Atoms

ATOMS Thomson’s Atomic Model : According to this model, an atom consists of a positively charged sphere in which entire mass & positive charge of the atom in uniformly distributed. Inside this sphere, the elctrons are embedded like seeds in a waternelon or like plums in a pudding. The number of elecrons is such that their negative charge is equal to positive charge. Thus, atoms is electically neutral.

_ + + +

+ _

+

Limitation of Thomson’s Model : (i) Could not explain the origin or spectral series of hydrogen & other atoms, (ii) Could not explain large angle scattering of - iparticles observed by Rutherford. Rutherford’s - scattering exsperiment : An - partifcle is He nucleus containing 2 protons & 2 neutrons. It has 4 unit of mass & 2 units of positive charge. Many radioactive elements emit - particles. S is a radioactive source contained in a lead cavity. The - particle emitted by the source are collimate into narrow beam with ehe help of collimated beam is allowed to fall on a thin gold fail of thickess = 105 m. - particles are scattered in different directions are observed though a rotatable detector consisting of ZnS screen & a microscope. The - particles produce bright flashes on ZnS screen. These are observed by the microscope & counted at different scattering angle .

Gold foil

ZnS screen

S Lead cavity

Collimator Observations : (i) Most of the - particle pass straigh through the gold foil or suffered very small angle fo deflections. (ii) A few - particle scatter through large angles (> 90°). (iii) Rarely, an - particle rebounces i.e., acattered through an angle of 180°. Explanation : (i) Since most of the - particles passed undeviated, the atom has a lot of empty space in it. (ii) To explain large scattering of - particles, Rutherfored suggested that all the positive charge & entire mass of the atom is confined to an extremely small central core called as nucleus. (iii) The scattering of - paticles through different angles was explained as : The - particles I, I’. Which pass through the atom at a large distance from the nucleus experience a small electrostatic force of repulsion & undergo a small defection. The - particle 2. 2’ which pass through the atom at a close distance from the nucleus suffer a larger defection. The - particle 3, which travels directly towards the nucleus down, comes to rest & is deflected through 180° and hence retraces its path.

1 2 3 2' 1'

+

The graph between scattering angle  & the number of - particles directly scatted N is as shown :

Number of scattered particle N

0°

Distance of closet Spproach : An - particle travelling towards the center of the nuclus slows down as it apparoaches the nucleus. At a certain distance, say r0 from the nucleus, the - particle comes to rest for a moment and then retraces it path. It inital kinetic energy is completely converted into elestrostatic pootential energy. This distance r0 is called the distance of closes approach. This distance gives an estimate of the size of the nucleus. Mathematically. +Ze

h0

1 1 2eZe  mv 2  2 4 f0 where 2 e is the charge on - particle and Ze is the charge on nuclus.

r0 

Ze 2e 1  4 0  mv2  2  

Impact Parameter : Impact parameter is defined as the perpendicular distance of the velocity vector of the - particle from the center of the nucleus, when it is far away from the nuclus of the atom. Rutherford derived the relation between impast parameter andde scattering angle, which is given by.

b Nucleus

1 Ze2 cot  / 2 1 4 0 mv 2 2 Rutherford’s Model : On the basis of - scattering experiment, Rutherford proposed the model of an atom as : 1. An atom consists of a small, and massive central core in which the entire positive charge and almost the whole mass are concentrated. This core is called a smaucleus. b

2.

The size of the nucleus is very small as compared to the size of atom.

3. The nucleus is surrounded by a number of electrons so that their total negative charge is equal to the total positive charge and the atom is electrically neutral. 4. The electrons revolve around the nucleus in various orbits. The censtripetal force required for their revolution is provided by the elctrostatic attracvtion between the elctrons and the nucellus. Drawbacks of Rutherford’s model of Atom : 1. If failed to explain the stability of the atom. The electrons revolving around the nuclus are accelerated towards the nucelus of the atom. Any chaged particle while accelerating ore retarding loses energy through electromagnetic radiations. As a rusult, the radius of the pathe rof the electron should go on decreasing and ultimately it sould fall into the nuclues by following spiral path. But, in fact, it nerer happens. It contradicts the stability of matter. 2. It failed to expain the spectrum of an atom. As electron can revolve around a nucleus in circular orbits of all possible radii, so atom should eemit continuous energy spectrum. But even the simplest atom i.e. hydrogen atom has line spectrum instead of continuous spectrum, which Rutherford’s model

could not explain. Bohr’s Model of Hydrogen Atom : 1. An atom consists of a central core called nucleus, in which entire positive charge and almost entire mass of the atom is concentrated. Electron revolve around the nucleus in circular orbits. The entripetal force required for revolution is provided by the electrostatic force attraction between the electron and the nucleus. If m is the mass of elctron moving with a velocity v in a circular orbit or radius

mv2 r Also, the electrostatic force of attraction between the nuclus of charge (+Ze) and electron of charge (e) is r , then the necessary centripetal force if F 

