb is the change in the area, AS, so A = B I AS, or A = I AO. The magnetic induction of a straight current-carrying conductor is B = \i
1
m 2nr '
where r is the shortest distance from the current-carrying conductor to the point at which the induction B is being determined. The magnetic induction of the field produced by a circular current is #
2r •
=
1
The magnetic induction of the field in a solenoid is n
f
0
)
l
'
where o> is the number of turns and I is the length of the solenoid. The magnetic field inside a long solenoid is uniform. Therefore, the magnetic flux in the solenoid is 0
=
B
S
=
] x
m
-
^
S
.
The magnetic field is also characterized by the magnetic field strength H which is related with the magnetic induction
158
Ch. II. Fundam entals of Electrodynam ics
through the form ula B = Hm#The force acting on an electric charge Q moving in a mag netic field is known as the Lorentz force F l : = BvQ sin a , where a is the angle between vectors B and v. The Lorentz force is alw ays perpendicular to the plane containing vectors B and v and hence does no work. W ithout changing the m ag nitude of the velocity of the charge, it only changes its d i rection and is responsible for centripetal acceleration. If a = 90°, we have
i.e. a particle of mass m and charge Q moves in a circle under the action of force F L. W orked Problems Problem 74. A direct current passes in the same direction thrpugh two parallel wires separated by a distance of 30 c m . The distance between the supports to which the wires are fixed is 50 m. The current in the wires is 150 A. D eterm ine the m agnitude and direction of the force w ith which the wires interact. Given: I x = I 2 = I = 150 A is the current in each w ire, a = 0.3 m is the separation between the wires, and I = 50 m is the distance between the supports. From tables, we find the m agnetic constant p,0 = 4n X 10"7 H /m and the perm eability of air, p, = 1. Find: the force F between the wires (its m agnitude and direction). Solution. Let us consider the distance between the su pports to be the active length of the wires. N oting th a t I » a, we can assume th a t the wires are infinitely long so th a t the following form ula is applicable for the force of interaction:
p F ~
4j i X lO"7 H /m X 1 (150 A )* 50 m _ n 7 r „ 2 n x 0 .3 m
§ 15. Electrom agnetism __________
159
In order to determ ine the direction of the force acting between the wires, we shall analyze Fig. 61. The m agnetic field between the wires is weaker (the m agnetic field lines are directed against one another). In the outer region, the m agnetic field lines have the same direction, and the m ag netic field is larger. C onsequently, wires carrying current in the same direction m ust a ttra c t each other. / X Answer. The force of attrac tio n J y ** between the wires is 0.75 N. Problem 75. W hen a 0.5-m long straig h t conductor carrying a curFig* 61 rent of 4 A is at right angles to a uniform m agnetic field, the field acts on it w ith a force of 2.8 N. W hat w ill the force exerted by the same field on the conductor be if the angle between them is 30°? Given: I = 0.5 m is the length of the conductor, / => 4 A is the current in it, F x = 2.8 N is the force corresponding to 90°, and a x = 90° and a 2 = 30° are the angles between the m agnetic field lines and the current in the first ahd the second case. Find: the force F 2 acting on the conductor. Solution. Using A m pere’s law Fx = B I l sin a lt we d e te r mine the m agnetic induction of the uniform m agnetic fie ld : 5 = -/f/ rsm ^— a ! .’
B = 4/ vA -x T0 . ?5 m—x lj- = 1.4 T.
Knowing the m agnetic induction, we can determ ine the force exerted by the m agnetic field on the current-carrying conductor when the angle between them is 30°: F 2 = B I l sin a 2,
F a = 1.4 T x 4 A x 0.5 m x 0.5 = 1.4 N.
Answer. The force acting on the conductor in the second case is 1.4 N. Problem 76. A 0.8-m long stra ig h t conductor moves in a m agnetic field of induction 5 X 10~2 T. The current in the conductor is 15 A. The conductor is at an angle of 30° w ith the m agnetic induction vector. Find the force acting on the conductor and the work done by the m agnetic field to move the conductor 1.8 m. Given: B = 5 x 10~2 T is the m agnetic induction of the field, I = 0.8 m is the length of the conductor, / = 15 A
160
Ch. II. Fundam entals of Electrodynam ics
is the current in the conductor, a = 30° is the angle be tween the direction of current I and vector B, and b = 1.8 m is the distance by which the conductor is moved. F in d : the force F acting on the conductor and the work A done by the m agnetic field to move the conductor. Solution. A current-carrying conductor in a m agnetic field is acted upon by the force F = B I l sin a , F = 5 x 10-2 T x 15 A x 0.8 m x 0.5 = 0.3 N. K now ing the force acting on the conductor and the dis tance by which it is moved, we can find the work: A = Fb, A = 0.3 N x 1.8 m = 0.54 J. Answer. The m agnetic field acts on th e current-carrying conductor w ith a force of 0.3 N and does a work of 0.54 J to displace it. Problem 77. A 30-cm long solenoid contains 3000 turns. The diam eter of each tu rn is 11 cm. D eterm ine the m agnetic induction of the field inside the solenoid for a current of 1.5 A and the m agnetic flux piercing each tu rn . W hat w ill the change in the m agnetic induction and in the m agnetic flux be if a carbon iron core is inserted into the solenoid? Given: I = 0.3 m is the length of the solenoid, N = 3000 is the num ber of turns in it, d = 11 X 10‘ 2 m is the di am eter of a tu rn , I = 1.5 A is the current; from tables, we find the m agnetic constant p 0 = 4 ji X 10”7 H /m and the perm eability of carbon iron, p = 3000. Find: the m agnetic induction B x of the field in the solenoid w ithout a core, the m agnetic induction B 2 of the field in the solenoid w ith the core, and the corresponding m agnetic fluxes and 2. Solution. Assum ing th a t the solenoid is long enough for the m agnetic field in it to be uniform and directed along the axis, the m agnetic induction can be determ ined from the form ula
In the absence of a core, p = 1 (for ,.ir). Then we have B x =■= 4 ji
x 10 7 H /m
1 .5 A X 3 0 0 0 0 .3 m
= 1.9 x 10-* T.
§ 15. Electrom agnetism
161
In the presence of the core, the m agnetic induction in creases by a factor of [i: B 2 = 3000 x 1.9 x lO '2 T = 57 T. The m agnetic flux can be determ ined from the form ula O = B S cos a. In the problem under consideration, a = 0 and cos a = 1, while S = nd 2!k. This gives
1 .9 X IQ-2 T X 3 .1 4 X 121 x 10~4 m 2 = l j 8 x 1 0 -'. W b
In the presence of the core, (D2 = 3000 x 1.8 x lO"4 W b = 0.54 W b. Answer. The m agnetic field induction in the solenoid w ithout a core is 1.9 x 10~2 T and 57 T w ith the core. The m agnetic fluxes in the solenoid are 1.8 x 10~4 and 0.54 W b respectively. Problem 78. A current-carrying conductor is looped into a circle of radius 10 cm. The m agnetic m om ent of the current loop becomes 0.314 A -m 2. D eterm ine the current in the loop and the m axim um torque exerted on it by a uniform m agnetic field of induction 5 X 10~3 T. Given: r = 0.1 m is the radius of the loop, p mag = 0.314 A -m 2 is the m agnetic m om ent of the loop, and B = 5 X 10-3 T is the m agnetic induction of the field. Find: the current / in the loop and the m axim um torque ^m ax-
Solution. K now ing the m agnetic m om ent of the loop, we can determ ine the current in it: p mag = I S , where S = n r 2, which gives p mag = I nr2, and hence T
Pmag
~
jir2
r
’
”
0 .3 1 4 A - n > 2 3 .1 4 X 10-2 m 2
in
\
The m axim um torque can be determ ined from the form ula ^m ax =
Pm ag^»
M mBx = 0.314 A -m 2 x 5 x 10"3 T ^ 1.6 x 10"3 N -m . Answer. The current in the loop is 10 A, and the m axim um torque is approxim ately 1.6 X 10“3 N -m . 11-0530
162___________ Ch. II. Fundam entals of E lectrodynam ics
Problem 79. A proton th a t has acquired a velocity while m oving across a potential difference of 1 kV enters a uniform m agnetic field of induction 0.2 T at right angles to the m ag netic field lines. D eterm ine the radius of the circle in which i t w ill move and the period of its revolution. Given: U = 1000 V is the accelerating poten tial difference, B = 0.2 T is the m agnetic induction of the field, a = 90° is the angle between vectors B and v. From tables, we find the mass of the proton m p = 1.67 X 10~27 kg and its charge Q = 1.6 x 10-19 C. Find: the radius r of the circle and the period T of revo lu tio n of the proton. Solution. An electric charge (in our case, the proton) m ov ing in a m agnetic field is acted upon by the Lorentz force F l = BvQ sin a , where a is the angle between vectors B and v. Since a = 90° and sin a = 1, we obtain F h = B vQ . Since the Lorentz force is always norm al to the plane con tain in g vectors B and v, it does no work, i.e. it does not change the kinetic energy of the m oving charge. The force only changes the direction of the velocity. Hence we can w rite BvQ = m pv2/r, whence *
BQ
In order to find the proton velocity, we can use the energy conservation law: the work done by the electric field U is equal to the kinetic energy acquired by the proton: Q U = m pv 2/2, whence -
2 X 4 .6 X 10~19 C X 40® V kg
4.67 X 40~27 v = 4.4 x 105 m/s.
W e can now determ ine the radius r of the circle: 4.67 x 40~27 kg x 4 .4 X 405 m /s r
=
0T
t
x
i
. 6 ' x
i
F
*
c
_
n n9Q w u
u
"
m
-
G iven the proton velocity and the radius of its o rb it, we can find its period: m
2jir
T = -2- * 3. y * 0- ° f m ~ 0.033 x 10"8 s ~ 0.3 |is. 4 .4 x 4 0 * m /s
r
§ 15. Electrom agnetism
Answer. The proton moves in a circle of radius 0.023 with period of revolution 0.3 .
163
m
jl is
Questions and Problems 15.1. A looped flexible conductor tends to expand into a circle when a current is passed through it. Why? 15.2. Using a d.c. voltm eter and a m agnetic needle on a pivot, one can determ ine the side on which a generator is located in a two-wire d.c. cable. How?
'FT Fig. 63
15.3. D eterm ine the polarity of the electrom agnets shown in Fig. 62. 15.4. D eterm ine the direction of the currents in the w ind ings shown in Fig. 63. 15.5. D eterm ine the p o larity and the direction of the m agnetic field lines for the generator shown in Fig. 64. 15.6. D eterm ine the direction of m otion of current-carry ing conductors in the m agnetic fields shown in Fig. 65. 15.7. W hat w ill happen to a perm anent bar m agnet if the current in the solenoid has the direction indicated in Fig. 66? 15.<8. W hat w ill the m otion of the coil be relativ e to the solenoid for the direction of the current indicated in Fig. 67? 15.9. W hat m ust the p o larity of the m agnet in Fig. 68 be for th e current loop to ro ta te clockwise? Through w hat angle will the loop be turned? u*
164
Ch. II. Fundam entals of Electrodynam ics
15.10. W hat is the force of the interaction between the wires of a d.c. trolleybus line over 30 m if the separation between the wires is 520 mm and the current in them is 200 A? 15.11. A d.c. transm ission line is rated for a current of 150 A. W hat is the separation between two wires if they
Ld 0
n F ig. 65
Fig. 66
in teract w ith a force of 2.8 X 10”1 N over a distance of 50 m? 15.12. Two parallel conductors carrying the same current are 25 cm apart. D eterm ine the current in each wire if it experiences a force of 2 mN per m etre. 15.13. Two parallel wires are 6 cm apart. The currents in the wires are 15 and 10 A. Over what segment of the wires will a force of 1.4 mN act? 15.14. A 0.5-m long stra ig h t conductor carrying a current of 5 A is in a uniform m agnetic field of induction 0.16 T.
§ 15. Electrom agnetism
165
D eterm ine the forces acting on the conductor when tike angles between the direction of current in it and the m agnetic induction vector are 90°, 30°, and 0°. 15.15. D eterm ine the m agnetic induction of a uniform m agnetic field in which a 0.7-m long stra ig h t conductor carrying a current of 10 A is acted upon by a force of 42 mN. The angle between the direction of current and the m agnetic induction vector is 30°. 15.16. Using a strong horse-shoe m agnet, one can deter mine w hether an incandescent lam p is connected to an a.c. or a d.c. source. How can this be done? 15.17. A force of 0.6 N acts on a 1-m long stra ig h t conduc tor placed in a uniform m agnetic field at right angles to the
A/ M / V ----------------IT'.
B
My
•
• • • • F ig. 69
'/V
Iz
Cf
Fig. 70
magnetic field lines. The current in the conductor is 12 A. W hat will the force exerted on th is conductor be if the angle between the direction of current in it and the m agnetic induction vector is 45°? Solve the problem using two different methods. 15.18. A conductor M N m ade of a m aterial w ith density p and having cross section S is suspended on two w eightless unstretchable strings in a uniform m agnetic field of induction B (Fig. 69). A t w hat current w ill the tension in the strin g be zero? W hat m ust the direction of this current be? 15.19. The m agnetic field stren g th 10 cm from a long cur rent-carrying stra ig h t conductor is 20 A/m. W hat force w ill act on each m etre of th is wire if it is placed in a uniform magnetic field w ith induction 2.5 T so th a t the angle between the direction of the current and the m agnetic induction vector is 30°? 15.20. D eterm ine the current th a t m ust be passed through
166___________ Ch. II. Fundam entals of Electrodynam ics
a long s tra ig h t wire to produce a m agnetic field of the same m agnitude 1 m away as the m agnetic field of the E a rth near its surface. The m agnetic induction of the m agnetic field of the E a rth should be taken to be 5.5 x 10-6 T. 15.21. D eterm ine the m agnetic field stren g th and the m agnetic induction of the field produced by a stra ig h t con ductor carrying a current of 7.8 A at a point 4.8 cm away. 15.22. A stra ig h t conductor carrying a current of 10 A produces a t a certain point a m agnetic field of strength 46 A/m. D eterm ine the m agnetic induction a t th is point and the distance between this point and the conductor. K 15.23. Two long current-carrying conductors are arranged at rig h t angles in the same plane (Fig. 70). D eterm ine the resu lta n t m agnetic induction of the field a t points M and N if I x = 10 A, / a = 6 A, MO = NO = 5 cm, and M C = N C X = 4 cm. 15.24. A current-carrying circular conductor produces a m agnetic field of strength 25 A/m at the centre. D eterm ine the m agnetic induction of the field and the radius of the loop if the current is 3.45 A. 15.25. The induction at the centre of a circular current of radius 4 cm is 1.57 x 10-4 T. D eterm ine the m agnetic field strength at the centre of the loop and the current. 15.26. The current passing through a stra ig h t long con ductor produces a m agnetic field of induction 0.8 x 10~4 T 4.4 cm away. The perm eability of the m edium is 1.1. De term ine the current in the conductor and the m agnetic field strength 16 cm away. 15.27. D eterm ine the m agnetic m oment of a circular loop carrying a current of 10 A and having a radius of 6.0 cm. 15.28. The m agnetic m om ent of a current-carrying w ire ring 15 cm in diaimeter is 4.2 x 10~2 A -m 2. D eterm ine the current in the ring and th e m agnetic field stren g th a t its centre. 15.29. A wire ring of rad iu s 5.0 cm carrying a current of 6 X 10~2 A is in a uniform m agnetic field of induction 1.2 X 10"2 T. The plane of the ring is parallel to the m agnet ic field lines. D eterm ine the m axim um m agnetic m oment exerted by the m agnetic field on the ring. 15.30. A solenoid w ithout a core is 100 cm long and con tains 600 turns. D eterm ine the m agnetic induction of the field in the solenoid for a current of 0.4 A.
§ 15. E lectrom agnetism
167
15.31. The current in a solenoid whose diam eter is sm all in com parison w ith its length is 6.5 A. The solenoid is 65 cm long and contains 750 turns. D eterm ine the m agnetic field strength and the m agnetic induction in the solenoid w ithout a core. 15.32. The m agnetic induction of a very long solenoid carrying a current of 3.0 A is 2.52 X 10~3 T. The solenoid is wound tig h tly in one row and has no core. D eterm ine the diam eter of the wire of which the solenoid is made. 15.33. D eterm ine the perm eability of soft steel in two m agnetic fields w ith strengths 1.5 and 5.0 kA/m respectively (see graph 22). How does the perm eability change w ith increasing m agnetic field strength during prim ary mag netization? W hy? 15.34. D eterm ine the m agnetic induction in a nickel core and the m agnetic flux if the strength of the uniform m agnetic field in the core is 2.0 X 103 A/m , the cross-sectional area of the core is 30 cm2, and the perm eability is 200. 15.35. The w inding of a solenoid w ithout a core is made of a wire whose diam eter is 1.0 X 10~3 m. The tu rn s have a radius of 1.0 cm and are wound tig h tly . D eterm ine the m ag netic flux in the solenoid for a current of 2.0 A. 15.36. A rectangular fram e m ade of a wire carrying a cur rent of 2.0 A is in a uniform m agnetic field of induction 0.10 T. The size of the fram e is 4 x 5 cm 2. In a certain po sition, the m agnetic flux through the fram e is 0.80 x 10~4 W b. D eterm ine the m axim um m agnetic flux in the frame when its plane is perpendicular to the m agnetic field lines. D eterm ine the work done in ro ta tin g the fram e to th is position. 15.37. The m agnetic induction in the core of an electro m agnet is 1.2 T and its cross-sectional area is 0.12 m 2. De term ine the m agnetic flux through the core. 15.38. A steel rod having a cross-sectional area of 4.5 cm 2 and a perm eability of 160 is placed in a uniform m agnetic field of strength 7970 A/m so th a t the m agnetic field lines coincide w ith the norm al to the cross section of the rod. De term ine the m agnetic flux piercing the rod. 15.39. The m agnetic field strength in a solenoid w ith an iron core is 1600 A/m. The cross-sectional area of the core is 10 cm2. Determ ine the m agnetic induction of the field and
Ch. H. Fundam entals of Electrodynam ics
the perm eability of iron if the m agnetic flux through the core is 2 X 10-4 W b. 15.40. An electron flies into a uniform m agnetic field as shown in Fig. 71. D eterm ine the direction of the force acting on the electron at the in itia l m oment. W hat w ill its tr a jectory be? 15.41. An electron moves in a uniform m agnetic field a t a velocity of 1.0 X 104 km /s at rig h t angles to the m agnetic induction vector. D eterm ine the force + + + + + acting on the electron for a mag+ + + + + netic field strength of 150 A/m. g x V 3 15.42. An electron flies into a uni+ + + + + form m agnetic field w ith induction + + + + + 9.1 x 10-5 T. The electron velocity is 1.9 X 107 m/s and at rig h t angles F ig. 71 to the m agnetic induction vector. De term ine the radius of the circle in which the electron will move, the period, and frequency of its revolution. 15.43. The Lorentz force exerted on an electron by crossed electric and m agnetic fields is determ ined by the form ula F l = eE + evB. W hat m ust the direction and m agnitude of the electron velocity be for it to move uniform ly in a straig h t line? 15.44. An electron flies into a uniform m agnetic field w ith induction 2.5 X 10~3 T and moves in a circle of radius 40 cm. The electron velocity vector forms an angle of 90° w ith the direction of the m agnetic field. D eterm ine the k i netic energy of the electron. 15.45. An electron having a velocity of 8.8 X 107 m /s flies into a uniform m agnetic field of induction 6.28 X 10”2 T. The angle between the velocity and m agnetic induc tion vectors is 30°. D eterm ine the radius and the lead of the helical trajectory of the electron. Use the charge-to-mass ratio of the electron, to the th ird significant digit. K 15.46. Two identical, singly charged ions fly at different velocities into a uniform m agnetic field. W hat w ill the periods of their revolution be? 15.47. An electron and a singly charged ion fly at the same velocity in to a uniform m agnetic field. W hat w ill the periods of their revolution be?
§ 16. E lectrom agnetic Induction
169
§ 16. ELECTROMAGNETIC INDUCTION
Basic Concepts and Formulas Electrom agnetic induction consists in the emergence of an induced emf and induced current in a closed loop if the m agnetic flux bounded by th is loop varies w ith tim e: e AO “ ~aT Let us suppose th a t a rectangular frame lying in a plane perpendicular to a m agnetic field moves at a velocity v and leaves the m agnetic field. Then the m agnetic flux piercing the fram e varies as AO = —B lv At. Therefore, the induced emf can be w ritten in the form g, = - B l v . If vectors v and B form an angle a , we have = —Blv sin a . An emf can also be induced in a statio n ary loop if the m agnetic induction of the field varies over tim e. The direction of the current induced in a closed loop can be determ ined using the right-hand rule or Lenz’s law. According to Lenz’s law , the direction of the induced current is such th a t the m agnetic field of this current opposes any change in the m agnetic field inducing the current. Self-inductance can be regarded as a special case of elec trom agnetic induction when a changing m agnetic flux is produced by a current as it varies. For exam ple, when a c ir cuit is closed, the self-inductance emf opposes (in accordance w ith Lenz’s law) the increase in the current. W hen a circuit is disconnected, the self-inductance emf opposes the decrease in the current. For th is reason, the induced current is d i rected against the m ain current when the circuit is connected and along the m ain current, when the circuit is disconnected. The self-inductance emf is proportional to the rate of change in the current in the circuit:
where L is the inductance of the circuit and is determ ined by the size and shape of the conductor in it and by the mag
170
Ch. II. Fundam entals of E lectrodynam ics
netic properties of the m edium containing the circuit. The SI u n it of inductance is the henry (H). Since it is a com ponent of electrom agnetic field, the mag netic field has energy which is given by ^ = 42_/=42- i / 2= 21 4,1 L W orked Problems Problem 80. A conductor A B is 0.6 m long and has a re sistance of 0.2 Q. I t can move along a copper busbars CD connected to a current source w ith an emf of 0.96 V and an in tern al resistance of 0.1 Q (Fig. 72). The resistance of the busbars is negligibly sm all. De +++ +++ + + term ine the current in the con + + + + + + -!- + ductor if it (a) is at rest, (b) moves + +8 T +++ — at a velocity of 0.5 m /s in a +++ ++ + + + uniform m agnetic field of induc ++ + ++ + + + tio n 1.6 T directed along the norc B m al to the plane of the figure Fiff 72 away from us. Given: Z=0.6 m is the length of the conductor, R = 0.2 Q is its resistance, % = 0.96 V is the emf of the current source, r = 0.1 Q is its in tern al resistance, v = 0.5 m /s is the velocity of the conductor, and B = 1.6 T is the m ag netic induction of the field. Find: (a) the current I x in the statio n ary conductor and (b) the current / 2 in the conductor moving in the m agnetic field. Solution. (a) If the conductor is at rest, the current in it is determ ined from Ohm ’s law for a circuit: 1
/? + r *
0.2Q + 0.1Q
(b) If the conductor moves a t a velocity v in a uniform m agnetic field, an emf £, is induced in it. If the conductor is a part of a closed circuit, then a current is induced in it w ith a direction determ ined by the right-hand rule. In the problem under consideration, the induced current is directed
§ 16. Electrom agnetic Induction
171
against the current I v Ohm’s law in this case has the form r _ 8 -# i hy *- - «+ r Sirtce Jf, = Blv, we obtain , _
% — Blv
R+r
Here we m ust note th a t the m agnetic induction vector and the velocity vector are at rig h t angles to each other. Hence, r
i2 _
0 .9 6 V — 1 .6 T X 0 .6 in X 0 .5 in /s A a v 0.2Q + 0.1Q - l . o A.
Answer. The current is 3.2 A in the statio n ary conductor, and 1.6 A when the conductor moves in the m agnetic field. Problem 81. A circular coil 10 cm in diam eter is placed in a uniform m agnetic field w ith induction 0.12 T so th a t the m agnetic induction vector is norm al to the plane of the coil (Fig. 73). D eterm ine the num ber of turns in the coil if an emf of 0.942 V is induced in it as a re sult of its ro tatio n through 180° in 0.14 s. Fig. 73 Given: B = 0.12 T is the m agnetic in duction of the field, d = 0.1 m is the diam eter of atu rn , a = 180° is the angle of ro tatio n of the coil, At = 0.14 s is the tim e in terv al, and {^ = 0.942 V is the induced emf. F in d : the num ber N of turns in the coil. Solution. W hen the coil is ro tated , the m agnetic flux boniided by the contour of th e coil changes, and the emf induced in it is
The problem statem ent indicates th a t before the coil is rotated the angle between its norm al and the m agnetic induction is a 0 = 0, and the m agnetic flux through the coil is O, = B S cos a 0. A fter the ro tatio n , the m agnetic flux becomes 0 2 = B S cos a , where a = 180°. The change in the m agnetic flux is 0 2 — = AO, AO = B S cos 180° — BS cos 0°, cos 180° = —1 and cos 0° = 1. Therefore, AO =
Ch. II. Fundam entals of Electrodynam ics
172
—B S — B S — —2BS. Since S = jic/2/4, we obtain A d)= S u b stitu tin g the change in the m agnetic flux into the expression for emf, we obtain ~ -7^ - N . Then the num ber of turns is v
2 x 0 . 9 4 2 V x 0 .1 4 s _ 0 .1 2 T X 3 .1 4 X 1 0 - 2 m 2
2exM
fljuf2
’
Answer. The num ber of turns is 70. Problem 82. D eterm ine the inductance of a coil in which the self-inductance emf induced by a change in the current from 5 to 10 A in 0.1 s is 10 V. W hat is the change in the energy of the m agnetic field of the coil in this case? Given: I x = 5 A is the in itia l current in the coil, / 2 = 10 A is the current after tim e At, Af — 0.1 s is the tim e during which the current changes, and £ s - 10 V is the self-inductance emf. F ind: the inductance L in the coil and the change A W in the m agnetic field energy of the coil. Solution. We w rite the form ula for the self-inductance emf %s = —L ^ and then determ ine the inductance of the coil: j
gs A/
~
$ s &t I2- I 1
10 V x 0 .1 s 10 A — 5 A
f
’
n o T-T
The m agnetic field energies for currents / , and w — LI* vv 1 — 2 ’
w — 2“
/ 2
are
2 *
The change in the energy is then
w = wt - Wi = - ^ L
_
_
=
J±. (II - If),
A W = - - - | H [(10 A)2 - (5 A)2] = 7.5 J. Answer. The inductance of the coil is 0.2 H, the energy of the m agnetic field increases by 7.5 J w ith current. Problem 83. A solenoid w ith a nickel core has 1000 tu rn s on 0.5 m, the cross-sectional area of a tu rn being 50 cm 2. Determ ine the m agnetic flux in the solenoid and the m agnet
§ 16. Electrom agnetic Induction
173
ic field energy if the current in the solenoid is 10 A and the perm eability of nickel is 200. Given: I = 0.5 m is the length of the solenoid, N = 1000 is the num ber of turns, S = 5 X 10‘3 m 2 is the crosssectional area of a turn, / = 10 A is the current, p = 200 is the perm eability of nickel. From tables, we find the m ag netic constant p 0 == 4jt X 10~7 H/m. Find: The m agnetic flux O in the solenoid and the energy W of the m agnetic field of the solenoid. Solution. The m agnetic flux can be calculated from the form ula O = B S . Since the m agnetic induction of the field / jy of a solenoid w ith a core is B = p 0p — , we have
O = 4ji x 10"7 H/m x 200
10 A X 1000 0 .5 m
5 x 10“3 m2
= 2.5 x lO-2 Wb. Given the m agnetic flux and the current, we can determ ine the energy of the m agnetic field in the solenoid: W =
= Y X 2.5 x 10-* W b x 10 A ~ 1.3 x 10"1 J.
Answer. The m agnetic flux in the solenoid is 2.5 x 10“2 W b, and the energy of the m agnetic field in the solenoid is 0.13 J . Questions and Problems 16.1. W ill an emf be induced in the conductors m oving as shown in Fig. 74?
Fig. 74
Fig. 75
16.2. W hen the poles of a horse-shoe m agnet are closed w ith an arm ature (Fig. 75), the pointer of the galvanom eter is deflected. W hy?
174
Ch. II. Fundam entals of Electrodynam ics
16.3. D eterm ine the direction of the emf induced in doctors m oving in uniform m agnetic fields shown in Fig. 76. 16.4. A rectangular frame in a translatory m otion <(a) enters a uniform m agnetic field, (b) moves in it, (c) leaver 10
tb)
M
B
+
+ N
Fig. 76
the field. The norm al to the fram e plane is directed aloMg the induction line. Is an emf induced in the fram e under these conditions? W hy? 16.5. A rectangular frame rotates in a uniform m agnetic field about an axis parallel to the m agnetic field lines. Is an emf induced in this case? (a)
(/>)
(cj
to
w w w w • - ! > cL Z .
Q 1 Z
I Z Z D
M M M H Fig. 77
16.6. D eterm ine the direction of the current induced in a frame (Fig. 77) ro ta tin g in a uniform m agnetic field in the direction indicated by the arrow. 16.7. Two identical steel rods are brought close to a cop per ring suspended in the vertical plane. In one case, the ring is repelled from the rod. W hy? 16JJ. D eterm ine the direction of the current induced £ solenoid if a bar m agnet is moved upwards in it (Fig. 78)..
§ 16. E lectrom agnetic Induction
175
16.9. A bar m agnet is caused to approach a copper ring as shown in Fig. 79. D eterm ine the direction of the current induced in the ring. 16.10. A perm anent bar m agnet is falling through a cop per cylinder. W ill the motion be free fall? 16.11. A conductor th at has been folded in two moves in a uniform m agnetic field crossing m agnetic field lines. Is an s
Fig. 78
Fig. 79
emf induced in the conductor? W hat would a galvanom eter attached to the ends of the m oving conductor indicate? 16.12. W hat is the value of the emf induced in a circuit if the m agnetic flux through it varies by 3.4 X 10-2 W b/s? 16.13. A conductor of active length 15 cm moves at 10 m /s at right angles to the lines of a uniform m agnetic field with induction 2.0 T. W hat current is induced in the conductor if it is short-circuited? The resistance of the circu it is 0.5 Q. 16.14. A linear conductor of active length 0.7 m crosses a uniform m agnetic field a t an angle of 30° and a t 10 m /s. Determ ine the m agnetic induction of the field if the emf induced in the conductor is 4.9 V. 16.15. D eterm ine the emf induced at the ends of an aero plane’s wings m oving horizontally at 900 km /h if the wing span is 36.5 m and the vertical com ponent of the E a rth ’s m agnetic field is 40 A/m. 16.16. An electrom agnet produces a m agnetic field of stren g th 4 x 105 A/m in a ir at its poles, Assuming th a t the field is uniform , determ ine the m inim um velocity of a con ductor of active length 10 cm required for an emf of 1 V to be induced in it if the angle between the vectors of m ag netic induction and velocity is 90°.
176
Ch. II. Fundam entals of Electrodynam ics
16.17. A m agnetic flux of 30 mWb piercing a closed circuit decreases to zero in 1.5 X 10~2 s. Determ ine the average emf and current induced in the circuit. The resistance of the circuit is 4 £2. 16.18. A straig h t conductor moves at a velocity of 4 m /s in a uniform m agnetic field of induction 6 mT. The active length of the conductor is 0.3 m. W hat is the angle between the velocity and m agnetic induction vectors if a potential difference of 3.6 X 10-3 V is induced across the ends of the conductor? 16.19. A solenoid contains 200 turns w ith a cross-sectional area of 80 cm2. The m agnetic induction in the solenoid increases over 0.1 s from 2 to 6 T. D eterm ine the average emf induced in the solenoid winding. 16.20. W hat is the change in the m agnetic flux in a coil over 0.05 s if the coil contains 1000 turns, and the emf in duced in it is 120 V? 16.21. D eterm ine the self-inductance emf in a coil of in ductance 0.5 H if the current in it decreases by 0.2 A over lO '3 s. 16.22. A voltage of 20 V is applied to a coil of inductance 1 H. How long does it take the current in the coil to reach 30 A? 16.23. The inductance of a circuit is 40 mH. W hat is the self-inductance emf emerging in the circuit if the current in it has changed by 0.2 A over 0.01 s? W hat is the change in the m agnetic flux in the circuit in this case? 16.24. The inductance of a circuit is 0.05 H. W hat is the m agnetic flux piercing the circuit if the current in it is 8 A? 16.25. Determ ine the inductance of a coil in which a self inductance emf of 0.5 V emerges as a result of a decrease of 0.2 A in the current in 0.04 s. 16.26. How can we increase the inductance of a solenoid? 16.27. W hy does a strong spark appear when a knifeswitch is used to disconnect a circuit containing a coil w ith a core? How can this effect be elim inated? 16.28. A m agnetic flux of 0.14 W b is coupled w ith a cir cuit whose inductance is 0.02 H. D eterm ine the current passing in the circuit. 16.29. At w hat current in a coil of inductance 40 mH is the m agnetic field energy equal to 0.15 J? 16.30. D eterm ine the energy stored in the m agnetic field
§ 16. E lectrom agnetic Induction
177
of a coil of inductance 85 mH if the current passing through the coil is 8 A. 16.31. A charge of 6 X 10~2 C for 0.01 s passes through the cross-section of a coil of inductance 12 mH for a long tim e. W hat are the m agnetic energy and the m agnetic flux in the coil? W hat will the self-inductance emf emerging at the moment the circuit is disconnected he if the m agnetic flux drops to zero in 0.05 s? 16.32. A circular loop of radius 5 cm is placed in a uniform m agnetic field of induction 1.2 X 10-2 T so th a t the norm al to the plane of the loop coincides w ith the direction of the field. The resistance of the loop is 3.1 Q. How much elec tric ity w ill pass through the loop if it is turned by an angle of 60°? 16.33. One tu rn of insulated wire in the form of a planar square frame w ith side I = 0.2 m is placed in a uniform m agnetic field perpendicular to the m agnetic field lines. D eterm ine the current passing through the turn if the mag netic field sta rts to decrease at a constant rate of 0.1 T/s. The resistance of the turn is 1 Q.
12 - 0 5 3 0
Chapter III
Oscillations and Waves § 17. MECHANICAL VIBRATIONS AND W AVES. SOUND AND ULTRASOUND
Basic Concepts and Formulas A v ib ratio n (oscillation) is a periodical m otion in which a body (particle) passes through its equilibrium position m oving altern ativ ely in opposite directions. The tim e taken for a v ib ratio n to be com pleted is called the period T , while the q u a n tity reciprocal to the period is known as the vibration frequency v: v = MT. The frequency is th num ber of vibrations th a t occur per second. The u n it of frequency is the hertz (Hz). The most im portant type of v ibration is the harm onic vibration (harm onic m otion) caused by a force proportional to the displacem ent x. The displacem ent of a point under going a harm onic m otion is given by the equation x = A sin (o)£ + cp0), where A is the am plitude of v ib ration, co£ -f
or
(o =
2JT./T.
The u n it of circular frequency is the radian per second (rad/s). R em ark. In the equation for a harm onic m otion, A is the m axim um displacem ent (:rmax) and can be denoted by X . By analogy, the am plitudes of velocity (umay) and acce leration (amax) can be denoted by V and A respectively. A sim ple pendulum (a particle suspended on a weightless unstretchable string) has the following period for small vibrations: T — 2n
VUg.
§ 17. Mechanical V ibrations and W aves
179
For a load of mass m vib ratin g on a spring w ith spring con sta n t k, the circular frequency and the period are given by a) = | / k/m , T == 2ji Y m / k . A v ib ratin g body has a potential and a kinetic energy. For a load v ibrating on a spring, the to ta l energy of the vibrations is given by
For an arb itra ry displacem ent x, the energy of the load is W = W p + W K= ^ f - + - ^
.
If a particle vibrates in a m edium , it causes neighbouring particles to vibrate, and a wave propagates in the m edium. The velocity v and the wavelength k of the wave are related thus v = k/ T, or v = A/v. The velocity of a wave does not depend on the frequency of vibrations and only depends on the properties of the me dium . Therefore a tran sitio n to a different m edium causes the velocity and w avelength to change, while the frequency rem ains unchanged. W aves having frequencies from 16 to 20000 Hz can be perceived by the hum an ear and are known as acoustic (sound) waves. The velocity of sound is usually denoted by c. W orked Problems Problem 84. A boy rocks up and down on a board. The tim e he takes to move from the upper position to the lower position is 1.5 s. W hat is the frequency, circular frequency, and period of the vibrations? Given: t = 1.5 s is the tim e during which the boy moves from the extrem e upper to the lower position. Find: the frequency v of the v ibrations of the boy on the board, the circular frequency (o, and the period T of the vibrations. Solution. In this case, the period of the v ibrations can be defined as the tim e elapsed between two successive upper-
12*
180
Ch. III. O scillations and W aves
most (or lowermost) positions of the boy. Thus, T
= 2t.
The frequency and the circular frequency can be deter mined from the form ulas v = l / 7 \ co — 2jiv . S ubstitu tin g in the num erical values, we obtain T = 2 x 1.5 s = 3 s, v = -^ —= 0 .3 3 s_1 = 0.33 Hz, os to = 2 x 3trad X 0.33 s"1 = 2.1 rad/s. Answer. The frequency of the vibrations is 0.33 Hz, the circular frequency is 2.1 rad/s, and the period is 3 s. Problem 85. The coordinates of a point are defined by the equation x = 1.2 cos ji (2£/3 -f 1/4). D eterm ine the am plitude, circular frequency, frequency, period, and the in itia l phase of the vibrations. D eterm ine the am plitudes of velocity and acceleration. W hat will the phase be in 0.375 s after the beginning of motion? Given: x = 1.2 cos n (2^/3 + 1/4) is the equation of mo tion of the vib ratin g point, t = 0.375 s is the tim e elapsed after the beginning of vibrations. Find: the am plitude X , the circular frequency co, the frequency v, the period 7\ the in itia l phase cp0, the am plitude V of velocity, the am plitude A of acceleration, and the phase cp at the in stan t t. Solution. Let us first transform the equation to a form containing a sine instead of a cosine. We know from trigo nom etry th a t cos a = sin (a + ji/2). Hence x = 1.2 cos ji (2£/3 + 1/4) = 1.2 sin [ji (2£/3 + 1/4) + jt/2] = 1.2 sin [2ji £/3 + n/4 + n/2] = 1.2 sin (2ji £/3 + 3 ji/4). Com paring this equation w ith the equation of harm onic m otion x = X sin (co£ +
§ 17. Mechanical Vibrations and W aves
181
The frequency, period, and phase can be determ ined from the form ulas v = (d/2ji, T = 1/v, (p = (at + (p0. In order to find the am plitude of the velocity, we m ust differentiate the equation of m otion. We obtain
Hence V = Xco. To find the am plitude of the acceleration, we m ust dif ferentiate the velocity equation — = - Fo) sin (
a = —A sin (a)t + (p0). Hence A = — Fa) = — .X co2. S u b stitu tin g in the num erical values, we obtain v
2ji rad/s 3 x 2 n rad
= 0.33 Hz,’ 7 = — = 3 s, V
2ji rad/s x 0.375 s ^ _ i 3
V = 1.2 m x A = - 1.2 m x
+ - ^ r a d = 3.14 rad,
s_1 = 2.5 m /s, s"2 = - 5.27 m /s2.
Answer. The am plitude of the vibrations is 1.2 m, the circular frequency is 2.1 rad/s, the frequency is 0.33 Hz, the period is 3 s, the in itia l phase is 2.36 rad, the velocity am plitude is 2.5 m/s, and the acceleration am plitude is -5.27 m /s2. The phase in 0.375 s after the beginning of mo tion is 3.14 rad. Problem 86. A load of m ass 100 g is fixed on a spring (see Pig. 85) w ith constant 100 N/m . I t is displaced 3 cm from the equilibrium position and receives a velocity of 10 cm/s. W hat are the po ten tial and kinetic energies of the load at
182
Ch. III. O scillations and W aves
the in itia l moment? W hat is the to tal energy of the load? W rite the equation of its m otion. Given: m = 100 g = 0.1 kg is the mass of the load, k = 100 N/m is the spring constant, x = 3 cm = 0.03 m is the in itia l displacem ent of the load from the equilibrium position, v = 10 cm/s = 0.1 m/s the in itia l velocity of the load. F ind: the potential energy E p0 of the load at the in itia l m om ent, the kinetic energy E k0 of the load at the in itia l m om ent, the total energy E of the load and the equation of motion. Solution. The potential and kinetic energies of the load can be determ ined from the form ulas r •^PO
kx2 o ?
ip kO
mv2 o
The to tal energy of the load is E = E v0 + E k0. In order to w rite the equation of m otion, we m ust find the am plitude A , the circular frequency co, and the in itia l phase of vibrations. I t should be noted th a t the in itia l dis placem ent of the load is not the am plitude since in addi tion to the in itia l displacem ent the load receives a veloci ty. However, the to tal energy can be expressed in term s of am plitude: E = k A 2/2, whence A = V 2 Ejk. The circular frequency can be found from the form ula (s)=z Y k / m . In order to determ ine the in itia l phase, we w rite the equa tion for harm onic m otion in the general form: X — A sin (at + %)• At tim e t = 0, the equation has the form x = A sin
§ 17. M echanical V ibrations and W aves
183
S ubstituting in the num erical values, we obtain
£ p0 = 100 N/m-* |2
0 .1 k g x O .O l m 2/s 2 = 5 x 1Q_4 j
£ = 4.55 x lO-2 J,
A _ y r2x 4.55x10-2 100 N /m
CO— •■/’ 100 ] V 0.1 kg
N /m
cp0 ~ arc sin
3.017
J
= 3.017 x lO-2
=31.6 rad/s, =arc sin 0.9945 = 1.4658 rad.
Answer. The potential and kinetic energies a t the in itia l moment are 4.5 X 10~2 J and 5 X 10“4 J respectively, the total energy is 4.55 X 10“2 J. The equation of m otion for the load has the form x = 3.017 X 10“2 x sin (31.6^ + 1.4658). Problem 87. W hat will the change in the period of oscillations of a pendulum be in a lift m oving upwards w ith an acceleration of 0.3 g? Given: a = 0.3g is the acceleration of the lift. Find: the period T of oscillations of the mg pendulum . Solution. Let us go over to a coordinate system fixed to the lift. Then the pendulum Fig. 80 will experience the action of an inertial force e q u a l to ma and directed against the accelera tion a. To clarify th is, let us consider the pendulum a t rest. It is acted upon by the tension F of the string and the force of gravity mg. Now let the pendulum move w ith the accel eration a together w ith the lift (Fig. 80). We w rite N ew ton’s second law F — mg = m a, or F = m (g + a). Thus, the weight of the load as if increases by the q u a n tity ma which is known as the in ertial force. We m ust then replace g in the form ula for the period of oscillations by g + a T = 2 n V l l ( g + a).
Ch. III. O scillations and W aves
184
In the absence of acceleration, we have D ividing these two equalities term wise, we obtain
S u b stitu tin g in the num erical values, we obtain
Answer. The period of oscillations of the pendulum will decrease by a factor of 1.14. Problem 88. Using vector diagram s, determ ine the am pli tude and phase of a v ib ration which is the resu ltan t of three harm onic vibrations x x = sin o)£, x 2 = 2 sin (to* + jt./2), and x 3 = 2.5 sin (cot + ji). W rite its equation. Given: x x = sin co£, x 2 = 2 sin (co£ + Jt/2), and x 3 = 2.5 sin (tot + n) are the component vibrations. F ind: the am plitude A and the in itia l phase cp0. W rite the equation of the resu ltan t vibration. Solution. From the given equations of harm onic m otions, we can determ ine the am plitudes and the in itia l phases: (a)
tb)
(c)
Fig. 81
A x = 1 m, A 2 — 2 m, and A 3 — 2.5 m, rp01 = 0,
§ 17. M echanical V ibrations and W aves
185
A 2, and A 3 and slopes 0, ji/2, and j i . Let us first compose the vectors lying on the same straig h t line (Fig. 81 b), and then find the required vector using the parallelogram rule (Fig. 81c). Thus we get A 3 — At = 1.5 m, A = Y A \ + (A 3 — A J 2, A = Y 4 m2 + 2.25 m2 = 2.5 m, sin a = -rpr- = 0.8,
a = 53°8',
186
Ch. III. O scillations and W aves
17.10. The angular velocity of the drive wheel of an au tom obile is 30 rad/s. W hat is the frequency and the period of m otion of the engine piston on direct gear?
\r=0 Fig. 82
17.11. A m erry-go-round turns through 90° in 0.5 s. De term ine its frequency, period, and circular frequency. 17.12. The m otion of a body is described by the equation
x = 4.25 sin (0.3f + 0.75). W hat is the am plitude, circular frequency, in itia l phase, and the period of oscillations? W hat is the phase at the in sta n t t = 0.5 s? 17.13. The equation of a harm onic m otion has the form x = 0.02 sin-^-^. D eterm ine the displacem ent of the body from the eq u ilib rium position at ti = 0, t 2 = 774, t 3 -- 772, and t 4 = 77712. 17.14. Using the conditions of Problem 17.13, determ ine the phase of vibrations for tx = T, t 2 = 772, and t 3 = 774. 17.15. The position of a v ib ratin g particle is determ ined by the equation x = 0.05 sin (ot. Determ ine the displacem ent of a particle if the vibration phase is ji /4. 17.16. The equation of a harm onically vib ratin g particle has the form
Determ ine the displacem ent of the particle at the instants t, = 0, t 2 = 77724, and t z = 117724.
