1.
Sub-Atomic particles
2.
Atomic models
3.
Planck’s quantum theory
4.
Photo electric effect
5.
Bohr’s model
6.
Hydrogen spectrum
7.
Limitation of Bohr’s theory
8.
Dual nature of electron
9.
Heisenberg’s uncertaintiy principle
10.
Quantum mechanical model of H-atom
11.
Quantum numbers
12.
Electronic configurations
... 1
Sub-Atomic Particles Experiments of Cathode Rays (Discovery of Electron) When electron discharges from a high potential source and passed through a discharge tube evacuated to a pressure of 0.01 mm or less, rays are emitted from the cathode. These are commonly known as cathode ray. ray. Cathode rays comprise of negatively-charged negatively-charge d particles called electrons. Gas at low pressure Cathode rays Cathode –
–
To vacu um pump Anode +
High voltage +
Discharge tube experiment-Production of cathode rays
Properties of cathode rays (i) (i) (ii) (ii) (iii) (iii)
They They trave travell in stra straigh ightt line lines s at at righ rightt angl angle e to the the cath cathod ode es sur urfa face ce.. They They are mate material rial part particle icles s as they they produ produce ce mecha mechanica nicall motion motion in a smal smalll paddle paddle whee wheell placed placed in their path. The name electron to these material particles particl es was given by Stony. They They are deflec deflected ted from from their their path by by electric electric and and magne magnetic tic fields fields as they they are elect electrica rically lly charg charged. ed. Direction of deflection indicates that they are negatively charged.
Explanation of discharge phenomenon: At low pressure the electrons can move through a considerably large distance before attaching to another molecule forming a negative ion which is not possible at high pressure. These free electrons and the positive ions are responsible for conduction of electricity. At very low pressure around 0.01 mm of Hg the walls wal ls of tube begin to glow this thi s is called florescence. The colour of glow depends on nature of glass. It is i s yellowish green for soda glass. If the pressure is still decreased the current through the gas gradually decreases and finally the tube stops conducting. In vacuum there are no charges and hence no conduction occurs.
Note:
Charge on electron was given by R. A. Mullikan ’s by an oil drop experiment. Its value was found to be 1.6 × 10 –19 coulombs which is taken as one unit negative charge.
... 2
e/m for electrons (cathode rays) is a fundamental property, property, i.e. it does not depend upon: (1) Metal us used as as ca cathode. (2) Gase ases use used d in in dis disch char arge ge tube tube.. Thomson determined determined e/m value as as 1.76 × 108 coulombs/g By knowing charge over an electron and e/m ratio, we can get the value of mass. e e/m
Mass of electron m =
=
=
1.6 × 10 −19 coulombs / el electron
1.76 × 108 coulombs / g 9.1 × 10− 28 g or 9.1 × 10 −31 kg
1 times the mass of a hydrogen atom. 1837 Mass of electron mov moving ing with a velocity v is given by
This is rest mass of electron which is
m=
mo 1−
v2 c2
(c = Velocity of light, v = Velocity of elelctron)
It is clear that with increase in velocity, mass of electron would increase but the velocity cannot equal or exceed the velocity of light.
Anode Rays (Discovery of proton) It is a well-known well -known fact that the atom is electrically electri cally neutral. The presence of negatively-charged electrons in the atom emphasized the presence of positively-charged particles. To detect the presence of positively-charged particles, the t he discharge tube experiment was carried out, in which a perforated cathode was used. Gas at low pressure was kept inside the tube. On passing high voltage between the electrodes it was observed that some rays were emitted from the side of the anode. These rays passed through the holes in the cathode and produced green fluorescence on the opposite glass wall coated with ZnS. These rays consist of positively-charged particles. Perforated cathode
Zn S Coating
+ –
H g a s in in s id id e at low pressure 2
For anode rays, e/m is not fundamental property as different gases used have different mass on C-12 scale.Highest e/m is for hydrogen gas. Mass of proton is found to be 1.672 x 10 –24 g or 1.0075 amu (1 amu= 1.66 x 10 –24 g)
... 3
Some of the well known fundamental particles particl es present in an atom are-protons, electrons and neutrons. Many others were discovered later viz positron, neutrinos etc. S u b at o m i c p ar t i c l e s
Sy m b o l
U n i t c h ar g e
U n i t m as s
p1
+1
1
n1
0
1
Proton
1
Neutron
0
Electron
1
–
e0
1
–
Negligible
C h ar g e i n Co u lo m b +1.60
×
Mas s in am u
19
10
–
0
1.008665
1.602× 10
–
1.007825
19
–
5.489×10
4
–
Chadwick in 1932 discovered the neutron by bombarding elements like beryllium with fast moving α -particles. He observed that some new particles were emitted emit ted which carried no charge and had mass equal to that of proton.
Atomic Models J.J. Thomson’s Model of an Atom An atom consisted of a uniform sphere of positive charge in which the electrons were embedded, so as to make the atom electrically neutral. This is known as plum-pudding model.
Positive sphere
Electron
Thomson's Atomic Atomic Mo del
... 4
Advantages of Thomson’s Model This model was quite attractive as it could explain several observations available at that time. 1. It could could explain explain why only only negat negative ively ly char charged ged particl particles es are are emitt emitted ed when when met metal al is hea heated ted and never never positive charged particles. 2. It coul could d also also explain explain the format formation ion of ions ions and and ionic ionic compou compounds nds of chemis chemistry try..
Disadvantages 1. 2.
Almost Almost free free passa passage ge of of catho cathode de rays rays throu through gh an an atom atom was was n not ot con consis sisten tentt with with Thoms Thomson on ’s model for that the atom should have a lot of empty space. The phenomenon of α - rays scattering could not be explained by this model.
Rutherford’s Experiment The experiment involved scattering of massive α-particles by very thin foils of metals metal s such as gold, silver etc., of thickness 0.0004 mm. The position of the particles after passing through the foil was ascertained by the flash produced on a ZnS screen.
Circular ZnS Circular screen
Mo st of particles strike here
Radium
-rays
Thin gold plate Block of lead Slit
Observations a. b. c.
Most of the α-particles passed through the metal foil without any deflection deflect ion in their path. Some α-particles were deflected through small angles. Very few α-particles (1 in 20,000) were deflected deflect ed by an angle greater than 90 °or even were deflected back completely.
Conclusions a. b. c.
Most ost o off the the spa space in the the atom atom is emp mpty ty.. The nucl nucleu eus s is surro surroun unded ded by nega negative tivelyly-cha charge rged d electr electron ons, s, which which revo revolve lve arou around nd the the nucl nucleu eus. s. An atom atom consist consists s of an extre extremel mely y smal smalll dense dense positiv positively ely-ch -charg arged ed nucleu nucleus s which which is pres present ent at the the 5 6 centre of the atom. Size of nucleus is 10 to 10 times less than the size of atom. This model is called Rutherford ’s model
... 5
Drawbacks of Rutherford’s Model According to Maxwell ’s equation of electro magnetism when a charged particle moves with acceleration, it continuously loses energy in the form of electromagnetic radiation. So S o should be the case with electrons, and as a result of loss of energy, its motion should slow down and it should eventually fall into the nucleus as it will no longer be able to withstand the attractive forces of the nucleus. As a result the atom should collapse. But we all know that an atom is quite stable. Rutherford was unable to account for the stability of the atom.