F

1 Zee  KZe2  2 4r0 r 2 r

1 where K  4 0

2. According to Bohr, electrons can revolve only is those certain discrete non radiating orbits, called steionary orbits, for which total angular momentum of the revolving electron is an intergral multiple of h/2, where h is Plank’s constant. Thus the angular momentum of the orbiting electron is quantized. While revolving in these stationary orbits, an electron does not radiate energy. For any permitted stationary orbit. nh 2 where n is any positive integer, 1,2,3 .................... It is called principle quantum number. mvr 

3. The emission/absorption of energy occurs only when an electron jumps from one stationary nonradiating orbit to another. The difference in the total energy of electron in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit i.e. absorbs photons, and emitted when electron jumps from outer to the inner orbit i.e., emits photons. If E1 is total energy o electron in an inner stationary orbit and E2 is its total energy in an outer stationary orbit, then hv = E2 - E1 Some quantities related to Bohr’s orbit : (a) Radii of Bohr’s stationary orbits : We know, mvr =

nh nh or v = 2 2mr

mv2 KZe2  2 Substituting v in the equation r r r

r r

n2h2 42m2r 2



n2h2 42mKZe2

KZe2 r2

for hydrogen atome Z = 1

n2h2

This shows that r  n2 i.e. radii of stationary orbits are in the ration 12 : 22 : 33 and 42mKe2 so on i.e. 1 : 4 : 9 .............. clearly the stationary orbits are not equally spaced. (b) Velocity of electron in BVohr’s stationary orbit : We have, r 

Therfore,

KZe2 mv 2

KZe2 mv 2



Also,

r

nh 2m

nh 2KZe2 or v  2mv nh

For hydrogen atom Z = 1

Therefore,

v

2Ke2 nh

(c) Total energy of electron : The energy of electron revolving in a stationary orbit is of two types : Kinetic energy is due to motion and potential energy is due to position of electron. We have

mv2 KZe2  2 r r

i.e., K.E. of electron =

1 KZe2 mv2  2 2r

Potential energy of electron =

KZe e KZe2  r r

Total energy of elecron in the orbit E = K.E.  P.E.  Substituting r =

n2h2

we get E 

42mKZe2

22mK 2Z2e 4

 E  

For hydrogen atom Z = 1

1 KZe2 KZe2 KZe2   2 r r 2r

n2h2 22mK 2e 4 n2h2

Substituting the standard values, we get E 

13.6

eV n2 Total energy of electron in a stationary orbit is negative, which means the electron is bound to the nucleus by means of electrostatic attraction. Origin of spectral lines : Suppose E1 = total energy of electron in the inner (n1 th) orbit and e2 = total energy of electron in the outer (n2 th) orbit. When an electron jumps from an outer to an inner orbit, the frequency of radiation emitted, according to Bohr’s third postulate is hv = E2 - E1

hv 

22m2e 4 n22h2



22mK2e 4 n12 h2

hc 22mK2e 4  1 1   2  2 2  h  n1 n2  E2 E1 +

1 22mK2e 4  1 1   2  2 2  h  n1 n2  Now, and



1   v the wave number of radiation emitted i.e., number of complete wae in unit lenght. 

22mK 2e 4 ch2

= R is a constant called Rydberg constant. R = 1.097 × 107 m-1

1  1 From above equation v  R  2  2   n1 n2 

This equation is called the Rydberg formula for the spectrum of hydrogen atom. Bohr’s explanation of Spectral of Hydrogen Atom : 1. Lyman Series : Bohr postulated that Lyman series of obtained when an electron jmps to the first orbit (n1 = 1) form any outer orbit (n2 = 2,3,4..................)

1  1 v  R  2  2  where n : ,2,3,4 ............... This series lies in the ultra violet region of the 2 n2  1  spectrum and agree well with the values of v observed experimentally by Lyman. 2. Balmer Series : According to Bohr, Balmer series is obtained when an electron jumps to the second orit (n\1 = 2) from any outer orbit (n2 = 3,4,5.........) Wave number of these spectral line is

1  1 v  R  2  2  where n = 3,4,5 ................. This series lie in the visisble part of 2 n2   2 the spectrum. 3. Paschen Series : According to Bohr Paschen series is obtained when an electron jumps to the 3rd orbit (n1 = 3) from any out of orbit (n2 = 4,5,6 ..........) Bohr calculated the wave number of spectral lines of paschen series from the relation.