§ 17. M echanical V ibrations and W aves
187
17.17. Using the conditions of Problem 17.16, determ ine the coordinate of the particle at tA = 177724, t 2 = 2 T , t 3 = 3778. Plot the graph x = f (t). 17.18. The position of a v ib rating body is described by the formula x = 8 sin tot. D eterm ine the displacem ent of the body if the vib ratio n phase is 30°. 17.19. D eterm ine the am plitude of the vibrations if the displacem ent of a particle at a phase of 45° is 10 cm. 17.20. Using the graph of the v ib ratio n al m otion of a particle, shown in Fig. 83, determ ine the am plitude, fre
quency, period, and the in itia l phase of the vibrations. W rite the equation of m otion. 17.21. The law of v ibration of a particle is represented graphically in Fig. 83. D eterm ine the displacem ent of the particle at t = 57712. 17.22. Given the equations of harm onic m otion of two bodies (a) x x = 2 s in 0 .5 n (2 1 + 1), x 2 = 3 sin (nt - f 1 . 5 j i ) , (b) x v = sin (tot + jc / 2 ), x 2 = cos tot. Indicate the phase difference in their vibrations. 17.23. Using the equations of vibrations of two bodies, determ ine their phase difference: (a) x { = 0.1 sin n (b) x x = 4 sin 1/4
, z2 = 0.4cos 1-5 jt ( * + 4 ”) * (co£-f 7 jt) ,
x 2
=
0.02 cos 1/3 (Q.75tot
— 3 ji) .
188
Ch. III. O scillations and W aves
17.24. W rite the (a) (o = it, x L = (b) x v = 0 for ti - 3 / 2 for f3 - 3. 17.25. W rite the
equation of a harm onic m otion if Y 3 for tx = 0, and x 2 = 1for t 2 = T ii. = 1, x 2 = Y~3/2 for t 2 = 2, and x 3 = equation of a harm onic m otion if
(a) A = 9 cm, v = 20 Hz, (b) A = 5 m, v = 0.5 Hz, cp0 corresponds to 778, (c) A = 1 m, f = 6 s, cp0 = — Jt/4. 17.26. A body floating on the surface of a liquid is immersed to a depth of 10 cm and then released. It sta rts to v ibrate a t a frequency of 2 Hz/s. How long does it take the body to move 5 cm from its equilibrium position? D eterm ine the tim e required for the body to traverse a distance of 5 cm from the point of m axim um deviation. W hat distance w ill the body cover in 10 s? The vibrations should be assum ed to be undam ped. 17.27. A pendulum consists of a ball suspended on a string. W hat forces cause it to oscillate? W hat is the direction of the restoring force? At w hat point is it at a m axim um ? At a m inim um ? 17.28. A ball suspended on a string performs sm all oscil lations about the equilibrium position. How does the veloc ity of the ball vary during one period? Can the m otion of the ball be defined as uniform ly variable? Why? 17.29. Two balls of the same size, one alum inium and one lead, are suspended on strings of equal length. The balls are deflected through the same sm all angle and released. W ill the periods of their oscillations be the same? 17.30. W ill the two balls in Problem 17.29 come to a h a lt at the same tim e? If not, which stops oscillating first? 17.31. At w hat position does a pendulum have the m axi mum and m inim um acceleration and velocity? 17.32. D eterm ine the restoring force acting on a sim ple pendulum of mass 10 g for a deflection angle of 45°. 17.33. D eterm ine the mass of a pendulum if the driving force is equal to 1 N at a deflection angle of 30°. 17.34. Through w hat angle is a copper ball of diam eter 2 cm deflected if the restoring force acting on it is 0.183 N? 17.35. W hy does a pendulum clock keep the correct tim e only at a certain latitu d e ?
§ 17. Mechanical Vibrations and W aves
189
17.36. The pendulum of a clock consists of a m etal rod and a load th a t can be fixed at any point along the rod. How w ill the pace of the clock change w ith decreasing room tem pera ture? How can the correct pace be restored? 17.37. W hat w ill the change in the period of oscillations of a pendulum be if its length is increased fourfold? 17.38. How should the length of the pendulum of a clock be changed for it to keep correct tim e on the Moon? 17.39. D eterm ine the oscillation periods of a 1-m long pendulum in Moscow (g = 9.816 m /s2) at the N orth pole (g = 9.832 m /s2), on the equator (g = 9.78 m /s2), and on the Moon (g = 1.63 m /s2). 17.40. The pendulum m ounted in the St. Isaac’s C athe dral in Leningrad is 98 m long and completes 181.5 oscilla tions per hour. Using these data, determ ine the free fall acceleration in Leningrad. 17.41. W hat w ill the speed of a pendulum clock be in (a) a uniform ly ascending lift? (b) a lift falling freely to the ground? W hat effect will the phase of the pendulum ’s oscil lations at the m oment the lift sta rts falling have? 17.42. A ball of mass m is suspended on a w eightless unstretchable string. The period of its oscillations is T 0. In addition to the force of grav ity , a force F acts on the ball along the vertical. W hat m ust the direction and m agnitude of this force (in com parison w ith the force of gravity) be for the oscillation period to be (a) 2 TV (b) 0 .8 7 V 17.43. W hat m ust the force F be (see Problem 17.42) to stop the pendulum ’s oscillations? 17.44. A horizontal force F acts on a ball (see Problem 17.42). W hat m ust the m agnitude of this force be (com pared to the force of gravity) for the oscillation period to be (a) T = 0.9467V (b) T = 0 .8 7 V W hat w ill be the eq u ilib rium position? Draw a diagram . 17.45. A pendulum consists of a m etal ball of mass 10 g and a silk thread. A charge of 10-6 C is supplied to the b all, and a horizontal electric field w ith strength 2 X 104 V/m is applied. How w ill the mode of the pendulum ’s oscilla tions change? W hat will the ratio of the oscillation periods bo in the two cases? 17.46. A seconds pendulum (viz. a pendulum w ith an os cillation period of 1 s) is in a tra in m oving w ith accelera
190
Ch. III. O scillations and W aves
tion a = 0.458g. Determ ine the period of oscillations of the pendulum . 17.47. A pendulum the period of whose sim ple oscilla tions is 0.5 s is in a lift descending w ith acceleration a = 0.19g. W hat is the frequency of its oscillations? 17.48. A load is fixed on a spring as shown in Fig. 84. W hen it is set vib ratin g along the vertical, the load is acted
Fig. 84
Fig. 85
upon by the elastic force of the spring and the force of g rav ity. W ill the load’s oscillations be harm onic? W hat effect does the force of g rav ity mg have? 17.49. One end of a spring w ith a constant of 50 N/m is fixed, and a 1-kg load is suspended from the other end. D eterm ine the vib ratio n frequency of such a pendulum . 17.50. The period of vibrations of a spring pendulum is 0.25 s. W hat is the spring constant if the mass of the load is 200 g? 17.51. A body of mass 0.5 kg fixed to a spring stretches it by 1 cm at rest. W hen it is displaced by 3 cm downwards and released, it sta rts v ib rating harm onically. D eterm ine the am plitude, circular frequency, period, and in itia l phase. W rite the equation of m otion. 17.52. A body of mass 800 g is fixed to a spring w ith con sta n t 40 N/m and vibrates as shown in Fig. 85. The am pli tude of vibrations is 2 cm. W hat is the energy of vibrations? D eterm ine the m axim um velocity and acceleration. 17.53. The m otion of a body whose mass is 2 kg is de scribed by the equation x = 0.8 sin | nt + - y ) . D eterm ine the energy of the v ib ratin g body. How does the energy depend on the in itia l phase? 17.54. A load of mass 1 kg is suspended from a spring w ith constant 1000 N/m and placed in a rocket th a t is
§ 17. M echanical Vibrations and W aves
191
launched w ith acceleration a = 2g. W hen the engine is switched off, the load starts vibrating. W hat is the energy of the vib ratin g load? W hat is its velocity when the load is 0.5 cm away from its equilibrium position? W hat are its kinetic and potential energies at this point? W rite the equa tion of m otion, assuming th a t the free fall acceleration g does not change during the flight. 17.55. The equation of m otion of a particle has the form x = 6 cos 0.2 sin n
cos
— 0.1 ji j
.
W rite the equation in a more convenient form and determ ine from it the am plitude, frequency, period, in itia l phase, and circular frequency of the vibrations. 17.56. A w atch w ith a spring and a pendulum clock are mounted on a rocket being launched v e rtica lly upw ards fly ing w ith an acceleration ax = g for 30 s, after which its engines are switched off. W hen it is 980 m from the surface of the E a rth , the rocket begins slow ing down and lands soft ly. W hat is the difference between the clock readings? 17.57. Plot the graphs of harm onic m otions described by the equations (1) x = 2 sin co£, (2) x = 3 sin (o£/2, and (3) x = 2 sin ( i o t + ji). 17.58. Figure 86 shows the graphs of five vib ratio n al mo tions. D eterm ine the am plitudes of vibrations I I and V. W hat is the am plitude of the resu lta n t vibration? Compose vibrations I I and V graphically. 17.59. Compose vibrations I and I I and / and V (Fig. 86) analy tically and graphically. Determ ine the am plitude of the resu ltan t vibrations. 17.60. Using vector diagram s determ ine the am plitude of the resu ltan t vibrations (see Fig. 86) for (a) I and I V , (b) I I , I V , and V, and (c) I , I I I , and I V . 17.61. Using vector diagram , determ ine the phase shift in the composition of vibrations I , I I I , and I V (see Fig. 86). 17.62. Using vector diagram s, determ ine the am plitude and phase of the v ibration obtained by subtracting V from I V (see Fig. 86). 17.63. Describe the m otion of a particle p articip atin g in a wave m otion.
192
Ch. III. O scillations and W aves
§ 17. M echanical V ibrations and W aves
193
17.64. Figure 87 shows the shape of a transverse wave. The wave propagates to the right. Use arrows to indicate the directions of the individual particle velocities. 17.65. D eterm ine the directions in which the waves in Figs. 88a and b propagate. The arrows indicate the direc tions of the velocities of the particles. 17.66. On w hat does the propagation velocity of a tra n s verse wave a t the boundary between two liquid m edia de pend? 17.67. A wave propagates over the surface of w ater at a velocity of 3 m /s. W hat will the velocities of the particles be on the crest of th e wave? 17.68. In southern countries, surfing (riding a board on the crests of waves) is very popular by the sea. At w hat velocity !n)
Fig. 87
Fig. 88
does a wave carry a surfer if the w avelength is 25 m and the wave particles v ib rate w ith a period of 1.5 s? 17.69. In which m edium is the velocity of sound higher: air or iron? Can sound propagate in vacuum ? 17.70. W hile opening a door, we som etim es hear a creak. W hat is the origin of th is sound? W hat is the role of the door? 17.71. W hich properties of a m edium determ ine the ve locity of sound? E xplain by tak in g air and w ater as examples. 17.72. W hat does the velocity of sound in a given m edium , say, air depend on? 17.73. W hy do people press an ear against the rails to know w hether a tra in is coming? 17.74. A m eteorite strikes the surface of the Moon. How long will it take for sensitive instrum ents on the E a rth to detect the sound of the explosion? 13-0530
194
Ch. III. O scillations and W aves
17.75. If a vib ratin g tuning fork is held against a table, the sound becomes much louder. Why? 17.76. A vib ratin g tuning fork is first held in the hand and then its end is brought in contact w ith a table. In which case does the sound cease sooner and why? 17.77. The hum an ear has its greatest sen sitiv ity in the frequency range 1.5-3 kHz. D eterm ine the w avelengths cor responding to this range if the velocity of sound is 340 m/s. 17.78. Sound propagates in air at 330 m /s for a v ib ration frequency of 0.5 kHz. Determ ine the shortest distance in th e propagation direction between two points v ib ratin g in phase. 17.79. If sound waves propagate in a m edium at 340 m/s and a frequency of 500 Hz, what is the phase difference of two particles of the m edium 17 cm apart? The particles lie on the line along which the wave propagates. 17.80. If we blow over the open end of the cap of a foun tain pen, a w histling sound is produced. E xplain its origin. 17.81. The cap in Problem 17.80 is 5.5 cm long. W hat is the frequency of the emerging sound? 17.82. The frequency of a tuning fork’s vibrations is 1.38 kHz. How long is the vibrating p art of the tuning fork? The velocity of sound is 332 m/s. 17.83. How will the frequency of the sound produced by a tuning fork change w ith decreasing tem perature? 17.84. The length of an acoustic wave of the same fre quency in air is a factor of 4.25 sm aller th an in water and a factor of 10.7 sm aller than in brick. Determ ine the velocity of sound in w ater and in brick assuming th a t the velocity of sound in air is 340 m/s. 17.85. W hat is the change in frequency, period of v ib ra tions, and w avelength of sound when it passes from air to steel? The velocity of sound in steel is 5000 m/s. 17.86. D eterm ine the depth of a sea if the response from an echo sounder is received in 1.6 s, and the velocity of sound in w ater is 1500 m/s. 17.87. U ltrasound is used to find defects in large bodies. At w hat depth is a defect in an alum inium com ponent de tected if the first reflected signal is received a fte r8 X 10_6s, and the second, after 2 X 10“5 s? W hat is the height of the component? The velocity of sound in alum inium is 510 m/s. 17.88. A person heard a clap of thunder 9 s after he saw the
§ 18. A lternating Current
195
flash of lightning. At w hat distance from the discharge was he? 17.89. A thunderstorm takes place 4.5 km away from an observer. W hat is the tim e in terv al between seeing the lightning and hearing the thunder? 17.90. A ship surveying the bottom of the ocean moves at 36 km /h. W hat is the percentage error in the m easured depth due to the m otion of the ship? W hat is the error for a depth of 3000 m? § 18. ALTERNATING CURRENT
Basic Concepts and Formulas An a ltern atin g current is one th a t periodically changes its direction in a circuit so th a t the current averaged over a pe riod T is zero. A lternating current is produced by vib ratio n al m otion of charge carriers, viz. electrons. W hen a fram e is rotated in a uniform m agnetic field, an emf is induced across its ends: e = £ max sin tot, where e is the instantaneous emf, £ max is !h e m axim um (am plitude) emf, and co is the angular velocity of ro ta tio n of the fram e and the circular frequency of a lte rn a tin g emf. The period and frequency are
The resistance R in which heat is liberated due to the pas sage of a current through it is known as ohmic resistance. If the fram e is connected to a load w ith an ohmic resistance, an altern atin g electric current will pass through it, and an a lternating voltage will appear across the term inals: i = / max sin (tit, u = U max sin (tit. Mere u = iR. An alte rn a tin g current (a.c.) is equivalent to a direct cur rent of the same power. The effective values of a ltern atin g voltage and current are jj
^m ax
]/2 13 *
j
’
__ f m a x
V2
196
Ch. III. O scillations and W aves
An a.c. circuit containing a inductive reactance XL = An a.c. circuit including a has a capacitive reactance: v __
coil of inductance L has an toL. capacitor of capacitance C
* (DC * The reactance of a circuit containing a capacitor, a re sistor, and an inductance coil in series is given by X = X L — Xc, while the im pedance is z = y m + ( x L - x cyi . The relation between Z, /?, and X is presented graphically by a triangle (Fig. 89), i.e. obeys the rules of geom etric com position. The angle (p in the figure is the phase difference between current and voltage. Using the am plitude and effective values, we can w rite Ohm ’s law for an a ltern atin g current, i.e. F1&* &9
jj
__
J
7
II
17
k'max — In the presence of a reactance, the current in the c irc u it is not in phase w ith the voltage. If we assume th a t the in i tial phase of the current is zero, the voltage across the induc tive reactance leads the current by Jt/2, while the voltage across the capacitive reactance lags behind the current by Jt/2. Therefore, only in an ohmic resistance do the current and voltage oscillate in phase. Hence, a lte rn a tin g currents and voltages cannot be added algebraically, as was the case for direct current. The com position is carried out using vector diagram s in which the emf, voltage, and current are depicted by vectors em erging from the origin and form ing w ith the abscissa axis an angle equal to the in itia l phase. The average power lib erated in an a.c. circuit is cos
§ 18. A lternating Current
197
ciated w ith the conversion of electrical energy into heat and is given by P = PR. The u nit of active power is the w att (W). The reactive power Q is given by Q
= P X L or Q = P X C.
I he u nit of reactive power is the var (volt-ainpere reactive). The to tal power S = PZ is connected w ith P and Q through the power triangle (Fig. 90). The u n it of to ta l power is volt-am pere (VA). The active power is the average pow er liberated in the circuit. The reactive power is associated w ith energy tra n s mission from the generator to a circuit and back (it is stored in inductance coils and capacitors). The reactive power a v eraged over a period is zero. The active, reactive, and to ta l power can be expressed in terms of the effective current and voltage in the circuit: P = I U cos (p, Q = I U sin (p, S = I U . Here cos
P
cos
198
Ch. III. O scillations and W aves
Given: R = 20 Q is the resistance, L = 0.0398 H = 3.98 X 10"2 H is the inductance of the coil, C = 159 p,F = 1.59 X 10~4 F is the capacitance of the capacitor, I = 2 A is the effective value of the current, and v == 100 Hz is the frequency of the altern atin g current. F in d : the voltages UR, U L, Uc , and U across the resis tor, the coil, the capacitor, and in the entire circuit respec tiv ely , the phase difference cp between the voltage and the current, and the im pedance Z of the circuit. Solution. The reactances of the coil and of the capacitor can be determ ined from the form ulas X L = and X c = i /n C . The im pedance is given by z=
V b 2+ ( x l - x c)2= V R 1 + (<*>£
-
i/ojC)2.
W hen the com ponents of the circuit are connected in se ries, the same current passes through it, while the to ta l vol tage is the sum of the voltages across the in d ividual compo nents. The voltage across the resistor is UR = I R , and is in phase w ith the current. The voltage across the coil is UL = /o L , and leads the current by jt/2. The voltage across the ca pacitor is Uc = / / (coC), and lags behind the current by n/2. The to ta l voltage can be found by constructing a vector diagram . We choose the origin and draw a horizontal vector, viz. the current axis relativ e to which the voltage vectors w ill be plotted (Fig. 91a). The vector of voltage UR across the resistor is plotted (from zero) along the current axis since its phase angle is zero. We now plot the vector of voltage UL across the inductance coil. P ositive angles are measured counterclockwise, so the voltage vector will be plotted upw ards at rig h t angles to the current axis. The vector of the voltage Uc across the capacitor is p lo t ted downwards from zero at rig h t angles to the curreivt axis.
§ 18. A lternating Current
199
The geom etric sum of the three vectors w ill give the to ta l voltage in the circuit. F irst we compose the vectors directed along the vertical, and as a result obtain a voltage triangle (Fig. 916) from which we can easily find the voltage and the phase difference: UX = UL - U C, v=
V U r + u \ = I V R* + (coL - 1/©C)2,
tan < p =
coZ,— 1/coC
g
=
Ux
If coL =» l/coC, the reactance is zero. This phenomenon is observed a t the circular frequency a) = co0 determ ined (a)
F ig. 91
from the form ula co^ = 1!(LC). Then the impedance of the circuit a tta in s its m inim um value and is equal to the resis tance: Z — R , while the current a tta in s its m axim um value. This phenomenon is known as voltage resonance, and o)0 is the resonance circular frequency. S ubstitu tin g in the num erical values, we obtain UR = 2 A x 20 £2 40 V, UL = 2 A x 2ji x 100 s '1 x 0.0398 H - 50 V, Uc = 2 A/(2ji x 100 s"1 x 159 x 10~4 F) = 20 V, U = V (40 V)2 + (30 V)2 = 50 V, X L - X c = 2ji x 100 s '1 x 0.398 H — (2n X 100 s‘ ‘ x 1.59 x 10-4 F )-‘ = 15 Q, tan
200
Ch. III. O scillations and W aves
Answer. The voltages across the resistor, the inductance coil, and the capacitor are 40, 50, and 20 V respectively, the to ta l voltage is 50 V, the phase difference is 36°52', the im
pedance is 25 £2. W hen the inductive and capacitive reac tances are equal, resonance is observed. Problem 90. A generator supplies an a.c. voltage of 80 V a t a frequency of 20 Hz and is connected to an induc tance coil and a capacitor of 750 p,F connected in parallel. The resistance of the coil is 1 £2 and the induc tance is 0.1 H (Fig. 92). D eterm ine the currents in the branches and the to ta l current in the circuit. W hat is the resonance condition for the circuit? D eterm ine the im pedance at resonance if co2L 2> /?2. Given: U = 80 V is the voltage in the circuit, v = 2 0 Hz is the fre quency, C = 750 p,F = 7 .5 x lO _4F is the capacitance of the capacitor, i = 0.1 H and R = 1 £2 are the inductance and the resistance of the coil. Find: I c , / L, and / , the currents in the branches and in the entire circuit and the im pedance Z at resonance. Solution. Let us first consider the branch containing the coil. We plot the vector diagram (Fig. 93) in the same way as in Problem 89.
§ 18. A lternating Current
201
Using Fig. 93, we can find U and (pL: o>L
U = I l V R 2 + <*2L \ tan(pL = - ^
For the parallel connection (see Fig. 92), the voltage U is the same in each branch, while the current is equal to the sum of the currents in the parallel branches. The currents in the branches have different phases, and hence they should be composed geom etrically, using vector diagram s. We plot the voltage axis (Fig. 94). The current in the branch w ith inductance has the value /
u
+
L
’
and lags behind the voltage by
C/coC,
and leads the voltage by j i / 2 . The com position of these vec tors results in a vector I a t an angle cp to the voltage axis. The current / can be determ ined by cosine law: J 2 = I 2l + I 2c — 21 1 I c cos a . We know from trigonom etry th a t sin (pL =
tan
V 1+ tan2
_
coL______
V R 2+ « 2£2
since tan L/i? (see Fig. 93); we have COS (pL :
1
Y 1+ tan2
R
\ f /?2+ (02L2
Since a = ji/2 — (pL, c o sa = sin (pL. Therefore, I ^ P L + I h - 2 I LI C
wL l/i? 2+ (o2L2
Angle (p can be determ ined from the equality of the projec tions of / and I L on the voltage axis: I cos (p = I L cos cpL, cos
2^2
Ch. III. O scillations and W aves
the origin. The next vector is plotted from the end of the preceding one. The resu ltan t vector then connects the ori gin of the first vector to the end of the last one. For our prob lem, such a construction yields a current triangle (Fig. 95), the calculations rem aining the same. If one of the param eters to, C, or L of the circuit is varied, the to ta l current and the phase angle (p w ill change. For
Fig. 95
Fig. 96
a certain relation between co, C, and L, the phase difference vanishes (Fig. 96). Then the current w ill have the m in i m um value, and hence the im pedance of the circuit, which becomes equal to the resistance, increases. This phenom e non is known as current resonance. The resonance condition can be determ ined from Fig. 96: I c = I L sin cpL. S ubstitu tin g / c , / L, and sin
UuC:
] / i ? 2 +
ioL co2 / , 2
j / / ? 2 +
cd
2L
2
The inductive reactance of inductance coils is usually much higher than their resistance (coL>> /?), the more so co2L 2^> R 2. Neglecting R 2 in the last equality and sim plifying, we obtain the condition for current resonance: n
9
1
® = wo =
'
Let us calculate the im pedance of the circu it. I t follows from Fig. 96 th a t / = / L cos
V
R
$ 18. A lternating Current
203
Since o)2L 2^ / ? 2, we obtain T
UR
URC
,
/ = -51Z r = —
rj
L
’ whence Z = ~RC
S u b stituting in the num erical values, we obtain I l ■-=
80 V
L
—
V ( 1 Q)* + ( 2 n x 2 0 s - ' x O . l H)2
= 6.35 A,
I c = 80 V x 2n x 20 s"‘ X 7.5 x 10"4 F = 7.54 A, COS a =
2 ji X 20 s-1 X 0.1 H = / (1 Q)2 + (2 ji X 20 s”1 X 0.1 H )2
q
ggy
1 = V (6.35 A)2 + (7.54 A)2— 2 x 6.35 A x 7.54 A x 0.997 = 1.28 A. The resonance frequency is 1 1 - 1 8 . 4 Hz. r
2jx V 0 A H X 7.5 X 10“ « F
2ji V L C
The impedance is Z
1
QX 7.5 XlO-4 F
"
133
Answer. The current through the coil and the capacitor is 0.35 A and 7.54 A respectively, the to ta l current is 1.28 A. Resonance sets in a t a frequency of 18.4 Hz, and the im pe dance of the circuit is then 133 £2. Problem 91. An electric bulb rated for 240 V and power of 200 W, a coil w ith inductance 0.15 H, and a capacitor are connected in series to a lighting circuit a t a voltage of 220 V. W hat is the capacitance of the capacitor if the active power liberated in the circuit is equal to half the to ta l power? D eterm ine the current, the voltages across the com ponents of the circuit, the active, reactive and to ta l powers, and the power factor. Given: U = 220 V is the voltage in the circuit, v = 50 Hz is the frequency of the a ltern atin g current, U h = 240 V and P h = 200 W are the rated voltage and power of the bulb, L = 0.15 H is the inductance of the coil, and P = 0.55 is the relation between the active and to ta l power. Find: the capacitance C of the capacitor, the current I in the circuit, the voltages UR, UL, and U c across the bulb, coil, and capacitor respectively, the active power P , reac tive power Q, to ta l power 5 , and the power factor cos (p.
204
Ch. III. O scillations and W aves
Solution. The resistance of the bulb can be determ ined from the form ula Pb = U \IR , P
whence
R = U l/P b =
= 288 £2.
The power factor can be determined from the relation = 0.55: 0.5 S
P
n .
cos
, R = 0.5Z = Q.5Vr R * + ( X c - X L) \
Squaring the last equality, we obtain 4R* = R* + (X c - X l )2, 3 R 2 = ( X c — X L)2. Hence
X C = R V 3 + X l , X c = 288 £2 ^ 3 + 47.1 £2 = 546 Q, 1
2n x 50 Hz X 546 Q
5.83 x 10"6 F = 5.83 pF.
The im pedance is Z = Y (288 Q)2 + (546 £2 — 47.1 £2)2 = 576 £2. Let us now find the current in the circuit
'- X .
' - S S - * * *
The voltage across all the com ponents can be found as follows: UR = IR, Uc = I X C, UL = I X L, UR = 0.38 A x 288 £2 = 109.4 V, Uc = 0.38 A x 546 £2 = 207.5 V, UL = 0.38 A x 47.1 £2 = 17.9 V. Let us now calculate the powers: to ta l power S = I U , S = 0.38 A X 220 V = 83.6 VA, active power P = 0.5S , P = 4 1 .8 W ,
§ 18. A lternating Curren
205 o
reactive power (? = £ sin = - ^ - 8 3 . 6 VA = 72.4 var. Answer. The capacitance of the capacitor is 5.83 pF, the current in the circuit is 0.38 A, the voltages across the bulb, the inductance coil, and the capacitor are 109.4, 17.9, and 207.5 V respectively, the active, reactive, and to ta l powers are 41.8 W , 72.4 var, and 83.6 VA respectively, and the power factor is 0.5. Questions and Problems 18.1. The standard frequency of the a ltern atin g current used in the USSR is 50 Hz. D eterm ine the period of the os cillations. How m any tim es per second does the direction of the charge carriers change? 18.2. An am m eter connected to an a.c. circu it indicates 3 A. W hat is the am plitude of the current? 18.3. The breakdown voltage of a capacitor is 250 V. Can it be connected to a lighting circu it a t a voltage of 220 V? 18.4. The emf in an a.c. circuit varies as e = 250 X sin IOOjiJ. D eterm ine the effective value of the emf and the circular frequency. 18.5. The tim e dependence of an altern atin g current is described by the equation i = 90 sin (314£ + ji/4). D eter mine the effective current and its phase, in itia l phase, and frequency. 18.6. The instantaneous voltage is u = 179 sin (ot. De term ine the instantaneous voltage a t tim es: 0, 0.0025, 0.005, 5/6, 4/3, 0.015, 0.0175, and 0.02 s. The frequency of the a l ternating current is 50 Hz. 18.7. The instantaneous current at 1/3 of the period is 2.6 A. W hat w ill the current be for a phase of (a) 1.5ic (b) 13ji /6? 18.8. A wire fram e rotates uniform ly in a uniform m ag netic field about an axis perpendicular to the m agnetic field lines. In w hat position of the fram e is the induced emf (a) zero, (b) equal to the am plitude value? 18.9. A rectangular fram e of length 10 cm and w idth 5 cm rotates uniform ly in a uniform m agnetic field of induction
206
Ch. III. O scillations and W aves
0.02 T. The speed of ro tation is 2865 rph. Determ ine the am p litude of the emf induced in the fram e. 18.10. A rectangular m etal frame of length 12 cm and w idth 4 cm rotates uniform ly in a uniform m agnetic field. W hat w ill the change in the emf induced in the fram e be if, w ithout changing the speed of its rotation, it can be deformed so th a t its length becomes 4 cm and w idth 12 cm? 18.11. A rectangular tates in a uniform m agnetic field at a constant speed about an axis per pendicular to the m agnetic field lines. The frame length is I = 14 cm Fig- 97 and its w idth is d = 6 cm. How w ill the emf induced in the fram e change if the frame is deformed such th a t it takes on the fol lowing dimensions: (a) I = 16 cm, d = 4 cm, (b) I = 10 cm, d = 10 cm, and (c) I = 8 cm, d = 12 cm. 18.12. How m any turns are wound on a fram e having an area of 367 cm 2 and uniform ly ro tatin g in a uniform m agne tic field w ith induction 0.115 T if a voltm eter connected to its term inals indicates 90 V? The ro tatio n period of the frame is 2.5 X 1 0 '2 s. 18.13. A frame containing 15 turns of area 200 cm 2 rotates uniform ly in a uniform m agnetic field a t 10 rps. W hen the fram e is a t an angle of 30° to the direction of the m agnetic field, the emf induced in the fram e is 0.75 V. W hat is the m ag netic field’s induction? 18.14. A wooden fram e having an area of 150 cm 2 and containing ten turns of wire rotates uniform ly in a uniform m agnetic field w ith induction 0.05 T. The position of the frame a t t = 0 is shown in Fig. 97. The frequency of the induced altern atin g current is 10 Hz. D eterm ine the am pli tude, circular frequency, the period, and the in itia l phase of the induced emf. W rite the equation for the instantaneous emf. 18.15. A conductor w ith inductance 2.55 X 10"3 H is switched from a circu it operating a t a frequency of 50 Hz to a circu it operating a t a frequency of 300 Hz. Does its im pedance change? If it does, in w hat proportion? W hat is its impedance in a d.c. circu it equal to?
§ 18. A lternating Current
207
18.16. A coil w ith inductance 8.42 X 10-3 H has a resis tance of 3 Q. W hat is its im pedance in an a.c. circuit w ith a current frequency of 50 Hz and in a d.c. circuit? 18.17. W hen a coil w ith inductance 0.3 H is connected to a d.c. circuit a t a voltage of 24 V, the current in it is 0.2 A. Determine the impedance of the coil when it is connected to an a.c. circuit of frequency 50 Hz. 18.18. W hat w ill the impedance of a 100-pF capacitor be m a d.c. circuit? 18.19. D eterm ine the impedance of a 100-p,F capacitor connected to a.c. circuits w ith current frequencies 50 and 3000 Hz. 18.20. A coil of inductance 0.15 H, a 500-Q resistor, and a 2 \iF capacitor are connected in series to an a.c. circu it oper ating at a frequency of 400 Hz. D eterm ine the im pedance of the circuit. 18.21. Determ ine the impedance of the circuit connected to the lighting m ains if the circuit consists of a 500-p,F and a 1000-|iF capacitors connected in series, a coil w ith a resistance of 10 £2, and inductance 0.08 H. 18.22. A 530-p,F capacitor and a potentiom eter w ith a resistance of 5 Q connected in series are connected to the mains which has a voltage of 220 V and a frequency of .30 Hz. Determ ine the current passing through the po ten tio meter. 18.23. A capacitor is connected to the m ains which has a voltage of 220 V and frequency of 50 Hz. The am plitude of t lie current in the circuit is 4.89 A. W hat is the capacitance of the capacitor? 18.24. A coil of inductance 0.2 H is connected to an a.c. circuit w ith a frequency of 100 Hz. The current in the coil is 1.01 A. W hat is the voltage across the coil? W hat is the am plitude of the voltage? 18.25. A voltage of 80 V is supplied to a variable capaci tor and a coil w ith a core connected in series to it. The current in the circuit was found to be 4 A. D eterm ine the voltage across the circuit elem ents if halving the capacitance of the capacitor and the inductance of the coil increases the current to 20 A. W hat is the frequency if the in itia l in ductance is 1.91 X 10-2 H? 18.26. An a.c. circuit contains a resistor. P lo t the am pli tude of the voltage across the resistor on a vector diagram
208
Ch. III. O scillations and W aves
based on the current axis. P lot the graphs of the current and voltage. 18.27. P lo t the voltage vector diagram if an a.c. circuit contains (a) a capacitor, (b) a solenoid. P lo t the graphs for the currents and the voltages. 18.28. P lo t the vector diagram for the voltages of the a.c. subcircuits operating at frequency 50 Hz and consisting of the following components connected in series (a) C = 1.33 x 103 pF, R = 2.4 £2, (b) L = 3.18 x lO"2 H, R = 17.3 £2. 18.29. P lo t the graphs of voltage and current in the com ponents of the circu it in Problem 18.28. 18.30. P lo t the vector diagram s of the voltage for the follow ing com ponents when connected in series (a) L = 0.03 H and C = 2 X 10~4 F, (b) L = 4.95 x 10“2 H, C = 398 pF and R = 7.55 £2. The altern atin g voltage frequency is 50 Hz. 18.31. P lo t the voltage graphs for the com ponents of the a.c. circu it in Problem 18.30. 18.32. Using vector diagram s, prove the form ula for the reactance of a subcircuit consisting of a capacitor and an inductance coil. 18.33. Using vector diagram s, prove the form ulas for the im pedances of series-connected R , C and L in an a lte r nating current. 18.34. An a.c. circuit a t voltage 120 V and frequency 50 Hz is connected to series-connected (a) R — 60 £2, L = 0.255 H, (b) R = 3.8 £2 and C = 2.27 X 10~3 F, and (c) L = 0.0764 H and C = 398 pF. De / term ine the impedance, current, and voltages on the com ponents of the L circuit. P lo t the vector diagram s. 18.35. An a.c. voltage of 220 V and frequency 50 Hz is applied to series-connected (a) R = 5 £2, L = F ig . 98 0.135 H , and C = 75 pF, (b) R = 30 £2, L = 0.2 H , and C = 97 pF. Determ ine im pedance, current, and voltage across the components of the circuit. P lot the vector diagram s. 18.36. An a.c. generator of voltage 36 V (Fig. 98) is con nected in parallel to (a) R = 3 £2, X L = 4 £2, (b) R = 1 £2 X c = 2 £2, and (c) X c = 2 £2, X L = 4 £2. The frequency of
§ 18. A lternating Current
209
the altern atin g current is 50 Hz. Determ ine the current in the branches, the total current, and cos
210
Ch. III. O scillations and W aves
to a d.c. circuit. W ill the bulb connected to the secondary w inding glow? 18.46. The prim ary of a transform er is connected to an a.c. circu it while the secondary is disconnected. Does the transform er consume energy from the circuit? 18.47. Two neighbouring turns of a transform er’s sec ondary are short circuited. W hat w ill happen? 18.48. W hen a transform er is connected to an a.c. supply w ith a voltage of 220 V, an emf of 110 V is induced in the secondary w inding. W hat is the trans£ form ation ratio? » j 18.49. The transform ation ratio of a transform er is 10. W h at is the ratio ~ of the num bers of turns in each windmg? Fig. 99 18.50. An emf of 600 V is induced in the secondary of a transform er con tain in g 1900 turns. How m any turns of the prim ary w ind ing are connected to a supply w ith voltage 220 V? 18.51. A transform er is connected to a supply w ith v o lt age 120 V. The prim ary w inding contains 300 turns. W hat m ust the num ber of turns in the secondary w inding be for a voltage of 6.4 V to be generated across it? 18.52. A voltage of 60 V at a frequency of 50 Hz is applied to a circu it containing a coil w ith inductance 0.1 H and a 20-|iF capacitor. W hat is the average power developed in the circuit during one period? W hat are the reactive and to ta l powers? 18.53. An a lte rn a tin g current passes through a seriesconnected resistor, capacitor, and coil whose resistances and reactances are 2, 3, and 5.67 Q respectively. D eterm ine the active, reactive, and to ta l power, the power factor (cos a ), and the voltage applied to the circuit if the current is 6 A. 18.54. The a ltern atin g voltage in a series-connected c ir cu it w ith a current of 2 A is 90 V. The circu it consists of a capacitor, a coil, and a resistor. The power factor is 0.6. W hat are the active and the reactive powers? W hat are the reactances and the resistance of the c irc u it’s com ponents if the reactance of the capacitor is twice th a t of the coil? 18.55. A source of an a.c. voltage u = 180 sin a t is connected to an unbranched circu it containing a resistance and two reactances, X L = 8 £2 and X c = 12 £2. Deter
§ 19. Electrom agnetic O scillations and W aves
211
mine the power factor, the active and reactive power if the current in the circuit is 3 A. 18.56. An a x . voltage of 100 V is applied to a 50-£2 resis tor and capacitor w ith reactance 55 £2 connected in p a ra l lel. D eterm ine the to ta l, active and reactive powers, and the power factor. 18.57. A source of a.c. voltage of 60 V and frequency f>0 Hz is connected in parallel to a resistance of 15 Q and an inductance coil w ith a negligible re L C sistance. The to ta l power in the cir cuit is 648 VA. W hat is the induc tance of the coil? W hat are the active and reactive powers? 18.58. An altern atin g current passes through a coil and a resistor connected in parallel. The voltage across the coil is 120 V, the current in the resistor is 2 A, and to ta l cur rent is 3.5 A. Find the resistance of the resistor and the coil. D eterm ine the active, reactive and total powers, and the power factor. 18.59. A coil, a capacitor, and a resistor are connected as shown in Fig. 100. The voltage of the source is 40 V. D eter mine the power factor and the to ta l, active, and reactive powers if (a) X L = 15 £2, X c = 15 £2, R = 10 £2, (b) X l = H £2, X c = 18 £2, and R = 13.3 £2. 18.60. A capacitor, a coil, and a resistor are connected in parallel to an a.c. circuit of a voltage 220 V. Their reac tances and resistance are 8 ,1 2 , and 6 £2 respectively. D eter mine the power factor, and the active and reactive powers. 18.61. The power in two parallel branches of an a.c. circuit is 200 var and 346 W . The current in the unbranched circuit is 8 A. W hat are the reactance and the resistance of the branches, the power factor, and the to ta l power? W hat is the impedance? § 19. ELECTROMAGNETIC OSCILLATIONS AND W AVES
llasic Concepts and Formulas Any varying m agnetic field generates an eddy electric field. In turn, an electric field varying over tim e generates an eddy magnetic field. This continuous transform ation of the eleci
Ch. III. O scillations and W aves
trie field’s energy into the energy of the m agnetic field and back is observed in an oscillatory circuit, viz. a circuit con tain in g a coil and a capacitor (Fig. 101). The resistance of the oscillatory circuit m ust be low, otherwise the energy of the electrom agnetic field w ill be converted into heat, and the oscillations in the circuit w ill soon cease. I Under these conditions, the period of the - i-f ^ 3 natu ral electrom agnetic oscillations is f
T = 2nVLC, and the frequency is
Fig. 101
v ——
1
T 2ji y i c An open oscillatory circuit em its waves of the w avelength X = c T , or X = c/v, where c is the velocity of electrom agnetic waves equal to the velocity of light in vacuum . The velocity of propagation of the electrom agnetic waves depends on the properties of the m edium , viz. — = / Lie, whence
u
v
=
-
V
vhere v is the velocity of propagation of the electrom agnetic waves in the m edium , and p and e are the perm eability and p e rm ittiv ity of the m edium . Since the perm eabilities of all dia- and param agnetic m edia differ only slightly from unity, we can assume th a t Vs ■ W hile using this form ula, it should be borne in m ind th a t the e known from electrostatics cannot always be used since the oscillation frequency affects the p e rm ittiv ity . W orked Problems Problem 92. An oscillatory circuit consists of a coil w ith inductance 20 pH and a capacitor whose capacitance can be varied from 2 X 10"8 to 10"8 F. W hat is the wavelength range of th is circuit? D eterm ine its frequency range.
§ 19. Electrom agnetic O scillations and W aves
213
Given: L = 2 X 10"5 H is the inductance of the coil, C{ = 2 X 10"8 F and C 2 = 10“8 F are the m axim um and m in imum capacitances of the capacitor. From tables, we take the velocity of light in vacuum c = 3 X 108 m/s. Find: the m axim um k { and m inim um k 2 w avelengths and the boundary frequencies vx and v2 of the oscillatory circuit. Solution. The wavelengths lie in the interval from k { = c T i for capacitance C i to k 2 = c T 2 for capacitance C2. The period of the c irc u it’s oscillations is determ ined by the form ulas 7\ = 2jx V L C , ,
T 2 =-- 2n V ~LC2,
T { = 2 x 3.14 V 2 x 10-5 H x 2 x 10-* F = 3.97 x 1 0 '8 s, 7 \ = 2 x 3.14 V 2 x 10~5 H x 10"8 F — 2.81
X
10"6 s.
Then the wavelengths are k, = 3 x 108 m/s
X
3.97
X
10*6 s = 1191 m,
k 2 = 3 x 108 m/s x 2.81 x l O '6 s = 843 m. Given the periods, we can determ ine the frequencies: = 1/77!, v2 = l/7 ’2, vt = 252 kHz, v2 = 356 kHz. Answer. The oscillatory circuit can operate in the w ave length range 843-1191 m a t frequencies 252-356 kH z. Problem 93. The m axim um voltage across the capacitor of an oscillatory circu it is 120 V. Determ ine the m axim um current if the inductance of the coil is 0.005 H and the c a pacitance of the capacitor is 10“5 F. The resistance of the circuit is assumed to be negligible. Given: t / max = 120 V is the m axim um voltage across the capacitor, L = 0.005 H is the inductance of the coil, and G = 10"5 F is the capacitance of the capacitor. Find: the m axim um current / max in the circuit. Solution. The problem can be solved by two m ethods. 1st method. At resonance, the reactance of the circu it is m inim al, while the current is a t its m axim um . The resonance frequency is <»0 = 1I V W , / max c * c j= l/(ce0C) = 1/ L I C , / maI c = UmAXl V L I C = Um&xV C I L . Argueing in the same way, we conclude th a t the current in
214
Ch. III. O scillations and W aves
the inductance coil can be calculated from the same form u la. S u b stitu tin g in the num erical values, we obtain An ax c = 120 V
= 5.37 A,
7may c = 7max.
2nd method. According to the energy conservation law, the m axim um energy of the m agnetic field m ust be equal to the m axim um energy of the electric field: L I ^ J l = CC/max/2. Hence /max = U mRXV C l L . Com paring the two solutions, we note th a t the second m ethod is much sim pler. Answer. In the absence of resistance, the m axim um cur ren t in the circu it is 5.37 A. Questions and Problems 19.1. How w ill the frequency of an oscillatory circu it change if the capacitance is reduced using a variable ca pacitor? 19.2. W ill the insertion of a ferrom agnetic core into the coil of the oscillatory circu it affect the period of the elec trom agnetic oscillations in th e circuit? 19.3. A tra n sm itte r operates a t a frequency of 600 kHz. W hat w avelength corresponds to this frequency? 19.4. An oscillatory circu it consists of a 200-pF capacitor and a coil of inductance 20 m H . W hat are the frequency and the period of n atu ral oscillations of the circuit? 19.5. W hen a m otor car enters a tunnel, the sound of its radio is attenuated if it does not vanish altogether. W hy? 19.6. The ionosphere reflects radiow aves. How then can com m unication w ith spaceships by radio be realized? 19.7. An oscillatory circu it consists of a coil of inductance 2 m H and a capacitor whose capacitance can be varied fiom 10~9 F to 40 pF. W hat is the w avelength range for this c ir cuit? 19.8. An oscillatory circu it consists of a coil of inductance 0.01 m H and a capacitor of capacitance 10~9 F. To w hat w avelength is this circu it tuned? W hat frequency corre sponds to this w avelength?