Note: Atomic number and mass number — Atom is electrically neutral hence it is necessary for the number of protons to be equal to the number of electrons in order to balance the charges. Atomic number (Z) = Number of protons present in the nucleus = Number of electrons present in the t he atom. ∴ Mass number of an element (A) = Number of protons + Number of neutrons.
Representation Representation of an element. An element is represented as
A ZY
where, Y is the symbol of element A is the mass number Z is the atomic number Illustrative
Example 1:
Example
Sodium is represented as
23 11 Na .
The mass number (A) of sodium is 23 and atomic
number (Z) is 11. Atomic weight of Na is 23 g, i.e., 1 mole or 6.023 × 1023 atoms of Na weigh 23 g. Then, weight of 1 atom of Na =
Where, 1 amu =
1 6.023 × 10
23
23 6.023 × 1023
= 23 amu
g
Electromagnetic Waves Electromagnetic waves are waves consisting of oscillating electric and magnetic fields. The T he basic source of electromagnetic waves is an accelerated charge, this t his produces changing electric magnetic field which constitute the wave. Both the components have same wavelength and frequency and travel in planes perpendicular to each other and also perpendicular to the direction of propagation of electromagnetic wave. y
λ
x
... 6
Characteristics Characteristics of a wave (i) Wave Wavele len ngth gth: It is repr repres ese ente nted by λ. Units are m, cm or Å. 1Å = 10 –8 cm = 10 –10 m (ii) (ii) (iii)
Frequen Frequency: cy: The num number ber of time times s a wave wave pass passes es through through a give given n poin pointt in 1 s s.. Velocity elocity:: The linear linear dista distance nce travel travelled led by by a crest crest or or a trough trough in 1 s. s. Its unit unit is cms –1.
(iv)
Wave Wave numbe number: r: The The numbe numberr of wave waves s prese present nt in 1 cm lengt length. h. It is is repres represent ented ed by by
1
λ
.
Electromagnetic Electromagnetic spectrum The arrangement of various electromagnetic radiations in order of their increasing (or decreasing) wavelengths (or frequencies) is known as electromagnetic spectrum. °
Electromagnetic radiation 1. Cosmic rays
Wavelength ( A ) Up to 0.01
2. Gamma( γ ) rays
0.0-0.1
3. X-rays
0.1-150
4. Ultraviolet (UV)
150-3800
5. Visible radiations
3800-7600
6. Infrared (IR) 7. Microwaves 8. Radiowaves
7600 – 6 ×106 3 × 106 – 3 × 109 3 × 107 – 3 × 1014
Origin Originate from outer space and penetrate the earth’s atmosphere to reach its surface. Produced from nuclei of radioactive substances during disintegration of nuclei. Produced by placing a metal plate in the path of fast moving cathode rays. Present in sun ’s rays. Lab sources include Hor Xe lamps etc. Produced from stars, arc lamps and any other source whose light is visible. Heat waves. Produced by incandescent bodies. Produced by special generators. Generate from alternating electric current of high frequencies.
Planck’s quantum theory The three key points of Planck ’s quantum theory are as follows. a. Radian Rad iantt energ energy y is abso absorbe rbed d or emitted emitted discont discontinu inuous ously ly in the form of smal smalll pa packe ckets ts of of energ energy y called ‘quanta’. In case of light energy these packets are called photons. photons. b. Energy Energy of each each pho photon ton is direc directly tly propor proportion tional al tto o the the frequ frequenc ency y of radiati radiation. on. This is math mathema ematica tically lly expressed as E ∝ ν E = h ν c
ν = λ h is called Planck’s constant h = 6.626 × 10 –34 Js. c.
Total otal amo amoun untt of ene energ rgy y emi emitt tted ed or abs absor orbe bed d is giv given en as as E = nhv
=
nhc
λ
n = 1, 2, 3, ... (no of photons)
... 7
Illustrative
Example 1:
Solution:
Example
It has been found that gaseous iodine molecules dissociate into separate atoms after absorption of light at wavelengths less than 4,995 Å. If each quantum is absorbed by one molecule of I2, what is the minimum energy in k cal/mole, needed to dissociate I 2 by this photochemical process? E(per mole) = NA.h υ E = NA
hc
λ
=
(6.023 × 10 23 mole −1) (6.626 × 10 −34 Js ) (3.0 × 10 8 m s − 1) 4995 × 10 −10 m
1 kcal 239.6 kJ / mole = 57.3 k cal/mole . kJ 4184 Photoelectric Effect Sir J.J. Thomson observed that electrons are emitted emitt ed instantaneously from a clean metal plate in vacuum when a beam of light falls on it. This is called the photoelectric photoelectric effect. effect. Usually such an effect is produced by a radiation in the UV region and also in some cases in the visible region. Photoelectric effect is an illustration of the particle nature of light. Photoelectric emissions are associated with the following facts. (a) Electro Electrons ns are emitte emitted d in insta stanta ntaneo neousl usly y from from a c clea lean n meta metall plate plate whe when n irrad irradiate iated d with with a rradi adiatio ation n of of frequency equal to or greater than some minimum frequency f requency,, called the threshold frequency. The energy corresponding to this frequency fr equency is known as the work function. (b)
Kinetic Kinetic ene energy rgy of the emitted emitted electr electrons ons dep depend ends s upon upon the frequen frequency cy of the the inciden incidentt radia radiation tion and not on its intensity. The kinetic energy increases linearly with the increase in the frequency of radiation.
(c) (c)
The The numb number er of of ele electr ctron ons s emitt emitted ed iis s prop propor ortio tiona nall to the the in inte tens nsity ity of of the incid inciden entt radia radiatio tion. n. Suppose, the threshold frequency frequency of light l ight required to eject electrons from a metal is ν 0 and hence the minimum energy required is h ν 0 . If the frequency of light is higher than ν 0 (Let ν ), then the excess energy will will be converted to the kinetic energy of the electron. h ν
= h ν 0 + K.E
h ν
= h ν 0 +
1 mv 2 2
1 mv 2 2
= h ν − h ν 0
... 8
M ax KE
M ax KE
ν
°
Illustrative
Example 1:
Solution:
f r e qu q u en en cy c y ( ν)
Intensity (I)
Example
Calculate the kinetic energy of a photoelectron emitted by Na surface when light of wavelength 400 nm strikes on it. (Given work function f unction of Na = 2.28 eV) Energy of photon = h υ =
hc
λ
=
6.626 × 10 −34 J s × 3 × 10 8 m s −1 400 × 10 −9 m
Work function = 2.28 eV = 2.28 × 1.602 × 10 –19 J (1 eV = 1.602 × 10-19J) = 3.653 × 10 –19 J Now, KE = h υ – Work function = 4.969 × 10-19 J – 3.653 = 1.316 × 10 –19 J
= 4.969 × 10-19 J
× 10-19 J
Types of Spectrum Spectra is classified into two types depending upon the source of radiation. (a) Emission sp spectra (b) Absorption sp spectra (a)
Emission spectra Emission spectrum is obtained when the radiations from a source (electric discharge through a gas at low pressure) are passed directly through a prism and received on a photographic plate. There are two types of emission spectra: (i) Continuous spectra
... 9
W hite light light
Beam
Prism Photographic plate When white light is passed through a prism, it splits into seven different colours from violet to red. The colour band is continuous and merge into other ot her colour bands before and after it. Thus, we get a continuous spectrum. (ii) Line spectra:
Prism Photograph ic plate plate
(b)
When an electric discharge is passed through a gas at low pressure, light is emitted. If this light is passed through a prism and on to a photographic plate, it is observed that lines separated from each other are obtained on the photographic plate. Thus, we get a line spectra. Absorption spectra When white light from fr om the sun or from a bulb is passed through the vapours or solution of a chemical substance, i.e. sodium vapours and then passed through a prism and on to a photographic film, some dark lines are observed in the continuous spectrum. The appearance of dark lines in the continuous spectrum is due to the absorption of certain wavelength by the sodium sodium vapours.