1  1 v  R 2  2  n2   3 where n2 = 4,5,6 ............. This series lies in the infrared region of the spectrum. 4. Brackett Series : According to Bohr, Brackett series is obtained when an electron jumps to the 4th orbit (n1 = 4) from any out orbit (n2 = 5,6,7................). This series lie in far infrared region.

1  1 v  R 2  2  n2   4

where n2 = 5,6,7 .....................

5. Pfund Series : According to Bohr, Pfund series is obtained when an electron jumps to 5th orbit (n1 = 5) from any outer orbit (n2 = 6,7,8..........). This series lies in far infrared region.

1  1 v  R 2  2  n2   5

where n2 = 6,7,8 ..............

Energy Level Diagram : A diagram which represents the total energies of electron in different sationary orbits of an atom are called the energy level diagram and are represented by parallel horizontal lines. Total energy in the nth orbit of hydrogen atom is given by E 

13.6

eV n2 Substituting n = 1,2,3 ............. we get the energies of electrons in various stationary orkbits as :

E1  E3  E5 

E7 

13.6 12 13.6 2

3

 13.6 eV

 1.51 eV

13.6 2

5

13.6

 0.54 eV

E2  E4  E6 

13.6 22 13.6 42 13.6 62

 3.4 eV  0.85 eV  0.37 eV

 0.28 eV 72 Clearly, as n increaes, En becomes less negative until at n =  , En = 0

n = 8 n = 7

n = 6

E = -0.54 eV

n = 5

Pfund Series

E = -0.85 eV

n = 4

Brackett Series

E = -1.51 eV

n = 3

Paschen Series

E = -3.4 eV

n = 2 Balmer Series

E = -13.6 eV

n = 1 Lyman Series

NUCLEI Limitations of Bohr’s Theory : 1. This theory is applicable only to simplest atom like hydrogen, with Z = 1. The theory fails in case atoms of other elements for which Z > 1 2. The theory does not explain why orbits of electrons are taken as circulr, while elliptial orbits are also possible. 3.

Bohr’s theory does not explain the fine structure of spectral lines oven in hydrogen atom.

4.

Bohr’s theory does not say anything about the relative intensities of spectral lines.

5.

Bohr’s theory does not take into account the wave properties of electrons.

Excitation : It is the process of absorption of energy by an electron of an atom when it jumps from lower energy state to higher energy states, First excitation energy of hydrogen is = E2 - E1 = -3.4 - (-13.6) = 10.2 eV Excitation potential : It is defined as the potential difference through which an electron in an atom must be accelerated so that it may go from the ground state to the excited state. First excitation potential of hydrogen = -3.4 - (-13.6)= 10.2 V Second excitation potential of hydrogen = -1.51 - (-13.6) = 12.09 V Isonisation : The process of knocking out an electron from the atom is called ionization. Ionisation energy : It is defined as the energy required to knock an electron completely out of the atom i.e. energy required to take an electron from its ground state to the outermost orbit (n =  ). Ionisation energy of hydrogen = E2 - E1 = 0 - (-13.6) = 13.6 eV Ionisation potential : The potential difference through which an electron of the atom is acceleratd so that it is knocked out of the atom. Th ionization potential of hydrogen atom = 0 - (-13.6) = 13.6 V Atomic Nucleus : Rutherford established that atomic nucleus is the central core of every atom, in which the entire prositive charge & almost entire mass of the atom are concentrated. The atomic nucleus is regraded as a tiny sphere of diameter ranging from 1015 m to 10-14 m. Most of the space around the nucleus in an atom is emply space. Thus, a nucleus of an atom consists of positively charged particle called rpotons and neutral particles callled neutrons. Atomic Number : The number of protons present inside the nucleus of an atom is called atomic number. It is equal to the number of electrons present in that atom. It is representd by Z. Mass Number : Mass number of an element is the total number of protons & neutrons present inside

the nucleus of the element. It is represented y A. Number of protons = NUmber of electrons = 2 Numer of nucleons = A Number of neutrons = A - Z An element is represented as

ZX

A,

where Z is the atomic number, A is the mass number & X is the

chemical symbol of that element, Example

23 n Na

Nuclear Size : It is experimentally found that volume of a nucleus is proprotional to its mass number A. 4 R 3  A 3

where R is the radius of the nucleus.

R  A1 / 3

 R  R 0 A1 / 3 where R 0 is called as empirical constant.