§ 19. Electrom agnetic O scillations and W aves
215
19.9. An oscillatory circuit em its electrom agnetic waves of length 500 m. Determ ine the capacitance of the capaci tor in the circuit if its inductance is 1.5 m H . 19.10. W hat is the distance between an aeroplane and a radar if a signal reflected from the plane was received 3 X 10"4 s after it had been em itted? 19.11. How does the power of electrom agnetic rad iatio n depend on the oscillation fre quency of the em itting circuit? 19.12. Figure 102 shows the circuit diagram of a crystal receiver. W hat component sm ooths the current pulses? 19.13. Program m es are per ceived from a crystal receiver using earphones since the received signals are very weak. How is a signal amplified in modern radios? 19.14. A rad ar em its electrom agnetic waves of length 10 cm a t a frequency of 2.25 GHz in a certain m edium . W hat is the propagation velocity of the waves in th is m edium and w hat would th e ir wavelength be in vacuum ? 19.15. An oscillatory circuit containing a 1.5-pF capaci tor should be tuned to a frequency of 1.5 kHz. W hat in ductance is required? 19.16. The tim e dependence of current in an oscillatory circuit is described by the equation i = 0.06 sin 106ji£. D eterm ine the frequency of the electrom agnetic oscillations and the inductance of the coil if the m axim um energy of the m agnetic field is 1.8 X 10‘ 4 J . 19.17. An oscillatory circu it tuned to a frequency of 20 MHz contains a coil w ith inductance 10~6 H and a p a ra l lel-plate m ica capacitor w ith a plate area of 20 cm 2. D eter m ine the thickness of the m ica if its p e rm ittiv ity is 6. 19.18. The m axim um current in an oscillatory circuit is 6.28 X 10 ^ A. and the m axim um charge on the capacitor is 10“8 C. D eterm ine the period of oscillations in the circuit. To w hat w avelength is it tuned? 19.19. The m axim um poten tial difference across the ca pacitor in an oscillatory circu it is 120 V. W hat w ill the
216
Ch. III. O scillations and W aves
m axim um current be if the capacitance of the capacitor is 9 (LtF and the coil has an inductance of 6 m H? 19.20. The tim e dependence of the current in an oscilla tory circu it is given by the equation i = 0.02 sin 500jt£. The inductance of the circuit is 0.1 H. D eterm ine the period of electrom agnetic oscillations, the capacitance of the c ir cuit, and the m axim um energies of the electric and m agnetic fields. 19.21. As the capacitance of a capacitor decreases by 100 pF, the resonance frequency of the oscillatory circuit increases from 0.2 to 0.25 MHz. W hat is the inductance of the circuit?
Chapter IV
Optics. Fundamentals of the Special Theory of Relativity § 20. GEOMETRICAL OPTICS
llasic Concepts and Formulas To analyze the phenomena associated w ith the propagation of visible radiation (light) in a homogeneous m edium or. Ibrough the boundary between two transparent m edia, we introduce the concept of a light beam. I n a homogeneous m edium, lig h t beams \ propagate in a straig h t line, which explains the form ation of the um bra \/ \/ and penum bra. If a reflecting surface is placed in the path of a light beam, light is reflected in Fig- 103 accordance w ith the following two laws: 1. The incident ray, the reflected ray, and the perpendicu lar to the reflecting surface a t the point of incidence of the ray lie in the same plane. 2. The angle of incidence e is equal to the angle of reflec tion e' (Fig. 103). If the angles of incidence of two rays on a plane surface are the same, i.e. if the incident rays are parallel, they will remain parallel after reflection. Diverging rays em itted by a source of light rem ain divergent upon reflection (Fig. 104). In this case, the image formed a t the intersection of the continuations of the reflected rays will be v irtu a l and sym m etric to the source about the m irror. If a plane m irror is placed in the path of a converging heam, the image formed by it will be real, the point of in te r section of the continuations of the incident rays w ill be a v ir tual image of the light source (Fig. 105), and S'O = S O . Light rays incident on a spherical surface are reflected from it according to the same laws. If the m irror (reflecting) surface is the inner surface of the p art of a sphere, the m ir
J
218__________ Ch. IV. Optics. Special Theory of R elativity
ror is called concave, and a m irror formed on the outer sur face of a sphere is called convex. The stra ig h t line passing through the centre of curvature C and point O (pole of a m irror) is known as the optical axis of the m irror (Fig. 106).
F ig. 104
F ig. 105
A beam of rays incident on a concave m irror parallel to the optical axis is reflected such th a t it intersects the axis a t a point F known as the focus. Concave m irrors have real fo-
F ig. 106
ci, while convex m irrors have v irtu a l foci. The distance FO is called the focal length and is denoted /: FO = / = RI2, where R is the radius of curvature of the m irror. The form ula for a spherical m irror has the form 1 1 ,1
§ 20. Geometrical Optics
219
Here a and a' are respectively the distances from the object and from its image to the m irror. The m inus sign indicates that the m irror is convex, and the plus sign corresponds to a concave m irror. The linear m agnification is given by
Here the m inus sign in the denom inator corresponds to a concave m irror, and the plus sign to a convex m irror. The image of an object formed by a spherical m irror is constructed by using (1) an incident ray parallel to the op tical axis, (2) a ray passing through the focus, and (3) a ray along the radius of the curvature. Light is refracted when a ray is incident on the interface between two different transparent m edia, and obeys the following laws: 1. The incident and the refracted rays lie in the same plane as the perpendicular a t the point of inci F ig. 107 dence to the interface between the two m edia. 2. For two given m edia, the ratio of the sines of the angles formed by the incident and refracted rays w ith the norm al is a constant and called the refractive index of the second m e dium relative to the first one (Fig. 107): s in e sin e'
*2, i-
In the case under consideration, the second m edium is op tically denser than the first one. As a ray passes from the first to the second m edium , the angle of refraction is always sm aller than the angle of incidence. If a ray passes from an optically denser medium to an optically rarer m edium , it is possible for the rays not to be refracted but reflected to ta lly , rem aining in the same m edium . This is called to ta l in te r nal reflection. I t occurs when a ray is incident on the in te r face a t an angle larger th an a critical angle, defined as the
220__________ Ch. IV. Optics. Special Theory of R elativity
angle for which the angle of refraction is equal to 90°: sin ecr . n;£ sin 90
1
.
1
= — * sin ecr = — . n C1 n
The relative refractive index can be expressed in term s of the velocities of light in two media: fl 2A = V1IV2The absolute refractive index is 71 =
d v ,
where c is the velocity of light in vacuum . T ransparent m edia bounded by convex or concave surfaces are called lenses. Convex lenses converge rays, while con
cave lenses diverge them , if the refractive index of the lens m aterial is larger th an the refractive index of the m edium surrounding it. The principal optical axis of a lens is the line passing through the centres of curvature of its two refracting sur faces (Fig. 108). The point O in a lens is called its optical cen tre. Rays passing through the optioal centre are not refracted. P oints F and F ' are known as the foci of a lens. Any straig h t line passing through the optical centre of a lens is called an auxiliary optical axis. In order to construct the image formed by a lens, three rays are norm ally used: (1) a ray parallel to the principal optical axis, (2) a ray passing through one of the foci, and (3) a ray passing through the optical centre.
§ 20. Geometrical Optics
221
As is the case of spherical m irrors, the following form ula is applicable to lenses:
The sign rule here is the same as for m irrors. The u n it of optical power O = 1/f is the dioptre (D). An image formed by a concave lens is always v irtu a l, while a convex lens forms a v irtu a l image when the object lies between the lens and a focus. Lenses are widely used in optical instrum ents to m agnify images of objects (projectors, magnifiers, and microscopes) or to increase the angle of view of exam ining objects a long distance from the observer (binocular and telescopes). Worked Problems Problem 94. W hat w ill happen if a plane m irror is placed in the path of a converging beam? Solution. Let us exam ine the path of the rays shown in Fig. 109. Rays 1, 2 , and 3 are incident on the plane m irror 5'
s Fig. 109
in a converging beam. After reflection, they w ill intersect at a point S ' which is a real image of point S . At this p oint the rays would converge in the absence of the m irror. Using
222__________ Ch. IV. Optics. Special Theory of R elativity
the laws of reflection of light, we can prove the equality of isosceles triangles M S ' N and M S N , whence it follows th a t S'O = SO. Answer. A converging beam of rays w ill form a real and sym m etric image of a v irtu a l bright point. Problem 95. A concave spherical m irror forms a threefold magnified image of an object. The distance from the object
Fig. 110
to the image is 2.6 m. W hat is the radius of curvature of the m irror? Given: jj = 3 is the linear m agnification of the image, I = 2.6 m is the distance from the object to the image. Find: the radius of curvature R of the m irror. Solution. We know th a t R — 2j. Consequently, the solu tion reduces to determ ining the focal length of the spherical m irror. The image formed by a concave m irror is magnified in two cases. L et us consider both. 1. The object lies between the focus and the centre of cur vature (Fig. 110). In order to construct the image A ' B ' we shall use two rays, one of which is incident on the m ir ror p arallel to the optical axis and the other is directed along the radius of curvature. After reflection, the rays w ill in te r sect a t point A ' . D ropping the perpendicular from A r onto the optical axis, we obtain point B ' . We have a real image of the object w ith linear m agnification a' = §a. The problem states th a t I = a 1 — a, and hence I = Pa — a. Therefore, a = Z/(p — 1) and a = pJ/(P — 1). We su b stitu te these values for a and a ' into the form ula for a
§ 20. Geometrical Optics
223
spherical m irror and determ ine the focal length: 1 /
1 ,1 a + a' X
f=
Bz
1_ ’ / “ j
T ’ /=
P - l | P —1 I + pz ’
3 X 2.6 m
9 -1
1 _ P2“ l / pi ’
r> n «r
= a 9 7 5 m-
Consequently, R = 2 X 0.975 m = 1.95 m. 2. The object lies between the m irror and the focus (Fig. 111). In order to construct the image, we shall use the same rays as above. I t can be seen from the figure th a t the image
of point A ' is not on the intersection of the rays them selves but on their continuations, i.e. the image is v irtu a l. In the formula for a spherical m irror, the q u a n tity 1la should be taken w ith a m inus sign. The subsequent reasoning is the same as before, the only difference being th a t I = a + a . The result w ill be the same. Answer. The radius of curvature of the spherical m irror is 1.95 m. Problem 96. The radius of curvature of a convex spherical m irror is 1.2 m. How far away from the m irror is an object of height 12 cm if the distance between its virtu al image and the m irror is 0.35 m? W hat is the height of the image? Given: R = 1.2 m is the radius of curvature of the m ir ror, a' = 0.35 m is the distance between the v irtu a l image and the m irror, and h = 0.12 m is the height of the object. Find: the distance a between the object and the m irror and the height h' of the image. Solution. Considering th a t the focus and the image for a convex m irror are v irtu a l, we p u t the m inus signs in front
Ch. IV. Optics. Special Theory of R elativity
224
of 1// and 1la in the form ula: 1
1
1
f
a
a'
/= -§ - —
m’
Let us determ ine the height of the image: h'
a'
~ ~ T _
, ,
’
ha'
“ “
0.12 m X 0.35 m 0.84 m
’ = 0.05 m,
h' — 0.05 m.
A nsw er. The object is 0.84 m from the m irror, and the image is 5 cm high. Problem 97. The Sun forms an angle of 60° w ith the h o ri zon. D eterm ine the length of the um bra at the bottom of an opaque vessel illum inated by sun light. The height of the vessel is 25 cm. W hat w ill the change in the length of the um bra be when w ater is poured into the vessel to a height of 20 cm (Fig. 112)? Given: H — 25 cm is the height of the vessel, cp = 60° is the a ltitu d e of the Sun F ig. 112 above the horizon, h = 20 cm is the height of the w ater colum n. From tables, we take the refractive index for w ater, n = 1.33. Find: the length l x of the um bra a t the bottom of the em pty vessel and the change AI of the length of the um bra in the vessel w ith w ater. Solution, W e use e to denote the angle formed by the d i rection of incident rays w ith the vertical wall of the vessel. I t can be seen from the figure th a t the angle of incidence e and the altitu d e of the Sun are related: e + (p = Jt/2, e = 90°—60° = 30°. The rays propagate in the em pty vessel along a straig h t line (along^4Af), and the length I of the um bra can be calcuH
A
B
C
D
§ 20. Geometrical Optics
225
la ted from the right-angled triangle A M D : l x = AD = H tan e, tan 30° = 0.577, l x = 25 cm x 0.577 ~ 14.4 cm. If the vessel contains w ater, the rays w ill be refracted a t the air-w ater interface since a light ray bends away from the rarer m edium into a denser m edium . In this case, the refracted ray is closer to the vertical, and the angle of re fraction e' is sm aller than the angle of incidence e. The length of the um bra in the vessel filled w ith w ater is l 2 = BD = BC + CD.
B ut CD = ON.
1 n the right-angled triangle O M N , we have ON = (H — h) x tan e, ON = 5 cm X 0.577 ~ 2.9 cm. From the triangle BOC, we find BC = h tan e '. In order to determ ine the an gle of refraction, we use the second law of refraction: sin e/sin e' = n , sin e' = sin e/n, sin e' = 0.5/1.33 = 0.3759, and e' ~ 22°. From tables, we take tan 22° = 0.404. Consequently, BC = 20 cm x 0.404 = 8.1 cm. The length of the um bra in the vessel w ith w ater is l 2 = BD = 8.1 cm + 2.9 cm = 11 cm. W hen the vessel is filled w ith w ater, the length of the um bra decreases by AI: AI = 14.4 cm — 11 cm = 3.4 cm. Answer. The length of the um bra in the em pty vessel is approxim ately 14.4 cm. The um bra in the vessel w ith w ater is 3.4 cm shorter. Problem 98. A light ray is incident at 45° on a glass slab. The slab is 3 cm thick, and the refractive index of the glass is 1.5. W hat w ill the displacem ent of the ray be as a result of its passage through the slab? At w hat angle w ill the ray emerge from the slab? Given: e = 45° is the angle of incidence of the ray, h = 0.03 m is the slab thickness, and n = 1.5 is the refractive index of the glass. Find: the displacem ent 6 of the ray and the angle at which the ray emerges from the slab. Solution. Figure 113 shows th a t the displacem ent of the ray is the shortest distance between the direction of the 15-0530
226__________Ch. IV. Optics. Special Theory of R elativity
incident ray and the ray emerging from the slab: 6 = CD. From the triangle A C D , we have 8 = i4C sin (e — e '), A C = ft/cos e '. Consequently, c
ft sin (e — e') c o se '
In order to determ ine the displacem ent, we m ust find the angle e \ Using the second law of refraction, we obtain n = sin e/sin e ', sin e' = sin e/n, sin e = ] / r 2/(2 X 1,5) = 0.47, and e' ~ 28°. This gives 6=
0.03 m X 0.29
0.88
~ 0.0099 m ex 9.9 mm.
Answer. After passing through the slab, the ray rem ains parallel to the incident ray but is displaced by 9.9 mm*, e ' = e. Problem 99. An object of height 6 cm is set a t rig h t angles to the optical axis of a double-convex lens of optical power 5 D and 25 cm away from the lens. Determ ine the focal length of the lens, the position of the image, the linear m ag nification of the lens, and the height of the image formed by it. Given: h = 0.06 m is the height of the object, CD = 5 D is the optical power of the lens, and a = 0.25 m is the dis tance between the object and F ig. 113 the lens. Find: the focal length / of the lens, the distance a from the image to the lens, the linear m agnification p, and the height h ' of the image. Solution . Given the optical power, we can determ ine the focal length of the lens: f - w - r D = °-2m Let us construct the image of the object (Fig. 114). By choosing an appropriate scale, we can determ ine the re quired q u antities from the diagram accurately enough.
227
§ 20. Geom etrical Optics
Using the form ula for a th in lens, we calculate a ': 1
1
I
1
/
fa a—f ’
— -----r** & /r — ---a 1 a ’
,
0.2 m X 0.25 m
A
a = --------0.05 ?r?vc------= 1 m. m
The linear m agnification p can be determ ined from the formula a*
P
q
“ T ’
1m 0.25 m
# *
(Consequently, the height of the image w ill be W = pfe,
h' = 4 x 0.06 m = 0.24 m.
Answer. The focal length of the lens is 20 cm, the image is 1 in from the lens, the linear m agnification of the lens is and the image is 24 cm high.
Fig. 114
Problem 100. An object 30 cm high stands v ertically 80 cm from a lens w ith an optical power of —5 D . D eterm ine the position of the image and its height. Given: h = 0.3 m is the height of the object, a = 0.8 m is the distance from the object to the lens, and O = —5 D is the optical power of the lens. Find: the distance a' from the lens to the image and the height h ’ of the image. Solution. The lens is concave since its optical power is negative. L et us determ ine its focal length: 1
1
228
Ch. IV. Optics. Special Theory of R elativity
We construct the image of the object formed by the lens (Fig. 115). The image is v irtu a l. From the form ula for a
th in lens, we determ ine the distance from the lens to the image: f _ 0 . 8 m ( — 0.2 m) JL = ± + - i ■, a = —a —r « a = ■ — 0.16 m. / a' ’ a— f ’ 0.8 m + 0.2 m In order to determ ine the image height, we shall use the form ula for m agnification: a ’
h
a
fe' = + ^ - 0 . 3 m = |0 .0 6 |m . 0.8 m
1
‘
Answer. The distance from the lens to the image is —16 cm (the m inus sign indicates th a t the image is virtu al) and it is 6 cm high. Problem 101. Given the principal optical axis M N , an object A B , and the image A ' of point A (Fig. 116). D eter.
•A f
Af
F ig. 116
m ine the positions of the lens and its foci. Is the lens con verging or diverging? C onstruct the image of the object. Solution. We draw a ray from A to A ' . I t intersects the principal optical axis a t the optical centre of the lens. We
§ 20. Geometrical Optics
229
direct the second ray from point A parallel to the principal optical axis. I t is refracted by the lens, to h it A \ intersec ting the optical axis a t a focus F' of the lens. We draw a vertical plane through the focus, which is known as the focal plane. In order to construct the image of the object, we need only construct the image of B . We draw a ray from I
B towards the lens at an arb itra ry angle to the optical axis Parallel to this ray, we draw an au x iliary optical axis, which must intersect the ray a t a point in the focal plane. Then point B ' is obtained where the ray intersects the principal op tical axis (Fig. 117).Connecting points A ' and B ', we obtain the image of the object. Problem 102. A microscope has a m agnification of 157.5. The focal length of the objective is 0.3 cm. An object is 0.31 cm from the objective. D eterm ine the m agnification of the eyepiece and its focal length. W hat is the length of the microscope draw tube if the image is 26.25 cm from the eye piece? Given: p = 157.5 is the m agnification of the microscope, h = 0.3 cm is the focal length of the objective, a i = 0.31 cm is the distance from the object to the objective, and a ' = 26.25 cm is the distance from the eyepiece to the final im age. Find: the m agnification p 2 of the eyepiece, the focal length / 2 of the eyepiece, and the length L of the draw tube. Solution. The m agnification of a microscope is d e te r mined by the m agnifications p2 and p 2 of the objective and eyepiece: p = p ^ . In order to determ ine P2, we m ust know the m agnifica tion px of the objective. L et us determ ine the position of
230__________ Ch. IV. Optics. Special Theory of R elativity
the image formed by the objective from the form ula J _ _ J
, J_
/l
«i a\ ’ 0.3 cm x 0.31 cm /ifli 9.3 cm. at 0.31 c m — 0.3 cm «i —/i Remark. To avoid cumbersome num bers, we shall express all q u a n titie s in centim etres: a x—
9.3 cm 0.31 cm
30.
We can now find the m agnification of the eyepiece: p2 = - ^ - = - ^ - = 5.25. 30
Eyepiece
F ig. 118
Let us determ ine the distance a 2 from the eyepiece to the image A[B[, which is the object relative to the eyepiece (Fig. 118):
§ 20. Geom etrical Optics
231
The eyepiece is arranged relativ e to the image A \B [ so th a t the image is as if magnified. The image A \ B \ w ill thus be v irtu a l, and while determ ining the focal length / 2 of the eyepiece we m ust put the m inus sign in front of 1/a': 1
_ 1
fi
1 a'2 »
1 /3
_ 1 5cm
1
4
o
26.25cm ’
The length of the draw tube of the microscope can be de term ined from the form ula L = a[ + 02>
— 9.3 cm + 5 cm = 14.3 cm.
Answer. The m agnification of the eyepiece is 5.25, the focal length of the eyepiece is 6.2 cm, and the length of the draw tube is 14.3 cm. Problem 103. A point is 35 cm along the optical axis from a spherical concave m irror having a focal length 25 cm. A t
w hat distance along the optical axis from the concave m ir ror should a plane m irror be placed for the image it forms to coincide w ith the point source? Given: f = 0.25 m is the focal length of the m irror and a = 0.35 m is the distance between the point source and the m irror. F ind: the distance I between the plane and concave m ir rors. Solution. W ithout the plane m irror, the image of point S would be formed a t a point S[ (Fig. 119). Converging rays are incident on the plane m irror, and hence point S\ should be treated as a v irtu a l source of light, its image being real (see Problem 94) and separated from the concave m irror by
232__________ Ch. IV. Optics. Special Theory of R elativity
a . The distance form ula ,
a to S ' can be determ ined
af a — f 9*
n
,
0 .3 5 m X 0.25 m n -n r -------- c\ OC----I) .ihmm——0.25 0 mm 0.35
from th e
A Q« r ~
O .o 7 0 m .
Then the distance between the m irrors is 7
i
®
®
l = a ^-------- ^—
i
7
a oc
i 0 .875m — 0.35 m
Z= 0 .3 5 n H --------- ^
~
A oj
0.61m .
Answer. The point source and its image formed by the plane m irror coincide when the distance between the m ir rors is approxim ately 61 cm. Questions and Problems Reflection of light. Plane and spherical mirrors 20.1. A light ray is incident on a plane m irror. By w hat angle w ill the reflected ray be deflected if the m irror is turned through 20°? 20.2. The angle of incidence of a light ray on a plane m ir ror is decreased by 15°. How much w ill the angle between the incident and the reflected ray be reduced? 20.3. Sun rays are incident at an angle of 24° to the h o ri zon. How can they be directed parallel to the horizon using a plane m irror? 20.4. The Sun is 46° above the horizon. W hat m ust the angle of incidence' of the sun rays on a plane m irror be so th a t the sun light is reflected (a) vertically downwards, (b) v ertically upwards? W hat w ill the angle between the reflected ray and the m irror be? Draw the light paths on a diagram . 20.5. A parallel beam propagates horizontally from a pro jector. How should a plane m irror be arranged so th a t the image of a slide is formed on the ceiling? 20.6. A m irror hangs vertically on a wall so th a t its u p per edge is level w ith the person’s head. The length of the m irror is 80 cm. W hat is the m inim um height of the per son to s till see him self full length? 20.7. A plane m irror is at 45° to the horizontal surface of a table on which a book lies. In w hat plane w ill the image of the book be formed?
§ 20. Geom etrical Optics
233
20.8. A beam is incident on a plane m irror (Fig. 120). Show graphically the light path. W hat type of image w ill be formed of a lam p and where? 20.9. Two plane m irrors form an angle of 120°. The dis tance between the two images of a point source formed in A
F ig. 120
F ig. 121
them is 20 cm. D eterm ine the distance from the light source to the point where the m irrors touch if it lies on the bisector of the angle formed by the m irrors. 20.10. How should two plane m irrors be arranged for a point source of light and its two images to lie a t the ver tices of an equilateral triangle? 20.11. How m any images of an object w ill be formed by two plane m irrors placed (a) a t rig h t angles to each other, (b) a t an angle of 60°? 20.12. How m any images w ill be formed by twro m irrors lying in parallel planes on both sides of an object? 20.13. How can we see the image of a large building formed by a sm all plane m irror? 20.14. On a sunny day, the focus of a concave m irror can be determ ined just by using a ruler. How can this be done? 20.15. Determ ine the radius of curvature and the focal length of a convex spherical m irror (Fig. 121) if M N = 3 cm and A B = 30 cm. 20.16. A point source of light is on the axis of a spherical m irror four focal lengths away from it. C onstruct its image. 20.17. The focal length of a concave spherical m irror is 0.6 m. At w hat distance from the pole m ust a point source
234
Ch. IV. Optics. Special Theory of R elativity
be located on the optical axis so th a t its image is formed a t the same point? 20.18. A concave spherical m irror forms a threefold m ag nified inverted image of an object. W hat are the principal focal length and the radius of curvature of the surface of the m irror if the distance from the object to the image is 28 cm? 20.19. The focal length of a concave spherical m irror is 25 cm. W here should an object be placed so th a t its v irtu a l image is formed 1 m from the m irro r’s pole? 20.20. W hen an object is 40 cm away from the pole of a concave spherical m irror, the size of its image is equal to th a t of the object. W hat is the focal length of the m irror? 20.21. A point source lies on the principal optical axis of a concave spherical m irror, and its v irtu al image is 40 cm away from the m irror. The radius of curvature of the m ir ror is 1.2 m. D eterm ine the distance from the source to the pole. W hat is the m agnification? 20.22. A concave spherical m irror forms a real image of an object w ith m agnification 3. W hen the object is moved a further 5 cm away from the m irror, the m agnification of the new real image becomes 2. Determ ine the radius of curva tu re of the m irror surface and the in itia l distance between the object and the m irror. 20.23. A converging beam of solar rays is incident on a concave spherical m irror whose radius of curvature is 0.8 m. D eterm ine the position of the point on the optical axis of the m irror where the reflected rays intersect if the extensions of the incident rays intersect the optical axis 40 cm from the m irro r’s pole. 20.24. How far from an object in front of a concave spheri cal m irror w ill a real image be formed if it is 42 cm from the pole and the radius of curvature of the m irror is 48 cm? 2 0 .2 5 . W hy are convex m irrors better than plane ones in m otor cars for view ing the road? 20.26. How far from a convex m irror does an object lie if its image is formed 1 m away from the m irror? The focal length of the m irror is 1.5 m. 20.27. W hen an object is 60 cm from the pole of a concave spherical m irror, its image is formed w ith a m agnification P = —5. W hat does the m inus sign of the m agnification in dicate? W hat is the radius of curvature of the m irror?
§ 20. Geom etrical Optics
235
20.28. W here w ill the image of an object be formed by a convex spherical m irror w ith a 1.2-m radius of curvature and w hat type of the image w ill it be if the object is 0.3 m away from the m irror? 20.29. Converging rays are incident on a convex spherical m irror so th a t their extensions intersect 30 cm behind the m irror on the optical axis. The reflected rays form a d i verging beam so th a t their extensions intersect the optical axis 1.2 m from the m irror. D eterm ine the focal length of the m irror. 20.30. Two spherical m irrors (convex and concave) having the same focal length of 36 cm are arranged so th a t their
F ig. 122
optical axes coincide (Fig. 122). The separation between the m irrors is 1 m. At w hat distance from the concave m irror should an object be placed so th a t its images formed by the concave and convex m irrors are identical? Refraction of light 20.31. W hat is the refractive index of flint glass if for a light rays incident a t an angle of 63° the angle of refraction is 29°40'? 20.32. A light ray is incident on the surface of w ater a t an angle of 50°. W hat is the angle of refraction in w ater? 20.33. At w hat angle should a light ray be incident on the surface of crown glass so th a t the angle of refraction is 27°? 20.34. D eterm ine the angle through which a light ray is deflected as it passes from air to w ater at an angle of inci dence of (a) 30°7 (b) 45°?
236__________Ch. IV. Optics. Special Theory of R elativity
20.35. W hen a light ray is incident on a quartz plate a t an angle of 44°, the angle of refraction is 27°. For w hat angle of incidence w ill the ray be refracted a t an angle of 30°? 20.36. W hy is it difficult to shoot a fish swimm ing in w ater? 20.37. A light ray is incident on an air-liq u id interface at 45° and is refracted a t 30°. W hat is the refractive index of the liquid? For w hat angle of incidence w ill the angle be tween the reflected and refracted rays be 90°? 20.38. A light ray passes through the interface between two transparent m edia. Under w hat condition w ill the angle of refraction be equal to the angle of incidence? 20.39. P arallel beams of light pass from air to w ater. The angle of incidence is 55°. D eterm ine the angle of refraction of the rays and the propagation velocity in w ater. 20.40. Determ ine the velocity of light in ice if, for an angle of incidence of 61°, the angle of refraction is 42°. 20.41. As a light ray passes from air to a transparent liquid a t an angle of incidence of 69°, the angle of refraction is
F ig. 123
F ig. 124
38°30'. W hat is the velocity of light in the liquid? W hat si the refractive index of the liquid? 20.42. Determ ine the angle of incidence of a light ray on the surface of acetone if the angle between the reflected and refracted rays is 120°. 20.43. The refractive index of rock crystal is 1.54. A t w hat angle m ust a light ray be incident from air to the crystal so th a t the reflected ray is perpendicular to tfie refracted ray? 20.44. A source of visible light a t the bottom of an aq u ar ium filled w ith w ater em its rays incident on the w ater-air interface (Fig. 123). Trace the light path afterw ards. 20.45. Determ ine the c ritic a l angle of incidence for a light ray passing from diam ond to w ater.
§ 20. Geometrical Optics
237
20.46. Determ ine the refractive index of benzene if the critical angle of incidence of rays on it is 42°. 20.47. For w hat angle of incidence of a light ray on a diam ond-water interface is to tal reflection observed? 20.48. If a bright m etal ball is first blackened w ith soot in a candle flam e and then immersed in w ater, it appears bright again. W hy? 20.49. Can you explain why a test tube imm ersed a t a certain angle in a tum bler of w ater appears to have a m ir ror surface for a certain viewing position? 20.50. The direction of an incident light ray is shown in Fig. 124. The sem icylinder is made of glass w ith refractive index 1.8. W hat is the path of the ray afterw ards? 20.51. A bulb is fixed at the bottom of an aquarium divided into two parts by a p artitio n (see Fig. 123). W hat is the m aximum height to which w ater m ay be poured if the bottom at the opposite face is to be illum inated and the aqua rium is 60 cm long? 20.52. Solar rays are incident on the surface of w ater at 45°. How long is the shadow of a pole erected a t the bottom of a pond if the pole is vertical and com pletely immersed in w ater? The length of the pole is 1.2 m. 20.53. Solve Problem 20.52 assum ing th a t 0.2 m of the pole is above the w ater surface. 20.54. A light ray is incident a t 45° on a plate m ade of crown glass w ith plane-parallel faces. D eterm ine the th ic k ness of the plate if a ray emerging from it is displaced by 1.5 cm. 20.55. D eterm ine the displacem ent of a light ray passing through a 2.1-cm thick plate m ade of crown glass w ith planeparallel faces if the angle of incidence is 30°. 20.56. A plate w ith plane-parallel faces, having refrac tive index 1.8, rests on a plane m irror. A light ray is in cident on the upper face of the plate a t 60°. How^far from the entry point w ill the ray emerge after the reflection by the m irror if the plate is 6-cm thick? 20.57. L ight rays passing through a plane-parallel plate are displaced. W hy do we not notice this through a window? Under w hat conditions does the displacem ent become no ticeable? 20.58. A light ray is incident on a glass plate w ith planeparallel faces a t 70°. The p late is 4-cm thick. The refractive
238
Ch. IV. Optics. Special Theory of R elativity
index of glass is 1.5. D eterm ine (1) the displacem ent of the ray and (2) the p ath length of the ray in the p late. 20.59. A light ray is incident a t rig h t angles on a rig h tangled glass prism (Fig. 125). W hat w ill the p a th of the ray be afterw ards? 20.60. How m ust two right-angled prism s be arranged in a periscope? Draw the lig h t p a th in it. 20.61.* Two parallel rays 1 and 2 are incident on a glass right-angled prism (Fig. 126). Trace the lig h t p ath after wards. 20.62. A lig h t beam is incident along the norm al on a lateral face of a prism w ith an angle of refraction of 30°
F ig. 125
F ig. 126
F ig. 127
ig. 127). D eterm ine the angular displacem ent of the beam as a result of its passage through the prism if its refractive index is 1.8. 20.63. A ray incident on the la te ra l face of a glass prism w ith an angle of refraction of 30° emerges from it a t 30°. The refractive index of glass is 1.5. Determ ine the angle of incidence of the ray. 20.64. A ray is incident on a late ra l face of a glass prism along the norm al and emerges from the prism at 25° from the direction of the incident ray. The refractive index of glass is 1.5. W hat is the angle of refraction of the prism? 20.65. The angle of refraction of a prism is 60°. A lig h t ray emerges from the prism a t the sam e angle as it is incident on it. The refractive index of the prism is 1.5. D eterm ine the angle by which the ray is deflected from its in itia l direction as a result of its passage through the prism . 20.66. The angle of refraction of a prism is A. A lig h t beam is deflected by an angle 6 upon passing through the prism .
§ 20. Geometrical Optics
239
D eterm ine the refractive index of the prism if the angle of incidence of the ray is equal to the angle of refraction as it emerges from the prism. Lenses. Optical instruments 20.67. D eterm ine the optical power of a double-convex lens whose focal length is (a) 12.5 cm, (b) 50 cm. 20.68. D eterm ine the optical powers of concave lenses w ith focal lengths of —25 cm and —0.4 m. 20.69. Given lenses w ith optical powers of 4, —5, and —2 D. D eterm ine their focal lengths. J
s
F
°
•
F'
(
M
to
00
> Fig Screen*
it
s
\ f
F ig . 129
20.70. In w hat case will a double-convex lens be d i verging? 20.71. Given an optical axis M N , a converging lens, its foci, and a point source S. on the optical axis (Fig. 128). Construct the image of the point source. 20.72. An opaque screen is placed between an object and a converging lens (Fig. 129). W hat will happen to the im age? Use a diagram to explain your answer. 20.73. W hat will the paths of the rays be after refraction in the lenses shown in Figs. 130a and fc? 20.74. Two converging lenses should be placed in the path of parallel rays so th a t the rays rem ain parallel after
240
Ch. IV. Optics. Special Theory of R elativity
passing through both lenses. How should the lenses be ar ranged? 20.75. The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. W here w ill the im age be formed and w hat kind of image is it?
F ig. 130
20.76. How far away w ill a converging lens w ith a focal length of 10 cm form the images of objects located 5, 15, 20, and 25 cm from the lens? 20.77. C onstruct the images of the objects in Problem 20.76 on a 1 : 10 scale and answer the following questions: (1) how does the height of the image change w ith the distance between the object ,yV and the lens? (2) w hat type of image is it when an ob ject is placed tw ice the foF ig. 131 cal length from the lens? (3) where should the object be placed to obtain a v irtu a l image? 20.78. W hen an object is placed 20 cm from a lens, the image is the same size as the object. W hat are the focal length and optical power of the lens? 20.79. Given an optical axis M N and the positions of an object and its image (Fig. 131), determ ine graphically the position of the lens (its optical centre O) and its foci. Is it a converging or diverging lens? Is the image real or v ir tual? 20.80. At w hat distance from a double-convex lens w ith focal length 0.42 m is an object placed if its v irtu a l image is formed 56 cm from the object? 20.81. A candle is 2 m from a w all. W hen a converging lens is placed 40 cm from the candle between the candle and
§ 20. Geometrical Optics
241
the w all, a sharp image of the candle is formed on the wall. Determ ine the optical power of the lens and its m agnifica tion. 20.82. A lens placed between a candle and a screen forms a real trip ly m agnified image of the candle on the screen. W hen the lens is moved away from the candle by 0.8 m w ithout changing the position of the candle, a real image one-third the size of the candle is formed on the screen. D eter mine the focal length of the lens. 20.83. An object 2 cm high is 15 cm from a double-con vex lens whose optical power is 10 D. D eterm ine the height of the image. 20.84. An object is 1.4/ (where / is the focal length) from a converging lens w ith an optical power of 2 D. W here is the image formed? W hat type of image is it? 20.85. An object is placed 0.8/ from a lens (see Problem 20.84). W here is the image formed? W hat type of image is it? 20.86. The focal length of a double-concave lens is 12 cm. W hat is the distance between the lens and an object if its image is 8 cm away from the lens? 20.87. An object is 20 cm from a double-concave lens with a focal length of 30 cm. W here is the image formed (rel ative to the lens) and how high is it if the object is 6 cm high? 20.88. A v irtu a l image is formed 8 cm from a diverging lens w ith a focal length of 0.12 m. How far is the object from its image? 20.89. W ill the focal length of a glass lens change if it is immersed in a m edium w ith the same optical density as the glass? 20.90. W hat will the m agnification of a converging lens w ith an optical power of 4 D be if an object is 1.4/ away from it? 20.91. A v irtu al image one-fifth of the size of an object is formed 6 cm from a diverging lens. Find the focal length and optical power of the lens and construct the image of the object formed by it. 20.92. The distance from a converging lens to an object is thrice the focal length. D eterm ine the m agnification. 20.93. The optical power of a lens is 2.5 D. For a certain position of an object relativ e to the lens, a real doubly mag 1 G—05 30
242
Ch. IV. Optics. Special Theory of R elativity
nified image is formed on a screen. W hat w ill the m agnifica tion be if the object is moved 0 . 1 m closer to the lens? 20.94. A converging lens forms a trip ly m agnified real image of an object. In order to obtain a v irtu a l im age w ith the same m agnification the lens is moved tow ards the object by 10 cm. W hat are the focal length and the optical power of the lens? 20.95. Two identical th in converging lenses brought in contact so th a t th eir axes coincide are placed 1 2 . 5 cm from an object. W hat is the optical power of the system and th a t of a single lens if the real image formed by the system of lenses is four tim es as large as the object? 20.96. A convex and a concave lens are brought in close contact along their optical axes. The focal length of the convex lens is 10 cm. W hen the system is placed a t 40 cm from an object, a sharp image of the object is formed on a screen on the other side of the system . D eterm ine the o p ti cal power of the concave lens if the distance I between the object and the screen is 1 .6 m. 20.97. The focal length of a projector’s objective is 15 cm. The distance between the slide and the objective is 15.5 cm. W hat is the linear m agnification of the projector? 20.98. D eterm ine the optical power of the projector’s objective if it produces a 24-fold m agnification when a slide is placed 2 0 . 8 cm from the objective. 20.99. The objective of a projector has a focal length of 12.5 cm. How far should the screen be placed away from the objective to obtain a tw entyfold m agnification? 20.100. The image of a 7 X 5 cm 2 slide is projected onto a 1.05 X 0.75 m 2 screen using projector at its m axim um m agnification. W hat is the optical power of the objective’s lens if the screen is 4 m away? 20.101. W hat purpose do diaphragm s serve in the objec tives of cameras? 20.102. The focal length of a cam era’s objective is 50 mm. The height of the image on the film of a building 80 m away from the cam era is 12 mm. W hat is the actual height of the building? 20.103. The image on the frosted glass of a camera is 13.5 mm high when the object is 8.5 m away and 6 cm when 2 m away. Determ ine the optical power of the objective.
§ 21. P hotom etry
243
20.104. The focal length of a microscope’s objective is 5 mm and the draw tube is 16 cm long. D eterm ine the m agni fication of the eyepiece for norm al sight if the m icroscope’s to tal m agnification is 2 0 0 . 20.105. The focal length of a m icroscope’s objective is 0.4 cm. A sam ple is placed 0.1 mm away from the focus. The m agnification of the microscope is 400. D eterm ine the focal length of the eyepiece and the length of the draw tube if the distance of best vision for an observer is 25 cm. § 21. PHOTOMETRY
Basic Concepts and Formulas In photom etry, the concept of a point source of visible ra d iation (light) is employed. A fraction of this rad ia tio n cor responding to the w avelength range from 400 to 760 nm acts on the retin a of th e eye and is perceived as light. A point source is assumed to be consid erably sm aller th an the distance over which its action is ana lyzed. Such a source is assumed to em it uniform ly in all directions. An exam ple of such a source is a star. The lum inous flux O is the power of visible rad ia tio n and is estim ated by its action on the retin a of a norm al eye. The u n it of a lum inous flux is the lum en (lm). The lum inous in ten sity (candle power) I of a source is defined as the ratio of the lum inous flux propagating from the source in a given direction w ithin a sm all solid angle to the m agnitude of this angle: T
AO) AQ ‘
The un it of solid angle is the steradian (sr). A steradian is th e solid angle bounded by a conic surface which cuts an area of a sphere’s surface AS equal to the square of the rad iu s (Fig. 132), the apex of the solid angle being a t the centre of the sphere. The solid angle em bracing the space around the 16*
244__________ Ch. IV. Optics. Special Theory of R elativity
point re s ts on the surface of the sphere and is equal to £2 =
= 4 ji sr.
The to ta l lum inous flux em itted by a lig h t source is O tot = JQ = 4 ji/ . T h e u n it of lum inous in ten sity is the candela (cd). A lu m in o u s flux incident on the surface of a body is par tia lly reflected, reaches the eye retin a, and causes the sen sation of lig h t—we see the things surrounding us. The illum inance E is defined as the ratio of the lum inous flux (D incident on a sm all surface elem ent to its area E = (D/5. The u n it of illu m in an ce is the lux (lx). The illum inance of a surface by norm al rays varies in in verse proportion to the squared distance from the lig h t source to the surface: E 0 = HR*. If rays are incident on a surface at an angle, the illu m i nance depends on the angle of incidence E = E 0 cose, or i? = -~ -c o se . The lum inous intensities of two sources can be compared using a photom eter. The two sources are placed on either side of the photom eter at distances such th a t the illum inances a t the photom eter are equal. B y m easuring the distances from the lig h t sources to the photom eter and given the lu m i nous in te n sity of one of the sources, we can determ ine the lu m inous in ten sity of the other source: E = -Rk\ - .’ _£l
R * ~
h-
i?! ’
2
E = -R \
’
or J ± - I L Ur
~
Rl
Worked Problems Problem 104. Two lam ps w ith lum inous inten sities of 200 and 300 cd are suspended from 3-m high poles (Fig. 133). The distance between the lam ps is 4 m. D eterm ine the illu m inance on the ground a t points A , B , and C.
§ 21. Photom etry
245
Given: h = 3 m is the height of the lam ps above the ground, I x = 200 cd and / 2 = 300 cd are the lum inous in tensities of the lam ps, and ! = 4 m is the distance between them . Find: the illum inances E A, E B, and E c a t points A , B , and C. Solution. A t points A, 5 , and C, the surface of the ground is illu m i nated by both the lam ps. Let us con sider each case separately. At point A, E a — E lA + E 2A, where E lA is the illum inance at point A from the first lam p and E 2A is th a t from the second lam p. The light rays from the first lam p are incident along the norm al to the surface at point A , and hence E\A — - j r , E iA —
200 cd 9 m2
=
2 2 .2
lx.
The illum inance produced at the same point by the second lam p is due to the rays incident at an angle e (see Fig. 133). Therefore, E 2A = E 0 cos e, where E 0 is the illum inance of the surface at point A due to rays incident along the nor m al, E 0 = I 2/r2. Since r = J/7 &2 + Z2, we have E0=
•If
can be h ..■■■■■■■ .
h cos e = — = r
seen
from
the
figure
th a t
\ / h 2+ l2
The second lam p produces a t point A an illum inance of n
i2 2A —
1,2 _I
ti 72
h2 -{-l2
,o 9 y, / Vh 2r- 7j -~ 12 ’
300 cd x o 3m ouuca m
t?
2A “
o r 25m2
_
X
r — "" 5 m
701-®—
'
IX .
The to ta l illum inance a t point A is E A = 22.2 lx + 7.2 lx = 29.4 lx. The illum inance a t the ground a t point B is found in the sam e way: E q = E ib + E 2b >
246__________ Ch. IV. Optics. Special Theory of R elativity
The angle of incidence of the rays from the first lam p re m ains the same, and j?
11
I
12
EB= -^ -c o s e + f c f ,
ri
200cd A C
£ fl = ^
300cd
QO 4 ,
r 0.6 + - ^ r - = 38.11x.
The illum inance at point C is E c — E lc + E 2C. Since point C is eq uidistant from the two lam ps, the rays from the lam ps form the same angle ex w ith the norm al. Therefore, Ec = -pf- cos ei + -^ ~ cos elt or E c = c° l ei ( /, + / 2). Here r\ = fe2+ (0.5Z)2= 1 3 m2, and c o s e ^ — = — 0.83. 4 rj o.om F in a lly , we have E c =■■ 1
(200 cd + 300 cd) = 31.9 lx.
Answer. The illum inance of the ground at points A , B , and C is 29.4, 38.1, and 31.9 lx respectively. Problem 105. Two incandescent lam ps w ith lum inous in tensities 25 and 225 cd are 1 m ap art (Fig. 134). Where should a screen be placed between sSreen them for it to be equally illuI . m inated from both sides? k I -..-o Given: I x = 25 cd and / 2 = p3 rfm* i ^ 225 cd are the lum inous in ten sities of the first and second F ig. 134 lam ps and I = 1 m is the sep aration between them . Find: the distance from the first lam p to the screen. Solution. We shall assume th a t the rays from the first and second lam ps are incident on the screen a t rig h t angles. Then the illum inances of the screen by the lam ps are Tp
11
-f1 r2
jp
rl
By hypothesis, E i = E 2, and hence l - r t.
Therefore,
ij. =
,
• or A
B ut =
r2 = .
§ 21. Photom etry
247
We shall solve th is equation for r<:
-^£=7^-, (/ - r,) v h = r t y r2) zy r,= r , (K/.+yr2>, whence r* =
i V h
1 m V 25~cd
A nK
- —*-=-, r t = — = 0.25 m. 1^/i + V /* j/25 cd + ]/225 cd
Answer. The screen should be placed 0.25 m from the first lam p. Problem 106. A 25-cd point source of light is a t the centre of a spherical surface of radius 0.5 m. D eterm ine the lu m i nous flux incident on the inner surface of the sphere over an area of 50 cm 2. Given: r = 0.5 m is the radius of the spherical surface, / = 25 cd is the lum inous in ten sity of the source, and S = 50 cm 2 = 5 X 10 " 3 m 2 is the area of the surface elem ent, iFind: the lum inous flux O . Solution. In order to determ ine the lum inous flux, we m ust know the area of the surface on which it is incident and its illum inance: (D - E S . The illum inance is produced by the rays incident along the norm al on the inner surface of the sphere. Therefore, E = / / r 2, and hence (D = - L s ,
5 .^ 10? ~ 3
0.25
sr = 0.5 lm.