... 10
Hydrogen Spectrum On supplying energy to the hydrogen atom, the electron in its ground state gets excited to higher energy levels. It then tends to return to its ground state by emitting the t he energy absorbed in the form of radiations which produce lines which collectively form the hydrogen spectrum. spectrum. Spectral lines of hydrogen Energy of an electron in Bohr ’s orbit =
∆E = hν =
−
2π2 Z2e4 mk 2 n2h2
hc
λ
∆E = Ef − Ei Ef = Energy in final orbit Ei = Energy in initial orbit 1
λ
=−
2πZ2 e 4mk 2 ch3
1 1 2 − 2 nf ni
For hydrogen Z = 1
∴
1
λ
=−
2πe4mk 2 e4
2πe4me 4k 2 3
ch3
1 1 − n2f ni2
= RH
ch RH is Rydberg’s constant RH = 1,09,678 cm –1 = 1.09 × 107m –1
∴ 1
λ
1 1 = RH 2 − 2 λ nf ni 1
=ν
= Wave number
Spectral series
Value of n1
Value of n2 Spectral region n2 = (n1+ 1), (n1+ 2) + ...
LYMAN
1
2, 3, 4, 5 ...
UV
BALMER
2
3, 4, 5, 6 ...
VISIBLE
PASCHEN
3
4, 5, 6, 7 ...
IR
BRACKETT
4
5, 6, 7, 8 ...
IR
PFUND
5
6, 7, 8, 9 ...
IR
... 11
Maximum number of lines produced when the electron elect ron jumps from the nth level to the ground state stat e =
n(n − 1) 2
Bohr ’s model of atom Neil Bohr on the basis of Planck ’s quantum theory overcame the drawbacks of Rutherford ’s model and suggested a modified model of an atom which retained the key features of Rutherford ’s model and introduced the concept of stationary orbits. Bohr’s postulates: i. An ato atom m comp comprise rises s of of a small small posit positive ively-c ly-char harged ged nucleu nucleus s in in the the centre centre in whic which h the the ent entire ire mass mass is concentrated. ii. Electron Electrons s revol revolve ve arou around nd the the nucl nucleus eus in circu circular lar orbits, orbits, the these se orbi orbits ts have have a fixed fixed value value of of energ energy y. They are popularly known as stationary orbits. When an electron revolves in a circular orbit, energy is neither emitted nor absorbed. This is i s known as quantization of energy. energy. iii. Electron Electron can revolv revolve e only only in tho those se orbit orbits s whose whose ang angula ularr momen momentum tum is an integra integrall multip multiple le of of the factor
h . 2π
i.e. mur
=
nh where n=1,2,3......... 2λ
What is angular momentum of electron in Bohr ’s second orbit? Answer: Angular momentum of electron in Bohr ’s second orbit
iv. iv.
= h / π
Energy Energy is emitt emitted ed or abso absorbe rbed d when when an an elect electron ron jumps jumps from from one orbit orbit to ano anothe therr.
∆E = E2 − E1 Bohr’s theory is used for a. Calcu Ca lcula latio tion n of of the the radi radius us of of the the orb orbit it in wh whic ich h the the el elec ectro tron n revo revolve lves. s. b. Velocity ity of of e ellectro tron in in th the or orbit. c. Energy of the electron
... 12
Calculation of radius of an electron in the orbit
m u2
n u c le u s +
r electron r
According to Coulomb’s law, the electrostatic force of attraction (acting inwards)
=
Fa
kZe 2 r
2
(k is a constant with value 9 × 109 Nm2 /C2)
mu2 Centrifugal force F c = (acting outwards) r In order to maintain the motion of the electron in the orbit Fa = Fc mu2 r
kZe2
=
r2
kZe2 mr According to Bohr’s postulates u2 =
mur = u= 2
u
nh 2π
nh 2πmr
=
n2h2 4π2m2r 2
n2h2 4π2m2r 2 r
=
=
kZe2 mr
n2h2
4π2mkZe2 For hydrogen Z = 1 For n = 1, Z = 1 r = 0.529 Å
... 13
Radius of an electron in the nth orbit of an atom with atomic number Z can be calculated as rn
= 0.529 ×
n2 Z
Å
Calculation of velocity of electron in an orbit m u2 r
=
k Ze 2
... (i)
r2
nh 2π Dividing (i) and (ii)
mur =
... (ii)
kZ2πe2 u= nh u=
2πe 2 u is in cm/s, for H atom nh
Z = 1, k in cgs unit, u =
2π e 2 nh
Velocity of an electron in nth orbit of an atom with atomic number Z is i s given by un
= 2.18 × 108 ×
Z cm / s n
Number of revolutions per second =
Velocity of electron in an orbit Circ Circum umfe fere renc nce e of orbi orbitt
Calculation of energy of electron in an orbit: Total energy = Potential Pot ential energy + Kinetic Ki netic energy TE = PE + KE PE =
−
kZe2 r
1 kZe2 2 KE = mu = (from (i)) 2 2r 1 TE = mu2 2
kZe2 − r
kZe2 ∴ TE = 2r
kZe2 − r
... 14
kZe2 TE = − 2r Putting value of r in the above expression TE =
−
2π2 Z2 e4 mk mk 2
n2h2 The expression for energy is given by
Z2 – 11 E = –2.18 × 10 × 2 erg/atom n
2
Z = − 2.18 × 10−18 × 2 J / atom n = –13.6
Z2
eV/atom n2 1 eV is defined as the amount of energy gained by an electron when it accelerates through a field fi eld of 1 V. Illustrative
Example 1:
Solution:
Example
Calculate the wavelength in Angstrom of the photon that is emitted when an electron in Bohr orbit n = 2 returns to the orbit n = 1 in the hydrogen atom. The ionization potential of hydrogen atom in ground state is 2.17 2. 17 × 10-11 erg per atom. The ionization potential of the ground state of hydrogen atom is 2.17 ×10 –11 erg. ∴ E1 = –2.17 × 10 –11 erg. Since En
=
E1
; E2
=−
2.17 × 10 −11
erg n 22 = –0.5425 × 10 10 –11 erg erg –2.17 ×10 –11) erg E2 – E1 = –0.5425 ×10 –11 erg – ( – ∆E
2
erg = 1.6275 × 10 −11 erg
∆E
= hυ =
hc
λ erg ∆E = 1.6275 × 10 –11 erg 6.626 × 10 10 −27 erg s × 3 × 10 10 10 cm s − 1 11 − ∴ 16 1.6275 × 10 erg =
λ
∴ λ=
6.626 × 10 10 −27 erg s × 3 × 10 10 10 cm s −1
= 12.2138
1.6275 × 10 −11 erg 10 –6 cm = 1221.4
10 –8 = 1,221.4 Å
... 15
Reduced mass According to Bohr, in an atom electron is revolving in a circular orbit around the stationary nucleus. But mass of o f electron 1 = nucleus will be stationary stationa ry only if, the mass of the nucleus is infinite and we know mass of proton 1836 so for lighter elements we can ’t assume nucleus as stationary. Actually the nucleus oscillates slightly about the center of gravity (nucleus ( nucleus and electron) so in Rydberg constant R R=
e 4m
8ε02h3C mass of electron (m) is replaced by reduced mass
µ
mM m+M where m = mass of electron M = mass of nucleus.