Nuclear Density : Nuclear Density is the ratio of mass of nucleus to its volume. If m is the average mas of a nucleon then mass of nucleus = mA Volum of nucleus = 

Nuclear density





4 4 R 3   R 0 A1 / 3 3 3



3



4 R 30 A 3

mA 4 R 30 A 3

3m 4R 30

It is of the order of 1017 Kg/m3

Atomic Mass Unit (amu or u) : One atomic mass unit is defined as 1/12th of the mass of an atom of 12 6C

isotope. 1 amu = 166 × 10-27 kg

Energy Equivalent of Atomic Mass unit : According to Einstein’s mass energy equivalent : E = mc2 m = 1 amu = 1.66 × 10-27 kg E = 1.66 × 10-27 × (3 × 108)2  E = 1.49 × 1010

E

1.49  10 10 1.6  10 10

MeV

E = 931.25 MeV Hence 1 amu = 931 MeV Isotopes : Isotopes of an element are the atoms of the element which have the same atomic number but mass number. For e.g. 1 2 1 H . 1H 2 He

3

. 1H3 are the isotopes of hydrogen.

, 2He 4 , 2H6 are the isotopes of helium.

10 11 6 C , 6C

, 6C12 , 6C14 are the isotopes of carbon.

Isobars : Isobars are the atom of different elements which have the same mass number, but different atomic number. The total number of nucleons is same. For e.g. Isotones :

3 1H

& 2He 4 are isotones

22 11 Na 16 8O

and

22 10 Na

are isobars.

& 6C14 are isotones.

Isomers : These are the nuclei with same atomic number and same mass number but existing in different energy states. Nuclear Forces : Nuclear forces are the strong attractive forces which hold together the nucleons (protons & neutrons) in the nucleus. Important characteristic of these force are : 1. Nuclear forces act between a pair of neutrons, a pair of protons & also between a neutron & a proton with the same strenght. This shows that these forces are independent of charge. 2. Nuclear forces are the strongest forces in nature. 3. Nuclear forces are the very short-range forces. 4. Each nucleon interacts with its immediate neighbors only.

5. Nuclear forces are non-central forces. The force between two nucleons does not act along the line joining their centres. 6. Nuclear forces are exchange forces. These forces are due to exchange of  mesons between the nucleons. Various of PE of a pair of nucleons with distance : Nuclear binding force dominates over the Coulomb repulsive force between protons inside the nucleus. So nuclear force is stronger than the Coulomb force or gravitational force between two charges.

PE

r0

r(fm)

(a) The PE is minimum at a distance r0 = 0.8 fm. For a distance greater than r0, the PE is negative which signifies that the nuclear force is attractive. It repidly decreases with distance and becomes negligible small at a distance of about 4 fm. (b) For a distance less than r0, the nuclear force is strongly repulsive, so the PE is positive. Mass Defect : The different beteen the sun of the masses of the nucleons in a nucleus & the rest mass of the nucleus is known as mass defect. It is denoted by m. The nucleus of an atom

zX

A,

contains Z

protons & (A-Z) neutrons. Mass of the ucleons in the nucleus = Zmp + (A – Z)mn where mp is the mss of proton and mn is the mass of a neutron. If ms is mass of the nucleus of by :



z XA

then mass defect is given



m  Zmp  A  Zmn  m

When a nucleus is formed from its nucleons, some of their mass is converted into energy whichbinds the nucleons together inside the nucleus. This energy is called binding energy and is equivalent to mass defect. Mass defect is measured in a.m.u. if atomic masses are in a.m.u and if atomic masses are in kg, then mass defect is meausred in kg. Binding energy : Binding energy of a nucleus is the nergy required to break up a nucleus into its constituent protons and neutrons and to separate the nucleons at infinit distance apart form the nucleus, so that they may not interact with each other. If m is the mass defect of the nucleus. Then according to Einstien’s mass energy relation :





Bonding energy = mc 2  Zmp  A  Zmn  m c2 If m is a.m.u. then Binding energy = Zmp  A  Z mn  m × 931.5 MeV Binding energy per nucleon : The binding energy per nucleon is the average energy required to extract one nucleon from the nucleus. It gives the measure of the stability of that nucleus. Greater the binding energy per nucleon, more stable is the nucleus.