Answer. A lum inous flux of 0.5 lm is incident face elem ent of area S .
on a sur
Questions and Problems 21.1. D eterm ine the to ta l lum inous flux em itted by a source of lum inous in ten sity 2 0 0 cd. 21.2. A lum inous flux of 2 lm is uniform ly distrib u ted w ithin a solid angle of 0.5 sr. W hat is the lum inous intens ity of the point source located a t the apex of the angle? 21.3. A 15-cd point source of light is placed a t the centre of a hollow sphere of radius 30 cm. D eterm ine the illu m i
248
Ch. IV. Optics. Special Theory of R elativity
nance of the inner surface of the sphere and the total lum i nous flux em itted by the source. 21.4. D eterm ine the lum inous flux passing through a su r face of area 20 cm 2 located 5 m from a 100-cd point source of light. The rays are incident normal to the surface. 21.5. The average illum inance in Leningrad during a sum m er night (when the Sun does not sink deep below the horizon) is 1 lx, while the illum inance during a m oonlit night is 0.1 lx. W hat are the lum inous fluxes incident on the Mars Field in Leningrad which covers an area of 0.1 km 2? 21.6. An incandescent lam p (w ithout a shade) of lum i nous in ten sity 25 cd is suspended 80 cm above a table. De term ine the illum inance on the table. 21.7. The lum inous in ten sity of the lam p in a photograph ic enlarger is 15 cd. Determ ine the illum inance on a piece of photographic paper if the enlarger is 30 cm above it and only 15% of the lum inous flux is used. 21.8. An incandescent lam p in a room produces an illu m inance of 28 lx at one wall and 7 lx on the opposite wall at the same level. W hat is the ratio of the distances between the lam p and the walls? 21.9. W hen w ill the illum inance under a lam p be higher: for a lum inous intensity of 120 cd 3 m away or for a lum i nous in ten sity of 25 cd 1.2 m away? 21.10. P arallel rays incident a t an angle of 25° produce an illum inance of 54 lx. A t w hat angle of incidence will the illum inance of the surface be 45 lx? 21.11. Before a sunset, sun light is incident on the sur face of the E a rth a t an angle of 81°. Compare the illu m i nances produced on the surface of the E arth and on a vertical w all facing the Sun. 21.12. The m axim um illum inance th a t can be created by sun light on the surface of the E arth is 1 0 8000lx. How far from the Sun is the planet Mars when the m axim um solar illum inance on its surface is 48000 lx? 21.13. W hy does snow m elt more quickly on sunlit slopes than on su n lit horizontal areas? 21.14. W hen parallel rays are incident on an object alo rg the norm al to its surface, the illum inance is 70 lx. W hat w ill the illum inance of the surface be if the object is turned so th a t the angle of incidence is 60°? 21.15. An electric bulb of lum inous in tensity 150 cd is
§ 21. Photom etry
249
suspended above a round table of diam eter 2 m. D eterm ine the m axim um and m inim um illum inances on the table if the distance of the cent e of the table from the lam p is 1.5 m. 21.16. A lam p w ithout a shade is suspended 1 m above the centre of a table w ith diam eter 1.2 m. D eterm ine the illum inance a t the edge of the table if the to tal lum inous flux from the lam p is 650 lm. 21.17. A bulb w ithout a shade is suspended 1 m above a table. The distance between the bulb and a book lying at the edge of the table is 2 m. W hat m ust lum inous in te n sity of the bulb be if the illum inance on the book is 25 lx? 2 1.18. The light from an electric bulb is in cident on a w orking space a t an angle of 45° and produces an illum inance of 141 lx. The lum inous in ten sity of the bulb is 200 cd. How far away is the bulb from the working space? How high above the working space is it suspended? 21.19. Two bulbs w ith the same lum inous in ten sity of 50 cd are suspended 1 m above a table. The distance between the bulbs is 140 cm. D eterm ine the illum inance on the table under each bulb. 21.20. Two lam ps of 200 cd each are fixed to a pole 2 and 3 m from the ground respectively. Determ ine the illu m i nance on the ground a t 1 m from the foot of the pole. 21.21. A lam p is suspended a t the top of a 10-m pole so th a t the illum inance on the ground 1 0 m from the base of the pole is 2.5 lx. W hat is the lum inous in ten sity of the lam p? 21.22. Two lam ps of 250 cd each are suspended from a height of 4 m on poles 5 m a p a rt. D eterm ine the illum inance on the ground m idw ay the poles. 21.23. A lam p of 800 cd is suspended 10 m above the ground. Over w hat area of the ground w ill the illum inance be a t least 1 lx? 21.24. Two incandescent lam ps w ith lum inous in te n sitie s 300 and 200 cd are suspended a t a height of 3 m. The d is tance between the lam ps is 4 m. D eterm ine the illum inance of the point on the ground between the lam ps, where rays from the first lam p are incident a t 4 5 °. 21.25. A lam p of lum inous in ten sity 32 cd which was sus pended 1 . 2 m above the m iddle of a table is replaced by another lam p whose lum inous in ten sity is 90 cd. How high
250
Ch. IV. Optics. Special Theory of R elativity
m ust the second lam p be suspended if the illum inance a t the m iddle of the table is to rem ain unchanged? 21.26. A lam p suspended from a height of 6 m illum inates a skating rin k . How far from the lam p and from the centre of the rin k w ill the illum inance o n /th e surface of ice be a factor of 3.4 less than th a t a t the centre? 21.27. A lam p w ith lum inous in ten sity 25 cd is placed 15 cm to the left of a photom eter. A nother lam p is placed 45 cm to the rig h t of the photom eter, the illum inance on both sides being the same. D eterm ine the lum inous intens ity of the second lam p. 21.28. Two lam ps w ith lum inous intensities 50 and 200 cd are 2.4 m ap art. W here should an opaque screen be placed between the lam ps so th a t it is equally illum inated on both sides? 21.29. The centre of a screen is illum inated by a light source w ith a lum inous intensity I th a t is placed a distance Screen
Fig. 135
I from the screen. W ill the illum inance change if the lum i nous in ten sity and the distance from the light source are both increased rc-fold? 21.30. A light source of 40 cd is placed between a screen and a plane m irror 0.5 m from the screen (Fig. 135). The distance between the screen and the m irror is 1.2 m. D eter m ine the illum inance of the screen where the lig h t ray is incident along the norm al. Assume th a t the m irror is per fectly reflecting. 21.31. A lam p used for p rin tin g photographs had lu m i nous in ten sity 50 cd and was placed 1.2 m from the photo graph. The exposure tim e was 3 s. W hen the lam p b urnt out, it was replaced by another lam p of 40 cd placed 1 m from the photograph. D eterm ine the exposure tim e for the new lam p.
§ 22. W ave Properties of Radiation
251
§ 22. PHENOMENA EXPLAINED BY THE WAVE PROPERTIES OF RADIATION. INTERFERENCE. DIFFRACTION
Basic Concepts and Formulas O ptical rad ia tio n occupies a sm all in terv al in the electro m agnetic wave spectrum and includes three regions, viz. the u ltra v io le t, visible, and infrared wave bands. U ltra v io le t rad iatio n corresponds to w avelengths from about 5 to 400 nm and it can induce m arked chem ical re actions. The rad iatio n causing the sensation of light is known as visible rad iatio n , or ju st light. The lower boundary of the spectrum of visible rad iatio n lies between 380 and 400 nm, and the upper boundary between 760 and 780 nm. Infrared rad iatio n corresponds to w avelengths from 780 nm (the upper boundary of visible radiation) to 1 mm. It has a w ell-pronounced therm al effect. O ptical rad iatio n is electrom agnetic in nature. The wave length is given by k = c T , or k = c/v, where c is the propagation velocity of electrom agnetic waves, T and v being their period and frequency. As light waves pass from vacuum to a m edium , the wave length changes (the frequency rem aining unchanged):
where n is the absolute refractive index and c and v are re spectively the propagation velocities of electrom agnetic waves in vacuum and in a m edium . Phenom ena which confirm the wave nature of visible ra diation include interference and diffraction. The interference of light is the enhancem ent or a tte n u a tion of light as a result of superposition of light waves. In order to observe an interference p attern , the waves m ust have the same w avelength and a constant phase difference, i.e. th ey m ust be coherent. Using Fresnel’s biprism , we can form two v irtu a l images S ' and S" of the same source S, which em it coherent rays (Fig. 136). W hen m onochrom atic rad iatio n is incident on the
252__________ Ch. IV. Optics. Special Theory of R elativity
biprism (i.e. the radiation w ith a single frequency), bright and dark fringes are formed on the screen. The m axim um brightness occurs when the optical path difference A is equal to an even num ber of half-waves: A = 2k ± ,
where k — 1 ,
2,
If, however, one wave lags behind another by a h a lf w avelength, i.e. the optical path difference is equal to an odd num ber of half-waves, the m axim um atten u atio n of light is observed: A = (2A -H ) A , where A: = 0, 1,
2
, ....
Interference can be observed in th in films or for light pas sing through a system formed by a plano-convex lens and a glass plate (Newton’s rings). If we know the radius of curvature R of the lens and m ea sure the radius rh of one of the dark rings (Fig. 137), we can
t
I I# I I I I
E
0 , ft ,
* . ,(Ljd
Fig. 137
determ ine the w avelength of the light illu m in atin g a device th a t yields N ew ton’s rings:
where ft = 0, 1, 2, . . . is the num ber of the dark ring. The form ula is valid when air, for which n = 1 , occupies the space between the plate and the lens.
§ 22. W ave Properties of Radiation
253
Diffraction is the bending of light round obstacles, and it occurs when the size of the obstacle is com m ensurate w ith the w avelength. A diffraction grating is formed by altern atin g transparent and opaque lines. The sum of the w idth of the line and of a separation between two lines is known as the g ratin g con s ta n t (Fig. 138): d — cl -|- by where d is the grating constant, b is the w idth of a line, and a is the distance between two lines. The form ula for a diffraction grating is d sin tp = kX, where k is the order of a m axim um . The brightest (zero-or der) m axim um lies on the screen opposite the centre of the g ratin g . The form ula of a diffraction grating shows th a t one m ust measure the angle cp to determ ine the wavelength using a diffraction grating. W hen w hite light is incident on a grating, a spectrum known as the norm al spectrum is formed on a screen. It contains all the coloured bands (from red to violet). Worked Problems Problem 107. The wavelength of light in glass is 450 n m . L ight propagates in glass at a velocity of 1.8 X 105 km /s. D eterm ine the frequency of light, the absolute refractive index of glass, and the w avelength of the light passing from glass into vacuum . Given: Xg = 450 nm = 4.5 X 10 ~7 m is the w avelength of light in glass, v = 1 . 8 x 1 0 6 km /s = 1 . 8 x 1 0 8 m/s is the velocity of light in glass. From tables, we take the velocity of lig ht in vacuum , c = 3 X 10 8 m/s. F in d : the frequency v of light, the absolute refractive in dex n of glass, and the w avelength X of light in vacuum . Solution. From the w avelength and the velocity of light in glass, we can determ ine its frequency: Xg = v/v, whence 1.8 X 108 m /s 4.5 X 10" 7 m
4.0
x 1014 Hz = 400 THz.
254__________Ch. IV. Optics. Special Theory of R elativity
As light passes from one m edium to another, its frequen cy rem ains unchanged, but its velocity and w avelength change. Consequently, we can determ ine the wavelength of light in vacuum: 3 X 108 m /s 4 .0 X 1014 Hz
:0 .7 5 x 10-6 m.
The optical density of a m edium is equal to the absolute refractive index: n= — ,
3 x 108 m /s ~ 1.8 X 108 m /s
1.7.
Answer. The frequency of light is 400 THz, the wavelength in vacuum is 750 nm, and the absolute refractive index is about 1.7. Problem 108. Two coherent light sources em it light of w avelength 550 nm which produces an interference pattern
F ig. 139
on a screen (Fig. 139). The sources are 2.2 mm a p a rt and 2.2 m from the screen. D eterm ine w hether the interference a t point O is constructive or destructive. Given: X = 550 nm = 5.5 X 10 ” 7 m is the w avelength of light, I = 2 . 2 m is the shortest distance from the first source to the screen, and d = 2.2 mm = 2.2 X 10 ' 3 m is the separation between the sources. Find: the path difference A of the rays. Solution. In order to answer the question, we m ust know the path difference for the rays. The optical path difference
§ 22. W ave Properties of Radiation
255
is equal to their geom etrical difference (since the rays pro pagate in the same m edium, viz. air): A = S 2D = S 20 -
5 ,0 , 5 ,0 = I .
From the tria n g le 5 t0 5 2, we determ ine S 20: S 20 = y W + d * = l V l + (dll)*. Since d/l is much sm aller th an Z, we can use approxim ation whence
+
Th<,n
. _
(2 .2 X 10"» m )a 2 x 2 .2 m
1.1
X 1 0 ”6 m .
There w ill be constructive interference a t O if the p ath difference contains integral num ber of waves, i.e. k = 1 , 2, 3, . . .: A l.lx lO “em K ~
5 .5 X 10-7 m
Answer. C onstructive interference (bright fringe) occurs a t O. Problem 109. A plan o co n v ex lens w ith a radius of cu rv a ture of 12 m is placed on a flat plate as shown in Fig. 140. M onochrom atic light is incident along the norm al to the plane face of the lens, and dark and bright rings are formed in the reflected light. D eterm ine the w avelength of the m o nochrom atic light if the radius of the six th dark ring is 7.2 X 10 - 3 m. Given: R = 12 m is the radius of curvature of the lens, k = 6 is the dark ring num ber, and r fl = 7.2 X 10 ” 3 m is the radius of the six th ring. Find: the w avelength k of the m onochrom atic light. Solution. The light wave incident on the plane surface of the lens is p a rtially reflected by the convex surface of the lens and p a rtially passes through the air gap d and then reflected from the flat plate, the path difference being in creased by k/2 by the reflection. Thus, the second wave pas ses through the gap d twice, and hence the p ath difference w ill be A = 2d -f k/2.
256______
Ch. IV. Optics. Special Theory of R elativity
For a dark fringe, we have A = A = Xk + A ,
X
(2k + 1), or 2d +
whence d = A - .
From triangle A B C , we obtain i ? 2 = (i? — d ) 2 + r^, 2/?d — d 2 = r\ . The q u a n tity d 2 can be neglected since d is sm all. This gives 2R d = r\, whence d = r\!2R. Com paring the obtained expressions for d, we obtain Xk_ _ r\ 2 ~ 2R ’
whence *=
kR ’
(7-2 x
l 0 ~3m )2
6 x 12 m
= 7.2 x 10~7 m.
The wavelength of light is 720 nm. Problem 110. The grating constant of a diffraction grating is 0.016 m m. The red line of the 2nd-order spectrum is 14.2 cm from the m iddle line. The dis tance from the grating to a screen is 1.5 m. Determ ine the w avelength of the red light and the w idth of t h e 2 nd-order spectrum . The wavelength of violet light is 4 X 10 -7 m (Fig. 141). Given: d = 0.016 mm = 1 .6 s ta n t, h T = 14.2 cm = 1.42 X 1 0 -1 m is the distance between the red region of the Fig. 141 2 nd-order spectrum and the m iddle line, k = 2 is the order of the spectrum , I = 1.5 m is the distance from the screen to the diffraction grating, and Xv = 4 X 10 ‘ 7 m is the w avelength of violet light. Find: the wavelength Xr of the red light and the w idth h of the 2 nd-order spectrum . Solution. The wavelength of the red light can be d eter m ined from the form ula for a diffraction grating: k k T = d sin (p. Here the angle
X
1
§ 22. W ave Properties of Radiation
257
put sin
d
tan
K = — k 1.6
«
’ or
dhi
*= r ~~n kl
'
x 10“5 m X 1.42 X 10"1 m
7.57 X lO " 7 m.
In order to determ ine the spectral w idth, we m ust know the distance from the m iddle line of the spectrum to the violet region h y . This can be determ ined from the form ula for .. .. , kvkl , 4XlO" 7 m x 2 x l .5 m a diffraction grating: h v = — , hy = i. 6 xlO-» m------- 7.5 x lO ' 2 m. Thence h = h T — h y = 14.2 X 10 ” 2 m — 7.5 X 10 " 2 m = 6.7 x 10 - 2 m.
Answer. The wavelength of the red light is approxim ately equal to 760 nm and the w idth of the 2nd-order spectrum is 6.7 cm. Questions and Problems 22.1. W hen would a bright spot be formed on a screen where two beams from a coherent m onochrom atic light source meet, and when would the spot be dark? Screen
22.2. Light from two coherent sources is incident a t points 1 and 2 on a screen (Fig. 142). W here will construc tive and destructive interference occur? 22.3. If crystals of common salt are placed in the flam e of a candle and then the flame is observed through a tra n sp a 17-0530
258
Ch. IV. Optics. Special Theory of R elativity
rent plate w ith parallel faces, altern atin g dark and yellow fringes can be seen against the background of the flame. The same pattern can be observed in reflected light (by placing the plate behind the flame). How can this effect be ex plained? 22.4. Oil spots on the surface of sunlit w ater are rainbow coloured. W hy? W ill the p attern change if the surface is illum inated by m onochrom atic light? 22.5. Rays from two coherent light sources of w avelength 0.5 |Lim and w ith a path difference of 0.5 mm arrive a t a certain point in space. W ill the interference be constructive or destructive at this point? 22.6. Red light of wavelength 760 nm from two coherent sources is incident on a screen, form ing an interference p a t tern of red and dark fringes. Determ ine the path difference of the rays if four half-waves fit into it. W hat type of fringe (red or dark) is formed for such a path difference? 22.7. A soap bubble displays all the colours of the ra in bow in sun light. W hy? 22.8. Determ ine the radius of the second dark Newton ring in reflected Ught if a plano-convex lens w ith the radius of curvature 8 m and a flat plate (Fig. 140) are illum inated by a m onochrom atic light w ith a wavelength of 640 nm. 22.9. An instrum ent for observing New ton’s rings is il lum inated by a m onochrom atic red light. The radius of the third dark ring is found to be 2.8 mm. D eterm ine the wave length of the red light if the radius of curvature of the plano convex lens is 4 m. 22.10. The air gap in an instrum ent for observing New to n ’s rings (see F ;g 140) is filled w ith water. W hat w ill the change in the radii of the interference rings be? 22.11. W hat is the antireflection coating of objectives based on? W hy do objectives have a bluish-violet tinge in reflected light? 22.12. The distance between two coherent m onochrom at ic point light sources is 1.5 cm. The sources are located at 36 m from a screen so th a t the line connecting them is p aral lel to the plane of the screen. Determ ine the w avelength of the light if the separation between adjacent interference fringes is 1 . 8 mm. 22.13. A plano-convex lens w ith a radius of curvature of 8 m is put on a flat transparent plate. W hen the system is
§ 22. W ave Properties of Radiation
259
illum inated w ith green light from thallium (wavelength 536 nm), N ew ton’s rings are formed in the reflected light. Determ ine the radius of the fifth dark ring. 22.14. The radius of the third dark Newton ring (see Problem 22.13) is found to be 2.8 mm when the system is illum inated by a m onochrom atic light. Determ ine the radius of curvature of the plano-convex lens if the w avelength of the m onochrom atic light is 720 nm. 22.15. If the light of a street lam p is viewed through one’s eyelashes, rainbow rings appear around the lam p. E xplain this effect. 22.16. W hat is the difference between a diffraction spec trum and the spectrum obtained w ith a prism? 22.17. In a practical experim ent on the w avelength of light using a diffraction grating, the first diffraction m axi mum was observed on a screen 30 cm from the central line. The grating co n stan t was 2 X 10 ~3 mm, and the distance from the screen to the grating was 1.5 m. Using these d ata, determ ine the wavelength. 22.18. Light from a gas-discharge tube is incident nor m ally on a diffraction grating whose constant is 2 X 10 -3 m m. The orange line in the lst-order spectrum is seen a t an angle of 18°, and the blue line is seen a t 14°. Determ ine the w ave lengths of this light. 22.19. Spectra formed by a diffraction grating are pro jected on a screen 3 m away. D eterm ine the w avelength of a m onochrom atic light if the distance from the central line to the lst-order spectrum is 2 2 . 8 cm and the grating constant is 0 . 0 1 mm. 22.20. M onochrom atic light corresponding to a sodium line of w avelength 5.89 X 10 ~ 7 m is incident on a diffraction grating. The angle a t which this line is seen in the lst-o rd er spectrum is 17° 18'. D eterm ine the grating constant. How m any lines does a centim etre of the grating contain? 22.21. M onochrom atic light em itted by a m ercury-vapour lam p and having a w avelength of 579 nm is incident on a diffraction grating w ith constant ? X 10 -6 m and forms a diffraction spectrum on a screen. The distance from the g ra t ing to the screen is 1.5 m. How far from the central line will the coloured line in the lst-o rd er spectrum be? 22.22. How m any lines per m illim etre w ill a diffraction grating have for the green line (wavelength 500 nm) in the 17*
260__________ Ch. IV. Optics. Special Theory of R elativity
3rd-order spectrum to be observed a t angle 48° 30'? 22.23. A m onochrom atic light of wavelength 500 nm is incident norm ally on a diffraction grating containing 500 lines per m illim etre. W hat is the highest order spectrum ob servable using this grating?
§ 23. RADIATION AND SPECTRA
Basic Concepts and Formulas W hile light is a composite. We can prove this w ith a trigo nal prism . After passing through the prism, a ray of w hite light splits into the coloured rays of the spectrum . Dispersion is the dependence of the refractive index on the wavelength of the transm itted light. The refractive index n r for red light is sm allest and for violet light n y, is largest. This means th a t red light propagates faster in a transparent medium than violet light: n T = cluT, n y = c/vy. Since n r < n y, clvT < c/vy, or vr > v y . The colour of a transparent body depends on the colour of the light rays it transm its. Glass is green if it only transm its the green light from w hite light. The colour of an opaque body is determ ined by the colour of the light it reflects. All bodies when hot enough produce emission spectra. 1 . Continuous spectra are em itted by heated liquids or solids. These spectra are the same for all bodies and consist of seven basic coloured bands which continuously merge into each other. 2 . Line spectra are obtained from heated gases or vapours. Each chem ical elem ent has its own line spectrum which differs from every other spectra in the num ber of lines, the colour, and the position of the lines against the background of a continuous spectrum . 3. Band spectra are obtained from molecules and consist of a num ber of bands. Absorption spectra appear when light passes through a m e dium which can selectively absorb light and has a lower tem perature. The emergence of absorption spectra obeys K irchhoff’s law: a substance m ainly absorbs light w ith the wave lengths which it can em it.
§ 23. Radiation and Spectra
261
The therm al rad iatio n of a body a t a given tem perature is determ ined by its rad ian t exitance R e which equals the ratio of the ra d ia n t flux em itted from a sm all surface ele m ent to the area of this surface. Bodies can absorb ra d ia tion incident on them . If a body com pletely absorbs the ra d ian t flux incident on it, it is called a blackbody. According to the law on therm al radiation, a blackbody has the m axi mum rad ia n t exitance. Stefan-Boltzmann's law. The exitance of a blackbody eh is proportional to the fourth power of its therm odynam ic tem perature: Re = oT\ where a is the Stefan-Boltzm ann constant. Wien's displacement law. The product of the wavelength corresponding to the m axim um of rad ian t energy and the therm odynam ic tem perature is constant for a blackbody: ^ma x T = b , where b is the W ien constant. Consequently, as the tem perature increases, the m axim um of the rad ia n t energy is displaced towards shorter waves. Worked Problems Problem 111. Assuming th a t the tem perature of the s u r face of the Sun is approxim ately 6000 K, determ ine the wavelength corresponding to the m axim um energy, consid ering th a t the Sun is a blackbody. Given: T = 6000 K is the tem perature of the S u n ’s su r face. From tables, we take the W ien constant: b = 2.89 X 10 - 3 m- K. Find: the w avelength Xmax corresponding to the m axi mum energy. Solution. The wavelength corresponding to the m axim um rad ian t energy in the spectrum of a blackbody a t a given tem perature can be found from W ien’s displacem ent law b — ^max^’ \ b n 2.89 X 10 ^ m •K. ^ qq .v/ ^ ^ ^max f » ^max — 6000 K 4.82 X 10 m. Answer. The m axim um of rad ia n t energy corresponds to Ihe wavelength 482 nm.
262
Ch. IV. Optics. Special Theory of R elativity
Questions and Problems 23.1. W here is the velocity of light higher: in diam ond or in water? 23.2. W hat is the ratio between the velocities of light in vacuum and in water? How long does it take for light to travel 225 km in w ater? 23.3. One standard m etre is as long as 1650763.73 wave lengths of 'the orange light em itted by krypton -8 6 atom s in vacuum . W hat is the frequency of this radiation? 23.4. D eterm ine the wavelengths for the red and violet light a t the edges of the visible spectrum if they correspond to frequencies 3.95 X 10 14 Hz and 7.5 X 1014 Hz. 23.5. W hat is the velocity of light in diam ond if the frequency 2.73 X 10 14 Hz corresponds in it to a wavelength of 450 nm? 23.6. The wavelength of blue light in vacuum is 500 nm. Determ ine the w avelength corresponding to blue light in w ater and its frequency. 23.7. Green light passes from air to water, its wavelength decreasing thereby. W hat colour w ill be perceived by a diver in the w ater? 23.8. W hat w ill the change in the wavelength of yellow light of frequency 5.3 X 10 14 Hz be as a result of its pas sage from glass to vacuum if its velocity in glass is 1.98 X 1 0 8 m/s? 23.9. D eterm ine the velocities of light in a transparent m edium for extrem e red (800 nm) and violet (400 nm) light if the refractive indices in this m edium for these wavelengths are 1.62 and 1.67 respectively. W hat are the frequencies and w avelengths of these types of light in the transparent m edium? 23.10. As light passes from w ater to vacuum , its wave length increases by 0.120 pm . D eterm ine the lig h t’s wave lengths in vacuum and in w ater. 23.11. As light passes from vacuum to a transparent m e dium its wavelength is reduced by a factor of 1.31. W hat is the m edium? 23.12. The word “light” is w ritten in green on a sheet of paper. Through which transparent medium would it be impossible io read the word?
§ 24. Quantum Properties of Radiation
263
23.13. The star Sirius has a surface tem perature of 104 K . D eterm ine the w avelength corresponding to the m axim um energy em itted by the star. 23.14. If one looks a t a bright red object for a while and then a t a w hite w all, a green silhouette of the object w ill be seen. How can this be explained? 23.15. W hat m ethod was used to study the chem ical composition of the Sun? 23.16. The to ta l energy em itted by the Sun per second is about E = 4 X 10 26 J . Assuming th a t the Sun is a black body, determ ine its surface tem perature. 23.17. The spectral analysis of a nebula has revealed th a t it has a continuous spectrum . W hat conclusion can be draw n from this? 23.18. The Sun em its about 4 X 10 26 J of rad ia n t energy per second. Some data indicate th a t the rad ian t exitance of the Sun has rem ained constant over the last 3 X 10 9 years (1017 s). D eterm ine the mass lost by the Sun per second in the form of rad iatio n and the mass th a t has been lost over the last 3 X 10 9 years. 23.19. W hat is the m elting point of tungsten if the w ave length of the rad iatio n corresponding to the rad ian t energy m axim um lies in the red spectral region of tungsten and is equal to 784 nm ? § 24. PHENOMENA EXPLAINED BY THE QUANTUM PROPERTIES OF RADIATION. PHOTOELECTRIC EFFECT
Basic Concepts and Form ulas In addition to its wave properties, visible radiation also exhibits particle properties. At low frequencies, the wave properties of electrom agnetic rad iatio n are more pronounced, while a t very high frequencies its particle properties dom i nate. According to P lan ck ’s theory, radiation is discrete. The units of electrom agnetic radiation are known as q uanta, and the particles carrying energy quanta are called photons. The m ain characteristics of a photon are its energy e and momentum p : e = hv = hc/X0,
264__________ Ch. IV. Optics. Special Theory of R elativity
where h is P lan ck ’s constant and k 0 is the w avelength of the rad iatio n in vacuum . Since photons, unlike other m icroparticles, move a t the velocity of light, the form ula for m om entum can be w ritten in the form p = me. According to E in ste in ’s form ula, m = e/c2, and hence p = e/c = hv/c or p = h / \ 0. The mass of a photon is m = hv/c2. Com paring the form ulas for energy, m om entum , and mass, we can arrive at the following conclusion: for a m onochro m atic lig h t, all photons of frequency v have the same ener gy, m om entum , and mass. Photons a t rest do not exist and therefore the rest mass of a photon is zero. The photoelectric effect arises during the interaction between lig h t and a substance. Electrons are detached from the surface of a substance a t the expense of the rad ia n t energy of photons to which the substance is exposed (the extrinsic photoeffect). The laws of the photoelectric effect are: 1. The satu ratio n photoelectric current is proportional to the lum inous flux incident on the surface of the substance. 2. The maximum kinetic energy of the electrons liberated from an irradiated surface does not depend on the ra d ia tio n ’s in ten sity and is given by m i |f ”L = e f / ,
where vmax is the m axim um velocity of the electrons, U the m inim um voltage at which no photoelectric current is ob served, and m and e are the mass and charge of the electron. E in ste in ’s relation for the photoelectric effect is hv = A + ^ k ^ where A is the work function defined as the kinetic energy which m ust be acquired by an electron in order to leave the surface of the substance.
§ 24. Quantum Properties o f Radiation
265
The m axim um wavelength (the m inim um frequency) at which a photoeffect can be observed is known as the photo electric threshold for a given substance. E q uating the kinetic energy of an electron in E in ste in ’s form ula to zero, we can determ ine the w avelength corre sponding to the photoelectric thresholds for different m ate rials:
The operation of photoelectric cells and photoresistors is based on the application of intrinsic and extrinsic photo electric effects. The photoelectric effect is w idely used in various fields. W orked Problems Problem 112. A surface elem ent of area 2 cm 2 is illu m i nated for 1 m in by a rad iatio n of energy 15 J. Determ ine the pressure exerted by the rad iatio n if the surface (a) com pletely absorbs the rad iatio n and (b) com pletely reflects it. Given: W = 15 J is the rad ia n t energy, t = 60 s is the tim e, S = 2 cm 2 = 2 X 10-4 m 2 is the area on which the ra diation is incident. From tables, we take the velocity of light in vacuum c = 3 x 108 m/s. Find: the pressure p t exerted by the radiation when it is com pletely absorbed and the pressure p 2 when it is com pletely reflected. Solution. The pressure arising as a result of the interac tion between rad iatio n and the substance can be found from the form ula p
= ^ - (i+ p )>
where W 0 is the rad ia n t energy per u n it tim e per u n it sur face area and p is the reflection coefficient. (a) W hen the rad iatio n is com pletely absorbed, p = 0 and
15 J Pl
Set ~
2 xx 10~4 10-« m2 m 2 X 3 x 108 m /s x 60 s
= 4.2 x 10-6 Pa.
Ch. IV. Optics. Special Theory of R elativity
266
(b) In the case of to tal reflection, p — 1, and Pz =
.
Pi = 8-4 x lO” 6 Pa.
Answer. The pressure produced by the rad iatio n in the first case is approxim ately 4.2 X 10 ~6 Pa, in the second case, it is double this value. Problem 113. The electron work function for zinc is 3.74 eV. Determ ine the rad iatio n threshold of photoelectric effect for zinc. C alculate the velocity of the electrons em it ted by zinc irradiated by u ltrav io let light w ith a wavelength of 2 0 0 nm. Given: A = 3.74 eV = 3.74 x 1.6 x 10 ~ 19 J is the elec tron work function, X = 200 nm = 2 x 10 “ 7 m is the w avelength of incident radiation. From tables, we take the velocity of light in vacuum c = 3 X 10 8 m /s, the Planck constant h = 6.62 X 10 ~34 J -s, and the mass of the electron, m e = 9,1 x 10 ~31 kg. F ind: the rad iatio n threshold Xth of the photoelectric effect for zinc, the m axim um electron velocity v. Solution. The rad iatio n threshold of the photoelectric effect is the m axim um w avelength which can cause the photoeffect. In this case the kinetic energy of an electron will be zero, and hence A — h v th = hc/Xih. Thus he
M h -— , . 6.62 X 1 0 -34 J -s X 3 X 108 m /s Q Q o„m = 3 -3 2 X 1 0 7 m = 3 3 2 n m *th = --------- 3.74 X l ie x i o ^ 9 J
The energy hv of u ltra v io let rad iatio n photons incident on the zinc plate is spent on the electron work function A and he
on the kinetic energy im parted to the electron: - j - = A + From this equation, we can determ ine the velocity acquired by the electrons in photoeffect: 2 (hc — %A)
mek
§ 24. Quantum Properties of Radiation
267
U =
/
2 (6.62 X 10-34 J .s x 3 X 108 m / s —2 x 10”7 m X 3.74 X 1.6 x 10~19 J) 9.1 X 10~31 kg X 2 X 10"7 m
= 9.3 x 105 m/s. Answer. The m axim um w avelength of the rad iatio n which may cause the photoeffect is 332 nm, the m axim um velocity of the knocked out electrons is 9.3 x 105 m/s.
Questions and Problems 24.1. W hat is the pressure exerted by light on 1 mm 2 of a black surface to which 500 J of rad ia n t energy are tra n s ferred per second? 24.2. The solar pressure on the surface of the E a rth is 4.7 X 10 ~4 Pa. Determ ine the energy of the rad iatio n inci dent per second on a square m etre of the E a rth ’s surface perpendicular to solar rays. 24.3. The num ber of X -ray photons of frequency 7 X 10 19 Hz incident per second on a square m etre of a black sur face is 2.5 X 1015. W hat pressure does this rad ia tio n pro duce? 24.4. The yellow light of sodium vapour has a w avelength of 530 nm. W hat is the energy of a quantum of th is light in joules and electron volts? 24.5. W hat is the ratio of the energy of a photon corre sponding to the y-radiation of frequency 3 X 1021 Hz to the energy of an X -ray photon of w avelength 2 x 10 " 10 m? 24.6. D eterm ine the energy of a quantum corresponding to w avelength 1 0 ” 7 m (in joules and electronvolts). 24.7. How m any photons m ake up 10"8‘J of rad ia tio n at w avelength 2 p,m? 24.8. D eterm ine the energy, mass, and m om entum of an u ltra v io let photon whose w avelength is 360 nm. 24.9. D eterm ine the energy, mass, and m om entum of X -ray photons of w avelength 4 X 10 ” 11 m. 24.10. A com et was observed in the sky after sunset. In which direction was its tail pointing? 24.11. The photoelectric threshold for tungsten corre sponds to a w avelength of 405 nm. Determ ine the work func tion for tungsten.
268
Ch. IV. Optics. Special Theory of R elativity
24.12. The work function for cesium is 1.9 eV. D eterm ine the m axim um wavelength of light at which a photoelectric eSect is observed. 24.13. D eterm ine the m axim um w avelength of light which m ay cause the extrinsic photoeffect from nickel if the work function for nickel is 4.5 eV. 24.14. The work function for platinum is 6.3 eV. W ill photoeffect be observed for a rad iatio n of w avelength 1 0 " 7 m? 24.15. L ight of w avelength 500 nm is incident on the sur face of silver. W ill the silver become charged or will it re m ain neutral? If it gets charged, w hat is the sign of the charge? The photoelectric threshold for silver is 261 nm. 24.16. The m axim um w avelength of rad ia tio n th a t can produce a photoelectric effect in platinum is 234 nm. De term ine the m axim um kinetic energy acquired by the elec trons due to rad iatio n of w avelength 2 0 0 nm. 24.17. W hat is the energy of electrons detached from the surface of copper irradiated by light of frequency 6 X 10 16 Hz if the work function for copper is 4.5 eV? 24.18. W hat is the velocity of the electrons knocked out of sodium irrad iated by light of w avelength 6 6 nm? The work function for sodium is 4 X 10 ” 19 J. 24.19. The photoelectric threshold for a certain m etal is 690 nm. D eterm ine the work function for th is m etal and the m axim um velocity acquired by its electrons due to ra diation of w avelength 190 nm. 24.2 0 .1 W hat is the m axim um velocity acquired by photoelectrons knocked out of m olybdenum by rad ia tio n of fre quency 3 x 1020 Hz? The work function for m olybdenum is 4.27 eV. Is the classical form ula applicable in this case? 24.21. If the surface of a m etal is successively exposed to rad iatio n of w avelengths 350 and 540 nm , the m axim um ve locities of the photoelectrons w ill differ by a factor of two. D eterm ine the m etal’s work function.
1 It would be useful to reconsider this problem after studying the section on special theory of relativity (§ 25).
§ 25. Special Theory of R elativity
269
§ 25. FUNDAMENTALS OF THE SPECIAL THEORY OF RELATIVITY
Basic Concepts and Formulas The special theory of re la tiv ity (STR) is based on two postulates. 1. All physical processes in in ertial frames proceed iden tically and do not depend on the choice of the reference frame. 2. The velocity of electrom agnetic waves in vacuum is the same for all in e rtia l frames. I t does not depend either on the velocity of the source or on the m otion of the observer (receiver of the light signal). The postulates of the special theory of re la tiv ity are in contradiction w ith the concepts of absolute tim e and space formed in N ew tonian classical m echanics. The STR im plies the following concepts. The relativity of lengths (distances): the length I of a rod (body) in a reference fram e relativ e to which it moves w ill, according to the STR , be less th an the length l 0 of the s ta tionary rod: i = i0 y i - v V c * . The transverse dim ensions of a m oving body will be the same in all in ertia l reference frames. The relativity of time intervals (relativ istic tim e dilatio n ): the tim e t m easured in a laboratory system in which an observer is a t rest and the in trin sic tim e t 0 m easured using a clock m oving together w ith the reference fram e are r e la t ed as T =
—
.
V i-v 't/c *
If the velocity of a m oving reference fram e is close to the velocity of lig h t, the m oving clock w ill lag behind the s ta tionary clock, i.e. tim e d ila tio n w ill occur. The mass of a body will depend on its velocity: the rest mass m 0 (intrinsic mass) and the mass m of the body m oving a t a velocity close to th a t of light are connected through the
270__________ Ch. IV. Optics. Special Theory of R elativity
following relation: m=
1/1 — v*/c*
G iven th a t the mass depends on the velocity, we can w rite the form ula for the m om entum of the body in the form pz= m v — — .
mnv u
j / l — v 2/ c 2
The relativistic law for velocity composition: the velocity v 2 of body M relative to a sta tio n a ry observer in frame K can be determ ined from the form ula l V i
«;. = ■v' + v ■
l + yiv/c2 9
K
where vx is the velocity of the body M relativ e to the refero , 7 F ence fram e K ' , v is the veloc ity of the reference fram e K ' relativ e to the sta tio n a ry ref erence fram e K (it is necesp. sary th a t the x-axis be the same for the two reference fram es, Fig. 143). If we assume th a t vx = v = c, the resu ltan t velocity can not exceed the velocity c of light. The equation relating mass and energy (E in stein ’s massenergy relation): for any interconversion of m atter into energy, the change in energy is proportional to the change in mass: AE = c2 Am. O '
Worked Problems Problem 114. A rocket moves w ith a velocity equal to of the velocity of light in vacuum relative to a sta tio n a ry pbserver. W hat is the change in the length of a 1-m long steel ruler and the density of the steel of which it is made in the rocket (in the direction of motion) w ith respect to the observer? How much tim e will elapse according to the clock of the statio n ary observer if six years have passed according to the clock in the m oving rocket? 0 .6
§ 25. Special Theory of R elativity
271
Given: v = 0.6c is the velocity of the rocket rela tiv e to the sta tio n a ry observer, l 0 = 1 m is the proper length of the ruler, £0 = 6 years is the intrinsic tim e. From tables we take the proper density of steel, p0 = 7.8 X 103 kg/m 3. Find: the length I of the ruler, the density p of steel, and the tim e t. Solution. The length of the ruler (along the trajecto ry of motion) for the statio n ary observer, relative to whom the rocket moves, can be determ ined from the form ula J = Z0 ] A - i > 2/c2,
/ = l m y/~ 1 — —
= 0.8 m.
The density of a substance is expressed in term s of mass and volume: p = m / V , but V = IS = l 0S ]A 1 — v2/c2. In this problem , the transverse dim ensions do not change, and hence n
___________________________
i
V 1 — V*/c* l0S ]/l — V*/C2
’
,
m °
___
l 0S
~
P °'
Therefore p=-r=fe-The tim e for the statio n ary observer w ill be dilated and is defined by the form ula t=
] / 1 — V2/ c 2 '
t=
= 7.5 years.
Answer. The length of the ruler for the statio n ary observer is 0.8 m, the density of steel is 1.2 X 104 kg/m 3, and the tim e period is 7.5 years. Problem 115. The rest energy of a proton is approxim ately 938 MeV. D eterm ine the rest mass of the proton and the mass and velocity of protons to which a kinetic energy of 70 GeV has been im parted in an accelerator. Given: E 0 = 938 MeV is the rest energy of the proton, E k = 70 GeV is the kinetic energy of a proton acquired in the accelerator. From tables, we take the velocity of light in vacuum c = 3 X 108 m/s. Find: the rest mass m Po of the proton, the mass m p of the proton after the acceleration, and the velocity v of the pro ton as a result of the acceleration.
272__________Ch. IV. Optics. Special Theory of R elativity
Solution. We first recalculate the p roton’s energy in SI u n its (joules) 1 eV = 1.6 X 10_19C x I V = 1.6 x 10“1 9 J , 1 MeV = 1.6 X 10 ~13 J . Consequently, the rest energy is E 0 = 938 x 1.6 x 10 - 13 J = 1.5 x 10 - 10 J. The kinetic energy of the proton after acceleration is
£ k = 7 x 1010 x 1.6 x 10-19 J = 1.12 x 10-8 J. In order to determ ine the rest mass of the proton, we shall use E in ste in ’s relation E 0 = m Poc2: 1.5 XlO"10 J 9 X 1016 m 2/s 2
1.67 x 10“ 27 kg.
The total energy of the proton can be determ ined from the form ula E = E k -\- E 0 = m pc2. D ividing the two sides of
— nro
_j_ i
whence
E°
1 .1 2 x 10 -®
I
1.5x lO - 1® J
+ 1) 1.67 x 10' 27 kg = 1.26 x 10” 25 kg.
H aving determ ined the mass of the proton after the accel eration, we can find its velocity:
whence
v =c
1.67x l O ' 27 kg 1.26 X 10-25 kg
) = 0.99c.
Answer. The rest mass of the proton is 1.67 X 10 “ 27 kg (which corresponds to the tab u lated value). The mass of the proton after the acceleration is nearly 75 tim es larger th an its rest mass, and the velocity is equal to 0.99 of the ve locity of light. Problem 116. Two rockets move tow ards each other each at a velocity of 0.8c relativ e to a statio n ary observer. De-
§ 25. Special Theory of R elativity
273
term ine the velocity w ith which the rockets approach each other according to (a) classical m echanics and (b) the theory of re la tiv ity . Given: vl = v 2 = 0.8c are the velocities of rockets relativ e to the sta tio n a ry observer on the E arth . Find: the velocity u cl at which the rockets approach each other according to classical m echanics and u rel, the veloc ity according to re la tiv ity , as well as the difference Au in these velocities. Solution. According to classical m echanics, we have u cl = vx +
u c\ ^
0 .8 c
+
0 .8 c
=
1 . 6 c.
According to relat.ivistic m echanics, we have w,.el =
V1+ V*
, urel = ----------------= 0.976c.
1 + -M L C2
i I ° -64c C2
The difference in these velocities is Au = 1.6c - 0.976c = 0.624c. These calculations show th a t the classical form ula for composing velocities cannot be used when bodies move at velocities close to the velocity of light since this contradicts the re la tiv ity ’s postulate th a t the velocity of lig h t in v a cuum cannot be exceeded. Questions and Problems 25.1. Compare the lengths of two m etre rods m oving in their longitudinal directions w ith velocities 0.5c and 0.75c relative to a sta tio n a ry observer. 25.2. At w hat velocity of a body relativ e to a sta tio n a ry observer is its length equal to 0 . 8 of its proper length? 25.3. A t w hat velocity should a body move rela tiv e to a sta tio n a ry observer for its size along the line of m otion to contract by a factor of two? W ill the size change to an ob server m oving together w ith the body? 25.4. A rocket moves at a velocity of 0.866c relativ e to a statio n ary observer. The w idth of a plane rectangle arranged along the line of the m otion of the body is equal to half its length. How w ill the rectangle appear to the statio n ary ob server? 1 8 -0 5 3 0
274__________ Ch. IV. Optics. Special Theory of R elativity
25.5. The radius of an electron at rest is 2 x 10 " 13 cm. W hat w ill the contraction in the electron’s radius be in the direction of m otion at 0 . 8 of the velocity of light? 25.6. The flight tim e measured in a rocket m oving a t a velocity 0.96c is one year. How much tim e w ill have elapsed for a terrestrial observer? 25.7. How m uch tim e w ill have elapsed on the E a rth if six years have passed in a rocket m oving at a velocity of 2.4 X 108 m/s relativ e to the E arth? 25.8. The distance to the sta r in the C entaurus constella tion nearest the Sun is traversed by light in approxim ately 4.25 years. Express this distance in astronom ical u n its and in kilom etres. D eterm ine the flight tim e (by a clock in a rocket) to th is sta r and back in the rocket if the velocity of the rocket relativ e to the observer is 0.99c, and 1 AU ~ 150 m illion kilom etres. 25.9. W hat w ill be the duration of a flight to a sta r by a rocket flying at a velocity of 0.9c for a cosm onaut and for an observer on the E a rth if the distance to the star is 40 light years? 25.10. One kilogram of water is heated by 80 K. W hat is the increase in the mass of w ater? 25.11. W hat m ust the velocity of a particle be a t which its kinetic energy is equal to its rest energy? 25.12. W hat is the velocity of an electron if its mass is four tim es its rest mass? 25.13. D eterm ine the mass of an electron moving at veloc ities of 50% and 90% of the velocity of light. 25.14. W hat m ust the velocity of a proton be in an ac celerator for its mass to be 5% greater? 25.15. A body w ith a rest mass of 5 kg and density of 7.8 X 103 kg/m 3 is in a rocket m oving relativ e to a terres tria l observer at a velocity of 2.4 X 105 km /s. D eterm ine the re la tiv istic mass and density of the body. 25.16. A rocket moves at a velocity close to th a t of light relativ e to a sta tio n a ry observer. By w hat factor w ill the mass and density of bodies in the rocket change for an ob server flying in it? 25.17. W hat m ust the velocity of a body be for its densi ty to be five tim es larger? 25.18. Express the rest energy of an electron (positron) in m egaelectronvolts. W hat will the rad ia n t energy em itted
__________________ § 25. Special Theory of R elativity______________ 275
as a result of the annihilation of an electron and a positron be? 25.19. Determ ine the m om entum of an electron moving at 0 . 6 of the velocity of light. 25.20. Two aeroplanes fly tow ards each other a t velocities 500 and 400 m/s. W hat is th eir relativ e velocity? 25.21. Two particles move tow ards each other at a ve locity (5/8)c each. W hat will their approach velocity be as calculated using the classical and rela tiv istic form ulas?