µ=
Facts in favour of Bohr’s theory (i) (i) (ii) (iii) (iv)
The The freq freque uenc ncies ies of the the spec spectra trall line line of the the hydr hydrog ogen en spec spectru trum, m, as calcu calcula late ted d usi using ng Bo Bohr hr ’s theory are in close agreement with their experimental values. The value of Rydberg’s constant as calculated from Bohr ’s theory is in full agreement with the value obtained from spectroscopic studies. The radii radii and and ene energy rgy of permis permissib sible le orbits orbits are consis consisten tentt with with the their ir experim experiment ental al values values.. The The abso absorp rptio tion n and and em emiss ission ion spec spectra tra of H and and H H-li -like ke ato atoms ms is is quite quite we wellll exp expla laine ined. d.
Drawbacks of Bohr’s theory 1. 2. 3.
4. 5.
The The theo theory ry exp explai lains ns the the s spe pect ctru rum m of only only hydr hydrog ogen en and and hyd hydro roge genn-lik like e spec species ies.. The The s spl plit itti ting ng of spe spect ctra rall line lines s in ma magn gnet etic ic fie field ld is is call called ed Zeeman effect, effect, and the splitting of lines in electric field is called Stark effect. effect. These phenomenon could not be explained by Bohr. It prov provide ides s no pro proof of for for postu postulat late e whic which h state states s quan quantiz tizat atio ion n of ang angula ularr mome moment ntum um nh mur = 2π It is not in acco accorda rdance nce with dua duall na nature ture of elec electron tron,, as acco accordin rding g to Bohr Bohr, electro electron n behav behaves es as as a particle and has no wave nature associated with it. It was unable to account for Heisenberg’s uncertainty principle. principle.
Dual nature of electrons de Broglie pointed out that matter has both particle and wave nature. The phenomenon of electron diffraction established the wave nature of electrons and the t he phenomenon of photoelectric effect and black body radiation established the particle nature of electrons. electr ons. de Broglie derived an expression for calculating calculati ng the wavelength of the wave associated with an electron. λ = h/p (p =momentum)
λ → Wavelength p → Momentum
... 16
The above equation is called de Broglie equation and the wavelength λ is called de Broglie’s wavelength. Let a photon of mass m moves with a velocity c around the nucleus and be associated with a wave of wavelength E = hv
λ . From Planck’s equation, we have
and Einstein’s equation gives E = mc2 Combining both equations, we have mc2 = hv mc2 = hc/ λ or
mc = h/ λ
or
λ = h/mc
= h/mass × Velocity = h/p
λ
or Illustrative
Examples
Calculate the momentum of a particle which has a de Broglie ’s wavelength of 3 Å.
Example 1:
(1 Å
= 10−10 m, h = 6.6 × 10−34 kg
m2 s−1)
6.6 × 10 −34 kg m 2s −1 λ 3 × 10−10 m Momentum = 2.2 × 10 −24 kg ms −1 mv
Solution:
=
h
=
Heisenberg’s Uncertainty Principle According to the principle, “It is impossible to determine exactly both the position and the momentum (or velocity) of an electron or any other microscopic moving particle at the same time.” Any attempt to locate an electron changes its momentum. If ∆x is the uncertainty in position positi on and ∆p is the uncertainty in momentum, then according to Heisenberg ’s principle, these two quantities are related as follows:
∆ x . ∆p ≥
h 4π
∆x . m∆v ≥ ∆x . ∆ v ≥
( a constant) h 4π
h 4πm
∆x → Uncertainty in position ∆v → Uncertainty in velocity
... 17
Explanation: Let Explanation: Let us try to measure both the position and momentum of an electron. To find the position of the electron we have to use light so that photon of light strikes the electron. As a result of lighting the position as well as the velocity of electron are disturbed but the accuracy with which the position of a particle can be measured depends upon wavelength of light used. The uncertainity in measurement of position using wavelength λ is ±λ. Therefore shorter the wavelength wavel ength the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. energy. When the high energy photon strikes stri kes the electron it changes its speed as well as direction. But this is not true for macroscopic moving particles. So Heisenberg’s uncertainity principle has no significance signifi cance in everyday life. Illustrative
Example
Example 1:
Calculate the uncertainty in position of an electron if the uncertainty in its velocity is 5.7 x 10 –5 ms –1. (h = 6.6 × 1034 Kg m2s –1 and m = 9.1 × 10 –31 kg)
Solution:
∆x × ∆p = ∆x = =
h 4π h 4π . ∆v . m 6.6 × 10−34 kgm2 s−1 4 × 3.14 × 57 × 105 ms−1 × 9.1× 10−31 kg
1 0 −10 m = 1× 10
Concept of Orbital Following Heisenberg’s uncertainty principle, we replace the concept of orbit (introduced by Bohr) by the concept of orbital. Orbital refers to three-dimensional space around the nucleus within which the probability of finding an electron of given energy is maximum.
Quantum or Wave Mechanical Model of Atom This new model of atom was put forward by Schrodinger taking into account the de Broglie concept of dual nature and Heisenberg ’s uncertainity principle. He described the motion of the electron in three-dimensional space in terms of a mathematical equation called Schrodinger wave equation.
∂2ψ ∂2ψ ∂ 2ψ 8π2m ( + 2 + 2 + 2 E − V )ψ = 0 2 h ∂x ∂y ∂ z Where ψ (psi) (psi) is the amplitude of the electron wave at a point with coordinates x, y and z. E is the total energy and V is the potential energy of the electron. is also called wave function. ψ 2 gives the probability of finding the electron at (x, y, z). The acceptable ψ is solutions of the above equation for the energy E are called eigen values and the corresponding wave function ψ are are called eigen functions. These T hese functions have to be single valued, continuous and finite. fini te.