2 B.E Zmp  A  Zmm  m c  A A Binding energy curve : The variation of average B.E. per nucleon with mass number A is shown as :

B. E. per nucleon =

10 8

O16

S32

Fe56

Mo100

l127

W184

U238

12

C He 4

6

N14

Li6 4 H3 2 H2 0

50

100

150

200

250

(i) Average B.E./nucleon for light nucel like 1H4 , 1H2 , 1H3 is small (ii) For mass number ranging from 2 to 20, there are sharply defined maxima peaks corresponding 2 He

4

, 6C12 , 8O16 . The peaks indicate that hese nuclei are relatively more table the nuclei in their neigh-

borhood like

6 10 14 3 Li , 5B , 7N

(iii) After mass number 20, binding energy per nucleon increases & for mass numbers between 40 & 120, the curve become more or less flat. Fro

56 26 Fe

, the B.E./nucleon in maximum & it is equal to 8.75

MeV. The average value of B.E./ nucleon is this region is about 8.5 MeV (iv) As the mass number increases, the B.E./ nucleon decreases gradually to about 7.6 MeV for

238 . 92 U

The decrease is due to Coulomb repulsion between the protons. The heavy nuclei are therefore, relativly less stable. (v) The B.E./nucleon has low value for both very light & very heavcy nuclei/ If order to attain higher value of B.E./ nueleon, the lighter nuclie may unit together (nuclear fusion) or a havy nucleus may split into higher nuclei (nuclear fission). Redioactivity : The spontaneous transformation of an element into another with the emission of some particles ofr radiations (-particles, - particles, -rays) is called natural radioactivity. The substances capable of emitting radiations are calld radioactivity was discovered by Becquerel, who noted that Uranium element gave out some invisible rays or radiations that can penetrate through several thick black papers and affect a photographic plate on the other side. Pierre Curie and Mariee Curie then confirmed the same phenomenon. They named these rays as Becqurerel rays. Rutherford and this associates investigated the nature of Becquerel rays. These are of three types : - rays, - rays & -rays. A smaple of radiactive element (Radius) is palced in small cavity drilled in a lead ablock. The radiations coming out of the cavity are subjected to an electric field providd by two plates as shown in the figure.

E

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Magnetic field

+ + + + + + + + + + .. . . . . .

. . .. . .. . . . .. . . . .

Radioactive Substance Lead Block

.. . . . . .

. . .. . .. . . . .. . . . .

The -rays are defected through smaller angles towards the negative plate. The - rays are deflected through larger angles towards the positive plate. The - rays or photons remain undeflected. So it was concluded the  - rays consist a stream of positively charged perticles, where as  - rays consists a stream of negatively cahrged particles. Since  - rays were undeflected they could be waves or uncharged particles. Same results were obtained when these radians were subjected to magnetic field applied are to plane of paper. Properties of - rays : 1.

An - particle is equivalent to a helium necleus

 He  consisting of two protons and two neu2

4

trons. 2.

It has a positive charge equal to +2e. where e = 1.6 × 10-19 C

3.

Mass of - particle is of the order of

1 th of the velocity of light. 10

4. Because of large mass, pentrating power of - particles is very small, - particles can be easily stopped by an AI shet, 0.02 mm thick. 5. Because of large mass, penetrating power of - particles is very small. - particles can be easily stopped by an Al sheet, 0.02 mm thick. 6.

Because of large mass, penetrating power of - particles have large ionizing power.

7.

- particles produce fluorescene in certain substances like sinc sulphide.

8.

They affect photographic plate.

9.

They are deflected by electric & magnetic fields.

10.

While passing through thin metal foils, they get scatted.

11.

They cause burns on human body.

Properties of - rays : 1.

A - particles is a fast moving electron



1 e

0



2. A - particle carries the charge of an electron i.e. 1.6 × 10-19 C of negative charge. 3. The rest mass of - particle is 9.1 × 1031 kg. same as that of electron. 4. The velocity of - particle range from 33% to 99% of velocity of light. So - particles are fast mvoing electrons. 5. Beacuse of small mass, the penetrating power of - particles is very large. They can easily pass through a few milimeter of Ai sheet. Their penetrating power is 100 times that of - particles. 6. The - particles ionize the gas through which they pass, but their ionizing power is 1/100th that of - particles. 7. They can also produce fluorescence in certain substance like Zinc Sulphide. 8. They affect photographic plate. 9. They are deflected by electric & magnetic fields. Properties of -rays : 1. They are electromagnetic waves which have wavelenght less than of X-rays. 2.

 - rays are not deflected by electric & magnetic field showing that they do not have any charge.

3.

 - rays travel with the speed of light. The rest mass of a  -photon is zero.

4.  - rays have very large penetrating power. Their penetration power is 100 greater than that of  - particles. They can pass through several cm. of iron & lead sheet. 5. They have small ionising power. Their ionizing power is about 1/100 times the ionizing power of  - particles. 6.