Chapter V
Physics of Atomic Nucleus § 26. STRUCTURE OF ATOMIC NUCLEUS. ATOMIC ENERGY AND ITS APPLICATION
Basic Concepts and Formulas According to the nuclear model of atom s proposed by R u therford and developed by Bohr, an atom consists of a pos itiv ely charged nucleus and electrons revolving about it. The nucleus contains protons and neutrons to which the general term “nucleons” is applied. A nucleus occupies a very sm all volum e as com pared to the atom . If we assume th a t the atom ic nucleus is a sphere, the radii of the nuclei of different elem ents can be determ ined by the form ula r = 1.4 x 1 0 - 15K ^> where A is the mass num ber of an elem ent. The density of m atter in a nucleus is of the order of 1.3 X 10 17 kg/m 3. The mean density of nuclear m atte r is calcu lated from the form ula mA
P — (473)^ ’ where m is the mass of a nucleon and r is the radius of the nucleus. The positive charge of a nucleus is determ ined by the product of the atom ic num ber Z of the elem ent in the Pe riodic T able and the elem entary charge, and hence depends on the num ber of protons in the nucleus. The mass of a nucleus depends on the num ber of nucleons in it. The u n it of mass of an atom , viz. atom ic mass un it (amu), is equal to 1 / 1 2 of the mass of the carbon atom !|C. The atom ic mass u n it can be expressed in SI units: 1 amu = 1.66057 x 10 " 27 kg.
§ 26. Structure of Atom ic N ucleus
277
In order to determ ine the mass of an atom , it is usually sufficient to use the mass num ber of the elem ent: ma = A x 1.66057 x 10 " 27 kg. The nuclei of atom s of chem ical elem ents are denoted by z X , where X is the symbol of an elem ent, A is the mass num ber, and Z is the atom ic num ber of the elem ent, which is equal to the num ber of protons in the nuclei. The particles are designated as follows: _\e is the electron, \e is the posi tron, is the neutron, \p is the proton (the nucleus of a hydrogen atom is JH), and JHe is the a-particle. The mass of an atom ic nucleus is less th an the sum of the masses of the nucleons by a q u a n tity known as the m ass defect: Am = Z m p + N m n — mx , where mx is the mass of the nucleus, m p is the mass of the proton, m n the mass of the neutron. The mass ma of the atom is taken from the Periodic Table. Then mx = ma — Z m d. Nucleons are confined in a nuclei by nuclear forces which considerably exceed the electrostatic (repulsive) forces. For this reason, when a nucleus is sp lit, a certain am ount of energy is required to overcome the nuclear forces. W hen nucleons combine to form a new nucleus, the energy known as the binding energy is liberated. The relatio n between the binding energy and mass defect is given by E in ste in ’s relatio n AE = Ame2. A mass defect of 1 amu corresponds to a binding energy of 931.5 MeV. R ad io activ ity (viz. the spontaneous disintegration of the nuclei of some isotopes w ith the emission of a - and {$-particles and Y“ra Ys) Is a phenomenon th a t confirms the com pound structure of nuclear atom s. Three principal types of rad ia tio n are em itted during ra dioactive decay (Fig. 144), a -ra d ia tio n which consists of helium nuclei *He (the mass num ber after an a-decay decreases by 4 and the charge by two units), ^ -radiation which consists of electrons whose velocities are close to the veloc
278
Ch. V. P hysics of Atomic N ucleus
ity of light (an electron is produced during (3-decay as a result of the transform ation of a neutron into a proton, and hence the positive charge of the nucleus increases by one, and the mass num ber remains unchanged), and y-radiation which accompanies a- and (3-decay. The emission of a y-quantum does not involve any change in the mass num ber or charge. The rules for the nuclei formed during radioactive decays are for an a-decay, for a
iX 2 + * H e ,
0 -decay,
z +JX 2 +
The law of radioactive decay states th a t the num ber A N of nuclei undergoing « radioactive decay during the tim e in terval from t to t + A M s propor tional to the num ber N of nuclei available at tim e t and to the length of the tim e interval At: AN = A*, where X is a decay constant which characterizes the rate of radioactive decay for a given species of nuclei. The m inus sign indicates th a t the num ber of nuclei decreases during F ig. 144 the decay process. In order to characterize the sta b ility of a nucleus, the concept of a half-life is introduced. The half-life T 1/2 is the tim e required for half the nuclei ca pable of decaying to do so: ^ 1
-
In 2 _
0.693
T ““
E
•
Worked Problems Problem 117. The radioactive waste from nuclear power plants contains the radioactive strontium isotope JJSr which has a half-life of 28 years. Over w hat period w ill the am ount of strontium decrease by a factor of four? Given: ™Sr is the strontium isotope, and T l/2 = 28 years is its half-life.
279
§ 26. Structure of Atom ic N ucleus
Find: th e tim e t during which the am ount of strontium decreases by a factor of four. Solution. The am ount of decaying nuclei as a function of the half-life is represented by the curve in Fig. 145. We conclude from this graph th a t the am ount of strontium nuclei w ill decrease to one quarter of the in itia l value in 2 X 28 years = 56 years. Problem 118. W hat fraction of radioactive cesium ^JCs w ith a half-life of 30 years will decay in a year? D eterm ine the decay constant. Given: ^ ’Cs is the radioactive cesium isotope, T 1/2 = 30 years is the half-life, and t = 1 year is the tim e. Find: the fraction A N ! N 0 of de F ig. 145 cayed nuclei and the decay constant. Solution: W e use N 0 to denote the in itia l num ber of atom s and N to denote the num ber of atom s rem aining after tim e t . Then we can w rite A N = N 0 — TV, where A N is the num ber of atom s th a t have decayed over the tim e t. According to the law for radioactive decay, N = N 0e(~u \ e = 2.718... being the base of N apierian logarithm . A ssum ing th a t the tim e t is sm all in com parison w ith the half-life T , we can use the approxim ation A N = - ± ^ - N 0t, or ATV _ TV0
0.693 4 T
’
AN N0
0.693 X 1 year 30 years
n noQ “
The decay constant X characterizes the rate of the rad io active decay and can be determ ined from the form ula 1
° -693 T
* ’
0.693 7 Q w 4A-10 - i 30 x 365 X 24 x 3600 s — S ‘
Answer. The fraction of cesium nuclei th a t decays over a year is approxim ately 2.3% , the decay constant is 7.3 X 1 0 “ 10 s - 1.
280
Ch. V. P h ysics of A tom ic N ucleus
Problem 119. Determ ine the com positions of lith iu m nuc lei, hydrogen isotopes w ith mass num bers 1 and 2 , and uranium isotopes w ith mass num bers 235 and 238. Given: JLi is the nucleus of a lith iu m atom , JH is the nu cleus of a hydrogen atom , 2H is the nucleus of a heavy hy drogen atom (deuteron), and 2^®U and are the nuclei of uranium isotopes. F ind : the num ber Z of protons and the num ber N of neu trons in the nuclei of the indicated elem ents. Solution. The num ber of protons in a nucleus is deter m ined by the charge num ber which is equal to the atom ic num ber of the elem ent in the Periodic Table. The charge num ber is the subscript to the left of the elem ent sym bol. The su perscript is the mass num ber A th a t is equal to the num ber of nucleons in the nucleus: A = Z + N . Therefore, the num ber of neutrons is N = A - Z. For lith iu m ’L i, Z = 3 and TV = 7 — 3 = 4. For ordinary hydrogen (}H), Z = 1, and for heavy hydro gen (;H ), Z = 1, N = 1. Fot 29328U, Z = 92, N = 146. For 23‘U, Z = 92, N = 143. In th is problem , hydrogen and uranium are represented by two isotopes. The isotopes of a chem ical elem ent have the same num ber of protons and different num ber of neutrons. Problem 120. The nucleus of the m agnesium isotope w ith a m ass num ber 25 is bom barded by protons. The nucleus of which elem ent is formed if the nuclear reaction is accom pa nied w ith the emission of a-particles? Solution. In the Periodic T able, we find the atom ic num ber of m agnesium (1 2 ) and w rite the nuclear reaction 25
12M g
+ [H
Since charge is conserved, the sum of the charge num bers on the left-hand side m ust be equal to the sum of the charge num bers on the right-hand side, viz. 1 2 - | - l = 2 + 2. Hence the charge num ber Z of the unknown elem ent is 11. The eleventh site in the Periodic Table is occupied by sodium . Since mass (the mass num ber) is conserved, the sums of the mass num bers on the left- and right-hand sides m ust be
§ 26, Structure of Atom ic N ucleus
281
equal (here this sum is 26). Consequently, we obtain a sodi um isotope w ith a mass num ber of 2 2 . In its final form , the reaction equation will be w ritten as “ Mg + J H - * ” N a + i H e .
Problem 121. As a result of bom bardm ent of alum inium by a-p articles, a new nucleus and a neutron are formed. W rite the nuclear reaction and identify the elem ent whose nucleus is formed. Solution. W e shall w rite the nuclear reaction. On the lefthand side, we w rite the in itia l q u antities, while the rig h thand side m ust contain the unknown nucleus and a neutron:
IJAI + J H e - ^ X + X E quating the charge and mass num bers on the left- and right-hand sides (as in Problem 120), we conclude th at the new nucleus is th a t of phosphorus. In its final form, the reac tion is « A i+ ;H e -;;p + > .
Problem 122. D eterm ine the mass defect and the binding energy of the nucleus of a nitrogen atom. W hat is the binding energy per nucleon? Given: 4*N is the nucleus of the nitrogen atom . From ta bles, we take the rest mass of a neutron m n = 1.00867 amu, th e mass of the nitrogen atom mN = 14.0067 amu, the mass of a hydrogen atom mH = 1.00797 amu, the velocity of light in vacuum c = 2.99792 X 108 m /s, and 1 amu = 1.66056 x 10 - 27 kg. Find: the mass defect Am, the binding energy AE of the nucleus of a nitrogen atom , and the binding energy per nu cleon AE /A . Solution. The mass defect is the difference between the sum of the rest masses of the free protons and the neutrons constituting a nucleus on one hand and the mass of the nucleus on the other, i.e. Am = Z m p + N m n — mx , where mx is the mass of the nucleus, mx = m^ — Z m €. The expression for the mass defect can be considerably simplified if instead of the sum of the masses of protons we
282
Ch. V. P hysics of Atomic N ucleus
take the sum of the masses of hydrogen atoms: Am = Zm u + N m n — mN. S u b stitu tin g in the num erical values, we obtain Am = 7 x 1.00797
amu + 7 x 1.00867 amu
— 14.0067 amu ~ 0.10978 amu, Am = 0.10978 x 1.66 x 10 - 27 kg ~ 1.822348 x 10 " 28 kg. The binding energy can be determ ined using E in ste in ’s relation A E = Ame2, AE = 1.822 x lO ' 28 kg (2.9979 m /s ) 2 = 1.638 x 10 " 11 J. In atom ic physics, the binding energy is expressed in megaelectronvolts: AE=
m« V ^ 102.4 MeV.
Considering th a t 1 amu corresponds to the energy approx im ately equal to 931.4 MeV, we can determ ine the binding energy in a sim pler way: A E = 0.10978 x 931.4 MeV ~ 102.25 MeV. The results of calculations clearly differ insignificantly. The binding energy per nucleon is AE /A = 102 MeV/14 ~ 7.3 MeV/nucleon. The result is in agreem ent w ith the tab u lated data. Answer. The mass defect is 0.1097 amu, the binding ener gy of the nitrogen nucleus is approxim ately 102 MeV, and the binding energy per nucleon is 7 MeV. Problem 123. The therm al power of each reactor installed at the Leningrad nuclear power plant is 3200 MW. The elec tric power is 1000 MW. The mass of the uranium charge in a reactor is 180 t. D eterm ine the efficiency of the u n it, the mass of the uranium -235 consumed by the reactor over a year of continuous operation at the to tal power. W hat fraction of the to tal charge does the mass of the consumed uranium constitute? Given: P t = 3.2 x 10° W is the therm al power of the re actor, P c = 109 W is the electric power of the u n it, t =
§ 26. Structure of Atom ic N ucleus
283
1 year = 3.15 X 107 s is the tim e for which the nuclear fuel consum ption is being determ ined, m = 1.8 X 105 kg is the mass of the uranium charge of a reactor. From tables, we take the Avogadro constant N A = 6.02 X 1023 m ol"1, the m olar mass of uranium M = 235 X 10 ~3 kg/m ol, and th e energy liberated as a resu lt of the fission of a 235U nucleus, E — 200 MeV. F ind: the efficiency q of the power u n it, the mass m 1 of uranium consumed and the fraction m j m of the consumed uranium . Solution. The efficiency of a un it is defined as the ra tio of the electric power of the un it to the therm al power of a reactor: r, = ^ 1 0 0 % ,
t) =
J x i S -w
100% = 31.3%.
In order to determ ine the mass of consumed uranium , we m ust find the num ber of 235U atom s entering into the nuclear reaction over the tim e of operation of the reactor. Since ab out 200 MeV of energy is liberated during the fission of a 235U nucleus and th a t 190 MeV are converted into heat, we can w rite N = P tt /E u where E x = 190 MeV. Given the num ber of atom s in the reaction and the mass of a uranium atom , m\j = M / N A, we can determ ine the mass of con sumed uranium : »r
mi = m v N , 1
or m i =
M
P it
-g — ,
2 3 5 x 1 0 -3 kg*m ol"1 X 3.2 X 10» W x 3.15 x 107 s _ 1 o q / . 6.02 X 1023 m ol"1 X 190 X 1.6 X 1 0 -13 J
Let us determ ine the fraction of the to ta l charge con sumed over a year: mi m
~
1294 kg 1.8 x 1 0 s kg
p. j-w-v-
Answer. The efficiency of the p lan t is about 31% , the am ount of uranium -235 consumed during a year of operation is about 1300 kg, which is less th an 1 % of the to ta l. Questions and Problems 26.1. W hat is a -ra d ia tio n ? W hy is it deflected less in a m agnetic field th an P-radiation? 26.2. W hat force deflects a - and p-rays in a m agnetic field?
284
Ch. V. P h ysics of Atom ic N ucleus
26.3. W hich of the three radiations (alpha, beta, and gamma) is not deflected by either m agnetic or electric fields? 26.4. W hat is y-radiation? W hat is the difference be tween th is rad iatio n and X -rays? 26.5. How m any electrons occupy the electron shell of a neutral atom whose nucleus contains six protons and six neutrons? 26.6. The atom ic nucleus of any chemical elem ent con sists of protons and neutrons. How can the em ission of 0-radiation be explained? 26.7. Determ ine the half-life of radon if 1.75 X 105 out of 106 atom s decay per day. W hat is the decay constant equal to? (Use the approxim ate form ula.) 26.8. The decay constants for bism uth-209 and poloni um-210 are 1.6 X 10 “ 6 s _1and 5.8 X 10~8 s ' 1 respectively. D eterm ine th eir half-lives. 26.9. W hat fraction of the radioactive nuclei of !JC de cays over 100 years if its half-life is 5570 years? 26.10. I t is known th a t 9.3 X 10 18 out of 2.51 X 10 19 available atom s of the sodium -24 isotope undergo P-decay. The half-life is 14.8 h. D eterm ine the decay tim e and con sta n t using the approxim ate form ula (see Problem 118). 26.11. The uranium isotope 2g®U of 1-g mass em its 1.24 X 104 a-p articles per second. D eterm ine the half-life and decay constant of the isotope. 26.12. D eterm ine the com position of hydrogen 8H, heli um 4He, alum inium JJAl, uranium 29®U, and neptunium 29sNp nuclei. W hat can be said about the neutron content of nuclei w ith increasing atom ic num ber? 26.13. W hat is the difference between the nuclei of the chlorine isotopes ®^C1 and 9’C1? How can you explain the fact th a t chlorine has a relativ e atom ic mass of 35.5 in the Periodic Table? 26.14. The nucleus of which elem ent contains 14 pro tons and the same num ber of neutrons? W hich of the first tw enty elem ents in the Periodic Table have nuclei contain ing equal num bers of protons and neutrons? 26.15. D eterm ine the charges of the lith iu m , copper, and uranium -238 nuclei in coulombs. 26.16. One gram of radium em its 3.7 X 10 10 a-p articles
§ 26. Structure of Atom ic N ucleus
285
per second. D eterm ine the charge of this rad ia tio n in cou lombs. 26.17. D eterm ine the radius and nuclear density of he lium and uranium -238 atom s. 26.18. W rite the reaction for the direct transform ation of actinium -227 into francium-223. W hat type of a radioactive decay is it? 26.19. W hat will happen to the uranium-237 isotope dur ing p-decay? W hat will the mass num ber of the new ele m ent be? To which side of the Periodic Table will the nuc leus be shifted? W rite the reaction equation. 26.20. A beryllium nucleus is formed by the reaction between a lith iu m nucleus and a deuteron. W hat particle is liberated in the process? W rite the nuclear reaction equa tion. 26.21. The “age” of the objects discovered during an arche ological dig is determ ined from the isotopes of a certain ele m ent the objects contain. D eterm ine the charge and the mass num ber and identify the elem ent from the following nuclear reaction:
*JN+ i » - * 4 x + lp. 26.22. W hat are the resu lta n t nuclei after the a - and P-decay of xenon? 26.23. Two y-quanta are formed by the an n ih ilatio n of an electron and a positron. Assum ing th a t the masses of the electron and the positron are the same, determ ine the energy of the y-radiation and its frequency. 26.24. The presence of explosives in the luggage of air passengers can be detected using nuclear physics. An explo sive norm ally contains nitrogen isotopes w ith mass num bers 14 and 15. As a result of bom bardm ent by neutrons, nitrogen isotopes w ith mass num bers 15 and 16 are formed. The la tte r isotope is radioactive and em its y-quanta th a t can be detected. W rite the equation of the nuclear reaction. 26.25. W hat m ust the energy of a y-quantum be for it to be convertable into an electron-positron pair? 26.26. I t is established th a t a proton is em itted during the bom bardm ent of an alum inium isotope by helium nuclei w ith the form ation of a new nucleus. W rite the equation for the nuclear reaction and id entify the new nucleus.
286
Ch. V. P h ysics of Atomic N ucleus
26.27. W hen boron “ B captures a fast proton, three alm ost identical tracks spreading in different directions are formed in a W ilson cloud cham ber where the process takes place. W hat particles are responsible for these tracks? 26.28. Identify the particle denoted by the question m ark in the nuclear reaction 2Li + ? “ ►‘JB + ‘n. 26.29. W hat nuclei and particles are formed as a result of the following reactions: « ;p „ + ;H e -* 2 x 4 -> , !h + v - ^ ; h + ? 26.30. The transform ation of phosphorus into silicon ™Si is accom panied by the emission of a positron. W hal changes occur in the nucleus? 26.31. D eterm ine the mass defect of a lith iu m nucleus in atom ic mass u n its and in kilogram s. 26.32. D eterm ine the mass defect of the boron nucleus ^B in atom ic m ass u n its and in energy units. 26.33. D eterm ine the mass defect and the binding energy for the uranium -238 nucleus. 26.34. Analyze th e following nuclear reactions and de term ine whether the energy is liberated or evolved: * H e-K H e-> 3 7L i-K H , ®Li + JH -► 2*He. 26.35. D eterm ine the energy liberated in the nuclear re action ILi + JH - > 42He + 2He (see Table 25). 26.36. In order to obtain 1 GW of electric power, 2 X 1 0 6 t of coal have to be burnt annually, which involves the discharge of 8 X 103 t of ash and tens of thousands of tons of sulphur dioxide into the atm osphere. How much uranium 235 is required to obtain the same power for the same ef ficiency? 26.37. The fission of a uranium -235 atom into two frag m ents is accom panied by the liberation of about 3 X 10 -11 J of energy. How m uch petrol has to be burnt to obtain the same energy as th a t liberated in a nuclear reaction in which 1 g of uranium is consumed? 26.38. C alculate the energy liberated by the com bustion of 1120 t of A-l grade coal, of 376 t of petroleum , of
§ 26. Structure of Atom ic N ucleus
287
5 X 105 m 3 of n atural gas and the energy generated by the fission of 260 g of uranium -235. 26.39. The efficiency of the reactors installed at the K olsk and Rovensk nuclear power plants is 32% . How much u ra nium 235 is consumed in a nuclear reactor per hour if its electric power is 440 MW? 26.40. How much energy is liberated in the nuclear re actors of the nuclear-powered icebreaker Lenin if the daily consum ption of uranium -235 is 62 g? 26.41. The therm al reactors installed at atom ic power plants (operating on slow neutrons) do not utilize the nuclear fuel very efficiently and cannot ensure the required scale of production of nuclear power. The situ atio n is different for reactors on fast neutrons. W hy? 26.42. W hich of the particles listed below are stable: a photon, an electron, a neutrino, a proton, a neutron, or a n-meson? 26.43. The an n ihilation of a proton and an antiproton generates y-radiation. C alculate the energy of the photons if the masses of the proton and the anti proton are the same and equal to 1.67 X 10 ~ 27 kg. 26.44. Controlled nuclear fusion would be a prodigious way of obtaining energy. As much energy is liberated by the fusion of the deuterium contained in one litre of ordinary water as by com bustion of 350 1 of petrol. C alculate this am ount of energy. 26.45. If 5 X 104 kg of hydrogen are fused into 49 644 kg of helium, how much energy is liberated?
Chapter VI
General Remarks on Astronomy § 27. FUNDAMENTALS OF ASTRONOMY
Basic Concepts and Formulas Astronom y facilitates a deeper understanding of the physics of the world and extends dialectical and m aterialistic ideas about m atter and various forms of its existence. The devel opm ent of cosmonautics has broadened the means for in vestigating the processes occurring in the U niverse and the way they influence the life on our planet. The celestial sphere is defined as having an a rb itra ry ra dius w ith the centre at the point of observation, onto which p
the celestial bodies are being projected. The celestial axis P P X is the axis of the apparent ro tation of the celestial sphere. The points where the celestial sphere intersects the celestial axis are known as celestial poles (Fig. 146). The diurnal ro tatio n of the celestial sphere makes it possible to determ ine the constellations which set at a g iv en geographical la titu d e and those which do not set. For
§ 27. Fundam entals of Astronom y
289
exam ple, the constellation of Ursa Minor does not set at the la titu d e of Moscow. The brightest sta r in this constella tion, Polaris, is very close to the north celestial pole. Since the geographical latitu d e of an observation point can be determ ined from the angular distance from the plane of the horizon to a celestial pole, the geographical la titu d e in the northern hem isphere can be determ ined from the a ltitu d e of Polaris. The ecliptic is the apparent annual path of the Sun on the celestial sphere. The ecliptic intersects the celestial equator at points known as vernal equinox and autum nal Z e n i t h
Z
\N
Z, Nadir Fig. 148
Fig. 149
equinox (Fig. 147). The most rem ote points from the equa tor are passed by the Sun on the 22nd of June and on 22nd of December, which are known as sum m er and w inter sol stices respectively. The ecliptic passes through tw elve con stellations of the Zodiac, nam ely Pisces, Aries, Taurus, G em ini, Cancer, Leo, Virgo, L ibra, Scorpio, S ag ittariu s, Capricorn, and A quarius. The position of a sta r on the celestial sphere is deter mined by two coordinates: its declination 6 and right ascen sion a (Fig. 148). The large circle of the celestial sphere passing through the celestial poles and a given sta r is known as the declination circle, while the angular distance from the celestial equator to the star, measured along the declination circle is known as the declination of the star. D eclinations are positive to 19-0530
290
Ch. VI. General Remarks on A stronom y
the north of the equator and negative to the south of it. The declination of the sta r is sim ilar to the geographical latitu d e . The right ascension is measured along the celestial equator from the vernal equinox to the declination circle passing through the given star. This coordinate is sim ilar to the geographical longitude and expressed in u n its of tim e. The passage of a star through the m eridian is known as its culm ination. Each star passes through the m eridian twice a day. The apparent noon is the upper culm ination of the cen tre of the solar disc, while the apparent m idnight corresponds to its lower culm ination. The altitu d e h of a sta r above the horizon at its upper cul m ination can be determ ined from the form ula (Fig. 149) h = 90° —
8,
where (p is the geographical latitu d e. In addition to the Sun and planets, the Solar System contains asteroids (small planets), comets, and m eteoric dust. Our star, the Sun, is an ordinary star in the M ilky W ay (Galaxy). Its diam eter is 1390000 km, i.e. 109 tim es the diam eter of the E a rth and its mass is 333000 tim es th a t of the E arth . The hot gaseous sphere of the Sun consists of 85% of hy drogen and 13% of helium . M any other elem ents are present in sm all am ounts. The tem perature at the centre of the Sun is about 2 X 107 K and the pressure is of the order of 2 X 107 G Pa. Under these conditions, fusion reactions take place, which are the source of the solar energy. The m otion of celestial bodies and their grav itatio n al a ttrac tio n obey the laws of physics. Kepler's first law states th a t all planets revolve around the Sun along elliptical orbits w ith the Sun as one of the foci. Kepler's second law: the radius vectors of a planet drawn from the Sun sweep equal areas in equal tim es (Fig. 150). Kepler's third law: the squares of the tim es taken by two planets to describe their orbits are proportional to the cubes of the sem im ajor axes of the orbits:
§ 27. Fundam entals of Astronom y
291
Newton refined K epler’s th ird law for com paring the masses of celestial bodies: m x + m % T j 2 __ R j 2 4
7*34
i?34 ’
where m m 2 and m 3, m 4 are the masses of two pairs of celes tia l bodies revolving one around the other, and ^ 12 > T 34, /?12, and i ? 34 are respectively the periods of revolution and the m ean distances between the bodies. Newton's law of universal gravitation: all bodies in the Universe are attracted to one another w ith a force propor-
S
A
tional to the product of th eir masses and inversely propor tional to the squared distance between their centres of mass: F = G mim2 where G is the g rav itatio n a l constant. The distances to ce lestial bodies are determ ined by parallax. The horizontal parallax p is the angle at which the radius of the E a rth per pendicular to the line of sight is seen from a celestial body (Fig. 151): s c = - Jsin ?— . p 19
292
Ch. VI. General Remarks on Astronom y
The annual parallax of a sta r is the angle a t which the ra dius of the E a rth orbit perpendicular to the line of sight D is seen from the star (Fig. 152): sin p
The unit of length in astronom y is the astronom ical (AU), 1 AU = 149.6 X 106 km is the mean distance tween the Sun and the E arth . W hen discussing stars galaxies, the units of length are the lig h t year and parsec
u n it be and (pc),
1 pc = 206265 AU = 30.86 x 10 12 km. One light year is the distance traversed by light in one year: 1 lig h t year = 0.3068 pc. W hen solving some of the problem s, the sta r chart should be used. To do this, trace the circle on the flyleaf of the book. It should be cut along the line corresponding to the geographical la titu d e of the site of observation. By apply ing the circle to the star chart on the front flyleaf, you will obtain the celestial map for the required m onth and hour of observation. For exam ple, to obtain the sta r chart at 10 pm, on October 10, it is necessary th a t 1 0 pm in the cut circle be m atched w ith the m ark for October 10 on the chart. In the slot of the circle you w ill have all the celestial bodies visible at th a t m oment. W orked Problems Problem 124. D eterm ine the ratio of the mass of th e Sun to the mass of the E a rth if the period of revolution of the Moon round the E a rth is 27.2 days and the m ean distance from the E a rth to the Moon is 384000 km.
§ 27. Fundam entals of Astronom y
293
Given: r M0on = 27.2 days is the period of revolution of the Moon round the E arth, aMoon = 3.84 X 105 km is the mean distance from the E a rth to the Moon, and r EarUl = 365 days is the period of revolution of the E a rth round the Sun. From tables, we take the mean distance from the E a rth to the Sun aEarth = 1.5 X 108 km. Find: the ratio between the mass of the Sun and the mass of the E a rth , mSun/^EarthSolution. We shall use the form ula for the refined veris i » xi ♦ i i sion ofp K epler’s th ird law:
m 'Sun^Earth^Earth — 2
a Earth
m Earth • m M o o i r M o o n
aMoon
- .
Since the mass of the E arth relative to th a t of the Sun and the mass of the Moon relative to the mass of the E arth are negligibly sm all, we can rew rite the fonuula as follows:
msun7^ arth - = 4 m Eartli* M o o n
^ aMoon
• Hence
x m Earth
flM o o n
1 Moon 1 Earth
S ubstitu tin g in the num erical values, we obtain m Sun m Earth
_ (1.5x10® km)3 (27.2 days)2 “
‘-1QOOOO
(3 *84 x 105 k m )3 (3 6 5 d a y s)2
Answer. The ratio between the masses of the Sun and E a rth is about 330000. Problem 125. Determ ine the mean distance between the E a rth and the Moon using the following data: (1) the hori zontal parallax of the Moon is p = 0.57', and (2) an elec trom agnetic signal sent to the Moon from the E a rth returns in 2.56 s. W hat is the m ean velocity of m otion of the Moon round the E a rth if the sidereal m onth is 27.3 days long? Given: p = 0.57' is the horizontal parallax of the Moon, t = 2.56 s is the tim e required for the electrom agnetic sig nal to traverse twice the distance from the E a rth to the Moon, T = 27.3 days = 27.3 X 3600 x 24 s is the sidere al m onth. From tables, we take the mean radius of the E arth /?Earth = 6370 km and the velocity of light c = 3 X 105 km /s. Find: the distance d from the E a rth to the Moon and the mean velocity v of the Moon in its orbit. Solution. 1. The horizontal p arallax of the Moon is defined as the angle at which the radius of the E a rth is seen from
294
Ch. VI. General Remark's on A stronom y
the Moon (see Fig. 151). Consequently, ^ 5 3 = • Here it is more convenient to express the parallax in degrees: p = 0.57' = 0.95°. Then 3600 X 6370 km 0.95° X 6.28
~ 384380 km.
2. A radio signal sent to the Moon w ill be reflected by its surface and retu rn to the E arth . We know the tim e in which the radio signal covers twice the distance from the E a rth to the Moon. Therefore, 300000 k m /s X 2.56 s
2
~ 384000 km.
In order to determ ine the mean o rbital velocity of the Moon around the E a rth we use the form ula v = 2ndIT, where T is the sidereal m onth, i.e. the tim e taken by the Moon to com plete one revolution around the E a rth relativ e to the stars. S u b stitu tin g the num erical values, we obtain
Answer. The approxim ate distance between the E a rth and the Moon given by the two m ethods are (1) 384380 km and (2) 384000 km. The m ean o rbital velocity of the Moon is 1 . 0 2 km /s. Questions and Problems 27.1. How long does it take light to travel from the Sun to the E arth ? The distance from the Sun to the E a rth is 149.6 X 106 km , and the velocity of light is 2.998 X 1 0 5 km /s. 27.2. D eterm ine the distance to a C entauri (the nearest sta r to the Solar System ) if its light takes 4.25 years to reach the E arth . Express the distance in kilom etres and parsecs. 27.3. The m ean distance from the Sun to the farth est planet P luto is 40 AU. How long does it take light to cover this distance? 27.4. I t takes 2 X 106 years for light to travel from the nearest to us galaxy in Androm eda. Express th is distance in parsecs. 27.5. L ist the brightest stars, whose m agnitudes are ex
§ 27. Fundam entals of Astronom y
295
pressed in negative num bers. 27.6. W hat constellation includes Sirius? W hat does its negative declination indicate? 27.7. The star Sirius is about 8.4 X 10 13 km from the E arth . How long does it take its light to reach the E arth ? 27.8. L ight takes over 100000 years to traverse the larger diam eter of our G alaxy. D eterm ine the approxim ate size of the G alaxy in parsecs. 27.9. W hat is the brightest sta r in the northern celestial hemisphere? W hat constellation does it belong to? 27.10. Name the points of the horizon lying on the celes tia l m eridian and celestial equator. 27.11. Name the equatorial constellation cut into two unequal parts by another constellation. 27.12. Compile a list of the nonsetting constellations in your locality by observation. Verify the correctness of your observations w ith star chart. 27.13. D eterm ine the position of the constellation Andro meda on the star chart at the m om ent of observation and locate it in the sky. The nearest G alaxy to us is a m isty spot in the vicin ity of this constellation. Observe this galaxy. 27.14. Does the Sun always rise exactly in the east and set exactly in the west? 27.15. W hich Zodiac constellations caanot be observed on the north pole? 27.16. W here m ust an observer be to see the north celes tia l pole at the zenith? 27.17. The angular distance of the celestial pole from the zenith is 34° 15' for an observer at Moscow. W hat is the geographical latitu d e of Moscow? 27.18. D eterm ine the a ltitu d e of Polaris for your locality. 27.19. The geographical la titu d e of Leningrad is 59° 56'. W hat is the angular distance between the zenith and the ce lestial pole for Leningrad? 27.20. Use the star chart to determ ine the constellation which is located on the northern horizon a t m idnight on May 15 at the la titu d e of Moscow. D eterm ine the constellation at th a t point and at the same tim e at your latitu d e. 27.21. By how m any degrees does the Sun move along the ecliptic per day? 27.22. W hat are the declination and the rig h t ascension
296
Ch. VI. General Remarks on A stronom y
of the Sun on the 22nd of March? 27.23. W hat are the declination and the rig h t ascension of the autum nal equinox? 27.24. W hat is the a ltitu d e of the Sun at noon in Moscow at the sum m er solstice? 27.25. W hat will the angular diam eter of the E a rth be for a cosm onaut on the Moon if the radius of the E a rth is 6370 km and the distance from the E arth to the Moon is 384000 km? 27.26. The angular diam eter of the Sun is 32'. Determ ine the d iam eter of the Sun. 27.27. The m axim um horizontal p arallax of Mars is 23". C alculate the m inim um distance between Mars and the E arth . 27.28. The annual p arallax of Vega is 0.121". C alculate the distance from the E a rth to Vega in parsecs. 27.29. C alculate the period of revolution of U ranus around the Sun if the m ean distance between it and the Sun is 19.19 AU. 27.30. S aturn orbits the Sun once in 29.46 years. D eter m ine the m ean distance between it and the Sun. 27.31. The period of revolution of J u p ite r round the Sun is 11.86 years. D eterm ine the m ean distance from it to the Sun in astronom ical u n its and in kilom etres. 27.32. How will the E a rth appear to an observer on the Moon at the in sta n t when full Moon is seen from the E arth ? 27.33. Venus is som etim es called the m orning sta r and evening star. For w hat latitu d e s is th is true? 27.34. W hat is the phase of Venus when it is observed as a m orning star? 27.35. The annual p arallax of our closest sta r, a C e n ta u ri, is 0.76". How long would it take for a spacecraft to fly there at a velocity of 17 km /s? 27.36. The distance to B arn ard ’s sta r is 1.83 pc. W hat is its annual parallax? 27.37. W hy do the seasons not change on Venus? 27.38. D eterm ine the m ean density of solar m atter. Take the required d ata from tables. 27.39. D eterm ine the linear velocity of the Vostok-1 spaceship, in which Y uri G agarin orbited the E a rth for the first tim e, assum ing th a t the o rb it was circular and its mean distance from the surface of the E a rth was 251 km.
Appendices
1.
B asic P h y sica l C onstants
V elocity of sound in air under norm al con d ition s V elo city of lig h t in vacuum G ravitation al constant Free fall acceleration A vogadro constant Molar volum e of ideal gas under normal con d ition s Loschm idt num ber Molar gas constant B oltzm ann constant Faraday constant E lectric con stan t M agnetic constant H ydrogen atom m ass E lectron rest m ass Proton rest m ass N eutron rest m ass E lem entary charge E lectron charge-to-m ass ratio R atio of proton and electron m asses A tom ic m ass u n it P lanck constant S tefan-B oltzm ann constant W ien constant R ydberg constant (for hydrogen) T riple point for w ater
2.
c c G g N
= = = = a
331.46 m /s 2.998 X 108 m /s 6.67 X 1 0 -n N *m2/k g 2 9.807 m /s2 = 6.022 X 102S m o l-i
Vm = 22.4 X 1 0 -3 m 3/m ol N r = 2.687 X 102* m -3 R = 8.314 J /(m o l -K) k = 1.381 X 1 0 -23 J/K F - 9.648 X 104 C/mol
e 0 = 8.85 X 1 0 -i2 F/m u0 = 4ji X 10*7 H /m = 1.257 X 10° H /m mH = 1.673 X 1 0 -27 kg m e = 9.109 X 1 0 -3i kg m„ = 1.673 X 1 0 -27 kg m i = 1.675 X 1 0 -27 kg e = 1.602 X 1 0 - in C e / m e = 1.759 X 10“ C/kg nip I m e = 1836.15 1 amu = 1.660 X 10~27 kg h = 6.626 X 1 0 -34 J -s o = 5.67 X 1 0 -8 W /(m 2 -K 4) h = 2.897 X 1 0 -3 m -K R 00 = 1.097 X 107 m -i T = 273.16 K (/ = 0.01°C)
D en sity of Som e Substances, p, k g /m 3 S o l i d s (at 293 K)
A lum inium Amber Brass Brick
2.7 1.1 8.5 1 .5
X 103 X 103 X 103 X 103
Cast iron gray w h ite
7 .0 X 7.5 X
298
Appendices
Concrete C onstantan Common salt Copper Cork D iam ond E b on ite Germ anium Gold G raphite Ice (0°C) Iridium Iron, steel Lead M anganin Mica
2.2 8.9 2.1 8.9 0.24 3.5 1.2 5.32 19.3 2.1 0.9 22.4 7.8 11.4 8.5 2.8
X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103
N ichrom e N ick el N ick elin e Oak Paraffin P latin u m Porcelain Rubber S ilv er Snow Tin T ungsten Uranium W indow g la ss Zinc
8.3 8.9 8.8 0.8 9.0 21.5 2.3 0.94 10.5 0.2 7.3 1.93 19.0 2.5 7.1
X 103 X 103 X 103 X 103 X 102 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103 X 103
L i q u i d s (at 293 K)
A cetone A n ilin e B enzene Carbon bisulphide Copper sulphate so lu tio n (saturated) E th y l (m ethyl) a l cohol E th y l ether G lycerine Kerosene Mercury (at 273 K)
0.8 1.02 0.85 1.26
X 103 X 103 X 103 X 103
1.15 X 103 0.79 0.71 1.26 0.8 13.6
X 103 X 103 X 103 X 10® X 103
N itrobenzene Oil castor (m ineral) transformer v egetab le Petrol Petroleum Turpentine W ater at 277 K heavy sea
1.2 X 103 0.92 0.89 0.93 0.7 0.9 0.87
X X X X X X
103 103 103 103 103 103
1.0 X 103 1.06 X 103 1.03 X 10s
Gases
(under normal conditions: p 0 = 1.013 X 105 Pa and T 0 = 273 K) A cetylen e A ir A m m onia Argon B utane Carbon d ioxid e Chlorine H eliu m
1.17 1.29 0.77 1.78 0.6 1.98 3.21 0.18
Hydrogen K rypton M ethane Neon N itrogen O xygen Propane X enon
0.09 3.74 0.72 0.9 1.25 1.43
2.01
5.85
3. Specific Heat Capacity of Some Substanc S o lid s
A lum inium Brass B rick
880 380 750
Cast iron Cement Concrete
550 800 920
A ppendices Copper G lass G old Ice (snow) Iron, ste e l, n ick el Lead N ap h th alen e Paraffin
380 840 125 2090 460 120 1300 3200
299
P la tin u m Sand S ilv er Sulphur T in W ood (spruce, pine) Zinc
125 970 250 712 250 2700 400
Liquids
E th y l alcohol E th y l eth er G lycerine Iron Kerosene
2430 2330 2430 830 2140
M achine o il Mercury Transform er o il Turpentine W ater
2100 125 2093 1760 4187
Gases (at co n sta n t pressure)
A ir A m m onia Carbon d iox id e H eliu m
1000 2100 880 5200
H ydrogen N itrogen O xygen W ater vapour
4. Specific H eat of Com bustion for Som e Sub stances,
14300 1000 920 2130 J /k g
Solid fuel
9.3 3.1 1.5 3.03 8.3 3.0
Brown coal Charcoal Chocks Coke Firewood Gun powder
X X X X X X
106 107 107 107 106 106
Coal A -l grade A-2 grade D onetsk E kibastuz Peat
2.05 3.03 2.55 1.63 1.5
X X X X X
Liq u id fuel
D iesel E th y l alcohol Fuel o il
4 .2 X 107 2.7 X 107 4 .0 X 107
Kerosene N aphtha P etrol, petroleum
4.4 X 107 4 .3 3 X 107 4 .6 X 107
Gaseous f u e l
(per cubic m etre under norm al con d ition s) B last-furnace gas Coke-oven gas N atural gas
3.7 X 106 1.64 X 107 3 ,5 5 x IQ7
Producer gas Town gas
5.5 X 106 2,1 X 107
300
Appendices 5. B oilin g P oint and Specific L atent Heat of V aporization S u b s ta n c e
A cetone Air A m m onia E th y l alcohol E th y l ether Freon-12 Iron Mercury P etrol Turpentine W ater heavy ordinary
T, K
t , °c
329.2 81 239.6 351 308 243.2 3023 630 423 433
56.2 — 192 —33.4 78 35 — 29.8 2750 357 150 160
374.43 373
J /k g r, <
5.2 2.1 1.37 8.57 3.52 1.68 5.8 2.85 3 .0 2.94
X X X X X X X X X X
10* 10* 106 105 10* 106 104 10* 10* 10*
2.06 X 106 2.26 X 10°
101.43 100
0. Pressure and D en sity of Saturated W ater Vapour a t V arious Temperatures t, °G
— 10 —5 -4 —3 —2 —1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
P, k P a
0.260 0.401 0.437 0.476 0.517 0.563 0.613 0.653 0.706 0.760 0.813 0.880 0.933 1.000
1.066 1.146 1.226 1.306 1.399 1.492 1.599 1.706
p x io ~ 3,
<, °c
k g /m 3
2.14 3.24 3.51 3.81 4.13 4.47 4.80 5.20 5.60 6.00 6.40 6.80 7.30 7.80 8 .3 0 8.80 9.40 10.0 10.7 11.4 12.1 12.8
'
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 80 100 120 160 200
p,
kPa
1.813 1.933 2.066 2.199 2.333 2.493 2.639 2.813 2.986 3.173 3.359 3.559 3.786 3.999 4.239 7.371 12.33 19.92 47.33 101.3 198.5 618.0 1554
pXlO"3, k g /m 3
13.6 14.5 15.4 16.3 17.3 18.3 19.4 20.6 21.8 23.0 24.4 25.8 27.2 28.7 30.3 51.2 83.0 130.0 293 598 1123 3259 7763
301
Appendices
7. Young’s modulus for Some Substances, E, GPa 70 110 28 90 20
A lu m iniu m B rass Brick C ast iron Concrete
Copper Iron Lead S teel
130 200 17 220
8. Boiling Point and Specific Latent Heat of Vaporization for Water under Various Pressures
t, °c
P, M Pa
10 100 151
0 .0 0 1
0.1 0.49
r,
M J /k g
<, &c
P.
2.47 2.26 2.11
197.4 346 347.15
1.47 15.7 22.1
M Pa
r , M J /k g
1.95 0.9 0
9. Boiling Point and Critical Parameters for Some Substances C r itic a l p a r a m e te r s S u b s ta n c e
Argon E ther E th y l alcohol H elium H ydrogen K rypton N eon N itrogen Oxygen W ater X enon
B o ilin g p o in t, *, °C
te m p e r a tu r e * c r’
-1 2 2 .4 193.8 243.1 — 267.9 — 241 —63.62 — 228.7 — 147.1 — 118.4 374.15 18.76
— 186 35 78 — 269 -2 5 3 — 193 — 246 — 196 — 183 100 — 108
p re s s u re Pc r x l U 5, P a
48 35.6 63 2.25 12.8 54.27 26.9 33.5 49.7 221.3 57.64
10. Melting Point and Specific Latent Heat of Fusion for Some Solids at Melting Point S u b s ta n c e
A lum inium Cast iron gray w hite
Tm ’
K
*m ’
°C
K J /k g
932
659
3 .8 X 105
1423 1473
1150 1200
9.7 X 104 1.3 X 105
A ppendices
302 10.