... 18
Schrodinger wave equation can be written as
∂2 ∂2 ∂2 8π2m ψ + 2 (E − V )ψ = 0 2 + 2 + 2 h ∂x ∂y ∂z 2
∇ ψ+
8π2m
Where,
2
h
(E − V ) ψ
=0
∂2 ∂ 2 ∂ 2 ∇ = 2+ 2+ 2 ∂x ∂y ∂z 2
is called laplacian laplaci an operator.
This equation can be rewritten as 2
∇ ψ=
−8π2m (
E − V )ψ
2
h
h2 2 − ∇ + ψ = Eψ V 2 8π m ˆ ψ = Eψ H ˆψ Where, H
=
−h2 2 ∇ + V is called Hamiltonian Hamilt onian operator. 2 8π m
In this operator, first term represents kinetic energy operator ( Tˆ ) and second term represents potential
ˆ = Tˆ + Vˆ . Hence Schrodinger wave equation can also be written as energy operator ( Vˆ ) i.e. H
(Tˆ + Vˆ ) ψ = Eψ Plot of radial wave function R The plot of radial wave function funct ion as a function of distance r from the nucleus gives the information i nformation about how the radial wave functi function on changes with distance r and about the t he presence of nodes where the change of sign of R occurs. The square of radial wave function R 2 for an orbital gives the probability density of finding the electron at a point along a particular radial line. Since Si nce the atoms have spherical symmetry, symmetry, it is more useful to discuss the probability of finding f inding the electron in a spherical shell between the spheres of radius (r + dr) and r. The volume of the shell is equal to 4 πr2 dr. dr. This probability which is independent of direction is called radial probability and is equal to 4 πr 2 drR2 . Radial probability function (= 4 πr2 R2) gives the probability of finding the elecron at a distance r from the nucleus regardless of direction. The plots of radial wave function R, radial probability density R 2 and radial probability function 4πr2 R2 for 1s, 2s and 2p atomic orbitals as a function of distance r from f rom the nucleus are shown in the figure.
... 19
1s
2s
R
2p
R
R Node
r
r
1s
R
r
2s
2
R
2p
2
R
2
Node
r
r 1s
r
2s
2
2p
2
R 2 r
R 2 r
4
4
π
2
R r
2
π
r
π
4
r
r
At node, the value of radial function changes from positive to negative. It has been found that ns-orbitals have (n – 1) nodes and np orbitals have n – 2 nodes. The radial probability function funct ion for the 1s orbital initially increases with increase in distance from the nucleus. It reaches a maximum at a distance very close to the nucleus and then decreases. The maxima in the curve corresponds to the distance at which the probability of finding the electron is maximum. This distance is called the radius of maximum probability. The radial probability function curve for f or 2s orbital shows two maxima, a smaller one near the nucleus and a bigger one at a larger distance. In between these two maxima it passes through a zero value indicating i ndicating that there is zero probability of finding the electron at that distance. The point at which the probability of finding the electron is zero is called a nodal point.
... 20
The distance of maximum probability for a 2p electron is i s slightly less than that for f or 2s electron. However, in contrast to 2p curve, there is a small additional maxima in the 2s curve. This indicates i ndicates that the electron in 2s orbital spends some of its time near the nucleus. In other words, the 2s electron penetrates a little closer to the nucleus and is therefore held more tightly than the 2p electron. So electron in 2s orbital is more stable and has lower energy than an electron in 2p orbital. Plots of angular wave function The plots of the angular wave function f unction ‘Θ, Φ’ and angular probability | Θ, Φ|2 for s and pz orbitals are shown below (fig a, b)
s
pz (a )
pz (b )
For an s-orbital, the angular part is independent of angle and is therefore of constant value. Hence the graph is circular or to be precise spherical. For the p z orbital, we get two tangent spheres The p x and py are identical in shape but are oriented along x and y axes respectively. In the angular wave function plots, the t he distance from the centre is proportional proporti onal to the numerical values of Φ’ in that direction and is not the distance from the centre of nucleus.
‘Θ,
The angular probability density plots can be obtained by squaring the angular function plots. For s-orbital, the squaring causes no change in shape. For F or both p and d orbitals, orbi tals, however, on squaring the plot tends to become more elongated (fig b). For a hydrogen atom, wave function of principal quantum number, n, there is a total of (n – 1) nodes that occur at finite values of the radial distance, r. The number of angular nodes is just equal to the angular momentum quantum number, l. Thus, we have angular nodes = l, radial nodes = n – l – 1. total nodes = n – 1, We should note that the total number of nodes in an atomic wave function is sometimes stated to be n rather than n – 1. In this case, the node that always occur at r = ∞ is being included in the count.
... 21
Shapes of Orbitals s-orbital:- s-orbitals s-orbital:- s-orbitals do not have directional character. They are spherically symmetrical. The s-orbital of higher energy levels are also spherically symmetrical. They are more diffused and have spherical shells within them where probability of finding the electron is zero. In the s-orbital, number of nodes is (n – 1) y
Node z
x
2s orbital p-orbital:- ‘p’ orbital has a dumb-bell shape and it has a directional directi onal character. The two lobes of a p-orbital are separated by a plane that t hat contains the nucleus and is perpendicular to the corresponding axis. Such plane is called a nodal plane because there is no probability of finding the electron. y y
y z
+
z
z +
–
+
x
x
x
– –
px
pz py
In the absence of an external electric or magnetic field, the three p-orbitals of a particular energy level have same energy and are degenerate. In the presence of an external magnetic field or electric field this degeneracy is removed.
... 22
d-orbitals:- For d-orbitals five orientations are possible viz., d xy, d yz, d xz, dx2 − y2 , d z 2 . All these five orbitals in the absence of magnetic field are equivalent in energy and are degenerate. The shapes of the orbitals are as follows: x
x
x
y
z
z
d xy
d yz
d xz
These three ‘d’ orbitals are similar. The maximum probability of finding the electron is in lobes which are directed in between the axes. Nodal region is along the axes. z
y
x
These two d-orbitals are similar. Probability of finding the electron is maximum along the axes and the nodal region is in between the axes.