They can produce fluorescence in certain substances.

7.

They can knock out electrons from the surface on which they fall.

8.

They affect a photographic plate more than  - particles.

9.

They can produce nuclear reactions.

Radioactive transformation : 1. Radioactivity is a spontaneous phenomenon i.e. it does not depend upon external factors like temperature pressure etc. 2. When a radioactive atom disintegrates, either an - particle of a - particle is emitted. Both are never emitted simultaneously. Also at a time, a radioactive atom can never emit more than one  particle or - particle. 3. The emission of - particle by a radioactive atom results in a new atom whose atomic number is 2 units less & mass number is 4 units less than the original atom ZX

A

  docay    z  2 Y A  4

4. The emission of - particle by a radioactive atom results in a new atom, whose atomic number is one unit more but mass number is same that of original atom. ZX

A

  docay    z 1 Y 4

Radioactive decay law : This law states that the number of nuclei disintegrated per second of a radioactive sample at any instant is directly proportional to the number of undecayed nuelipresent in the sample at that insant. Mathemeatical from of decay law : Let N0 be the total no. of nuclei present originally in a sample at t = 0. Let N be the total no. of nuclei left in the sample at time t and dN be the total no. of nuclei that dN Here (–) ve sign indicates that no. of nuclei left dt undecayed decrease with time. Accoriding to radioactive decay law :

disintegrate in time dt. Rate of disintegration R 

dN dN N  R   N ....(1) dt dt where is called as distegration constant or decay constant. Eq. (1) can be written as : 

dN  dt N N

Integrating both sides :



N0

dN  N

N

 dt

N0

loge N  loge N0  t loge

N  t N0



N  e  t N0

N0

number of atoms left (N)

0

t

N  Nee 11 Graph of radioactive decay shows that number of active nuclei in the sample decreases exponentially with the time.

Dacay of Disintegration Constant : We have N= Nee-11 N0 1  0.368 N0 we get, N = Nee-1 =  e The disintegration constant of a adioactive element is the reciprocal of time at the end of which, the

Substituting = t 

1 number of active nuclei left undecayed in a radioactive substance reduces to   to,es pr 36.8% of its e -1 initial numbe its SI unit is as s .

Half Life of Radioactive Element : Hair life of radioactive element is defined as the time during which half the number of radioactive nuceli present initally in the sample of the element decay. It is represented by T1/2. When t = T1/2, N = N0/2 Substitution in N = Nee-lt, we get

N0 T  N0e 1 / 2 2

1 T  e 1 /2 2



 2e

T1 / 2

Takein lig of both sides :  T1 / 2 loge e  loge 2  2.303 log10 2  T1 / 2  1  loge 2  2.303  0.3010 T1 / 2 

0.693 Thus half life is inversely proportional to decay constant and independent of N0. 

N0 2

When T  T1 / 2 N 

After 2 half lives, N 

After 3 half lives, N 

1  N0  2  4

 1   N0   2 

1  N0  2  2

 1   N0   2 

2

3

n

1 1 After n half lives, N  N0    N0   2   2

.................................................... T1 / 2

Where t = n × T1/2 is the total time of n half-lives.

Average life of mean life of Radio effective Element : The average time for which the nuceli of radioactive sample exist is called mean life or average life of that sample. Average life of radioatie element can be determined by calculating the total life time of all the nuclei & deviding it by the total number of nucler pesent initially in the sample. Total life of dN nuclei = tdN N0

 tdN

Total life time of all the Ne nuclei =

e

n0

Averrage life T  av We know

Therefore,

 tdN 0

N0



1 N0

N0

 tdN 0





dN   Ndt    N0e  t dt

Tav

1   N0

N0

  N e 0

 t

dt  t

0





Tav   te  t dt 0

Integrating by parts, we get Tav 

1 

when N = 0, t =  when N = N0, t = 0

0.693 Ths average life is the reciprocal of decay constant. As   T 1/2

T1 / 2

Therefore, Tav

0.693

 1.44T1 / 2

Decay rate of activity of a radioactive sample : The activity of a radioactive sample is defined as the number of radioactive disintegration taking place per second in the sample. If a radioactive sample contains N radionucleir at any time t, then activity R is given by R 

dN Here (–) sign indicates that activity of the sample decreases with the passage of time dt

 t We know N  N0 e

R  t 

so, R 





dN d  N0 e  t  N0e  t dt dt

0.693N T1 / 2 More is the half life, less is the activity of the substance.