M elting P oin t and Specific L aten t H eat of F u sion for Som e Solid s a t M eltin g P oint (cont.) Substance
Copper Gold Iron Lead M ercury N aphthalene Silver S teel Sulphur T in T ungsten W ater, ice W ater, heavy W ood’s m etal* Zinc
K J /k g
1356 1337 1803 600 234 353 1233 1673 385.8 505 3683 273 276.82 341 692
1083 1064 1530 327 -3 9 80 960 1400 112 .8 232 3410 0 3.82 68 419
1.8 6.6 2.7 2.5 1.25 1.51 8.8 2.1 5.5 5.8 2.6 3.35 3.16 3.2 1.18
105 104 105 104 104 105 104 105 104 104 104 105 105 104 105
X X X X X X X X X X X X X X X
* W o o d ’s metal with thi9 melting point contains 5 0 % bismuth, 2 5 % lead, 1 2 . 5 % tin, and 1 2 . 5 % cadmium.
11.
Surface T ension for Som e S u b stances, a , N /m (a t 293 K)
A cetone Castor o il Copper su lp h ate so lu tio n E tner E th y l alcohol G lycerin e Kerosene
0.024 0.033 0.074 0.017 0.022 0.059 0.024
0.470 0.045 0.029 0.040 0.027 0.072
Mercury M ilk Petrol Soap solu tion T urpentine W ater
12. Coefficient of Linear E xpansion for Som e S o lid s, a , A lu m in iu m , duralum in 2 .3 B rass 1.9 Bronze 1.8 Cast iron 1.0 C oncrete, cem ent (1-1.4) Copper 1.7 E bonite 7 .0 G lass quartz 6 w indow 9
X X X X X X X
X X
10-5 10 - 5 1 0- 6 io -5 10- 6 1 0- 5 1 0- 5 10- 7 1 0- 6
* Invar contains 6 4 % iron a nd 3 6 % nickel.
Gold Invar* Iron, steel Lead N ick el P latin u m P la tin ite Tin T ungsten Zinc
1.4 6 1.2 2.9 1.28 9 9 2.1 4 2.9
X X X X X X X X X X
Appendices
303
13. C oefficient of Volume E xpansion for Som e L iq u id s, p, K" A cetone E th yl alcohol E th y l ether G lycerine Kerosene Mercury Petrol Petroleum
1.2 1.1 1.6 5.0 1.0 1.8 1.0 1.0
X x X X x X x X
1 0- 3 i o -3 10- 3 10- 4 io -3 10- 4 io -3 1 0- 3
Sulphuric acid Transform er o il W ater at 5-10°C 10-20°C 2Q-40°C 40-60°C 60-80°C 80-100°C
5.7 X 10- 4 6 .0 X 10- 4 5 .3 1.5 3.02 4 .5 8 5.87 7.02
X X X X X X
10- 5 10-4 10-4 10-4 10-4 10-4
14. P e r m ittiv ity e for Som e S ubstances A ir at 0.1 MPa at 10 MPa Amber A n ilin e E b on ite E poxy resin G lass G lycerine H ydrogen Ice at — 18°C Kerosene Marble Mica
1.0006 1.055 2.8 84 2.7 3.7 5-10 39 1.0003 3.2 2.0 8-9 6-9
Paraffin Petrol P orcelain Rubber R u tile Sh ellac Sulphur Transform er o il Vacuum W ater W ater at 0°C W ax W axed paper
2.2 2 .3 4-7 2-3 130 3.6 3 .6 -4 .3 2 .2- 2 .5 1 81 88 5.8 2.0
15. R e s is tiv ity of Som e M aterials, p, fl-m A lum inium 2.7 B rass 6.3 Coal (4.0-5.0) Constantan 4.7 Copper 1.68 Ferro-alum inium h igh-resistance a llo y 1.1 Gold 2.2 Iron 9.9 Lead 2.07 Manganin 3.9 Mercury 9.54
X X X X X
10“8 10- 8 1 0 ~5 10- 7 IO" 8
x X X X X X
io -6 IQ- 8 10- 8 1 0- 7 10- 7 10"7
N ick el N ick elin e Nichrom e Osmium P latin u m R heotan S ilv er S teel T in T ungsten Zinc
7.3 4 .2 1.05 9.5 1.05 4 .5 1.58 1.2 1.13 5 .3 5.95
X X X X X X X X X X X
10- 8 10- 7 1 0- 6 10- 8 10- 7 1 0- 7 10- 8 10- 7 1 0- 7 10- 8 10- 8
16. Tem perature R esistan ce Coefficient for Som e S u b stances, a , K -1 Cast iron C onstantan Manganin N ichrom e and ferro-alum inium h igh-resistan ce a llo y
0.002 0.000005 0.000008 0 .Q002
N ick elin e R heotan S teel T ungsten
0.0001 0.0004 0.006 0.0050
304
A ppendices 17. E lectrochem ical E q u ivalen t for Some Substances, fc, kg/C
A lum inium Calcium Chlorine Chromium (bivalent) Copper bivalen t m onovalent Gold H ydrogen Iron bivalen t trivalen t
9.32 X 10-8 2.077 X 10"7 3.67 X 10-7 2.79 X 10-7 3.29 X 10-7 6.6 X 10-7 6.81 X 10-7 1.045 X 10-8 2.89 X 10-7 1.93 X 10-7
1.074 X 10-6 1.26 X io - 7 2.072 X 10-6
Lead M agnesium Mercury N ick el b ivalen t triv a len t Oxygen P otassium S ilv er Sodium Zinc
3.04 X io - 7 2.03 X io - 7 8.29 X 10-8 4.052 X io - 7 1.118 X 10-6 2.383 X i o - 7 3.388 X io - 7
18. R efractive Index for Some M aterials, n A cetone A ir A n ilin e Benzene Carbon b isulp hide Carbon tetrachloride D iam ond E th yl alcohol G lass crown flint
1.36 1.0003 1.59 1.50 1.63 1.46 2.42 1.36
G lycerine Ice M ethyl alcohol Quartz Rock sa lt Sugar S y lv ite Turpentine W ater
1.47 1.31 1.33 1.54 1.54 1.56 1.49 1.51 1.33
1.5 1 .6- 1.8
19. Mass of Soine Isotopps, amu* Element
Hydrogen
H elium L ithium B erylliu m Boron Carbon
Isotope
:h ?h ?H ?He £He ?Li 7Li ?Be JBe *|C
Mass
1.00783 2.01410 3.01605 3.01603 4.00260 6.01513 7.01601 8.00531 9.01219 11.00930 12.00000
Element
Carbon Oxygen Fluorine A lum inium Phosphorus Radon R adium Uranium N ep tu n iu m P lu ton iu m
* 1 a m u is equal to 1/12 of the mass of an a t o m of
Mass
Isotope
ic 10 t9i? s*1 13A 1
2!e2Rn 2i§R a
2i}|u %u 2B N p 29®Pu
13.00335 15.99491 18.99843 26.98153 29.97867 222.01922 226.02435 235.04299 238.05006 237.04706 239.05122
isotope.
Appendices
305
20. Psychrometric Table Dry-bulb temperature K
273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303
°G
0 1 2 3 4 5
6
7 8
9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29 30
Difference between dry-bulb and wet--bulb temperatures 1
0
100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100
21.
82 83 84 84 85
86 86
87 87
88 88 88 89 89 90 90 90 90 91 91 91 91 92 92 92 92 92 92 93 93 93
2 0 -0 5 3 0
3
45 48 68 51 69 54 70 56 72 58 73 60 74 61 75 63 76 64 76 65 77 66 78 68 79 69 79 70 80 71 81 71 81 72 82 73 82 74 83 74 83 75 83 76 84 76 84 77 84 77 85 78 85 78 85 78 86 79 86 79 63 65
4
5
28 32 35 39 42 45 47 49 51 53 54 56 57 59 60 61 62 64 64 65
6
7
8
9
10
11
11 16
20
24 10 28 14 32 19 35 23 37 26 40 28 42 31 44 34 46 36 48 38 49 40 51 42 52 44 54 45 55 47 56 48 58 50 66 59 51 67 60 52 68 61 54 69 61 55 69 62 56 70 63 57 71 64 58 71 65 59 72 65 59 72 66 60 73 67 61
6 10 14 18
7
21 11 24 26 29 31 33 36 37 39 41 43 44 46 47 48 49 50 51 52 53 54 55
14 17
4
8 20 11
23 14 25 17 27 20 30 22 32 24 34 26 35 29 37 30 39 32 40 34 42 36 43 37 44 38 45 40 47 41 48 42 49 43 50 44
6 9
12
5
15 17
8
10 20 13 22 15
24 18 26 20 28 22 30 24 31 26 33 27 34 29 36 30 37 32 38 33 39 34
Permeability of Some Materials, ji
P a ra m a g n etics (p >
Air A lu m in iu m O xygen liq u id Tungsten
2
1)
1.00000038 1.000023 1.0000019 1.003400 1.000176
D ia m a g n e tic s
B ism u th Copper G lass H ydrogen W ater
(p <
1)
0.999824 0.999990 0.999987 0.99999993? 0.999991
306
A ppendices F e rrom ag ne tic s
(m axim um perm eability) Cast iron Iron carbon soft transform er P erm alloy (N i-F e a llo y for transform er cores)
2000 3000 8000 15000 80000
R e m a r k . The m axim um p erm eab ilities of ferrom agnetics are given sin ce the param eter depends on the external m agn etic field stren gth .
22.
Magnetic Induction as a Function of Magnetic Field Strength for Soft Steel during Primary Magnetization
23. Some Astronomical Quantities (mean values) R ad iu s of the Earth M ass of th e E arth D en sity of the Earth V elo city of orbital m otion of the E arth R ad iu s of the Sun M ass of the Sun R adius of the Moon M ass of the Moon D istan ce betw een the cen tres of the Earth and th e Sun D istan ce betw een the cen tres of the Earth and th e Moon Period of revolution of th e Moon around the E arth In clin ation of the eclip tic to th e equator
6.37 5.9 8 5.52
106 m 1024 kg X 103 k g /m 3 105 km /h 6.95 X 108 m 1.98 X 1030 kg 1.74 X 10° m 7.33 X 1022 kg X
X
1.49
X
1011 m
3.84
X
108 m
27.3 d ays 23.5°
Appendices
307
24. Param eters of Som e B right S tars Magni tude m
S ta r
8 co. 8 8 8 8 8 8
Tauri (Aldebaran) O rionis (R igel) A urigae (Capella) O rionis (B etelgeuse) C anis M ajoris (Sirius) Gem inorum (Castor) Lyrae (Vega) C ygni (Deneb)
1.06 0.34 0.21 0.92 — 1.58 1.99 0.14 1.38
Declina tion 0
R ig h t a sc e n sio n a
4 5 5 5 6 7 18 20
h h h h h h h h
31 m in 54 s 12.1 m in 13 m in 52.5 m in 42.9 m in 31.4 m in 35 min 39 m in
+ 45°57' —8°15' + 45°57' + 7°24' — 16°39' + 32° + 38°41/ + 45°06'
25. B in d in g Energy of A to m ic N uclei Nucleus
?H ?H iH e £He °3Li ’ Li |B e 4!B ‘|C *3C 14N ifO ll 0
?IA1 iJs i ?°5P 2I?Rn 22|B a 2932BU 23|U 293JPu
2 9*
E b, M e V
E h/ A , M e V
2.2 8 .5 7.7 28.3 32.0 39.2 58.2 64.7 76.2 92.2 97.1 104.7 127.6 131.8 225.0 255.2 250.6 1708.2 1731.6 1783.8 1801.7 1806.9
1.1 2.83 2.57 7.075 5.33 5.60 6.47 6.47 6.93 7.68 7.47 7.47 7.975 7.75 8 .33 8.51 8.35 7.69 7.66 7.59 7.57 7.56
Appendices
308
26. Prefixes Used w ith SI U n its Notation
Prefix
Factor
lOl8
hexa peta tera
109
giga
1Q15 101? 10° 103 103
H P T
G M k h da d c m
mega kilo hecto deca deci cen ti m illi micro nano pico fem to atto
101 io - 1 1 0 -3
io - 3 1 0 -6 1 0 -9 10-12 10-15 10-18
P
n P f a
27. S in es and T angents of A ngles from 0 to 90° Angle degrees
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
radians
0 0.0175 0.0349 0.0524 0.0698 0.0873 0.1047 0.1222 0.1396 0.1571 0.1745 0.1920 0.2094 0.2269 0.2443 0.2618 0.2793 0.2967 0.3142 0.3316 0.3491
Sine
0.0000 0.0175 0.0349 0.0523 0.0698 0.0872 0.1045 0.1219 0.1392 0.1564 0.1736 0.1908 0.2079 0.2250 0.2419 0.2588 0.2756 0.2924 0.3090 0.3256 0.3420
Tangent
0.0000 0.0175 0.0349 0.0524 0.0699 0.0875 0.1051 0.1228 0.1405 0.1584 0.1763 0 1944 0.2126 0.2309 0.2493 0.2679 0.2867 0.3057 0.3249 0.3443 0.3640
Appendices
309
27. Sines and Tangents of Angles from 0 to 90° (cont.) A n g le d e g re e s
r a d ia n s
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65
0.3665 0.3840 0.4014 0.4189 0.4363 0.4538 0.4712 0.4887 0.5061 0.5236 0.5411 0.5585 0.5760 0.5934 0.6109 0.6283 0.6458 0.6632 0.6807 0.6981 0.7156 0.7330 0.7505 0.7679 0.7854 0.8029 0.8203 0.8378 0.8552 0.8727 0.8901 0.9076 0.9250 0.9425 0.9599 0.9774 0.9948 1.0123 1.0297 1.0472 1.0647 1.0821 1.0996 1.1170 1.1345 1.1519
66
S in e
T angent
0.3584 0.3746 0.3907 0.4067 0.4226 0.4384 0.4540 0.4695 0.4848 0.5000 0.5150 0.5299 0.5446 0.5592 0.5736 0.5878 0.6018 0.6157 0.6293 0.6428 0.6561 0.6691 0.6820 0.6947 0.7071 0.7193 0.7314 0.7431 0.7574 0.7660 0.7771 0.7880 0.7986 0.8090 0.8192 0.9774 0.8387 0.8480 0.8572 0.8660 0.8746 0.8829 0.8910 0.8988 0.9063 0.9135
0.3839 0.4040 0.4245 0.4452 0.4663 0.4877 0.5095 0.5317 0.5543 0.5774 0.6009 0.6249 0.6494 0.6745 0.7002 0.7265 0.67536 0.7813 0.8098 0.8391 0.8693 0.9004 0.9325 0.9657 1.0000 1.036 1.072 1.111 1.150 1.192 1.235 1.280 1.327 1.376 1.428 1.483 1.540 1.600 1.664 1.732 1.804 1.881 1.963 2.050 2.145 2.246
*
310
Appendices 27. S in es and T angents of A ngles from 0 to 90° (cont.) A n g le d e g re e s
67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
r a d ia n s
1.1694 1.1868 1.2043 1.2217 1.2392 1.2566 1.2741 1.2915 1.3090 1.3265 1.3439 1.3614 1.3788 1.3963 1.4137 1.4312 1.4486 1.4661 1.4835 1.5010 1.5184 1.5359 1.5533 1.5708
S in e
0.9205 0.9272 0.9336 0.9397 0.9455 0.9511 0.9563 0.9631 0.9659 0.9703 0.9744 0.9781 0.9816 0.9848 0.9877 0.9903 0.9925 0.9945 0.9962 0.9976 0.9986 0.9994 0.9998 1.000
T angent
2.356 2.475 2.605 2.747 2.904 3.078 3.271 3.487 3.732 4.011 4.331 4.705 5.145 5.671 6.314 7.115 8.144 9.514 11.43 14.30 19.08 28.64 57.29 oo
28. B asic Form ulas in P h y sics F o r m u la
Q u a n titie s a p p e a rin g in f o r m u la s
U n its
Kinematics r = r (<)
x = x (<). y = y (t) s = s(t)
r is th e disp lacem en t of a p article t is the tim e in terval x and y are the coordinates of a particle s is th e d istan ce covered by the particle
m s m m
Uniform motion r — r „ = vf
x — x 0 = uxt y
— y0= V
r — r0 is the increm ent of d isp lacem en t x 0 ana y0 are the in itia l coordinates of a particle
m m
A ppendices
311
28. B asic Form ulas in P h y sics (cont.) Q u a n titie s a p p e a r i n g in f o r m u la s
F o r m u la
| r | = x = s =
s is the path len g th (the coordinate
vt
U n its
m
a x is co in cid es w ith the d irection of displacem ent)
Uniformly variable motion -•
Ar At
y
v = lim ^
is the m otion
m ean
v e lo c ity
of variab le
v is the in stan tan eou s v e lo c ity
m /s
m /s
A f-0
In vector form a = const
a is the acceleration
= V0 +
V
tit
m /s2
v0 is the in itia l v e lo c ity
m /s
v is the in sta n ta n eo u s v e lo c ity
m /s
In scalar form
vx and vy are the p rojections of v e lo c ity m /s
= "o* + =
Vy
X —
Xq
on the coord in ate axes
+ ayt
V0 y
= v0x +
_ jO 2
ax and ay are the projections of a ccel m /s2 eration on the coordinate axes
ayt2
y — y0 = voy + “ 2“ i>* — yj = 2 as
Free fall =
%
«
g is the free fa ll acceleration
gt
is the h eig h t from w hich the b od y fa lls
m /s2 m
- !4
H
II
v
v is the v e lo c ity at th e end of th e free m /s fall
Motion of a body thrown upwards V =
v„ —
h=
"o *
-
gt
T
II *
H
is the m axim um h eigh t of ascent
m
Appendices
312
28. B asic Form ulas in P h y sics (cont.) F o r m u la
Q u a n titie s a p p e a r in g in fo r m u la s
U n its
Curvilinear motion v is the lin ear v e lo c ity R is the radius of rotation
v = 2nR f
/ is the rotation al frequency o) is the angular v e lo c ity
co = 2ji/ to = v/R
ac = vVR ac = co2/?
ac is the cen trip etal acceleration
p = mlV
p is the d en sity m is the m ass V is th e volum e
m /s m H z, s _1 rad/s m /s2
Dynamics k g/m 3
H m3
N ew ton ’s second law
a = Ftm Ft = A mv
F is the (resultant) force Ft is the im p u lse of force mv is the m om entum
N N *s k g * m /s
C onservation of m om entum mv = con st N ew ton ’s third law —
7712 ^ 2
P = mg
P is the force of g ra v ity
N
Law of u niversal gravitation
F — G mim* r2
F is the g ra v ita tio n a l force
N
G is the g ra v ita tio n a l con stan t N *m2/k g 2 r is the d ista n ce betw een the cen tres m of m ass
Fc = mv2/R Fc = mco2/? Fei = kx Ff
= ilN
Fc is the cen trip eta l force Fe] is the e la stic force
k is the sp rin g con stan t F f is th e force of friction
H is the frictio n coefficient N is the force of norm al pressure Work, power, energy
N N N /m N N
A = FS cos a A = FS for a = 0
J A is the m echanical work a is the an gle betw een the direction deg
P = A lt P = Fv
P is the power P is the m ean power "v is the m ean v e lo c ity
of force and displacem ent W W m /s
Appendices
313
28. B asic Form ulas in P h y sics (cont.) F o r m u la
= mv2/2 E p = mgh
Q u a n titie s a p p e a r i n g in f o r m u la s
U n its
E k is the k in etic energy Z?p is the p o ten tia l energy
J J
A u.seful x 1 0 0 % n is the efficiency
1spent-
A c -
is the u sefu l work A spent Is the spent work
^ useful
Molecular Physics. Heat n
O
-
n° ~ M ~P
N \ is the A vogadro con stan t
no = * x /V m
M is the m olar m ass n0 is num ber d en sity of m olecu les Vm is the m olar volum e
m0 = M /N jl
m 0 is the m ass of a m olecule
F0 = J * 0 n aP
V0 is the v olu m e of a m olecule 3M
- v . 4npATA X = viz
m o l-1 k g /m o l
m~3
m3/ m ol
hg m3
r0 is the radius of a m olecule
m
X is the mean free path of a m olecule
v is the (arithm etic) mean v e lo c ity of
m m /s
m olecu les
X=
y 2nd2, ettn o
z is th e num ber of c o llisio n s per second
s _1
deff is
m
the
effectiv e
diam eter
of
a
m olecule P =
3",lo
p is the pressure
Pa
yrms is the root-m ean-square v e lo c ity
m /s J /K K
p = n0kT
k is the B oltzm an n con stan t T is the therm odynam ic tem perature
E\r == ~ kT
Ek is the k in etic energy of tran slation al
m otion of m olecu les E q u ation s of state of an id eal gas R is the m olar gas con stan t
p v = w RT P iV i
Ti
^ |^ = c o n st
*2
J
J/(m ol*K )
m /M = v is the am ount of substance Flt and Tx are the param eters of the first sta te of a gas p 2, V2J and T 2 are the param eters of the second sta te of a gas
m ol
Appendices
3i4
28. B asic Form ulas in P h y sics ( cont.) F o r m u la
Q u a n titie s a p p e a r in g in fo r m u la s
U n its
D a lto n 's law p is th e pressure of a gas m ixtu re p = p[ + p'2 + . . . p [ and p'2 are p artial pressures Q = c m ( T 2 — 7 \) Q is the h eat required to h eat a body
Pa Pa J
or ffiven off as it co o ls c is the specific heat J/(kg»K ) Q is the h eat g iv en off as fu el burns J q is the specific heat of com b u stion J/k g
Q = qm
of fuel __ Quseful 1 0 0 % vspent
■— Psat
B = --^ a
^sat
p o= —
kg
tj is the efficien cy of the heater
%
(?usefui is the useful heat J
Q = rm
B =
m is th e m ass of burnt fuel
100%
B is the rela tiv e h u m id ity of air
pa is the ab solu te h u m id ity of air psat is the saturated vapour d en sity 100%
p a is th e pressure of vapour con tain ed in air Psat Is the saturated vapour pressure
a is the surface ten sion F is th e force of surface ten sion I is th e len g th of the boundary of free
% k g/m s k g /m 3 Pa Pa N /m N m
surface A
a = AS
A is the work of m olecular forces
J
A S is th e decrease in the area of free
m2
surface 2a h = ——
h is the h eig h t to w hich a w ettin g liq u id
pL = 2 o / r
r ise s (n on w ettin g liq u id fa lls) in a ca p illa ry r is th e radius of a ca p illa ry p L is th e L aplacian pressure for a spherical surface
m m Pa
Appendices
315
28. B asic Form ulas in P h y sics (cont.) Q u a n titie s a p p e a r i n g in f o r m u la s
F o r m u la
Q = km
U n its
Q is the heat required for fusion or
liberated
during
c ry sta lliza tio n
k is the specific la ten t h eat of fusion
J /k g
D eform a tion s
a =-
o is the m echanical stress
Pa
F is the force of e la stic deform ation S is the c r o ss-sectio n a l area of a de-
N m2
form ed body H ooke’s law o = E -j-
e = M /l E p = F M /2
E is the Y ou n g’s m odulus
Pa
AZ is the a b solu te deform ation (elongation) I is the in itia l len g th e is the strain E p is the p o ten tia l energy of e la stic d eform ation
m m J
T h e r m a l e x p a n s i o n o f bodies
SI l0S T
l 0S T
l 0 is the len g th o f a body a t 273 K I is the len g th of the body at an y tern-
I = l 0 (1 + a A T )
S = S 0 (1 + 2aA T)
V -V 0
a is the coefficien t of lin ear exp an sion K " 1
perature AT* is the change in tem perature S 0 is the surface area o f th e body a t 273 K S is th e surface area of the body a t an y tem perature
m m K ma m2
AV
ft — y A f "” A T $ *s coeft*c *en t volu m e exp an sion K ~l V = V 0 (1 + ft AT) V is th e vo lu m e of th e body a t any tern- m3 perature V 0 is the volu m e of the body at 273 K m3
p
_
Po
1 +f t AT
p is the d e n sity of a su bstance a t
k g/m 3
any tem perature p0 is th e d e n sity of the substance at 273 K
k g /m 3
316
A ppendices 28. B asic Form ulas in P h ysics ( co nt .) Q u a n titie s a p p e a rin g in f o r m u la s
F o r m u la
U n its
E lectrostatic s
Charge con servation law Q i + Qz + • • • + Q n = const
Q ly Q 2j . . n
Q n are
electric
charges
C
the num ber of electric charges in a closed sy stem
C oulom b’s law
1
I Qi J J.ff2 y f
F = -t~ — ^
4ne0
E = F /Q t Q
4jte0er2 (p =
W /Q t
(P = 4Jie~er~
er2
i s the force of in teraction betw een N
p o in t electric charges (?i» Qz are electric charges C r is the separation betw een the charges m e0 is the electric co n sta n t C/(N *m2) e is th e p e r m ittiv ity of the m edium E is the electric field stren gth N/C Q i is a test charge in the field
C
Q is the charge producing the field (p is the electric field p o te n tia l
C V
^
Po te n t^a^ ener£y ° f the te st
J
U is the v o lta g e is the work done by th e electric field to m ove a charge from one p oin t in the field to another
V J
IS
charge ^7 =
A
E = - ^ j-
E is the stren gth of a uniform field U is the p o ten tia l difference betw een tw o
V /m , N/C V
p o in ts in th e field C = Qltp C = Q /U C=
G GtS —
W = C U 2!2
d is the d ista n ce betw een the tw o p o in ts
m
a lo n g th e field lin e C is the cap acitan ce
F
C is th e cap acitan ce of a (p arallel-p late)
F
cap acitor S is the area of a p la te d is th e th ick n ess of a d ielectric W is th e energy of the cap acitor
m2 m J
Appendices
317
28. B asic Form ulas in P h y sics (cont.) Q u a n titie s a p p e a r i n g in fo r m u la s
F o r m u la
U n its
Electric current in metals is the electric current A is the charge of an electron (ion) C is the num ber d en sity of charge ra-3 carriers v is the mean v e lo c ity of the directed m /s m otion of charge carriers
/ = Q it
I = envS I — enuES
I e n
u — v/E
u is the m o b ility of the charge carriers m2/(V *s)
I i~ ~ s
j is the current d en sity
S
i
is the c ro ss-sectio n a l area
R is the resistan ce of the conductor
R = p -j-
p is the r e s is tiv ity of the conductor m aterial I is the conductor len g th S is the c ro ss-sectio n a l area of the conductor R t = R 0 (1 + a A 7*) R t is the resistan ce of a conductor at an y tem perature R 0 is the resistan ce of the conductor at 273 K a is the tem perature resistan ce coefficient Ohm ’s law for a conductor / = -g -
A /m 2 m2 Q Q*m m m2 Q Q K _1
U is the vo lta g e across the conductor
V
is the resistan ce of the conductor
Q
R
Series connection of conductors /?eq = /?! + R 2 /?eq is the eq u iv a len t resistan ce of conQ + . . . + Rn ductors R eq = 7?j ft for R x ~ R 2 = . . . ft is the num ber of the conductors = Rn R 2» . . R n are the resista n ces of Q the con d u ctors Parallel connection of conductors
1
1
— = -ft ,
1
1
/? eq is the eq u iv a len t resistan ce
Q
318
A ppendices 28. B asic Form ulas in P h y sics (cont.) Q u a n titie s a p p e a r in g in fo r m u la s
F o r m u la
Ran =
——
**
n is the
n
for R x = = ...
num ber of resistors
R lt R 2, /?3, . . = /?n
U n its
R n are
the
resis- 12
tances
Ohm ’s law for a closed circu it I =
%
p i— "** r
% is the electro m o tiv e force
of a cur- V rent source R is the resistan ce of the ex tern al cir- 12 cu it r is the resistan ce of the internal c irc u it 12 Ohm’s law for a sub circu it co n ta in in g an em f
U _
I = -----5 ----Jti A = IU t A = I*Rt A =
U2
- 5— t
it P = IU
P = PR P = U*!R P q = 1% P\ = P 0 — P r
% is the counter em f
V
A is the work done by electric current
J
t is the tim e
s
P is th e power of an electric current
W
P 0 is the to ta l power developed by a cur-
W
rent source P \ is the power of the load in a closed circu it
W
Electric current in electrolytes Faraday’s law m = k Q or m = k i t
m is the m ass of a substance dep osited kg
on electrod es G eneralized law 1 M m = Q
r
n
k
is
the electroch em ical eq u iv a le n t
M is the m olar m ass F is th e Faraday con stan t
UqU /./ a/
P = — nZ 2na
kg/C kg/m ol C/mol
n is the valen cy Electromagnetism F is the force of in teraction betw een
N p arallel current-carrying conductors / j and 1 2 are the currents in the conduc- A tors jx is the p erm eab ility p 0 is the m agnetic con stan t H /m a is the separation betw een the con- m ductors
A ppendices
319
28. B asic Form ulas in P h ysics (conf.) F o n n u la
F x = B I l sin a Fa = B Il for sin a = 1
A = IB AS A = I AO
* s tr =
W
~2n
I B clr =
27
n B sol
/o) ~
=
^sol
HoH
fiSol5
= no(i/r FL = B v e sin a
b
Q u a n titie s a p p e a r in g in f o r m u la s
U n its
F a is A m pere’s force (actin g on a cur N rent-carrying conductor) B is the m agnetic in d u ction T I is the a ctiv e len g th of a conductor m a is the an gle betw een the d irection of d eg the current and the m agnetic in d u ction vector A is the work done in m ovin g a cur rent-carryin g conductor in a m agnetic field AS is the change in the area em braced m* b y the current loop AO is the change in the m agnetic flux Wb B st r is the m agnetic in d u ction of a straigh t current r is the d istan ce from the conductor to the p oin t where the in d u ction is bein g determ ined
T m
B clr is the m agnetic in d u ction at the
T
centre of a circular current r is the radius of the circular current
m
2?S0l is the m agnetic in d u ction of the T field in a solen oid co is the num ber of turns I is the len gth of the solen oid m O sol is the m agnetic flux in the so len o id Wb S is the cro ss-sectio n a l area of the m 2 solenoid H is the m agnetic field strength A /m Fl is the Lorentz force (acting on a N charged p article in a m agnetic field) e is the charge of a p article C v is the v e lo c ity of tne p article m /s a is the an gle betw een vectors B and v deg
Electromagnetic induction
S = -
AO
At
% is the induced em f
AO is the change in the m agnetic flux Af is the tim e in terval over w hich the m agnetic flux changes
Wb s
320
A ppendices 28. B asic Form ulas in P h ysics (cont.) Q u a n titie s a p p e a r i n g in f o r m u la s
F o r m u la
AO At # s t r = B l v sin a = — CO
A/ = — L At L I2
W =
* S0l is the em f induced in a solen oid i s the em f induced in a stra ig h t conductor during it s m otion I is the a ctiv e len gth of the conductor
$ s tr
Units
V V m
£ s is the self-in d u ctan ce em f
V
L is the in ductance of th e circu it
H
W is th e energy of the m agnetic field
J
of t h e . circu it O s c illa tio n s a n d W aves
^
n
v = co =
1I T 2jt/ T = 2jiv cot cot +
T is th e period of o sc illa tio n s n is the num ber of com p lete o sc illa tio n s
v is the o scilla tio n frequency co is the circular frequency
E quation of harm onic m otion x is the d isp lacem en t A is the am p litu d e of v ib ra tio n s
x = A sin (cot +
-
2* Y -
T = 2 k V m !k
co~ V f — m k = vT
s Hz rad/s rad rad m m
T is the period of o sc illa tio n s of a s sim p le pendulum Iis the len g th of the pendulum m g is th e free fa ll acceleration m /s2 T is the period of ela stic v ib ra tio n s s
co is the circular frequency
rad/s
k is the spring con stan t N /m m is th e m ass of the load kg X is the w avelen gth m v is the v e lo c ity of propagation of w ave m /s
A l t e r n a t i n g cu rre n t
e = #<> sin(©* + u =
q>o) e *s th 0 in stan tan eou s em f in the circu it # 0 is the am p litu d e of the em f U 0sin (co/ +
V V V V
A ppendices
321
28. B asic Form ulas in P h y sics ( con t .) Q u a n titie s a p p e a r i n g in fo r m u la s
F o r m u la
i = 7 0 sin (cof +
7 = 70/ | / 2 U = U 0I | / 2 # = /2
1
1
coc
zjivc
X c = —7 r = o - „ r
i is the in stan tan eou s current 7 0 is the a m p litu d e of the current 7 is the effective current U is the effective v o lta g e % is the effective em f
Units A A A V V
*s ca p a citiv e reactance of th e Q circu it v is th e a ltern a tin g current frequency Hz X L = 0)7/ = 2jxvL X l is the in d u ctiv e reactance of the Q ________________ circu it Z = Y R 2- \ - ( X l — X c )2 Z is the im pedance of the circu it w ith Q series-con n ected com ponents R is the resistan ce of the circ u it Q cos
Electromagnetic oscillations and waves T = 2ji Y L C k = cT
T is the period of natural o sc illa tio n s
s of an o sc illa to r y circu it k is the len g th of the electrom agn etic m w aves c is the v e lo c ity of lig h t in vacuum m /s
Geometrical optics n =
cfv
n2*1 =
2 1 -0 5 3 0
sin e' "
n is the ab solu te refractive in d ex v is the v e lo c ity of lig h t in a m edium
m /s
I *s re^a t^ve refractive in d ex e is the angle of in cid en ce e' is the angle of refraction
deg deg
322
Appendices 28. Basic Form ulas in P h ysics (cont.) F o r m u la
” 2,1 =
m /s m edium v2 is th e v e lo c ity of lig h t in the second m /s m edium icr is the critica l angle of to ta l internal deg reflection
V2
— 1/ n
1
1 , 1
is th e focal len g th of a len s or mirror
m
a is the d istan ce from the object to the
m
~T~~a+ ~ar
len s (mirror) a ' is the d istan ce from the im age to the len s (mirror) O is th e o p tica l power of a len s
= 1/ / ID t
f
U n its
vx is the v e lo c ity of lig h t in the first
~
sin
Q u a n titie s a p p e a r i n g in f o r m u la s
~T
m D
/ is the focal len g th of a mirror
m
R is the radius of curvature of the
m
mirror ) is the lin ear m agnification
1
P = x
H is th e h eig h t of the im age h is the h eig h t of the object
m m
Photometry oII
e
O is the lu m in ou s flux
Q is the lu m in ou s energy cDtot = 4 j t / / = CD/co
I is the lu m in o u s in te n sity
E = 0 /5
E is the illu m in a n ce
O tot is the to ta l lu m in ou s flux co is the so lid angle
lm J lm cd sr lx
Laws of illu m in a tio n
E =
//r2
E ~ E 0 cos a
r is the d istan ce from a source of lig h t to the surface being illu m in a ted E 0 is the illu m in a n ce due to the norm al rays
m lx
or
i? =
cos a
a is the an gle of incidence
deg
Wave and quantum properties of radiation kX — d sin (p
m d is the g ratin g con stan t
A ppendices
323
28. B asic Form ulas in P h y sics (c o n t.) Q u a n titie s a p p e a r in g in f o r m u la s
F o r m u la
•st
>
II
CO
e is the energy of a photon
e
h
h
h is the P lanck con stan t
J J *S
m is the m ass of a photon
kg
me is the m om entum of a photon
me= x
U n its
k g*m /s
II
o.
- I' ll
p is the pressure of lig h t
Pa
N is the num ber of photons I
is the ra d ia tio n in te n sity
J /(m 2 -s)
E in stein s equ ation for photoelectric effect hv = A +
m v2 —| —
A is the electron work function
J
29. Som e M ath em atical Form ulas A Igebra 1.
(fl
b)2 =
a 2 ”}“ 2 a b -j- 6 2,
(a — b )2 = a 2 — 2a& + 62,
(a -f- ^)3 = 03 + 3a26 + 3a62 + b3y (a — 6)3 = a3 — 3a26 + 3a62 — b 3, a2 — b2 = (a — b) (a + b), a3 — b3 = (a — b) (a2 + ab + 62), a3 + b3 = (a + 6) (a2 — a& + 62). 2. The follow in g in eq u a lity m eans th at the arith m etic mean is larger than or equal to the geom etric mean
r
a~^b >
l/a6, a > 0, 6 > 0.
The eq u a lity arises when a = b. 3. For a given quadratic equation a x 2 + bx + c = 0 ,
its tw o solu tion s can be found from the form ula _ 21*
— b dh V b2 — 4ac
Appendices
324 or
a
The la tter is more con ven ien t when b is even. For a reduced quadratic equation x2 + px
+ 9 = 0,
its two solu tion s are given by
4. knowns:
Suppose we are given a sy stem of tw o eq u ation s in two u n b xy — Ci, c2.
{a 2x - \ - b 2y =
In order to find its so lu tio n , one unknown from one of the eq u a tio n s (say, the first) m ust be expressed in term s of the other unknown ( for instance, y in term s of x) and su b stitu ted in to the other equation:
S o lv in g th is equation , we ob tain ci&g— c2b 1 a xb2 — a2b i S u b stitu tin g x into the exp ression for y, w e find th at
fljCj— V~ “1*2—
*
The sam e m ethod can be used to so lv e a sy ste m w ith a larger n u m ber of equations: we exp ress one unknow n from one of th e eq u a tio n s of the sy stem and su b stitu te it in to th e rem aining eq u a tio n s. T h is reduces th e num ber of eq u a tio n s and unknow ns by on e. S im ila r ly , w e elim in a te in turn the other unknow ns u n til w e are le ft w ith one eq u a tion in one unknown, w h ich can be so lv ed . The rem ain in g unknow ns are determ ined in reverse order.
5. Some formulas for approximate calculations.
Appendices
325
If e < 1 (at le a st b y a factor of 10), we have
1
1 +8
= 1 — e,
1 1 —e = 1+ 8, ( l + s ) * = l + 2e,
(1 —e)2= 1 — 2e,
(l + e)3 = l + 3e, (1 — e)3= 1 — 3e,
Y T + l= i+ ± ., v — < T hese form ulas y ield q uite accurate resu lts. For exam p le, le t us calcu late Y 1/3.96:
i /J L ^ V
3.96
1
!
iY/ kI — T 0.04
°-04 j
j / j (j
1 2 11
0.01 j
1
1
1
2
1 — 0.005
2 V 1 — 0.01
= - i - (1 + 0.005) = 0.5025.
Geometry 1. G iven a right trian gle w ith sid es a and b and h y p oth en u se c. Then
a2 + b2 = c2#
2. The area of a trian gle is
where h is the h eigh t dropped on sid e a from th e op p osite vertex . The area of a rectan gle w ith sid es a and b is S = ab.
The area of a trapezium w ith b ases a and b and h eig h t h is
326
A ppendices The area of a circle of radius r is
S = Jir2. 3.
The area of the surface and the volum e of a sphere of radius r
a re
S ~ 4nr2,
4
V = — nr3.
The area of the lateral surface of a right cy lin d er w ith radius r and len gth h is S = 2 nrh. The to ta l surface area and the volum e of the right cy lin d er are
S = 2nr (r + h),
V = nr2h.
Trigonometry 1 . The trigonom etric fu n ction s of angles are sin (a + P) = sin a sin (a — P) = sin a cos (a + P) = cos a cos (a — P) = cos a
the sum s and differences of cos cos cos cos
p + cos a sin P, P —cos a sin p, p —sin a sin P, p + sin a sin p.
2 . The fu n ction s of double the angle are sin 2a = 2 sin a cos a cos 2a = co s2 a — sin 2 a . 3. The fu n ction s of h a lf the angle are a 2 sin 2 — = 1 — cos a ,
a
2 cos2 — = 1 + cos a . 4. The sum and the difference of the fu n ction s are • Q • a + P cos — a — sm a +I sin p = 2o sin —
a - T
si n a — si n p = 2 cos —
6
a
si n —
—
P
B
> »
a a + P a —p cos a + cos P = 2 cos — cos — »
. a+ P . a —P cos a — cos p = — 2 sm — s m — ^— .
A ppendices
327
5. The trigonom etric fu n ction s in term s o f the tan gen t are sin a =
tan a y f 1 -j* tan2 a ’
cos a =
1
----] / 1 + tan 2 a *
6 . Cosine theorem : for an arbitrary trian gle w ith sid es a , 6 , and c , w e have a2 = b2 + c2 — 26c cos A , where A is the angle opposing sid e a. The sin e theorem : sin A
sin B
sin C
where A , B , and C are the an gles opposing sid es a , 6, and c resp e c tiv e ly . 7. For 8 < 1 , w e have sin e = e, cos e = 1 , tan e = e. 8 . Som e d eriv a tiv es and integrals: ( x n ) ' =
n
i n - 1 ,
n
(i)'(sin x )' = cos x , (cos x)' = — sin x,
S
~n+i *” d x = T + T '
JifL-m ui,
J
sin x d x = — cos x,
s
cos x dx = sin x.
P e r io d s
PERIODIC TABLE cc
1
I
2
II
G r o u p s
s o
I
II
o f
V
IV
111
(H )
L i
3
6.94, L it h iu m
N a III
3
B e
2s* i
2
2 *? 2 2
M g
12
i
24.305
3s1 a 2
K
19 , Ca
IV 3 9 .0 9 8 .
V
VI
3 2p* 2
13
A l
2
3
f
2698154
8 2
2
A lu m in iu m
8 8 2
4*2
C a lc iu m
21
Ga
Z in c
ie 87.62
85.467a 5s'
8 2
R u b id iu m
8 4p1 2
88.9059
ii
5*2 8 S t r o n t iu m 2
G a lliu m
Y ttr iu m
5 A g V II is
Cs
,1 7 9 IX
,i
6*>
8 2
A U 196.9665
Fr X
(2231
2
Hg
8 0
18
•
2
200.59 Ssa
M e rc u ry
87 j Ra 226.0254
8 8 ,s
tl
F r a n c iu m
2
114.82
8 5p’ 2
. d ^ iu m In
R a d iu m
32
7s
2 ’«2
L a n th a n u m
,8 8 6p*
2
.3 2
,
2
5 0 8 5 P2 2
A c iin iu m
82
5 6P’
, Ku
8 9 1 2
4
4 8
£
Sn T in
33
749 216
a 4p3 2
A r s e n ic
Nb
4 * ,;
929064
p b
is
4045s* 8 N io b iu m 2 ,|5 1
Sb
!S
1217s
8 5P# * a2 A n t im o n y
Ta
a 2
a 2
A S
1!
7 2 » la 180.9479
5076*7 H a f n iu m
i
23
, ,n
3 0 34 *2 V a n a d iu m
Ge
H8 6 9
18
«
12271
2
4d*5*2 a Zirconium 2
2 04 37
p
, 30.97376
50.9415
91-22
Tl _ T h a lliu m
Ac**
...
e S ilic o n
18 72.59 a 4p2 2 G e r m a n iu m
Hf
,2 81
15 s „
a 3p3 2 P h o s p h o ru s
2 2 } V
57 2 56 j is 138.905s Is 178.4, 50*6*7 8 6*7 8
B a r iu m
i
a 2
La*
Ba
5 M ’06s' Gold
7
2
^ . C a d m iu m
56 i S 137.33
C e s iu m
6
8 5 s?
1!
14.0067 5 2p3 2 N itr o g e n
S i 28.085s
4790 T it a n iu m
In
2 4 8
11241 !s
!!
Silver
V III 1 329054
C d
2 4 8
107.868
8 5 *1 2
14 4
39 2 Zr 40*5 *2
12.011
C a rb o n
T i
69.72
18
8 4»* 2
^
N
7
C
4 2p* 2
8 3p2 2
4 4 9 5 5 9 3d*4*2 J S c a n d iu m 2
,29 C U ,30 Zn ,3i 65.38 , ’! 63546 is Rb 37, Sr 38 2 Y 3d"4»' 2 Copper
8
B o ro n
«), Sc ,
, s 40.08
4s1 8 P o t a s s iu m 2
4
3*7 M a g n e s iu m
_ S o d iu m
B 10.81
„ B e r y lliu m
11
22.90977
8
4
9.01218
7 3 « ”
5 036*2 a T a n t a lu m 2
.1 8 3
B i
2072
ii
2089804
Lead
2
B is m u t h
104
86p3
s
r,' aa. 1 0 5 ,? 32 32
6 037*2
1 26,1 6 d > 7 ,> » K u r c h a to v iu m 2
»8 2
LANTHANIDES Pr 59 1 Nd 6oi Pm6 1 1 Sm62i Eu63! Gd84? Ce 5 8 1 140.9077 J 151.96s , 5 4f 6 s 8 Shies’ 'I 4f3 6s2 8 ,442<;-6,’ -s Gadolinium 2 Cerium
2 Praseodymium 2 Neodymium 2 Promethium 2 Samarium
2 Europium
2
** ACTINIDEJ |
Jh904 Pa
U
91 j
9 2 j
2 3 2 .0 3 8 1 32 2 3 1 .0 3 5 9 32 5f26dVs 2 60?7s2 8 Thorium 2 Protactinium ? Uranium
1
2
NP93J Pu 94 J Am95j Cm96j 1244|,rs^| ( 2 4 3 | 32 2
2 37 048232 5f 4 6d'7s2 '5 Neptunium j Plutonium
5f 7s
Americium
2 Curium
2
OF THE ELEMENTS elements VI
VII
VIII H
1
8
0
2
O xygen
16
8 3p s 2
24
5 1 .9 9 6
t
8 2
Mo
42
9 5 9 4
t
5 p 4 TT e llu r iu m 2
8
7 4 « i8
5d46s 2 8 T u n g s te n 2
jl i 84
Po 209 P o lo n iu m
18 0
Fe 5 5 .0 4
Co r
£ 3de4s7 8 2
Ir o n
3d74s2 8 C o b a lt 2
Br 43
T e c h n e tiu m
836
,! 5 4 S 8
131.30
4d05s1 R h o d iu m 2
75
76j2 lr 192.2?