Quantum Numbers These are used to determine the region of probability of finding a particular electron in an atom. (a) (a)
Prin Princi cipa pall quan quantu tum m num number ber (n): (n): This denotes the energy level or level or the principal or main shell to which an electron belongs. It can have only integral values 1, 2, 3 etc. The letters K, L, M ... are also used to designate the value of n. Thus, an electron in the K shell has n = 1, that is L shell has n = 2 and so on. Example 1: The 1: The principal quantum number of 2s-electron is ___. Solution: n Solution: n = 2
... 23
(b)
Azim Azimut utha hall quan quantu tum m numb number ers s (): This denotes the orbital (Sub-level ( Sub-level)) to which an electron belongs. It gives an idea about the shape of the orbital . ( n – 1), for a given value of n, can have any value from 0 to (n i.e. = 0, 1, 2, ... (n – 1) Value of Subshell
0
1
2
3
s
p
d
f
The value of of orbital angular momentum momentum of the electron for a given value of ‘’ is
(
+1
h . 2π
Question: What will be the value of angular momentum of d orbitals? orbit als? Answer: Angular Answer: Angular momentum of d-orbital (c)
=
h h 2(2 + 1) (For d-orbital value of = 2) 6 2) = 2π 2π
Magnet Magn etic ic quan quantu tum m numb number er (m): (m): It gives us the idea about the orientations an orbital can have in space in the presence of magnetic field. The values of ‘m’ depend on ‘’ orbital quantum number. number. Total number of alue of m=(2 +1) and it varries from – to +. For example, for = 0 the value of magnetic quantum number m is also equal to zero, i.e. ‘s’-orbital can have only one orientation in space in presence of magnetic field. fiel d.
(d)
Spin pin quant antum numb umber (s): s): The electron while moving round the nucleus in an orbit also rotates or spins about its own axis either in a clockwise direction or in an anticlockwise direction. Its value is +
1 1 or – corresponding 2 2
to clockwise or anticlockwise spin.
h 2 (2) The spin spin magnetic moment moment of electron electron (excluding orbital magnetic momentum) momentum) is given
(1) The value value of spin angular momentum for a given value of s is
by
µeffective =
s(s 1)
n(n + 2)BM (Whe (Where n = Numberof unpairede dellectrons)
... 24
Distribution of Electrons in an Atom:- Electronic configuration The filling up of orbitals with wi th electrons takes place according to certain rules which are given below: (i) (i)
The The max maximu imum m num numbe berr of of ele electr ctron ons s in in a ma main in shel shelll is eq equa uall to to 2n 2n 2, where n is the principal quantum number.
(ii) (ii)
The maximu maximum m nu numbe mberr of electro electrons ns in in a subsub-she shellll like like s, p, d, d, f is equ equal al to to 2(2 2(2 + 1), where is the azimuthal quantum number for the respective orbitals. Thus s, p, d, f can have a maximum of 2, 6, 10 and 14 electrons respectively r espectively..
(a)
Aufbau Pr Principle According to this principle, “Electrons are added progressively to the various orbitals orbital s in the order of increasing energy. ”
What does the word ‘Aufbau’ mean? Aufbau is a German term which means building up or construction. The energy of various orbitals increase in the order given below: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s ...
1s 2s 2p 3s 3p 3d
4s 4p 5s
4d 5p
4f
6s
5d 6p 7s
... 25
(i) (i) A new elect electron ron ente enters rs the orbital orbitals s for whic which h (n + ) is minimum, e.g. if we consider 3d and 4s orbitals, the electron will first fi rst enter 4s-orbitals in preference to 3d. This is because the value of (n + ) for 4s-orbitals is less (4 + 0 = 4) than that for 3d-orbital (3 + 2 = 5) (ii) (ii) In case case wh wher ere e ( n + ) values are the same, the new electron enters the orbital for which ‘n’ is minimum, e.g. in a choice between 3d and 4p for which (n + ) values are same (3 + 2 = 5, 4 + 1 = 5), the electron will prefer to go to the 3d-orbital, since n is lower for this orbital. (b)
Pauli’s Exclusion Principle It states that it is impossible for two electrons in a given atom to have same set of all four quantum numbers. OR An orbital can have a maximum of two electrons and that too with opposite spin.
Example: (a) n = 2, n = 2, (b) n = 2, n = 2, n = 2, n = 2, n = 2, n = 2, (c)
= = = = = = = =
0, 0, 1, 1, 1, 1, 1, 1,
m = 0, s = +1/2 m = 0, s = –1/2 m = 0, s = +1/2 m = 0, s = – 1/2 m = +1, s = +1/2 m = +1, s = –1/2 m = –1, s = +1/2 +1/2 m = –1, s = –1/2
Hund’s Rule of Maximum Multiplicity According to this rule, electrons enter the orbitals (e.g. s, p x, py, pz ...) in the same sub-level in such a way as to give maximum number of unpaired electrons. In other ot her words it means that pairing begins with the introduction of the fourth electron sixth electron in d subsheel and so on in p d subshell.
What is the electronic configuration of Cu (Z = 29)? 1s 2 2s 2 2p 6 3s2 3p6 3d10 4s 1
Exceptional Electronic Configuration Some atoms such as copper and chromium exhibit exceptional electronic configuration. For example: Cr(z = 24) has an electronic configuration 1s2 2s2 2p6 3s 2 3p6 3d5 4s 1 It is because of the extra stability associated with the half-filled and completely filled orbitals.
... 26
1.
Account for the following. Limits your answer to two sentences. “Atomic weights of most of the elements are fractional.” (1979)
2.
Nitrogen atom has atomic number 7 and oxygen has atomic number 8, then the total number of electrons in azide ion is (a) 20 (b) 21 (c) 22 (d) 19
3.
The increasing order (lowest first) for the value of e/m (charge/mass) for electron (e), proton(p), neutron (n) and alpha particle ( α ) is: (1984) (a) e, p, n, α (b) n, p, e, α (c) n, p, α , e (d) n, α , p, e
4.
Rutherford’s scattering experiment is related to the size of the : (a) nucleus (b) atom (c) electron
(1983)
(d) neutron
5.
With what velocity should an α -particle travel towards the nucleus of a copper atom so as to arrive at a distance 10 –13 meter from the nucleus of the copper atom? (1997)
6.
The number of photons emitted in 10 hours by a 60 W sodium lamp is (λphoton = 5893 A ) (a) 6.40 × 1023 (c) 6.40 × 1025
(b) 6.40 × 1024 (d) 5.40 × 1024
*7.
Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005 nm.
*8.
A photon of 300 nm is absorbed by a gas and thus re-emits re-emi ts two photons. One re-emitted re- emitted photon has wavelength 496 nm. Calculate the energy of other photon re-emitted out.
*9.
What accelerating potential is needed to produce an electron beam with wit h an effective wavelength of 0.09 A ? (a) 100 kV (b) 18.6 kV (c) 1 kV (d) 12.2 kV
10.
The minimum energy required to overcome the attractive attracti ve forces between electron and the surface of 19 – Ag metal is 7.52 × 10 J. What will be the maximum kinetic energy of electron ejected out from Ag which is being exposed to UV light of λ = 360 A°?
11.
The photo electric emission requires a threshold frequency v 0. For a certain metal
λ1 = 2200 Å and
λ 2 = 1900 Å produce electrons with a maximum kinetic energy KE 1 and KE2. If KE2 = 2KE1, calculate v0 and corresponding λ 0 .
... 27
12.
An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 A .
(1997)
13.
A photon of wavelength 4000 Å strikes a metal surface, the work function of metal being 2.13eV. Calculate, the energy of Photon in eV, (ii) the kinetic energy of emitted photoelectron, and (iii) the velocity of photoelectron.
14.