Units of Activity : 1. Becquered (Bq) : The activity of a radioactive substance is said to be 1 bequered if it has 1 disintegration per second. 1 Becquerel = 1 disintegration/second. 2. Curie (Cr) : The activity of a radioactive substance is said to be 1 Curie if it has 3.7×1010 disintegration per second. 1 Curie = 3.7 × 1010 disintegration/second = 3.7 × 1010 Bq 3. Rutherfored (rd) : The activity of a radioactive substance is said to be 1 Rutherford if it has 106 disintegration per second. 1 Rutehrford = 106 disintegration/second. Alpha Decay : Alpha decay is a process in which an unstable nucleus transforms into a new nucleus by emitting an - particle. When a nucelus emits an - particle its mass number decreases by 4 & atomic number decreases by 2.

  z  2 Y A  4  2He 4  Q where Q is the energy released in the decay. Q is calculated by Einstein mass energy relation E = mc2 Q = (mX - mY - mHe) c2 1X

A

238 92 U

90 Th234  2 He 4  Q Beta Decay : It is the penomenon of spontaneous emission of an electron (e-) or a position (e4) from a radioactive nucleus. e.g. :

Beta minus decay : in - decay, the mass number remain uncharged but its atomic number increases  by one. An electron and a new particle antineutrion ( v ) are emitted from the nucieus. zX

A

 z 1 Y A  1 e0  v or

zX

A

 z 1 Y A  0  v

10 N22 e  e' v Note : A nucleus does non contain any electron, position neutriono and antineutrino, yet it can eject these particles. In beta-minus deay, a neutron transforms into a proton inside the nucleus via the reaction. e.g. :

22 11 Na

n  n  e  v It shows that a beta decay procss involves the conversion of a neutron into a proton or vice-versa. These nucleons have equal mass. That is why mass number A under-goin beta decay does not change. The graph between energy of - particle is continuous i.e. - particles can carry all possible energies from zero to maximum.

end point energy

Gamma Decay : It is phenomenon of emission of gamma ray photon from a radioactive nucleus. It occurs when an excited nucleus makes a transition to a state of lower energy.

 zX A   After an - or - decay, the nucleus is usually in an excited state & it attains ground state by emitting one or more - rays photons . e.g. : The - decay of 27Co60 transforms it into an excited 28Ni60 nucleus. This reaches the ground sate by emission of - rays of energies 1.17 MeV & 1.33 MeV. The energy level diagram is as shown. zZ

27 CO

60 

60   28 Ni 60   28 Ni

A

 28Ni60   1 e0

 28Ni60   E 1.17 MeV   28Ni60   E 1.33 MeV 

Nuclear Reaction : It represents the transformation of stable nucleus of one element into stable nucleus of another element, Rutherford bombarded nitrogen with -particle given by 14 7N

 2 He 4  8O17 1 H1 In any nuclear reaction the quantities conserved are : Momentum, Nucleon, Charge and Energy. Nuclear Fission : The phenomenon of splitting of a heavy nucleus into two or more lighter nuclei.  0 n1  56Ba141  36 Kr 92  30 n1  Q Mass defect in this reaction is 0.2153 amu. e.g. :

235 92 U

Chain Reaction : In the above reaction neutrons produced may bring the fission of three more nuceli & produce 9 neutrons, which in turn bring fission of nine

235 92 U

235 92 U

nuclei & so on. Thus a continuour

reactin called chain reaction would start & a huge amount of energy is released in a short time. It is of two types : (i) Uncontrolled chain reaction : If the fissionable material has a mass greater than the critical mass, then the reaction will accelerated at such a rapid rate that the whole material will xplode within a microsecond, liberating a hage amount of energy. Such a chain reaction is called uncontrolled chain reaction. If forms the underiying principle of atom bomds.

Kr2 Kr

U

U n

Ba

Ba

U

U Ba

U U

(ii) Controlled chain reaction : If the chain reaction is controlled an maintained at steady state by absorbing a suitable number of neutrons at each stage, then the reaction is called as controlled chaing reaction. here the energy released does not get out of control. A nuclear reactor works on this principle. Multiplication Factor : If is define as the ratio of rate of production of neutrons to the rate of loss of neutrons.

k

rate or production of neutrons rate of of neutrons

(i) If k=1, the chain reaction will be steady. Th size of the fissionable material is aid to the critical size and its mass, the critical mass. (ii) If k > 1, the neutron population increases exponentially with time and the chain reaction accelerates resulting in an explosion. The size of the material in this case is said to besupercritical.