J OS 1 9 0 .2
2
5dfi6 s 2 8 O s m iu m 2
77 j
, 5d 6s i r id iu m
2 8 2
At
Xe
2
Uranium
2
..
78» 18
Rn 12221 ^
R adon
Atomic mosses conform with fhe Intem cfionol table ol 1977 the ctcurocy of fhe Icsi significant digit )S t | or ±3 if if is set in smcil type the numbers
922f-l■Distribution n.
Xenon
5d*6s2 B P la t in u m 2
» 86
Atomic num ber
Atomic moss
by unfilled ond following-
Pt 1 9 5 .0 9
«8
12101
8 6p5 2 A s ta tin e
completed sublevels
w e i s ,° i P a lla d iu m 2
2
I o d in e
5 d 56 » 2 8 R h e n iu m 2
\] U
K ry p to n
5p®
1 8 6 .2 0 7
,285 1
Kr
8 3 .0 0
48°
4 d 75s’ 8 R u t h e n iu m 2
|
Re
28 3 4 .4 ,1 2
Rh 4s.SPd 44; f S 1029055
1269045 8 5p* 2
N ic k e l
8 4p6 2
] Ru
2
70
58
is
ia 1 01.07
Ar 3 9 .9 4 s A rg o n
Ni
27
5 8 .9 3 3 2
7 9 .9 0 4
4 d 5 ss2 e
Distribution of electrons
N eon
2
8 4 PS 2 B r o m in e
127 6 0
2
13
Te ,53
,s 5 2
1 0 3 .8 s
,
Sd^s2 8 M anganese 2
\i. Tc 989062
4d 5s1 8 M o ly b d e n u m 2
w
25
5 4 .9 3 0 0
Se is,35
634
£ A 7 0 .9 6 8 4P 2 S e l e n iu m
2p«
0 3p® C h lo r in e
Mn 13
3 d s4s' C h r o m iu m
Ne 20.17c
C l 35 453
7
„ S u lp h u r
Cr
10 8 2
F l u o r in e
17
3206 3p 4
2
He 4 .0 0 2 8 0 H e liu m
18998403 7 2p*
§
6 8 2
2
F
9
1 5 .9 9 9 4
, 5 is2
H y d ro g e n
1
6 2p«
2
1 .0 0 7 9
1s*
.
of electrons by levels
m brockets ore the moss numbers of the most stable isotopes the nomes ond symbols of elem ents m parentheses ore not generally adopted.
Yb 7 0 1 Lu 7 1 5 T b 65j Dy66 1Ho 6 7 1 Er 88j Tm69l 9 34 3 S173.0, 158.9254 2 162.56 2 174.907 2 167.2e 2 ,6 8 4 6s 0 5d'6s 1 Thulium Lutetium Ytterbium Hofmium Erbium Terbium Dysprosium 4 f 96 s 2 0 2
4 f ‘° 6 s 2
2
2
4 f ,26 s 2 8 2
4f
2
8 2
2
Bk97j Cf 98 j Es 99 j Fm100i Md101,?(No)102j (Lriwaj [25%rSl 5%*%(Lawrencium) ,28W * PStU-a 7 i2 (Nobelium) Fermium Berkelium ?Californium Einsteinium 2
2
2 Menrlelevium
2
2
Answers
1.1 . Brow nian m ovem ent and diffusion. 1 .2 . M olecular ad h esive forces in liq u id s are much stronger than in gases. 1 .3 . D uring p a in tin g , the p a rticles of a p ain t diffuse into pores (interm olecular spaces) of the surface being painted . 1 .4 . In the region where the sa lt d isso lv e s a h igh sa lt concentration is created. A s a resu lt, diffusion tak es place to regions w ith a low er con cen tration . 1 .5 . The end faces of Johansson b lock s are h ig h ly polished; for th is reason, m olecular ad h esive forces com e in to p lay. 1 .6 . G lu in g is based on m olecular ad h esive forces and w ettin g . 1 .7 . A bout 316 m ol. 1 .8 . 24 kg, 28 kg, and 4 kg. 1 .9 . 1.6 kg. 1 .1 0 . 6.02 X 1026 and 6.02 X 1023. 1 .1 1 . A bout 2600 m3. 1 .1 2 . A bout 0.14 m ol, 8.2 X 1022. 1 .1 3 . A bout 5.7 X 10~? m3. 1 .1 4 . T he m ass of 1 m3 of brass is 8500 kg. 1 .1 5 . 105 m ol. 1 .1 6 . 5.32 X IO" 26 and 2.66 X IO" 26 kg, 4 .6 5 X 10~26 and 2.33 X 1 0 ~26 kg, 6.64 X 10- 27 kg. 1 .1 7 . 16 X 10~3 k g/m ol, 2.66 X IO-26 kg. 1 .1 8 . 1.46 X 1020. 1 .1 9 . A bout 1.54 X 1016. 1 .2 0 . The ratio is a p p roxim ately 2 X 1014. 1 .21. The ratio is 14.3 (considering th at equal v olu m es of gases under normal co n d itio n s con tain the sam e num ber of m olecu les). 1 . 2 2 . 7.3 X 1 0 - 26 kg, 1.97 k g/m 3. 1.23. 9.9 X IO-26 kg. 1 .2 4 . 2 X IO" 3 k g/m ol. 1 .25. 4 .2 x 10-° m. 1 .2 6 . 1.26 X 10~25 kg, 1.01 X IO" 28 m3, 5.8 X 1 0 - 10 m. 1 .2 7 . 9 X 109 m, 225 tim es. 1 .2 8 . 16 X 1 0 ~3 k g/m ol. 1.29. 1.4 X 10~5/ F 0, or 6.3% , w here V 0 is the m olar volum e of the gas under norm al con d itio n s. 1 .3 0 . 4 .0 X 10“ 8 m. 1 .3 1 . 1.12 X 10“4 s - 1. 1 .3 2 . 550 m /s. 1 .3 3 . 6.5 X 1 0~8 m. 1 .3 4 . 9 .3 X IO' 8 m. 1 .3 5 . A bout 2 X 1 0- 6 m. 1 .3 6 . 2.1 X 1 0“23 k g .m /s . 1 .3 7 . A bout 3.9 X IO" 8 m. 2 .1 . 470 m /s and 510 m /s for air and 450 m /s and 480 m /s for o x y g en 2 .2 . The mean k in etic energies of tra n slation al m otion of heliu m and neon are the sam e. 2 .3 . A t 322 K. 2 .4 . 2.25. 2 .5 . A bout 631 and 8640 m /s. 2 .6 . 6.6 X 1 0 ~22 J, 1.2 X IO"19 J. 2 .7 . A bout 3860 K. 2 .8 . 1.25 X 104 J, 6.2 X 105 J. 2 .9 . 290 K. 2 .1 0 . 9.9 X 10-11 Pa, high vacuum . 2 .1 1 . 1.45 X 1025. 2 .1 2 . A bout 6.5 X IO" 21 J, 9 .3 X 102&. 2 .1 3 . 1.3 X IO"6 Pa. 2 .1 4 . 1350 m /s, 6.07 X IO" 21 J. 2 .1 5 . 3.6 X 1012 m olecu les/cm 3. 2 .1 6 . 105 Pa. 2 .1 7 . 1.3 X 1027. 2 .1 8 . 2.1 X 10- 21 J. 2 .1 9 . 3.6 X 1021. 2 .2 0 . B y 1.7% . 2 .2 1 . 4 .4 6 X 1025, 148 g. 2 .2 2 . The therm odynam ic tem perature and the k in etic energy have increased fourfold.
Answers
331
3 .1 . A bout 1.2 1. 3 .2 . A bout 64.4 kP a. 3 .3 . 323 K. 3 .4 . 614 1. 3 .5 . 1.82 m3. 3 .6 . 44 X IO-3 kg/m ol, carbon d ioxid e. 3 .7 . 1.2 m ol. 3 .8 . About 2 kg. 3 .9 . 1.4 m ol. 3 .1 0 . 4.7 X IO-3 m3. 3 .1 1 . 2.6 tim es.
T
0
Fig. 153 3 .1 2 . A bout 3.5 kP a. 3 .1 3 . 513 K. 3 .1 4 . 1.62 m ol, 195 kP a. 3 .1 5 . 3 MPa. 3 .1 6 . 58 X 10-3 k g/m ol, 2.2 X 1025. 3 .1 7 . 932 K . 3 .1 8 . H yd ro gen. 3 .1 9 . I l l kg. 3 .2 0 . 606 K. 3 .2 1 . 73.3 M Pa. 3 .2 2 . A bout 105 Pa. 3 .2 3 . 1.2 k g/m 3. 3 .2 4 . 0.46 kg/m 3, 9 .6 X 1021 m -3, 444 m /s. 3 .2 5 . 28 X P IO-3 kg/m ol, nitrogen. 3 .2 6 . 1.66 kg. 3 .2 7 . 0.7 M Pa. 3 .2 8 . 2 X 105 Pa. 3 .2 9 . See F ig. 153. 3 .3 0 . 53. 3 .3 1 . See F ig. 154; (1) in joules, (2) isotherm 2 for the second equation w ill lie further from the F -ax is since it corresponds to a higher tem per ature. 3 .3 2 . 296, 963, and 17.6 g. 3 .3 3 . See Fig. 155. 3 .3 4 . Isobar 1 corresponds to the process occurring at a higher pres sure. An arbitrarily chosen tem perature T 1 corresponds to tw o va lu es Vx and V2 0 V of the volum e, and hence to two v a lu es p x and p 2 of pressure. B ut the product of F l£* pressure and volum e at a con stan t tem perature is con stan t, i.e . pxV1 = p 2Va. Therefore, volu m e V1 on isobar 1 corresponds to a higher pressure. 3 .3 5 . The gas lea k s. 3 .3 6 . 8.5 1. 3 .3 7 . To 225 K. 3 .3 8 . B y 82 K. 3 .3 9 . See F ig. 156. 3 .4 0 . 38.3 kN . 4 .1 . 250 J. 4 .2 . 380 J. 4 .3 . A lu m in iu m , 1.9. 4 .4 . 4187 J, 460 J. 4 .5 . The specific heat of copper is larger than th at of alu m in iu m , the sp e cific heat of lead is the sm a llest of the three. 4 .6 . Copper w eig h t w ill be heated to a higher tem perature. N o, it does n ot. 4 .7 . The energy required to heat a gas at con stan t pressure is spent both to increase the internal energy of the gas ana to do work of exp an sion . 4 .8 . 25.5 MJ. 4 .9 . 1 kg. 4 .1 0 . 1.73 X 108 J. 4 .1 1 . 1026 J. 4 .1 2 . 86 1.
332
Answers
4 .2 6 . 24 K . 4 .2 7 . 0.7 K . 4 .2 8 . 1.9 X 106 m3. 4 .2 9 . 225 MJ. 4 .3 0 . 187 K . 4 .3 1 . To about 78 m. 4 .3 2 . A bout 0.2 K . 4 .3 3 . 0.11 K. 4 .3 4 . 25. 4 .3 5 . 19 K . 4 .3 6 . 174 m /s. 4 .3 7 . 250 K . 4 .3 8 . 29% . 4 .3 9 . 30% . 4 .4 0 . 1029 kg. 4 .4 1 . A bout 24% . 4 .4 2 . A bout 17 kW . 4 .4 3 . A bout 550 km . 4 .4 4 . 4.2 8 kJ. 4 .4 5 . 26.4 k J . 4 .4 6 . 831 J. 4 .4 7 . 7462 J, 1662 J . 4 .4 8 . 48.1 k J. 4 .4 9 . 32 X 1 0-3 k g/m ol, oxygen . 4 .5 0 . 416 J. 4 .5 1 .
(a)
IV
Ic)
Fig. 155
106 J. 4 .5 2 . 1.8 kJ. 4 .5 3 . 8.31 J; no, it does not. 4 .5 4 . 0.31 kg. 4 .5 5 . Y es, it can, for exam ple, during ad iab atic expansion or com p ression . 4 .5 6 . The pressure w ill change more during the ad iab atic com pression. 5 .1 . T h is can be done if the pressure is reduced som ehow . 5 .2 . No, th ey are not; the internal energy of steam is higher. 5 .3 . In such vessels, evaporation takes place both from the w ater surface and through the v essel w a lls, w hich con tain pores to fa c ilita te co o lin g . 5 .4 . The pressure gauge in d icates the ex cess pressure over th e atm ospheric pressure. 5 .5 . The num ber d en sity of m olecu les in creases. 5 .6 . A t n igh t, when the air tem perature fa lls con sid erab ly, th e w ater vapour in the air saturates it and p a rtia lly con d en ses to form w ater drops, i.e . m ist. 5 .7 . T hese substan ces h ave different b o ilin g p o in ts. 5 .8 . B o ilin g w ater quenches fire more q u ick ly . 5 .9 . N o, it is n o t. 5 .1 0 . The work of expan sion of the gas is done a t the expense of a decrease in the internal energy of the gas, as a resu lt of w hich the tem perature of the released gas drops, ana the v a lv e is covered by frost. 5 .1 1 . The differ ence betw een the h u m id ities of air in the tw o clim a te s determ ines different evaporation rates from th e sk in . 5 .1 2 . N o, it is n ot. In the
Answers
333
critica l sta te, the specific la ten t heat of vap orization is zero for a ll liq u id s. 5 .1 3 . 300 J. 5 .1 4 . The energy liberated during the conden sation of w ater vapour is about eigh t tim es more than th a t lib erated during the condensation of mercury va pour. 5 .1 5 . 11.3 MJ, 2.1 MJ. 5 .1 6 . 4.0 k J. 5 .1 7 . 2.6 X 105 J. 5 .1 8 . 100 g. 5 .1 9 . 36.6°C. 5 .2 0 . 53 kg. 5 .2 1 . A bout 10°C. 5 .2 2 . 71.6°G. 5 .2 3 . A bout 3.3 kg, see F ig. 157. 5 .2 4 . 84 kJ. 5 .2 5 . 2.25 X 106 J/kg, 0.01 X 106 J /k g, 0.44% . 5 .2 6 . 334 g. 5 .2 7 . 2.32 kJ. 5 .2 8 . 0.04 m3. 5 .2 9 . 17 m in. 5 .3 0 . 51.8 kg. 5 .3 1 . 6.5 kg. 5 .3 2 . 35% . 5 .3 3 . 10~2 k g/m 3, 78% . 5 .3 4 . 19.4 X IO" 3 kg/m 3, 22°C. 5 .3 5 . 17.3 X 10- 3 kg. 5 .3 6 . 1°C. 5 .3 7 . At about 7°C. 5 .3 8 . 9.4 mg, 65% . 5 .3 9 . To 284.5 K. 5 .4 0 . 9.4 X 1 0~3 k g/m 3, 48% . 5 .4 1 . A t 4°C. 5 .4 2 . 18°C. 5 .4 3 . 60% , about 11 g. 5 .4 4 . 17°G, 13°C. 5 .4 5 . 19°C, the relative h u m id ity increases. 5 .4 6 . 0.4 m3. 5 .4 7 . It decreases to 54% . 5 .4 8 . 1.21 kg, 271 g. 5 .4 9 . 2.33 kPa. H i n t . p s = p a/ £ , p a can be determ ined from the eq uation of state: p a = m R T / M V , p s = m R T / M V B . 5 .5 0 . 12.3 X 10 ~3 k g/m 3, about 48% . H i n t . Solve the problem by using the equation of state. 6 . 1 . The surface tension changes: it decreases when soap is placed in w ater and increases when sugar is placed. 6 .2 . Under the action o f the forces of surface tension, a p ellet acquires the shape for w hich the surface area is at a m inim um . 6 .3 . Y es, it w ill. 6 .4 . W ater is a n on w ettin g liq u id for a surface covered w ith grease. 6 .5 . See answer to P roblem 6.4 . 6 . 6 . T h is is done to create c o n d itio n s in w h ich the solder is w ettin g. 6 .7 . A n o n w ettin g liq u id . 6 . 8 . T h is is done in order to prevent m oisture from risin g in the ca p illa ries of the b u ild in g w a lls 6 .9 . The presence of ca p illa ries in caked so il m akes it p o ssib le for w ater to rise closer to the surface. 6 .1 0 . See answer to Problem 6 .9 . 6 .1 1 . 9.6 X IO*3 N . 6 .1 2 . 8.4 X 1 0" 3 N . 6 .1 3 . 1.37 X 1 0 -3 and 0.12 N .6 .1 4 . 1.56 X 1 0 ~3 N , tow ards the w ater. 6 .1 5 . 0.105 N . 6 .1 6 . 1.44 X 10- 4 J. 6 .1 7 . 7.2 X 10"4 J. 6 .1 8 . 3.15 X IO" 2 N /m . 6 .1 9 . 0.022 N /m , e th y l alcoh ol. 6 .2 0 . 2.9 cm . 6 .2 1 . 2.56 mg. 6 .2 2 . 15 mg. 6 .2 3 . In the test tube of pure w ater. 6 .2 4 . 1.14 X 103 k g /m 3. 6 .2 5 . 0.7 cm . 7 .1 . Strong h eatin g m ay recry sta llize ste e l, w hich m akes its m echan ical properties w orse. 7 .2 . The shape of the sin g le cry sta l w ill change. 7.3. Glass is b r ittle . 7.4. T h e m eltin g p o in t of lead is considerably low er than the m eltin g p o in ts of other m etals. 7 .5 . Lead can be m elted by h eating it in a h erm etically sealed vessel w ith w ater. The temper ature of the w ater can then be raised con sid erab ly above the m eltin g p oin t of lead. 7 .6 . In sum m er, the tem perature of the la y ers of air c lo se to the surface of the E arth is above 0°C, and sm all ice c ry sta ls formed in the upper cold la y ers of the atm osphere m elt on th eir w ay
334
A nswers
to the ground. Larger cr y sta ls have no tim e to m elt and reach the ground in the form of h a il. 7 .7 . The m eltin g p oin t of the m ixture of snow and com m on sa lt is low er than 0°C, and hence snow m elts even below 0°C. 7 .8 . 0.4 cm . 7 .9 . 1.5 mm, 7.5 X IO-4. 7 .1 0 . 4 mm, about 1.4 J. 7 .1 1 . It w ill decrease by a factor of four. 7 .1 2 . 9 mm. 7 .1 3 . 3 X 107 Pa. 7 .1 4 . 1.07 X 1011 Pa. 7 .1 5 . 5 X IO" 3 m. 7 .1 6 . 9.4 kN . 7 .1 7 . 25. 7 .1 8 . 7.5 cm 2. 7 .1 9 . A bout 4 .2 km . 7 .2 0 . 3 J. 7 .2 1 . 0 .3 mm.
7 .2 2 . N o, it w ill not. 7 .2 3 . A lcohol has a low er freezin g p o in t. 7 .2 4 . A bout 159 kJ. 7 .2 5 . A bout 380 kJ. 7 .2 6 . 25.1°C. 7 .2 7 . Q ] : Q t = pl^i : ptA,t = 0.67. 7 .2 8 . 320 K, see F ig. 158. 7 .2 9 . A bout 12 g, see F ig. 159 . 7 .3 0 . 3.5 X 105 J/k g. 7 .3 1 . 19 kg. 7 .3 2 . 227 kg. 7 .3 3 . 17% . 7 .3 4 . A bout 80 kg. 7 .3 5 . 4.5 W. 7 .3 6 . 20% 7 .3 7 . 20 g. 7 .3 8 . 0.9 kg. 7 .3 9 . 0 .8 . 7 .4 0 . 3.35 X 105 J/k g. 7 .4 1 . 97 g. 8 .1 . The a llo y has a very sm all coefficient of lin ear exp an sion and it is used to ensure th at the m echanism is accurate in d ep en d en tly of the therm al con d ition s. 8 . 2 . The o val hole a llo w s for lin ear ex p a n sion during w hich the gap betw een the ra ils is reduced. 8 .3 . D ifferent coefficients of linear exp an sion w ould weaken the tu b e during it s operation (when it is heated); no, it cannot. 8 .4 . It w ill in crease. 8 .5 . The sleeve should be h eated . 8 . 6 . The relation is approxim ate. H i n t . We w rite the expression for a u n it volum e after h ea tin g a body through I K : 1 + 0 = (1 + a )3. A fter raisin g the right-hand sid e to the third power, we can n eglect the term s 3 a 2 and a 3 in view of their sm alln ess, w hich g iv es P = 3 a . 8 .7 . T h is is due to the r e la tiv e ly sm all therm al expansion coefficien t. 8 . 8 . Therm al e x p a n siv ity should be taken in to account. I ts v alu e for liq u id s is m uch larger than th at for the m aterial of w hich the tan k s are m ade. ,8 .9 . D uring h eatin g, stresses m ay em erge in teeth , w hich cau ses crack s in the en am el. 8 .1 0 . The strip w ill bend tow ards the m etal w ith the sm aller co efficien t of linear expansion; such strip s are used in therm al rela y s used to control tem peratures a u to m a tica lly . 8 .1 1 . B y 12 mm . 8 .1 2 . 10 K. 8 .1 3 . T ungsten. 8 .1 4 . 533.128 and 532.872 m. 8 .1 5 . A bout 903°C. 8 .1 6 . B y 417 K. 8 .1 7 . To 391 K . 8 .1 8 . The a lu m in iu m com ponent is 330 m longer than the steel com p on en t. 8 .1 9 . 1.99 m. 8 .2 0 . 739.2 k N . 8 .2 1 . 15 K . 8 .2 2 . 79.2 k N . 8 .2 3 . 500QC. 8 .2 4 . A bout 4 k J . 8 .2 5 . 37 k N . 8 .2 6 . 16.2 cm 3. 8 .2 7 . T h is relation is ap p roxim ate. 8 .2 8 .
A nswers
335
0.306 m2. 8 .2 9 . 46 K. 8 .3 0 . 0.09 1, 0.087 1. 8 .3 1 . 0 .08 1. 8 .3 2 . 1.85 X 103 kg/m 3. 8 .3 3 . 400 K. 8 .3 4 . A bout 3.6 kg. 8 .3 5 . 299 K. 8 .3 6 . 9.48 1, 7.49 kg. 8 .3 7 . 9.49 l t 7.5 kg. 8 .3 8 . 57.8 m3, about 40.5 t. 8 .3 9 . 2.14 m 3. 8 .4 0 . A bout 3.6 m3. 8 .4 1 . 1.8 X IO"4 K - 1. 8 .4 2 . 42.5 kJ.
K
9 .1 . Two. 9 .2 . + 4 . 8 X 1 0 ~19 C. 9 .3 . N o, it can n ot. 9 .4 . T h is can be done if the end of the rod held in the hand is coated w ith an in su la t in g m aterial. 9 .5 . The sphere should be brought in co n ta ct w ith tw o sim ilar neutral spheres. 9 .6 . The sta tic charge th at m ay be formed as the car m oves is conducted to the earth through the ch ain . 9 .7 . Y es, it can. 9 .8 . The surface charge d en sity w ill increase. 9 .9 . A t first the b all w ill touch the e le c trica lly charged rod and then w ill be re pelled from it. 9 .1 0 . No, it w ill n ot. 9 .1 1 . 0, 2. 9 .1 2 . 1.44 m N , by a factor of 81. 9 .1 3 . 2.1 X IO" 8 G. 9 .1 4 . 9 G N, 4 .5 G N . 9 .1 5 . 95 k m .
9 .1 6 . (1) 6-9, (2) in kerosene and w ax paper, (3) to w ater. 9 .1 7 . Yes; it w ill decrease by a factor of 3.2. 9 .1 8 . The drops w ill approach each other w ith an in itia l acceleration of 3.4 m /s2, w hich w ill increase further as the drops approach each other. Y es, it w ill. 9 .1 9 . 1.84 X 10- 7 N . 9 .2 0 . 3.45 X 1011, about 0.01 N, yes, it w ill; see F igs. 160a and b. 9 .2 1 . 2.25 X IO"4 N . 9 .2 2 . 5 X 1 0 1? C/m2. 9 .2 3 . 4 X IO" 6 C. 9 .2 4 . 4 X 10~7 and 2 X 10~7 C, 0.42 m. 9 .2 5 . 4 .2 X 1042. 9 .2 6 . 1.77 X 10’ 11 F/m , 5.2 X 10" 7 C. 9 .2 7 . No, th ey do n ot. 9 .2 8 . T h is is done for electrostatic sh ield in g . 9 .2 9 . (a) yes, th ey can, (b) no, th ey can n ot. 9 .3 0 . The field strength w ill be greatest where the curvature of the surface is the largest, nam ely, at a p o in t. For th is reason, charge m ay leak from the p oin t, form ing an “electric w in d ”. 9 .3 1 . See the answ er to Problem 9.30. 9 .3 2 . 0 . 9 .3 3 . It m eans th at the electric field acts on a charge of one coulom b placed at th is p oin t w ith a force of 300 N. 9 .3 4 . 7 X 103 V /m . 9 .3 5 . 1.6 X IO" 5 N . 9 .3 6 . 3 .2 X 10~14 N . 9 .3 7 . 3 X 104 V /m . 9 .3 8 . 2.2. 9 .3 9 . 1.44 X 105 V/m ; the field stren gth w ill be doubled. 9 .4 0 . 3.75 X 104 V /m , 42 nC. 9 .4 1 . 104 electrons. 9 .4 2 . C oncentric spherical surfaces. 9 .4 3 . N o, it w ill n ot. 9 .4 4 . 9 X 103 V /m , 2.7 X 104 V /m . 9 .4 5 . 1.71 X 104 V /m . 9 .4 6 . A bout 3.6 X 105 V /m . 9 .4 7 . 1.13 kV /m . 9 .4 8 . (a) 0 , (b) 0 , (c) o /e 0e * 9 .4 9 . 0.1 mg. 9 .5 0 . 9.93 m /s2. 9 .5 1 . 0, ] / 2
336
Answers
9 .5 5 . 38 V. 9 .5 6 . 3 X 10~4 J. 9 .5 7 . 6 kV, 6 k J. 9 .5 8 . 1.44 X 106 V. 9 .5 9 . 2.4 X 10“5 J, 150 V. 9 .6 0 . 2.56 kV, 4.1 X 10 ~16 J. 9 .6 1 . 3 .5 X 104 V /m , 2.4 kV. 9 .6 2 . 36 V . 9 .6 3 . 1.8 kV. 9 .6 4 . 8 cm , 1 0 “7 s. 9 .6 5 . Y es, it can be changed by bringing another conductor clo se to the charged one. 9 .6 6 . 2 C. 9 .6 7 . 9 X 106 km, 7?s p h / ^ E a r t h ^ 1400. 9 .6 8 . The capacitance, charge, and energy of the cap acitor w ill in crease, w h ile its voltage w ill rem ain unchanged. 9 .6 9 . 1.1 X 10~u F, 1.1 X 10~5 pF, 11 pF. 9 .7 0 . A bout 711 pF. 9 .7 1 . 4.5 cm , 3 X 1 0" 7 C/m 2. 9 .7 2 . The oxide film used as the dielectric in th ese cap acitors is very th in . 9 .7 3 . A bout 455 and 1786 V; the charge from the sm aller sphere w ill be transferred to the larger sphere u n til the p o ten tia ls becom e equ al. 9 .7 4 . A bout 100 pF. 9 .7 5 . 1.8 mm . 9 .7 6 . 6.6 X 1 0" 10 C. 9 .7 7 . 1.1 X IO-7 C; it w ill decrease by a factor of six . 9 .7 8 . 6 mm. 9 .7 9 . 5 X IO" 9 F. 9 .8 0 . 3.8 X 10" 5 J. 9 .8 1 . (1) 1.09 pF, 2.2 X IO" 2 J, (2) 12 pF, 0.24 J. 9 .8 2 . 0.6 pF. 9 .8 3 . 0.24 pF, 2.4 X 10~5 C. 9 .8 4 . 300 V . 9 .8 5 . 100 V, IO" 2 C, 0.5 X 10~2 C. 9 .8 6 . 1 pF. 9 .8 7 . The ca pacitor w ith the higher capacitance. 9 .8 8 . 36 J, 13 kW . 9 .8 9 . 4.8 pF, 9.6 X 10 ~4 C.
9 .9 0 .
In the second case.
4 Tf F
F r2
9 .9 1 . C = — ~ — ; Ar
for
vacu u m , 555 pF. 9 .9 2 . 0.215 J /m 3. 1 0 .1 . 7.5 X 1016. 10.2 . 6.3 X 1021. 10.3. 103 A. 10.4 . 3 X 1023 cm"3. 1 0 .5 . 1.1 X 10-4 m /s. 1 0 .6 . 3.6 X IO" 2 V /m , 518 m. 1 0 .7 . 5.4 X 10-4 m 2/(V -s). 1 0.8. 2.5 X 104 A /m 2; y es, it is. 10.9. 2 A /m 2. 10.1 0 . A bout 23 mm. 10.11. 6.2 A /m m 2. 10.12. 0.054 Q; it w ill be half. 1 0.13. N ichrom e w ire, 2.5 tim es. 10.14. The current in the tungsten w ire is 1.87 tim es larger than that in the steel w ire provided th at the internal resistance of the accum ulator can be neglected in v iew of it being sm all. 10.15. The resistance decreased by a factor of four. 10.16. 12 pF. 10.17. 1.2 A, about 42 Q. 10.18. 272 m. 10.19. 60 Q. 10.20. 736 m. 10.21. 2.7 mA. 1 0 .2 2 . 3 V. 10 .2 3 . 2.7 V, 2.67 A /m m 2. 1 0.24. 15 mm2. 10.25 . 7.5 k N . 10.26. 0.017 Q. 10 .2 7 . The resistance of the nichrom e w ire is larger by a factor of 3.125. 1 0 .2 8 . S teel. 10.29. W ill increase by 3.24 Q for ca st iron and by 0.34 Q for the ferroalum in ium h igh -resistan ce a llo y . 10.30. The tem perature resistance coefficient m ust he as large a sp o ssib le . 10.31. 230 Q. 10.3 2 . 25 K. 10.33. 6 Q. 10.34. 250 K. 10.3 5 . 558 V. 10.36. 146 m m 2. 10.37. A bout 1260 kV. 10.38. / 2 > / j , the resistan ce w ill increase; the current in the circu it w ill decrease. The readings on am m eters A x and A 2 w ill be the sam e. 10.39. 4 Q. 1 0 .4 0 . 0.033 Q. 10.41. 22 Q, 10 A, *7, = 50 V, U 2 = 70 V, U 3 = 100 V. 1 0 .4 2 . 20Q. 10.43. 24Q, 3Q. 10 .4 4 . 3.24, 2.12, 1.2, and 0.37 A. 10.45. 0.03Q . 10.46. Into tw o parts; in parallel. 10.47. 25 Q. 10.48. 44 Q, 5 A and 0.5 A. 10.49. 0.64 Q, 3.2, 1.6, and 0.2 A. 10.50. 1.8 Q and 1.2 Q. 10.51. 0.55 A, 0.275 A, 1.65 A, and 0.275 A. 10.52. 6 9 , / 1 = 2 A , I 3 = / 4 = 1 A. 10.53. 2.1 Q, 21 V, U x = 1 2 V, I 3 = 7.5 A, U 3 = 9 V, I 2 = / 4 = / 5 = 2.5 A, U 2 = U* = U b = 3 V. 10.54. The resistan ce of the cold filam ent is low er than that of the hot filam ent. C onsequently, the sw itch-on current is larg er than the sw itch -off current. 10.55. 4 Q, 3 A. 10.56. 508 V, 488 V, 475 V. 10.57. 27 V. 10.58. 4 .5 Q, 2 Q, 1.5 Q, 0.67 Q, 0.46 Q, 0. 10.59. 0.78 Q. 10.60. 2.6 Q. 10.61. 1 A, 0.5 and 0.5 A. 1 0 .62 . 440 Q, 110 V, 0.25 A. 10.63. 146 V, 73 V , 0 .33 A, 0.17 A. 10.6 4 . 0.014 Q. 10.65.
A nswers
337
A bout 2.3 Q, 52.5 A, I x = 22.5 A, / 2 = 30 A, / 3 = / 4 = A = 7.5 A. 1 0.66. 20 A, / j = 5 A, / 2 = 15 A, / 3 = / 4 = 2.5 A, / 5 = 1 7 = 10 A, /« = 7.5 A. 10.67. 2.5 X 1 0 -3 Q. 1 0 .6 8 . 60 A. 1 0 .6 9 . 400 Q. 1 0 .7 0 . 10 kQ. 10.71. 250 Q, 225 V . 10.72. 17 m. 1 0 .7 3 . 5. 1 0 .7 4 . I t m easures the voltage of the extern al circ u it. 10.75. 1.5 V, 0.24 V. 1 0 .7 6 . 1.93 V, 0.25 Q. 1 0.77. 0.29 A, 1.02 V, 70% . 1 0 .7 8 . 230 V , 229 V . 1 0 .7 9 . 10 A. 1 0.80. 0.1 Q. 10.81. IO" 5 C. 10 .8 2 . 1.5 X 1 0“ 6 C. 10 .8 3 . 12 V , 2 Q. 10 .8 4 . 147 V, 120 V . 1 0 .8 5 . 0.6 A, 3 A. 1 0 .8 6 . 225 V . 1 0 .8 7 . 236 V , 220 V . 1 0.88. 222 V, 21 m m 2. 10 .8 9 . 535 V. 1 0 .9 0 . 230 V , 195 V. 10 .9 1 . 2 A, 540 V . 1 0.92 . (a) 1.2 A, (b) 17.5 A, (c) 4.7 A. 1 0 .9 3 . 0 .29 A, 0 .8 A. 10.94. A bout 3.6 A. 10.9 5 . 0 .6 Q. 1 0 .9 6 . The internal resistan ce of acid accum ulators is very sm all, and therefore a large current em erg in g in a circu it m ay cause the p la te to break down. 1 0 .9 7 . 2.5 A. 10.98. 1.25 A, 0.75 A. 10.99. / p a r s e r = 3. 10.100. T hey should be connected in tw o parallel groups of three series-con n ected c e lls. 10.101. T hey sh ould be con n ected in three parallel groups of four se ries-connected ce lls. 10.102. 1.3 V, the em f w ill be reduced by a factor of seven. 1 0.103. S eries connection. 10.1 0 4 . It is more ad van tageou s when the external resistan ce is higher than the internal one. 1 0 .1 0 5 . 52 V, 50 V. 10.106. 3.42 V, 3.36 V. 1 0.107. 0.4 Q. 1 0.108. 0.77 A, I .2 6 V . 10.109. / =
- r ri ~ t - 2-r — r
rir2 + R ('i + r2)
= 1.24 A .
I I .1 . 18 C, 216 J. l i . 2 . 317 MJ, 88 kW h. 1 1 .3 . 1125 kW h, 5 ru b les 76 kopecks. 1 1 .4 . 91 %, 1 ruble 6 kopecks. 1 1 .5 . 800 MW, 7 X 10» kW h. 1 1 .6 . 176 kW h. 1 1 .7 . 234 M J, 65 kW h. 1 1 .8 . 48 W . 1 1 .9 . 4 Q, U c D = 0, / , = 0, 144 W. 11 .1 0 . 54.5 A. 1 1 .1 1 . 8.4 kW . 1 1 .1 2 . 1 .35 MW. 1 1 .1 3 . 2.7 MW. 11 .1 4 . (a) 13.8 MJ, (b) 8 .64 X 10s J, (c) 3.46 MJ. 11 .1 5 . 82.5 kW , 74.25 kW . 11 .1 6 . 102 kW , 34 kW . 1 1 .1 7 . 125 kW , 110 kW , 0.8 8 . 11 .1 8 . 550 W, 5.5 kW h. 1 1 .1 9 .2 .8 m m 2. 1 1 .2 0 . The power is higher for the p arallel con n ection . 1 1 . 2 1 . 16. 1 1 . 2 2 . 0.16 m /s, 12.5 s. 11.23. 63.4 km /h. 1 1 .2 4 . 462 A. 11.2 5 . The m eltin g p o in t of lead is rela tiv ely sm all (327°C). 11 .2 6 . The fu ses have different cro sssection al areas. 1 1 .2 7 . The co il of the 127-V hot p la te is th ick er. 1 1.28. 570 MJ. 11.29. (a) 110 V, 60 W, 0.22 MJ, (b) 88 V , 132 V; the 40-W bulb is under a v o lta g e higher than the nom inal value; 38.4 W, 57.5 W, 0.14 MJ, (c) 138 V, 82 V; the 60-W bulb is under a v o lta g e higher than the nom inal value; 94 W, 56 W, 3.38 X 105 J, 0 .2 MJ. 11.30. 26 g. 11.31. 79% , 0 .8 kopecks. 1 1 .3 2 . 2 Q. 1 1 .3 3 . 24 V, 0.1 Q. 11.34. 2.64 kW , 90% . 11.35. The w ire should be shortened to 3 m. 11.36. 20 or 80 Q. 11.37 . 15 kg. 1 2 .1 . N o, we cannot. 12.2 . N o, it can n ot. O therw ise, the o x id e film w ould be destroyed by e lectro ly sis. 1 2 .3 . Current reversal m eans th a t its direction changes. W hen the a rticle becom es an anode during the current reversal, the uneven regions (protrusions) are d isso lv ed more v igorou sly and the edges or surfaces are sm oothed over; y es, it can. 1 2 .4 . 9.32 X 10- 8 kg, 1.118 X IO 6 kg, 0.33 X IO" 6 kg, 6.25 X 1018. 1 2 .5 . 283 m g. 1 2.6. 362 g. 1 2 .7 . 3.3 X 1 0-7 kg, 6.6 X 10" 7 kg. 1 2 .8 . 2 .96 X IO*4 kg. 12.9. 3.33 X 10~7 kg/C , 0.004 X IO" 6 kg/C, 1.2% . 12 .1 0 . Copper. 12.11. S ilv er, 1. 12.12. 1.6 X 10~19 C, 3.2 X IO" 19 C, 4.8 X 10 - 19 C. 12.13. 1.118 X 10~6 kg/C , 6.8 X 10~7 kg/C. 12.14.
\
22 —0 530
338
Answers
2.24 X 1022. 12.15. 4.13 X 1 0~3 kg, 1.88 X 1022, 1.26 X 1022. 12.16. 11.8 kg. 12 .1 7 . 64.5 G J, 0.12 Q. 12 .1 8 . 278 A /m 2. 1 2 .1 9 . 9.6 5 , 18.4 g. Iron is liberated at the cathode and chlorine a t the anode. P o sitiv e m etal ion s m ove to the cath od e, w h ile n eg a tiv e ch lorin e ions m ove to the anode. 12.20. 9.4 h. 12.2 1 . 52 m in. 1 2 .2 2 . 2.6 X 104 kW h, 520 rubles. 12 .2 3 . 4 .5 kg. 1 2 .2 4 . 12.5 h. 1 2 .2 5 . 2.7 X 10“9 m /s. 12 .2 6 . 312 K . 12.27. N o, it w as n ot. A correction of 0.1 A is required. 12 .2 8 . 1.045 X IO' 8 kg/C, 2.38 X 10‘ 7 kg/C, 1.26 X 10" 7 kg/C. 1 2 .2 9 . 50 C, 56 m g. 1 2 .3 0 . A bout 29 W . 1 2 .3 1 . 369 m g. H i n t . See F ig. 161. The area of the trapezium is equal to the am ount of e le c tr ic ity p a ssin g through th e electro ly te from the m om ent w hen the current sta rts changing. 1 2 .3 2 . A bout 28.2 A /m 2. 12 .3 3 . 1.9 X 1022. 1 2 .3 4 . » PoI =
U -
£
R = 0.8 V.