Following data were collected for the photoelectric emission of an electron from an element X,
λ(in nm)
KE(in eV)
254
1.93
313
0 .9
365
0 .5
405
0 .2
Calculate the wave function of an an element X and also the value of Planck ’s constant. 15.
Iodine molecule dissociates into atoms after absorbing light to 4500 A . If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of iodine atoms. (Bond energy of I 2 = 240 kJ mol –1) (1995)
16.
In hydrogen atom, electrons are excited to 4th energy level. The number of lines that may appear in the spectrum will be (a) 4 (b) 6 (c) 10 (d) 12
17.
The velocity of an electron in the first Bohr orbit of a hydrogen atom is 6.32 × 104 m/s. Its velocity in the second orbit would be (a) 2.16 × 104 (b) 1.5 × 104 (c) 3.16 × 104 (d) 3.16 × 108 21.7 × 10 −12
18.
The electron energy in hydrogen atom is given by E = −
19.
Estimate the difference in energy between 1st and 2nd Bohr ’s orbit for a hydrogen atom. At what minimum atomic number, a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with λ = 3.0 × 10−8 m ? Which hydrogen atom-like species does this atomic number corre-
ergs . Calculate the energy n2 required to remove an electron completely from the n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition? (1984 )
spond to? 20.
(1993)
Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr ’s orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionisation potential of the ground (1982) state hydrogen atom is 2.17 × 10 –11 erg per atom.
... 28
21.
An electron in a hydrogen atom absorbs 1.5 times as much energy as the minimum required for its escape (i.e., 13.6 eV) from the atom. Calculate Cal culate the wavelength of the emitted electron.
22.
Calculate the Bohr radius of third orbit of He + ion.
23.
What is the velocity of electron electr on in Bohr’s second orbit of H-atom?
24.
Calculate the wave number of the shortest wavelength transition in Balmer series of atomic hydrogen.
25.
Electromagnetic radiations of wavelength 242 nm is just sufficient to ionize sodium atom. Calculate the ionization energy of sodium in kJ/mol.
26.
The energy of the electron in the second and third Bohr orbits of the hydrogen atom is –5.42 × 10 –12 ergs and –2.41 × 10 –12 ergs respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third t hird to second orbit.
27.
Calculate the frequency of the spectral line emitted when the electron in n = 3 in hydrogen atom is de-excited to the ground state (Rydberg constant = 109, 737 cm -1).
28.
Find the number of waves made by a Bohr electron in one complete revolution in its third orbit.
29.
The ionization potential of hydrogen atom is 13.6 eV. eV. The energy required to remove an electron el ectron from n = 2 state of the hydrogen atom is (a) 3.4 eV (b) 6.8 eV (c) 13.6 eV (d) 27.2 eV
30.
If the wavelength of the first fi rst line of the t he Balmer series of hydrogen atom is 656.1 nm, the wavelength of the second line of this series would be (a) 218.7 nm (b) 328.0 nm (c) 486.0 nm (d) 640.0 nm
31.
The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a hydrogen-like species is (a)
32.
(b)
−
1 2
(c) 1
(d) –1
The ratio of kinetic energy and total energy of an electron in a Bohr orbit of a hydrogen-like species is (a)
33.
1 2
1 2
(b)
(c) 1
(d) –1
The ratio of potential energy and total energy of an electron in a Bohr orbit of a hydrogen-like species is (a) 2 (b) –2 (c) 1 (d) –1
... 29
34.
What is the wavelength emitted during the transition of electron in between two levels of He + ion whose sum is 5 and difference is 3? (a) 200 A° (b) 250 A° (c) 244 A° (d) 240 A°
35.
Consider the hydrogen atom to be a proton embedded in a cavity of radius a 0 (Bohr’s radius) whose charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron electr on in its ground state in a hydrogen atom as the work done in the above neutralisation process. Also, if the t he magnitude of the average kinetic energy is half the magnitude of the average potential energy, energy, find the average potential energy. (1996)
*36.
1.8 g of hydrogen atoms are excited by the radiations. The study of spectra indicates that 27% of the atoms are in third energy level and 15% of atoms in second energy level and rest in ground state. Ionization energy of H atom is i s 13.6 eV. eV. Calculate (i) the number of atoms present present in third and second energy levels, (ii) the total energy evolved when all the atoms returned to ground state.
*37.
The ionization energy of H is i s 13.6 eV, it is exposed to electromagnetic waves of wavelength 1028 Å and gives out induced radiations. Find the wavelength of these induced radiations.
38.
The angular momentum of an electron in a Bohr ’s orbit of H-atom is 4.2178 × 10 –34 kg m2 /s. Calculate the spectral line emitted when electron falls fall s from this level to next lower level. l evel.
39.
Which electronic level allows the hydrogen atom to absorb a photon but not emit photon? (a) 3s (b) 1s (c) 2p (d) 3d
*40.
The temperature at which the de Broglie wavelength of helium heli um atom is 0.62 A ° will be (Atomic mass of helium = 4.04) (a) 710.5 K (b) 750.4 K (c) 410.75 K (d) 570.4 K
41.
A dust particle has mass equal to 10 –11 g, diameter of 10 –4 cm and velocity 10 –4 cm s –1. The error in measurement of velocity is 0.1 –1. Calculate the uncertainty in its position. Comment on the result.
42.
What is the maximum precision with which the momentum of an electron may be known if the position is determined within ± 0.0001 Å.
43.
(a) The Schrodinger wave equation for hydrogen atom is: /
− / ψ = − / (π ) Where ao is Bohr’s radius. Let the radial node in i n 2s be at r o. Then find ro in terms of ao.
(b) A base ball having mass 100 g moves with velocity 100 m/sec. Find out the value of wavelength of base ball.
... 30
44.
The principal quantum number of an atom is related to the : (a) size of the orbital (b) spin angular momentum (c) or orientation of th the orbital in in space (d) orbital angular momentum
45.
The electrons, identified by quantum numbers n and (i) n = 5, = 1 (ii) n = 5, = 0 (iii) n = 4, = 1 (iv) n = 4, = 2, can be placed in order of increasing energy, energy, from the lowest to highest, as (a) (ii) < (iii) < (iv) < (i) (b) (iii) < (ii) < (iv) < (i) (c) (i) > (iv) > (ii) > (iii) (d) (i) > (iv) > (iii) > (ii)
46.
Which of the following sets of quantum numbers is not allowed? (a) n = 3, = 1, m = +2 (b) n = 3, = 1, m = +1 (c) n = 3, = 0, m = 0 (d) n = 3, = 2, m ± 2 Which of the following sets of quantum number is/are not allowed.
47.
(I) n = 4, = 3, m = - 1, s = +
1 2
(III) n = 3, = 0, m = + 1, s = + (a) only I 48.
(II) n = 2, = 3, m = + 1, s = 1 2
(IV) n = 2,
(b) II, III and IV
1 2
(d) only IV
Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is : (1984) (a) 5, 0, 0,
+
1 2
(b) 5, 1, 0,
+
1 2
(c) 5, 1, 1,
+
1 2
(d) 6, 0, 0,
49.