(iii) If k < 1, the neutron population decreases exponentially with time and the cahin reaction gradually dies out. The size of the material in this case is said to be subcritical. Nuclear Reactor : The controlled nuclear Fission chain reactions for peaceful purposes can be carried out in nuclear reactors. It words on the principle of controlled chain reaction and provides energy at a constant rate. Construction : Its essential parts are : 1. Nuclear fuel : Uranium isotope (U235), Thorium isotope (Th232), etc are the most commonly used fuels in the reactor. 2. Moderator : Doderator is used to slow down the fast moving nutrons. Most commonly used moderator are grapohite, heavy water etc. When heavy water is used as maderator, then non-enriched U ca be used as fuel because heavy water has more neutrons to produce dission. 3. Control Rods : They are used to control the chain reaction and to maintain a stabel rate of reaction. It controls the number of neuttrons availibale from the fission. Cadmium rods are inserte into the core of the reaction because they can absorb the neutrons. The neutrons available for fission are controlled by moving the cadmium rods in or out the core of the reactor. 4. Coolant : It is a cooling material that removes the heat generated du4e to fission in the reactor, Commonly used coolants are water, liquid sodium etc. 5. Protective Sheilds : Protective shilds in the form of acconcrete thick wall surrounds the core of the reactr to save the persion working around the reactor form the hzazardous radiations. Working : A few 92U235 Nuclie undergo fisison liberating fast neutrons. These sfas neutrons are slowed down by surrounding modeator. The cadmium rods are used to control the chain reaction. The fission rpdocues hat in the nuclear core. The coolant transfers this heat from the core to the exat exchange, where steam is formed. This steadm produced at very high-pressire runs a turbine and the electricity is obtained at the generator. The dead steam from the turbine condenses into water and is returned to the heat exchange. The process repeats and combinuous supply of elec5tricla energy is obtained. In addition to the production of electricity, nuclar reactor is used to produce radioactive isotopes. Nuclear Fusion : The phenomenon of fusing two or more lighter nuclie to form a single heavy nucleus. The mass of product nuclear is slightly less than the sun of the mass of the higher nucleif fusing together. This difference in masses results in teh release of tremendous amount of energy. e.g.

(i)

1 1H

 1H1  1H2  e     0.42 MeV

(ii)

2 1H

 1H2  2H3  n  3.27 MeV

 1H2  1H3 1 H1  4.03 MeV In (i) two protons combine to form a deutron & a positorn with release of 0.42 MeV In (ii) two deutron combine to form a light isotop of He & a neutron with release of 3.27 MeV In (iii) two deutrons combine to form triton & a proton with release of 4.3 MeV The essential conditions for carrying out neclear fusion are : (iii)

2 1H

(i) High temperature is necessary for the light nuclei to have sufficient kinetic energy so that htey can overcome nutual Coulombic repulsions and come closer than th range of nuclar force. This process is alled thermonucler fusin. (ii) High density or pressure increases the rate of fution. Energy source of start : Protons are most abundat in the body of sun and stars. At extremely high temperatures protons fuse together to form helium nuclei, liberating a huge amount of energy. This fusion takes place via two different cycles: (i) Proton-proton cycle (P-P cycle) (ii) Carbon - Nitrogen cycle (C-N cycle) (i) P-P cycle

1 1H

 1H1  1H2  e   0.42 MeV

e   e       1.2ev 2 1H

 1H1  2He3    5.49 MeV

 2He3  2He 4  1H1  12.86MeV For the fourth reaction to occur, the first three reactions must occur twice. Thus the net reaction will be: 2He

3

(ii) C-N cycle :

12 6C

1 H1  7H13    1.93 MeV

13 7H

 6 C13  e     1.20 MeV

13 8C

1 H2  7H14    7.6MeV

14 7C

1 H1  6H15    7.39MeV

15 8O

 7N15  e   v  1.71MeV

15 7O

1 N1  6C12  2 He 4  4.99MeV The overall reaction is : 1 1H

 6 C12  2He 4  6 C12  2e   2  3  24.8MeV Thus four protons combine to form a helium nucleus, gamma rays and neutrons to liberte about 25 MeV of energy. Both P - P and C - NB cycles participate almost equally in the generation of energy in the sun. Starts hotter than the sun get their energy from the C -N cycle, while those cooler than sun get their energy from P - P cycle. Continuously and can be used fro generation of electrical power. The energy produced by fusion is clean and is not polluted by any radioactive waste. Moreover, the fuel is avialable in plenty is sea water. But t is very difficult to set up a sustained and controllable source of energy because fusion requires a very high temperature to 106 - 107 K. This temeperature is attained by causeing explosing throgh a fission process. Moreover, no solid containeer can withstand such a high temperature.