1 3 .1 . A s a result of recom bination, charged p articles are transform ed in to neutral atom s, and the gas becom es an in su lator. 13.2 . OA is the region of dynam ic eq u ilib riu m betw een ion iza tio n and recom bi nation (the region where Ohm ’s law is ap p ro x im a tely v a lid ), A B is the region of the saturation current (the current depends o n ly on the in ten sity of the ionizer), and B C is the region of im pact ion ization (self-su sta in ed discharge). 1 3 .3 . (1) Charge carriers in a gas are formed by an ionizer, w h ile in so lu tio n s th ey are form ed due to the effect of a so lv en t w ith therm al m otion taken in to ac count. (2) G ases ex h ib it electron and ion con d u c tiv itie s, w h ile so lu tio n s o n ly have ion c o n d u c tiv ity . (3) C o n d u ctiv ity in a gas does 300 t , s n o t ob ey Ohm ’s law . 1 3 .4 . The tem perature should be increased. 1 3 .5 . N o, it is n ot. For F ig. 161 a sm all mean free p ath , the electric field stren gth should be n igh . 1 3 .6 . The large cu rren t in the spark reduces the volta g e across the electrod es, w hich interrupts the discharge. 1 3 .7 . (a) Coronas formed around h ig h -v o lta g e power lin es; (b) the corona d ischarge can be used as a filtre for the purification of flue gas. 1 3 .8 . G low discharge; electrons, gas ion s, and m ercury vapour. 1 3 .9 . When the gas is being rarefied, the mean free path increases, and hence the energy W \ = e E k required for ion ization can be acquired at a low er electric field strength. 13.10. The a ctiv a tio n of the cath od e (coatin g it w ith a la y er of a more activ e m etal) reduces the work fu n ction of cathode electron s and saves energy. 13.11. The horizontal lin e s of the graph correspond to saturation cur rents for different tem peratures of the filam ent. A s the tem perature of the filam ent increases, the em ission of electrons is increased. 13.1 2 . The electron beam can be con trolled by u sin g electric or m agnetic fields. In CRT tubes, th is is done u sin g p a ra llel-p la te cap acitors. Current-carrying c o ils on the neck of the tube are used to produce m agnetic fields. 1 3.13 . In a gas under norm al c o n d itio n s, charge carriers are absent, or th eir num ber is n e g lig ib ly sm all. Plasm a is a m ixture of elec trica lly charged p articles, such that the to tal n egative
A nswers
339
charge of the p articles is equal in m agnitude to the to ta l p o sitiv e charge. The presence of charged p a rticles in the plasm a ensures its electrical co n d u ctiv ity . T h is cannot be said of a gas. 13.14. 4 .3 V. 13.15. It w ill increase by a factor of 1.2, 1.6 X 107 m /s, 1.9 X 107 m /s. 13.16. 1.03 X 108 m /s. 1 3 .1 7 . 1.03 X 107 m /s, 9.7 X 10 ~10 s. 13.18. 7.95 X 105 m /s. 13.1 9 . 2.18 X 106 m /s. 13 .2 0 . 6.4 X 1 0 -14A /m 2. 13.21. 3 pm . 13.22. 1.26 X 1013 m '3. 1 3 .2 3 . 1.46 X 10~4 m2/(V -s). 1 4.1. Curve 1. 1 4 .2 . E lectrons and h o les. 1 4 .3 . The num ber of h oles is equal to the num ber of electrons; no, it is n ot. 1 4 .4 . ra-type (p h os phorus is pentavalent); p -ty p e (alu m in iu m is triv a len t). 1 4 .5 . It w ill decrease. 1 4 .6 . It w ill increase by a factor of 2.4 X 104. 1 4 .7 . b = 2.5 X 10 ~3 m 2/( V -s ), n = 1018 m -3. 1 4 .8 . It is a sem icon d u ctor w ith term in als to be connected to a c ir c u it. For th ese th erm istors, the current does not depend lin ea rly on the applied v o lta g e. 1 4 .9 . T her m istors are based on the tem perature dependence of resistan ce, w h ile photoresistors make use of the dependence of resistan ce on illu m i nance. 1 4.10. A tran sistor is a sem icon d u ctor d ev ice in the form of a cry sta l co n ta in in g tw o p - n ju n ction s. T hese regions are the base, the co llector, and the em itter. 1 4 .1 1 . If the th ick n ess of the base is much sm aller than the mean free path, m in o rity charge carriers g e ttin g in to it have no tim e to recom bine. 1 4 .1 2 . / em = A> + ^col* 14.1 3 . The ap p lication of sem iconductor d ev ices in radio en gin eerin g saves energy and m akes ap pliances more com pact and durable. 15.1. If we m en ta lly d iv id e the cu rrent-carrying con d u ctor in to a large num ber of elem en ts, for each elem en t there e x is ts an elem en t in w hich the current has the op p osite d irection , and hence th ese tw o elem en ts repel each other. W ith such in teraction betw een the elem en ts, the conductor w ill assum e the circular shape. 1 5 .2 . C onnecting the
F ig. 162
F ig. 163
voltm eter to p o in ts 1 and 2 (F ig. 162), we can determ ine th e p oin t w ith a higher p o ten tia l. Then we can bring a m agnetic n eed le to one of the conductors and determ ine the direction of current from the deflection of the north p ole. 1 5 .3 . See F ig. 163. 1 5 .4 . See F ig. 164. 15 .5 . See F ig. 165. 1 5 .6 . See F ig. 166. 1 5 .7 . The m agnet w ill be re p elled by the solenoid and w ill rise. 1 5 .8 . The c o il w ill be deflected to the righ t. 1 5 .9 . For the in d icated d irection of the current, the loop w ill be rotated clock w ise if the north pole is on the right; through 90°. 15.10. 0.46 N . 15.11. 0 .8 m. 1 5 .1 2 . 50 A. 1 5 .1 3 . 2.8 m. 1 5 .1 4 . 0.4 N , 0.2 N , 0. 15.15. 1.2 X IO-2 T. 15.1 6 . We m ust bring the m agnet to the lam p. The filam ent in the a .c. cir c u it w ill o sc illa te and w ill
340
Answers
appear as blurred. 15.17 . 0 .42 N . 15.18. / = p S g / B ; from N to M . 15.19. 15.7 N. 15.20. 275 A. 15.21. 26 A /m , 3.25 X 1 0 '“ T. 1 5 .2 2 . 5 X IO' 5 T , 4 cm . 1 5 .2 3 . 7 X IO' 5 and 1 X IO' 6 T. 1 5 .2 4 . 3 .14 X IO' 5 T, 6 .9 cm . 15 .2 5 . 125 A /m , 10 A. 15.2 6 . 16 A, 16 A /m . 1 5 .2 7 . 1.1 X 10~l A *m 2. 15.28. A bout 2.4 A, 16 A /m . 1 5 .2 9 . 5.7 X 1 0 -6 N -m . 1 5 .30. 3 X 10- 4 T. 15.31. 7.5 X 103 A /m , 9.4 X IO' 3 T. 1 5 .3 2 .
s [
81
> >
Fig. 164 1.5 mm . 15.33. 800, 280. The p erm eab ility of steel decreases. When the m agnetization of a ferrom agnetic a tta in s saturation, the m agnetic in duction increases on ly due to increasing m agnetic field strength. 1 5 .34. 0.5 T, 1.5 mW b. 1 5.35. 7.9 X IO" 7 Wb. 1 5 .3 6 . 2 X IO"4 Wb, 2.4 X 10- 4 J. 15.37. 144 m W b. 15.38. 7.2 X 10"4 W b. 15.3 9 . 0 .2 T, fa)
lb)
Ld Id FI Fig. 165
FI
Fig. 166
100. 15.40. A t the in itia l m om ent, the force is directed v er tic a lly downwards; a circle. 15.41. 3 .0 X IO-16 N. 15.42. 1.2 m, 4 .0 X 10" 7 s, 2.5 M Hz. 15.43. The v e lo c ity vector must be perpendicular to the plane co n ta in in g vectors E and B. B y h y p o th esis, the electron m oves u n iform ly in a straigh t lin e, and hence F h = 0, v = E l B . 15.4 4 . 1.4 X IO*14 J. 15.45. 3 .9 m m , 4 .4 cm . 1 5 .4 6 . T %1 = T i2. 15.47. Ti > T c. 1 6 .1 . An em f w ill be induced in th e case (a). 1 6 .2 . W hen the p oles o f the m agnet are connected (disconnected), the m agnetic field in duction changes, w hich lea d s to the em ergence of an induced current. 1 6 .3 . From N to M \ from M to N . 1 6 .4 . An em f is induced when the fram e enters the m agnetic field or lea v es it since in th ese cases the m agnetic flux piercing the fram e changes. 1 6 .5 . N o. 1 6 .6 . The current induced in the upper part of the frame is directed aw ay from us in the cases (a) and (d) and tow ards us in the cases (b) and (c). 16.7. The
A nswers
341
second rod is non m agnetized. 1 6 .8 . The current through the g a lvan o m eter flows in the upward direction. 1 6 .9 . The induced current has the counterclock w ise direction. 16.1 0 . W hen the m agnet en ters the c y l inder and em erges out of it, the m agnetic field produced by the current induced in the cylin d er decelerates the m otion. If the cy lin d er is lon g, a = g in it. 16.11. Y es, it is. The galvan om eter reading is 0 sin ce the em f’s induced in each h alf of the conductor are equal in m agnitude and have op p osite d irection s. 1 6 .1 2 . 3.4 X 1 0~2 V. 16.1 3 . 6 A. 1 6 .14. 1.4 T. 16.15. 0.46 V. 1 6 .1 6 . A bout 20 m /s. 1 6 .1 7 . 2 V, 0.5 A. 1 6 .18. 30°. 16.1 9 . 64 V. 1 6 .2 0 . 6 m W b. 1 6 .2 1 . 100 V . 1 6 .2 2 . 1.5 s. 1 6 .2 3 . 0 .8 V, 8 mW b. 1 6 .2 4 . 0.4 W b. 1 6 .2 5 . 0.1 H . 1 6 .2 6 . A core m ade of a ferrom agnetic m ust be inserted in to the solen oid . 16.27. A self-in d u ctan ce em f em erging during the d iscon n ection of the circuit has the sam e d irection as the em f of the source, and their jo in t action cau ses sparkling. A cap acitor connected to the sw itch elim in a tes th is effect. 1 6 .2 8 . 7 A. 1 6 .2 9 . A bout 2.7 A. 1 6 .3 0 . A bout 2.7 J. 16.31. 0.216 J, 0.036 W b, 0 .7 2 V. 16.3 2 . 1.5 X IO' 5 C. 1 6 .3 3 . 4 mA. H i n t . I = fc//?, % = — AO/A*, A
17.28. v = A © cos (©* +
342
Answers
variable since the restoring force varies all the tim e. 17.2 9 . Y es, they w ill. 17.30. The alum in iu m ball w ill com e to a h a lt first. 17.31. The acceleration for the am p litu d e value of the d isp lacem en t is m axim al and in the equilibriu m p osition is m in im al. The v e lo c ity has the m axim um valu e in the eq u ilib riu m p osition . 17.32. F = m g sin a = 6.93 X IO" 2 N . 17.33. 204 g. 17.34 . 303. 17.35. The period of o sc illa tion s of a pendulum depends on the free fall acceleration w hich is different at different la titu d es. 17.36. The len gth of the pendulum becom es sm aller w ith decreasing tem perature and hence tne period changes (the clock w ill be fast). The correct pace can be restored by changing the len gth of the pendulum by d isp la cin g its load . 17.3 7 . The period w ill double. 17.38. The len gth of the pendulum should be reduced by a factor of 6 .0 5 . H i n t . B y h y p o th esis, the periods of the pendulum on the Earth and on the Moon m ust be the sam e. E xp ressin g the p eriods by the form ula T = 2 n l /g for the co n d itio n s on the
Moon and on the Earth, we obtain ( / E a r t h ^ M o o n ) ( ? M o o n / * E a r t h ) = 1 . ^ E a r th /^ M o o n ^ £ E a r t h / £ M o o n = 6,05, ^M oon = ^ E a r t h / 6 . 05, 1 7 .3 9 , 2.004 s, 2.003 s, 2.008 s, 4 .9 2 s. 17.4 0 . 9 .82 m /s2. 1 7 .4 1 . (a) The clock w ill keep correct tim e, (b) In the freely fa llin g lift, w eig h tlessn ess sets in, and the restoring force v a n ish es. If the pendulum is in an extrem e p o sitio n , no o sc illa tio n w ill take place, w h ile in the e q u ilib riu m po sitio n the pendulum w ill u n iform ly rotate in the vertical plane. 1 7.42. (a) The force m ust be directed upw ards and equal to 0.75m g; (b) the force m ust be directed dow nwards and equal to 0.5625m g. H i n t . If the period increases under the action of force F , the acceleration im parted by th is force h as the m in u s sign . T 0 = 2 n Y l / g , 2 f 0 = 2 n Y l ! ( g — *j), 1/2 = y (g — g j / g , g = 4 (g — g j), g , = 0.75g. 17.43. G = mg, it is directed upw ards. 17.4 4 . (a) 0 .5 m g , a = 0.464 rad or a = 26°30', (b) 1.2m g, a = 0.876 rad or a = 5 0 ° 1 2 \ S o l u t i o n . The resultant force actin g on the ball can be found from the rightangled triangle (F ig. 168): F x = m a x ] / m 2g 2 + F 2, w hence a = V g 2 + F 2/ m 2. S o lv in g th is eq u ation together w ith the tw o eq u ations for the periods of o sc illa tio n s T 0 = 2 j i V l / g and k T 0 = 2 j i where k = T / T Q, we ob tain F = m g Y \ / { k * — 1). The equilibrium
Answers
343
p osition corresponds to tan a = F /( m g ) . 17.4 5 . T = 0 .9 9 T 0 (see solu tion of Problem 17.44). 17.4 6 . 1.9 s (see solu tion to Problem 17.44). 17.47. 0.9 Hz (see solu tion to Problem 17.44). 1 7 .4 8 . Y es, th ey w ill be harm onic; the force of g ra v ity changes the p osition of eq u ilib riu m . 17.49. 1.13 H z. 17.50. 126 N /m . 1 7 .5 1 . 0 .0 3 m, 31.3 rad /s, 0 .2 s, — n / 2 rad, * = 0.03 sin (31.3* — n / 2 ) . 17.5 2 . 8 X IO" 3 J. 0.14 m /s, 1 m /s2. 17.53. 6.32 J, the energy does not depend on the in itia l phase. 17.54. 0.192 J, 0 .6 m /s, 0.1795 J, 0.0125 J, x = 1.96 X 1 0 _2sin (31.6* — 0 .5 ji). H i n t . W hen the engine is sw itched off, the spring is stretched to the a m p litu d e q2 of the d isp lacem ent. In th is case, m>2g = k A . 17.55. x = 3 cos 0 .2 X sin(n* — 0.2ji), 3 c o s 0 .2 m , 0.5 H z, 2 s, —0.2ji rad/s, n rad/s. H i n t . B y u sin g the form ula for the sine of double angle sin 2a = 2 sin a cos a , the equa tion can be transform ed to th eco n v e n tion al form. 17.56. The pendulum clock is 47.6 s slow . S o l u t i o n . The m otion ^ '9 of the rocket can be divid ed into three parts (Fig. 169). On the first segt, m ent, the acceleration is directed upw ards, and a x = g; on the second segm ent, w hich la s ts u n til the engine Fig. 169 is sw itched off, the rocket fa lls free ly , and on the third segm ent of dece leration, the acceleration is directed upw ards. Let us first d eterm in e *2 and *3. In order to find *2, we sh all use the form ula for a u n iform ly accelerated m otion: h = h 0 + v0t + g t 2/2 . In the case under co n sid eration, h = H 2, h Q = H ^ and v 0 = vx. T aking in to account the sign of the acceleration and the q u a n tities vx = g t x and H i — g t \ l 2 , we can w rite the eq uation H 2 = g t \ / 2 + g t xt 2—g t \ / 2, w hence *2 = *i ~\r V 2 (*i — H J g ) = 70 s. Let us now consider the third segm ent of the path. Here v2 = vi — g t 2 = g (*2 — *2) = — 392 m /s, a 3 = — v y ( 2 H 2) = —8g, *3 = v 2/ a 3 = 5 s. On the first segm en t, the period of the pendulum clock is T = 2 n l /2 g = T j \ 2. The clock is = ^ i(l/ 2 — 1 ) = 12.4 s fa st. On the second segm en t, the pendu lum clock has zero speed, and on the third segm ent it is fast: T = T J 3, A*3 = *3 (3 — 1) = 10 s. In to ta l, A* = A*! + A*2 + A*3 = 12.4 s — 70 s + 10 s = —47.6 s. 17.5 7 . See Fig. 170. 1 7 .5 8 . 2 m, 3 m, 5 m (Fig. 171). 1 7 .5 9 . x = 0, x = — sin co*, 0, — 1 m (F ig. 172). 17.60. (a) 2 1 /2 m, (b) 5.4 m, (c) 5.4 m (F ig. 173). 17.61. (p^arctan 0.4 = 0.38 rad = 21°48' (Fig. 173c). 1 7 .6 2 . 3.6 m, 30°27' (Fig. 174). 17.63. The p article vib rates about its eq u ilib riu m p o sitio n . 17.6 4 . See F ig. 175. 17.65. (a) To the righ t, (b) to the le ft. 1 7 .6 6 . The pro pagation v e lo c ity depends on surface tension. 1 7 .6 7 . 0. 17.68. 16.7 m /s. 17.69. The v e lo c ity of sound is higher in iron; no, it cannot. 17.70. The sound em erges as a resu lt of friction betw een m etal sur faces in the hinge. H a v in g a larger surface, the door augm ents the
t
Answers
344
(c)
F ig. 174
_/v Fig. 175
345
Answers
vib ration s. 17.71. E lastic forces. 17.72. The v elo c ity of propagation of sound depends on tem perature and pressure. 17.73. An acou stic w ave propagating in the rail is p a rtia lly reflected from the surface of the rail inw ards, and hence the in te n sity of the sound does not atten uate as rap id ly as in the spherical w ave p ropagating in air. 17.74. The sound w ill not be detected. 17.75. The v ib ra tio n s of the tu n in g fork are transm itted through the surface of the ta b le w hich is m uch larger than the surface of the tu n in g fork. 17.7 6 . The sound from the tu n in g fork in con tact w ith the tab le w ill die aw ay sooner since the energy of the tu n in g fork is spent on the vib ra tio n of the tab le. 1 7.77. 22.7 cm -11.3 cm . 17.7 8 . 0.66 m. 17.7 9 . 0.5ji rad. 17.80. The air colum n in the cap w ill start vib ra tin g . 17 .8 1 . 1545 H z. 1 7 .8 2 . 6 cm . 17.83. The frequency of the sound w ill increase. 17.84. 1445 m /s, 3638 m /s. 17.85. The v alu es of v and T w ill not change, w h ile k w ill increase by a factor of 14.7. 17.86. 1200 m . 17.87. 2.04 mm, 5.1 mm. 17.88. 3 km . 17 .8 9 . 13.2 s. 17.90. 2 X IO" 3 %, 6 cm . 18 .1 . 0.02 s, 100 tim es. 1 8 .2 . 4.24 A. 1 8 .3 . N o, it can n ot. 1 8 .4 . 177 V, 314 rad/s. 1 8 .5 . 63.8 A, (314* + j i / 4 ) rad, j i / 4 rad, 50 H z. 1 8 .6 . 0, 126.6 V, 179 V, 89.5 V. - 1 5 5 V, — 179 V , — 126.6 V, 0. 1 8 .7 . - 3 A, 1.5 A. 1 8 .8 . (a) If the plane of the fram e is perpendicular to the m ag netic field lin es, (b) if tne plane of the fram e is p arallel to th e m agnetic field lin es. 1 8 .9 . 5 X IO-4 V. 18.10. The induced em f w ill not ch an ge. 18.11. (a) The induced em f decreases by a factor of 1.3125; (b) increases by a factor of 1.19; (c) increases by a factor of 1.143. 1 8 .1 2 . 120. 18.13. 0.046 T. Hint. The begin n in g of o sc illa tio n s of the em f corre sponds to the p osition of the fram e in w hich its plane is perpendicular to the m agnetic field lin es. 18.14. 0 . 1 5 j i V, 2 0 j i rad/s, 0.1 s, — j i / 4 rad, e = 0 . 1 5 j i sin j i (20* — 0 .25). 18.1 5 . The im pedance of the conductor increases sixfold; the im pedance is equal to the resistan ce of the con ductor. 18.16. 4 Q, 3 Q. 18.1 7 . 152.6 Q. 18.1 8 . The resistan ce w ill be in fin itely large. 1 8.19. 31.8 Q, 0.53 Q. 1 8 .2 0 . 531 Q. 1 8 .2 1 . 18.5 Q. 18.22. 28.2 A. 1 8 .2 3 . 50 jxF. 1 8 .2 4 . 127 V, 179 V. 1 8 .2 5 . U c = 16 V, U L = 96 V, 200 H z. Hint. The answ er is obtained by so lv in g the two equations: X L — X c = U / I 1 and X L /2 — 2 X C = U H 2. 1 8 .2 6 . See Fig. 176. 1 8 .2 7 . (a) See F igs. 177a and 6 , (b) see F igs. 177c and d. 18.28. (a) See F ig. 178a, (b) see Fig. 178 h. 18.29. (a) See F igs. 179a and 6, (b) see Fig. 179c. 1 8 .3 0 . (a) See F ig. 180a, (b) see F ig. 1806. 18.31. (a) See F ig. 181a, (b) see F igs. 1816 and c. 18.3 2 . See Fig. 180a to Problem 18.30. 18.3 3 . See F ig. 1806 to Problem 18.30. 18.3 4 . (a) 100 Q, 1.2 A, 72 V, 96 V, see F ig. 182a, (b) 4 Q, 30 A, 114 V , 42 V, see F ig. 1826, (c) 16 Q, 7.5 A, 180 V, 60 V, see Fig. 182c. 1 8 .3 5 . (a) 5 Q, 44 A, 220 V, 1866 V, 1866 V , see Fig. 183a, (b) 42.4 Q, 5.2 A, 156 V, 327 V, 171 V, see F ig . 1836. 18.36. (a) 12 A, 9 A, 15 A, 0 .8 , see F ig. 184a, (b) 36'A , 18 A, 40.25 A, 0.894, see F ig. 1846, (c) 18 A, 9 A, 20 A, 0, see Fig. 1,84c. 1 8 .3 7 . (a) 6 A, 12 A, 12 A, 6 A, 1, see F ig. 185a, (b) 4 A, 3.6 A, 7,5 A, 6 A, 0.6 7 , see F ig. 1856, (c) 9 A, 18 A, 6 A, 15 A, 0 .6 , see Fig. l£5c. 18 .3 8 . The inductance should be increased by 0.184 H; 120 V / H i n t .
From the con d itio n
— = = = = =
j/ / ? 2 + o)2/,2 we obtain 2 3 -0 5 3 0
——
= iC ? 2
eq u a lity of the phase differences lead s to
Answers
346
.(b)
(a) ^ m a x~ Irn at R
Fig. 176
(a)
/-K/2
Uc =/(Lc
Fig. 177 /
C2
(b )
ULsIcuL
O^er1_____ » . Uk-IR I
Fig. 178
347
A nswers
(b)
■9^ u*
Fig. 179
(a)
(b)
fy'/cul
ur u.
Uc'f-r LUC Fig. 180
(bj
(c)
(a) n U*UL -UC
Fig. 182
348
Answers
R + Ml
the equation V ( /? +
A /? ) 2 +
R
to2 ( L +
A L )2
VR* +
H ence co2Z/2.
L 1 ' A L = & R ~ = A/? — TT? • 1 8 .3 9 . 17.3 Q, 0.04 H . 18.40. U R = R 0) (/ o 32 V , U L = 48 V , U c = 32 V , tft ot = 36 V , 26°34', see Fig. 186. 1 8 .4 1 . 1594 fiF. 18.42. H yd rau lic turbines cannot ensure the required (a)
(V
4
/j"
Uf=U
Uc F ig. 183
(0
r< Ic-h
1*
1'
F ig. 184
(o)
•W
A F ig. 185
speed of rotation of the generator’s rotor. In order to ob tain the stan dard frequency (50 H z), m u ltip o le generators are used. An increase in th e num ber of pole p airs is eq u iv a len t to an increase in the speed of rotation of the rotor. 1 8 .4 3 . 20. H i n t . The frequency v of a lte m a t-
A nsw ers
349
in g current and the speed of rotation n of the rotor are connected through the relation v = p n , where p is the num ber of pole pairs. 18.44. 100 H z. 18.45. N o, it w ill n o t. 1 8 .4 6 . The transform er consu m es an in sign ifican t am ount of energy. 1 8.47. The huge current p a ssin g through u. a sh ort-circuited turn lead s to h ea tin g of the transform er and m ay dam age it. 18.48. 2. 18.49. 10. 18.5 0 . 696. 18.51. 16. 18 .52. 0, 28.2 var, 28.2 V A . 18.53. 72 W, 96 var, 120 V A, 0 .6 , 20 V. 18.54. 108 W, 144 var, R = 27 Q, X L = 36 Q, X c = 72 Q. 18.55. 0.995, 380 W , 36 var. 18.56. 270 V A, 200 W, 182 var, 0.74. H i n t . For the p arallel con n ection , the power factor can be determ ined by / two m ethods: cos
i / / j 2 _ l i / X c wh*c h fo llo w s from the vector diagram for currents. 1 8 .5 7 . 0.019 H, 240 W, 602 var. 1 8 .5 8 . 60 Q, 41 .8 Q, 240 W, 344.7 var, 420 VA, 0.5 7 . F ig. 186 18.59. (a) 1, 160 W, 0; (b) 0 .6 , 120 W, 160 var. 1 8 .6 0 . 0.97, 8067 W, 2020 var. 18.61. X = 12.5 Q, R = 7.2 Q, 0.865, 400 V A , 6.25
Q.
1 9 .1 . The frequency increases. 1 9 .2 . Y es, it w ill; the period w ill in crease. 1 9 .3 . 500 m . 1 9 .4 . 7.94 X 104 H z, 1.26 X 10-5 s. 1 9 .5 . The reflection of electrom agn etic w aves tak es place. 1 9 .6 . U sin g ultrashort w aves p en etratin g through the ionosphere. 1 9 .7 . 53 and 2640 m. 19 .8 . A bout 200 m, 1.6 M Hz. 1 9 .9 . 70 pF. 19.1 0 . 45 km . 1 9 .1 1 . The power is proportional to the fourth power of frequency. 1 9 .1 2 . C x. 19.13. The signal is am plified at the exp en se of the source of elec tric energy. 19.14. 2.25 X 108 m /s, 0.13 m. 1 9 .1 5 . 7.5 m H . 1 9 .1 6 . 0 .5 M Hz, 0.1 H . 1 9.17. 1.6 mm. 19 .1 8 . 1 0~5 s, 300 m. 19 .1 9 . 4 .65 A. 1 9 .2 0 . 4 X IO’ 3 s, 4 pF, E mag = E el = 2 X 10‘ 5 J. 1 9 .2 1 . L = 4n 2A(7vJv|
= 2.3 m H .
2 0 .1 . B y 40°. 2 0 .2 . B y 30°. 2 0 .3 . The mirror should be placed on the path of the rays at an an gle of 78 or 12° to the horizon tal. 2 0 .4 . (a) 68°, 22°; (b )22°, 68°; see F ig. 187. 2 0 .5 . The mirror should be placed at an an gle of 45° to th e h orizon tal w ith the mirror surface facin g up w ards. 2 0 .6 . 160 cm . 2 0 .7 . In the v ertica l plane. 2 0 .8 . The im age of the lam p w ill be virtu a l and sym m etric about the mirror, see F ig . 188. 2 0 .9 . 11.5 cm , see F ig . 189. 2 0 .1 0 . The mirrors should be placed at righ t angles to the lateral sid es of the trian gle S ,iS S 2 at equal d is tan ces from the v ertices, see F ig. 190. 2 0 .1 1 . (a) 3, (b) 5. 2 0 .1 2 . As a result of m u ltip le reflections by the m irrors, a large num ber of im ages w ill be form ed. 2 0 .1 3 . B y m aking the d istan ce to the b u ild in g large. 2 0 .1 4 . The rays em itted by the Sun can be treated as p arallel, there fore, the p oin t of th eir convergence w ill be the focus. 2 0 .1 5 . 39 cm ,
350
Answers
F ig. 190
Fig. 192
Answers
351
19.5 cm . 2 0 .1 6 . See F ig. 191. 2Q.17. 1.2 m . 2 0 .1 8 . 10.5 cm , 21 cm . 2 0 .1 9 . 20 cm from the mirror. 2 0 .2 0 . 20 cm . 2 0 .2 1 . 24 cm , — 1 .7 . 2 0 .2 2 . 60 cm , 40 cm . 2 0 .2 3 . 0 .2 m from the mirror. 2 0 .2 4 . 14 cm . 2 0 .2 5 . The field of v isio n can be increased by u sin g c o n v ex m irrors.
F ig. 193
F ig. 195
Fig. 197
Fig. 194
Fig. 196
F ig. 198
2 0 .2 6 . 3 m . 2 0 .2 7 . 1.5 m. 2 0 .2 8 . — 0 .2 m, the im age is v irtu a l and reduced. 2 0 .2 9 . 0.24 m, see F ig. 192. 2 0 .3 0 . 86 cm . 2 0 .3 1 . 1 .8 . 2 0 .3 2 . 35°. 2 0 .3 3 . 43°. 2 0 .3 4 . (a) A bout 8°, (b) 1 2 ° 2 0 \ 2 0 .3 5 . 50°. 2 0 .3 6 . The im age of the fish in w ater is v irtu a l and d isp laced . 2 0 .3 7 . 1.41, 54°40'. 2 0 .3 8 . (1) For n j = n2, (2) when e = 0. 2 0 .3 9 . A bout 38°, 225000 k m /s. 2 0 .4 0 . 2.3 X 105 k m /s. 2 0 .4 1 . 2 X 106 k m /s, 1.5. 2 0 .4 2 . 35°. 2 0 .4 3 . 57°. 2 0 .4 4 . See F ig. 193. 2 0 .4 5 . A bout 24°. 2 0 .4 6 . 1.5. 2 0 .4 7 . 3 3 °2 (/. 2 0 .4 8 . A ir bubbles on the surface o f the blackened b a ll create the
352
Answers
c o n d itio n s of the total reflection at the w ater-air interface. 2 0 .4 9 . The effect is exp lained by the to ta l reflection. 2 0 .5 0 . See F ig. 194. 2 0 .5 1 . 26 cm . 2 0 .5 2 . 0.75 m, see F ig. 195a. 2 0 .5 3 . 82.5 cm , see F ig. 1956. 2 0 .5 4 . 4 .5 cm . 2 0 .5 5 . 0.4 cm . 2 0 .5 6 . M N = 6.6 cm , see F ig. 196. 2 0 .5 7 . A ll p oin ts are displaced sim ila rly ; the disp lacem en t is n o tice-
F ig. 201 able if the th ick n ess of the g la ss is nonuniform . 2 0 .5 8 . 2.65 cm , 5 cm . 2 0 .5 9 . See F ig. 197. 2 0 .6 0 . See F ig. 198. 2 0 .6 1 . See F ig. 199. 2 0 .6 2 . 34°. 2 0 .6 3 . 17°. 2 0 .6 4 . 3 5 ° 3 0 \ 2 0 .6 5 . 38°. 2 0 .6 6 . n ^ s i n "4 ^ 6- :
A
sin — . 2 0 .6 7 . 8 D, 2 D. 2 0 .6 8 . —4 D, —2.5 D. 2 0 .6 9 . 25 cm , —20 cm , —50 cm . 2 0 .7 0 . It w ill be d ivergin g if the transparent m edium in w hich the len s is placed is o p tic a lly denser than the m aterial of the len s. 2 0 .7 1 . See F ig. 200. 2 0 .7 2 . The brightness of the im age w ill be reduced, see F ig. 201. 2 0 .7 3 . See F ig. 202. 2 0 .7 4 . T hey should be arranged so th at the foci of the len se s coin cid e. 2 0 .7 5 . 30 cm from the len s, th e im age w ill be real and reduced by a factor of tw o. 2 0 .7 6 . — 10 cm , 30 cm , 20 cm , and about 17 cm . 2 0 .7 7 . (1) The h eigh t of the im age decreases, (2) the im age is real, inverted, and fu ll-sized , (3) the object should be betw een the focu s and the lens. 2 0 .7 8 . 10 cm , 10 D. 2 0 .7 9 . See F ig. 203. In order to determ ine the op tica l centre 0 , we
Answers
353
draw a ray from point A to A ' . It w ill in tersect the o p tica l a x is at the op tical centre. Then we draw another ray from p o in t A parallel to the o p tica l a x is, w hich after the refraction by the le n s w ill p a ss through the focus F ' and w ill arrive at p oin t A '. In th is problem , th e con verging le n s produces a real im age. 2 0 .8 0 . 24 cm . 2 0 .8 1 . A bout
Fig. 2 0 2
F ig. 2 0 3
D, 4 . 2 0 .8 2 . 0 . 3 m . 2 0 .8 3 . 4 cm . 2 0 .8 4 . 1 . 7 5 m from the len s, the im age is magnified X 2 . 5 . 2 0 .8 5 . — 2 m, the im age is v irtu a l, erect, and m agnified X 5 . 2 0 .8 6 . 2 4 cm . 2 0 .8 7 . — 1 2 cm , 3 . 6 cm . 2 0 .8 8 . 1 6 cm . 2 0 .8 9 . Y es, it w ill, / = o o . 2 0 .9 0 . 2 . 5 . 2 0 .9 1 . — 7 . 5 cm , about — 1 3 D; see F ig. 2 0 4 . 2 0 .9 2 . 0 . 5 . 2 0 .9 3 . 4 . 2 0 .9 4 . 0 . 1 5 m, about 7 D. 2 0 .9 5 . 1 0 D , 5 D . H i n t . The o p tica l power O of the sy stem is the sum of the op tical pow ers of the len ses th at are in con tact and form the 3 .1
Answers
354
sy stem .
2 0 .9 6 .
A bout
6.7 D , ^ con cave — ^ ^
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2 0 .9 7 . 30. 2 0 .9 8 . 5 D . 2 0 .9 9 . 2.63 m. 2 0 .1 0 0 . 4 D. 2 0 .1 0 1 . For im prov in g the sharpness of the picture. 2 0 .1 0 2 . 19 m . 2 0 .1 0 3 . A bout 8.9 D. 2 0 .1 0 4 . 7.25. 2 0 .1 0 5 . 2.5 cm , 16 cm . 2 1 .1 . 2.51 X 103 lm . 2 1 .2 . 2.4 cd. 2 1 .3 . 3.167 lx , 188.5 lm . 2 1 .4 . 8 X 10”3 lm . 2 1 .5 . 105 lm , 104 lm . 2 1 .6 . 39 lx . 2 1 .7 . 25 lx . 2 1 .8 . 2. 2 1 .9 . E x ~ 13.3 lx , E 2 = 17.4 lx , in the la tter case the illu m in an ce is higher. 2 1 .1 0 . A t about 41°. 2 1 .1 1 . The illu m in a n ce of the vertica l w all is 6.33 tim es higher. 2 1 .1 2 . 224 X 106 km . 2 1 .1 3 . S olar rays are in cid en t on the slop es at a sm aller angle to their surface, co n seq u en tly, the sam e surface area absorbs a higher energy per u n it tim e. 2 1 .1 4 . 35 lx . 2 1 .1 5 . 67 lx , about 38 lx . 2 1 .1 6 . 32.3 lx . 2 1 .1 7 . 200 cd. 2 1 .1 8 . 1 m, 0.71 m. 2 1 .1 9 . The illu m in an ce under each bulb is the sam e and equal to 60 lx . 2 1 .2 0 . 55 lx . 2 1 .2 1 . A bout 700 cd. 2 1 .2 2 . 19 lx . 2 1 .2 3 . Over the area sm aller than 942 m2. 2 1 .2 4 . A bout 31 lx . 2 1 .2 5 . 2 m. 2 1 .2 6 . A bout 9 and 6.7 m . 2 1 .2 7 . 225 cd. 2 1 .2 8 . 0 .8 m from the lam p w ith a low er lum inous in te n sity . 2 1 .2 9 . Y es, it decreases by a factor of n. 2 1 .3 0 . 123 lx . H i n t . The illu m in an ce on the screen is determ ined as the sum of the illu m in a n ces produced by the lam p and its im age formed by the plane mirror. 2 1 .3 1 . 2.6 s. H i n t . In order to obtain photographs of the sam e q u a lity , the fo llo w in g co n d ition m ust be fulfilled during printing: the sam e am ount of energy m ust be supplied to the photographic paper: W x = W 2. Since W = t = E S t , we obtain E 1S t l = E 2S t 2. U sin g th is eq u a lity and h a v in g determ ined the illu m in a n ces, we calcu la te the tim e t 2. 2 2 .1 . A bright spot (enhancem ent of ligh t) is obtained when the op tica l path difference of the w aves is equal to an even num ber of half-w aves. A dark spot (attenuation of lig h t) is obtained when the path difference of the w aves is equal to an odd num ber of h alf-w aves. 2 2 .2 . C onstruc tiv e interference w ill occur at p oin t 1 and d estru ctiv e interference at point 2. 2 2 .3 . The effect is exp lain ed bv interference. 2 2 .4 . The ph e nom enon appears due to interference of lig h t. Y es, it w ill; a ltern a tin g dark and bright fringes are form ed, w hose colour is determ ined by the colour of rays in cident on the surface of w ater. 2 2 .5 . The interference w ill be con stru ctive (bright fringe). 2 2 .6 . 1.52 X IO-6 m, red. 2 2 .7 . When, the soap film is illu m in a ted by w h ite lig h t, interference occurs: in various regions, depending on the film th ick n ess, path difference is created such th at it cau ses enhancem ent of som e w a v elen g th s and atten u ation of others. 2 2 .8 . 3.2 mm. 2 2 .9 . 652 nm . 2 2 .1 0 . The radii w ill decrease. 2 2 .1 1 . It is based on interference: the film c o a tin g the ob jective is chosen so th at the w aves in the m edium part of the sp ec trum are m ain ly atten u ated . Red and v io le t rays are atten u ated in sig n ifica n tly and produce a b lu ish -v io le t tin ge. 2 2 .1 2 . k = d h [L = 750 nm , where d is the separation betw een the lig h t sources, h is the d istan ce betw een tw o adjacent interference frin ges and L is the d is tance from the screen to the lin e con n ectin g the lig h t’s sources. 2 2 .1 3 . 4 .6 mm . 2 2 .1 4 . A bout 3.6 m. 2 2 .1 5 . T h is is exp lain ed by diffractiop of lig h t. E yelash es p lay the role of a diffraction grating. 2 2 .1 6 . U n lik e the spectrum obtained w ith a prism , a diffraction spectrum is uniform
A nswers
355
ly extended in a ll regions. B esid es, spectra of several orders are ob ta in ed to the right and to the le ft of the central bright lin e. 2 2 .1 7 . 0.4 pm . H i n t . For sm all angles, sin es can be replaced by tangents; k k = d sin (p. 2 2 .1 8 . 018 nm , 484 nm . 2 2 .1 9 . 0 .76 pm . 2 2 .2 0 . 2 X 10-° m, 5 X 103 c m -1. 2 2 .2 1 . 4.34 cm . 2 2 .2 2 . 500. 2 2 .2 3 . 4. H i n t , k k = d sin cp, k = d sin A.th . 2 4 .1 6 . 1.44 X 1 0- 19 J. 2 4 .1 7 . A bout 3.93 X IO-17 J. 2 4 .1 8 . 2.4 X 105 m /s. 2 4 .1 9 . 2.9 X 10~19 J, 1.3 X 106 m /s. 2 4 .2 0 . 2.85 X 108 m /s. H i n t . The electron v e lo c ity obtained in d ica tes th a t the re l a t i v i s t s form ula has to be used. S in ce the electron work fu n ction for m olybdenum is n eg lig ib le in com parison w ith the energy of a photon, it can be n eglected in ca lcu la tio n s, and then the form ula hv= A 4-E * I , — l ) becomes h v = E 0 ( —— ■ * r —1 ' ] / 1 — y2/c 2 ' ' " |/1 — u2/c 2 ) where £,0 = 0.51 MeV is the rest energy of the electron . 24.21. 2.84 x in
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356
A nswers
2 5 .1 . A I = 0.87 in — 0.66 m = 0.21 m. 2 5 .2 . v = 0.6c. 2 5 .3 . 2.61 X 108 m/s; no, it w ill not. 2 5 .4 . In the form of a square. 2 5 .5 . 1.2 X IO-13 cm , by a factor of 1.67. 2 5 .6 . A bout 3.57 years. 2 5 .7 . 10 years. 2 5 .8 . A bout 40.2 X 1012 km; about 2.68 X 105 AU; about 1.2 years. 2 5 .9 . 44.4 years for the observer and 19.4 years for the cosm onaut. 2 5 .1 0 . 3.73 X 1 0 - 13 kg. 2 5 .1 1 . 0.866c. 2 5 .1 2 . 0.968c. 2 5 .1 3 . 1.05 X 10- 30 kg, 2.09 X IO"30 kg. 2 5 .1 4 . 9.2 X 107 in/s. 2 5 .1 5 . 8.33 k g, 2.2 X 104 k g/m 3. 2 5 .1 6 . The m ass and d en sity w ill not change for the observer in the rocket. 2 5 .1 7 . 0.89c. 2 5 .1 8 . 0.51 MeV, 1.02 MeV. 2 5 .1 9 . 2.05 X 10“22 k g -m /s. 2 5 .2 0 . 900 m /s. 2 5 .2 1 . 1.25c, w hich con tradicts the postu late of the theory of r e la tiv ity on the im p o ssib ility to exceed the v e lo c ity of ligh t; ^ 0 .9 c . 2 6 .1 . It is the flow of doubly ionized h elium atom s. T h iir m ass is a lm ost 8000 tim es as large as the electron m ass. 2 6 .2 . The L oren tz force does. 2 6 .3 . y-rad iation . 2 6 .4 . y-rad iation is elec tro m a g netic w aves follow in g X -rays in the electrom agnetic spectrum and differing from them in a higher frequency (of the order of 1020 Hz) and hence in a higher energy. 2 6 .5 . 6 . 2 6 .6 . An electron is born as a result of conversion of a neutron in to a proton. 2 6 .7 . A bout 4 d ays, 2 X 10~° s _1. 2 6 .8 . TBi = 5 d ays, T Po = 138 days. 2 6 .9 . About 1.2% . 2 6 .1 0 . 7.9 h, 1.3 X 10~5 s~l . 2 6 .1 1 . 4.5 X 109 years, about 5 X IO-18 s _1. 2 6 .1 2 . The atom of JH con tain s one proton and two neutrons, h eliu m atom has tw o protons and two neutrons, alum inium atom has 13 protons and 14 neutrons, uranium atom has 92 protons and 146 neutrons, neptunium atom has 93 protons and 144 neutrons. The num ber of neutrons is increasing. 2 6 .1 3 . T hey differ in the num ber of neutrons, tabu lated chlorine c o n sists of 75% chlorine w ith the m ass num ber of 35 and 25% chlorine w ith the m ass number 37. 2 6 .1 4 . The nucleus of silico n atom ; the n u clei of h eliu m , carbon, nitrogen, o x y gen, neon, m agnesium , sulphur, and calciu m . 2 6 .1 5 . 4 .8 X 1 0 ~19 C, 4.64 X IO"18 C, 1.47 X IO-17 C. 2 6 .1 6 . 1.2 X 1 0~8 C. 2 6 .1 7 . R He = 2.2 X 10- 15 m, R v = 8.7 X 1 0~15 m, pHe = 1.44 X 1017 k g /m 3, Pu = 1.4 X 1017 k g/m 3. 2 6 .1 8 . 2f 7Ac -► 223Fr + 2H e, ct-decay. 2 6 .1 9 . It w ill be converted into neptunium : 2i$U 2£JNp + J{e. The m ass num ber does not change. Tne n u cleu s w ill be sh ifted to the righ t. 2 6 .2 0 . A neutron. 7Li + 2H 8Be + Jn. 2 6 .2 1 . Z = 6 , A = 14, carbon isotope: + h i -*■ 4JC + \ p . 2 6 .2 2 . It is an iod in e atom 1JJI. 2 6 .2 3 . 1.02 MeV, 2.5 X 1020 H z. 2 6 .2 4 . ‘4N + + 2J» -► *fN + 19N + y. 2 6 .2 5 . The energy m ust be h igher than the sum of the rest energies of an electron and a positron, i.e . larger than 1.02 MeV. 2 6 .2 6 . 2JA1 + £He ?2Si + Jp, the n u cleu s of silico n iso to p e. 2 6 .2 7 . Id en tical tracks are le ft by id en tica l p a rticles, v iz . n uclei of h eliu m atom s formed in the reaction !£B + \ p 3 £He. 2 6 .2 8 . The m issin g particle is an a -p a rticle . 2 6 .2 9 . Curium 2£SCm, neutron. 2 6 .3 0 . A pro ton is converted into a neutron. 2 6 .3 1 . 0.12 amu or 1.99 X IO" 28 kg. 2 6 .3 2 . 0.07 am u, about 65.2 MeV. 2 6 .3 3 . 1.97 amu or 3.27 X 10“27 kg, 1833 MeV or about 2.9 X 10~10 J. 2 6 .3 4 . (1) Energy is absorbed, (2) lib erated , (3) liberated. 2 6 .3 5 . 4 .0 MeV or 6.4 X IO-13 J. 2 6 .3 6 . 770 kg. 2 6 .3 7 . 1.67 t. 2 6 .3 8 . In a ll ca ses, the energy is ap p roxim ately equal to 2 X 1013 J. 2 6 .3 9 . A bout 64 g. 2 6 .4 0 . 4 .8 X 1012 J. 2 6 .4 1 . In fast neutron reactors, natural uranium is u tilized w ith a higher
Answers
357
efficiency, and besides the nuclear fuel (plutonium ) is reproduced. 2 6 .4 2 . G am m a-quantuin, electron, neutrino, and proton. 2 6 .4 3 . 3 X 1 0 -11 J. 2 6 .4 4 . A bout 1.13 X 1010 J. 2 6 .4 5 . A bout 3.2 X 1019 J. 2 7 .1 . 8 min 19 s. 2 7 .2 . 4.02 X lO13 km , 1.31 pc. 2 7 .3 . 5.54 h. 2 7 .4 . 6.2 X 105 pc. 2 7 .5 . Sirius, Canopus. 2 7 .6 . C anis Major; th is c o n ste l lation is in the southern part of the c e le stia l sphere. 2 7 .7 . 8 .9 years. 2 7 .8 . 3.04 X 104 pc. 2 7 .9 . Vega; Lyra. 2 7 .1 0 . The north and south p o in ts on the cele stia l m eridian and p oin ts of east and w est on the ce le stia l equator. 2 7 .1 1 . Serpens co n stella tio n . 2 7 .1 2 . The lis t of c o n ste lla tio n s m ay change depending on geographical la titu d e . 2 7 .1 3 . See the answ er to Problem 27.12. 2 7 .1 4 . The Sun rises e x a c tly at the east and se ts ex a ctly at the w est o n ly during the vernal and autum nal eq u in oxes. 27 .1 5 . A quarius, Capricorn, S a g itta riu s, Scorpius, and Libra. 2 7 .1 6 . A t the poles. 2 7 .1 7 . A bout 55°. 2 7 .1 9 .3 0 ° . 2 7 .2 1 . B y 1°. 2 7 .2 2 . 0, 0. 2 7 .2 3 . 0, 12 h. 2 7 .2 4 . 5 7 ° 4 2 \ 2 7 .2 5 . 1.9°. 2 7 .2 6 . 1.392 X 106 km . 2 7 .2 7 . 5.71 X 107 km . 2 7 .2 8 . 8.2 pc. 2 7 .2 9 . 84.01 'y ea rs. 2 7 .3 0 . 9.539 A U . 2 7 .3 1 . 5.203 A U , 778 X 106 km . 2 7 .3 2 . In the form of a narrow crescent. 2 7 .3 3 . T h is is v a lid for any la titu d e. 2 7 .3 4 . In the first quarter. 2 7 .3 5 . 7.5 X 104 years. 2 7 .3 6 . 0.545". 2 7 .3 7 . The in c li nation of the V enus a x is to the plane of it s orb it is clo se to 90°. 2 7 .3 8 . 1400 kg/m 3. 2 7 .3 9 . 7.7 X 103 m /s.
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HIGHER MATH FOR BEGINNERS (MOSTLY PHYSICISTS AND ENGINEERS)
Ya.B. Zeldovich, Mem. USSR Acad. Sc., and I.M. Yaglom, D.Sc. (Phys.-M ath.) This took has been w ritten as an introduction into higher m athe m atics. Aside from such traditional topics as analytic geom etry and differential and integral calculus, the book introduces the^notions of power and trigonom etric series, studies sim ple differential equations, and discusses som e special topics in physics. It is intended for high-school students, freshm en at universities and technical colleges, and anyone w ish in g to brush up on, the elem ents of higher m athem atics.
A COLLECTION OF QUESTIONS AND PROBLEMS IN PHYSICS L.A. Sena The C o lle c tio n contains more than 400 questions and problems covering all the sections of the physics course. A ll questions and problem s have detailed answ ers and solutions. For th is reason the tw o m ain sections of the book, Q uestions and Problems and A ns w ers and Solutions, have identical headings and num bering: each chapter in the first section has a corresponding chapter in the second, and the num bering of answ ers corresponds to the number-* ing of problems. A special feature of the C o lle c tio n is the draw ings and diagram s for m ost of the questions and answers. The diagram s use a variety of scales: linear, sem ilog, log-log, and quadratic. A rrangem ent of the m aterial in this C o lle c tio n corresponds to the structure m ost com m only used in college physics textbooks. One exception is the questions and problems involving the special theory of relativity. T hese are placed in different chapters, starting from the one dealing w ith m echanics. The C o lle c tio n is intended for the self-instruction of students of technical colleges.
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Vernal equinox Autumnal
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