Which of the following atoms would be expected to be most paramagnetic? (a) 3Li (b) 4Be (c) 5B (d) 6C
50.
The ratio of magnetic moments of Cr, Cu, Fe is
51.
(1983)
1 2
= 2, m = + 1, s = +
(c) only II and III
(a) 4 : 1 : 8
(b) 6 : 1 : 4
(c) 2 : 1 : 8
(d) 8 : 1 : 8
+
1 2
A 3d orbital has (a) zero radial and two angular nodes (b) two radial and two angular nodes. (c) three radial and three angular nodes (d) two radial and zero angular nodes
... 31
52.
53.
Ground state electronic configuration of oxygen atom can be represented by (a)
(b)
(c)
(d)
Give reason why the ground state outermost electronic configuration of silicon is : 3s
3p
3s
(1985)
3p
and not
54.
Which of the following statement(s) is(are) correct? (a) Oxygen molecule is diamagnetic (b) The electronic configuration of Au is 5d106s1 (c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type (d) The oxidation state of nitrogen is N 3H is –3
55.
What is the maximum number of electrons that may be present in all the atomic orbitals with prin princip cipal al qu quan antu tum m num numbe berr 3 an and d the the azim azimut utha hall qua quant ntum um nu numb mber er 2? (1985)
56.
A compound of vanadium has a magnetic moment of 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound.
(1997)
57.
Write electonic configuration of the following: C(z = 6), Ar(z = 18), Br(z = 35), Cs (z = 55)
58.
Write electronic configuration of Mn2+ ion and calculate (i) number of unpaired electrons, (ii) magnetic moment and find whether the ion is paramagnetic or diamagnetic. (a) Statement – – I is True, Statement – –II is True; Statement – – II is a correct explanation for Statement – – I (b) Statement – – I is True, Statement – – II is True; Statement – –II is not a correct explanation for Statement – – I (c) Statement – –I is True, Statement – –II is False (d) Statement – – I is False, Statement – –II is True.
59.
Statement 1: - An orbital cannot have more than t han two electrons Statement 2:- The two electrons in an orbital create opposing magnetic field.
60.
Statement 1:- On increasing the intensity of radiation, the number of photoelectrons ejected and their KE increases. Statement 2:- Greater intensity means, greater energy which in turn means greater frequency of radiation
61.
Statement 1:- The configuration of B atom cannot be 1s 2 2s2 Statement 2:- Hund’s rule demands that configuration should display maximum multiplicity.
... 32
62.
Statement 1:- Angular momentum of e – in 1s, 2s, 3s etc. is zero Statement 2:- 1s, 2s 3s ... all have spherical shape
63.
Statement 1:- In Rutherford’s gold foil experiment, very few Statement 2:- Nucleus present inside the t he atom is heavy. heavy.
64.
–I : Band gap in germanium is small. Statement – Because –II: The energy spread of each germanium atomic energy level is infinitesimally small. Statement – (2007)
65.
Match the following Column I (i) Aufbau principle (ii) de Broglie (iii) Hund’s rule (iv) Balmer series
Column II P. Line spectrum invisible region Q. Orientation Q. Orientation of e – in an orbital R. λ = h = m ν S. Electronic S. Electronic configuration
66.
Column I (i) Thomson (ii) Pauli (iii) Bohr (iv) Chadwick
Column II P. Exclusion principle Q. Atomic Q. Atomic model R. Cathode R. Cathode rays S. Neutron S. Neutron
67.
Column I (i) Cathode rays (ii) Alpha particles (iii) X-rays (iv) P-orbitals
Column II P. Helium nuclei Q. Dumbell Q. Dumbell R. electrons R. electrons S. Electro S. Electro magnetic radiation
68.
According to Bohr ’s theory En = total energy Kn = Kinetic energy Vn = Potential energy rn = Radius of nth orbit
(2006)
Match the following: Column I (A) Vn / Kn = ?
Column II (P) 0
=?
(Q) –1
(C) An Angular mo momentum in in lo lowest or orbital
(R) –2
(B) If radius radius of nth oribit
(D)
∝ , = ?
α -particles are deflected back.
∝ ,
(S) 1
... 33
Passage Based upon the information given below, answer the questions that follow: According to Bohr ’
s theory for single electron species. The radius of orbit of an e – r
n
the velocity of e – in an orbit,
En
= − 1 3 .6 ×
νn = 2.18 × 106 ×
n2 = 0 .5 2 9 × Z
Å,
z m / s and, the Energy of e – in an orbit n
Z2
eV / atom . n2 Where n is the orbit in which e – revolves and Z is the atomic number of species
69.
The radius of second orbit of He + ion is: (a) 0.529 Å (b) 0.529 × 2 Å (c) 0.529 × 4 Å
70.
1 2
Å
The ratio of velocities of electron in i n third orbit of H-atom to that of electron elect ron in second orbit of Li 2+ ion is: (a)
71.
(d) 0.529 ×
1 2
(b)
3 2
(c)
2 9
(d)
1 3
The energy required to remove an electron from 1st excited state of H-atom is: (a) + 13.6 eV (b) + 3.4 eV (c) + 1,9 eV (d) zero
... 34
Atomic Structure 2. c
3. d
4. a
5. 6.34×104 ms –1 8. 2.63 × 10 –19 J
6. b 9. b 11. 1.1483×1015 ms –1 , 2.612 × 10 –7 m
7. 32.85 V 10. 47.68 × 10 –19 J 12. 63.57 V
13. 5.87 × 105 m / s
14. 6.38 × 10 −34 Js
15. 2.16×10 –20 J
16. b
17. c
18. 3.67×10 –5 cm
19. 16.35×10 –19 J, 2, He+
20. 1220 A
21. 4.7 × 10−10 m
22. 2.38 × 10−10 m.
23. 1.09 × 106 m / s 26. 6.6 26. 6.6 × 10 –7 m
24. 274 27419. 19.5 5 cm−1
25. 494.5 kJ/mole 28. 3 31. b 34. c
29. a
27. 2.92 × 1015 s –1 30. c 30. c
32. d
33. a 33. a
35.
− − , πε πε
36. (i) 1.62 × 1023 atoms, (ii) 830.50 J 37. 1.028 × 10 –7 m, 1.216 × 10 –7 m, 6.5688 × 10 –7 m 38. 1.8 × 10 –4 cm 41. 5.27 × 10 –6
39. b
40. c
42. 5.28 × 10 −22 Kg ms −1
43. (a) 2ao (b) 6.626×10 –25 A
44. a
45. b
46. a 46. a 49. d 52. d 56. 1s2 2s2 2p6 3s2 3p6 3d1 4s0 61. b
47. b 50. a 54. b, c 59. b 62. b
48. a 48. a 51. a 55. 10 60. d 63. b
64. c
65. (i) → S, (ii) → R, (iii) → Q, (iv ) → P
66.((i) → R, (ii) → P, (iii) → Q, (iv ) → S 66.
67. (i) → R, (ii) → P, (iii) → S, (iv ) → Q
68. A
→ R, B → Q, C → P, D → S
69. b
70. c 70. c
71. b 71. b
... 35
