ATOMS

Thomson’s Model : Atom is a sphere of +vely charged matter with e-s embedded in it. The +ve charge is
uniformly distributed over it. This arrangement is similar to that of seeds in a water melon or plums in a pudding.
This model is famous as Plum – Pudding model.
Atom as a whole is neutral and stable.
Thomson’s model was able to explain with some success processes like chemical reaction and radioactive
disintegration.
Failure :
i) It could not explain the origin of spectral lines in case of hydrogen and other atoms.
ii) It failed to explain large scale scattering of -particles in Rutherford experiment.
RUTHERFORD’S -  Particle Scattering Experiment

EXPERIMENTAL ARRANGEMENT
A radioactive source of -particle is enclosed in a thick lead block provided with a narrow opening. The -particles from
this source are collimated into a narrow beam through a narrow slit. The beam is allowed to fall on a thin gold foil. The -
particles scattered in different directions are observed and counted with the help of a rotatable detector which consists of
ZnS screen and a microscope. The whole apparatus is enclosed in an evacuated chamber to avoid scattering of -particles
by air molecules.

OBSERVATION
1. Most of the -particles pass through the gold foil or suffer only small
N
deflections.
2. A few -particles about 1 in 8,000 get deflected through 900 or more.
3. Occasionally an -particle gets rebounded from the gold foil suffering a
deflection of 1800. Scattering Angle
SIGNIFICANCE OF OBSERVATION
i) As most of -particles pass straight through the foil so most of the space within atoms must be empty.
ii) To explain large angle scattering, Rutherford suggested that all the positive charge and mass of the atomic
is concentrated in a very small region called nucleus of atom.
iii) The nucleus is surrounded by e-s whose –ve charge is equal to the +ve charge of nucleus, so atom is
neutral.
Scattering of  particles on Rutherford Model
Scattering of -particles is due to Coulombic respulsion
between positively charged -particles and nuclei.
a) an - particle like (1) passing atom at large
distance from nucleus experiences small repulsion
and passes almost undeflected.
b) particles like (2/2’) & (3) which pass closer to the
nucleus experience large repulsive force and hence scatter through large angles.
c) Very rarely an - particle (like 4) travels head-on collision towards the nucleus. The strong repulsive
force slows down the -particle, which is finally repelled back along the original path.
Distance of Closest Approach – Estimation of Nuclear Size
As -particle. of mass ‘m’, initial velocity ‘v’ moves directly towards the centre of the nucleus, it experiences
Coulombic repulsion and its K.E gets progressively converted into electrical energy.
At r0, the  particle stops for a moment, then retraces its path i.e. scattered through 1800 & entire K.E is converted
into electrostatic P.E at r0.
i.e. K.E = P.E.
1 kq q
m 2 = 1 2 q1 = +2e ( particle)
2 r0

q2= +Ze (Nucleus of atomic no. ‘Z’

1 2 2kZe 2
 mv =
2 r0

4kZe 2 2kZe 2
 r0 =  r = (1)
m 2
0
K .E 

Radius of nucleus must be smaller than r0.

The Rutherford expt.
K.E| = 5.5 MeV, Zgold = 79
So putting in (1) r0 = 41.3 Fermi

So it was predicted that size of nucleus is of the order of

1 fm = 10-15 m
IMPACT-PARAMETER
It is defined as the perpendicular distance of the
velocity vector of the -particle from the centre of
nucleus, when it is far away from the atom. It is
denotated by ‘b’.
1. For large impact parameters, the repulsive force experiment by the -particle is weak and the -particle
passes almost undeflected.
2. For small impact parameter, the repulsive force is large and so the -particle is scattered through large
angle.
3. For a head on collision, when the -particle just aims at the centre of the nucleus, b = 0, scattering angle
 = 1800.
kZe 2 cot  / 2
b= ; shape of trajectory of scattered -particles depends on the impact parameter and
E
nature of potential field.
Rutherford’s Model of an Atom
1. An atom consists of a small and massive central core in which the entire positive charge and almost the
whole mass of the atom are concentrated. Core is called the nucleus.
2. The size of the nucleus (~ 10-15m) is very small as compared to size of atom (~10-10m).
3. The nucleus is surrounded by a suitable number of e-s that their total negative charge is equal to the total
positive charge on the nucleus and the atom as whole is electrically neutral.
4. The e- revolve around the nucleus in various orbits just as planets revolve around the sun. The centripetal
force required for their revolution is provided by the electrostatic attraction between the e-s & the nucleus.
LIMITATIONS TO RUTHERFORD’S MODEL

1) According to e.m. theory, an accelerated charge particle must radiate electromagnetic energy. An e- which
is revolving around the nucleus under continuous acceleration towards the centre, should continuously
ease energy and move in orbits of gradually decreasing radii. The e- should follow a spiral path and finally
it should collapse into the nucleus.  It could not explain the stability of an atom :
2. As e-s are continuously accelerated and so it should emit a continuous spectrum. But an atom like
hydrogen always emits a discrete line spectrum.
BOHR’S Quantisation Condition of Angular Momentum
An e- orbiting in a circular orbit of radius ‘r’, can be taken to be a stationary
energy state only if it contains an integral number of de-Broglie wavelength ,
i.e. 2r = n
nh
De-Broglie Wavelength  = h / mv  2r =
mv
Thus, angular momentum ‘L’ of the e- must be
 nh
L = mvr = n = 1,2,3
2
Bohr’s quantisation condition.
Thus, only those circular orbits can be allowed stationary states of an e- in which its angular momentum is an
integral multiple of h/2.
BOHR’S ATOMIC MODEL
Postulates
1. Nuclear concept : An atom consists of a small and massive central core, called nucleus around which
planetary electrons revolve. The centripetal force required for their rotation is provided by electrostatic
attraction between the e-s & nucleus.
2. Quantum Condition ; e-s are permitted only in those orbits in which the angular momentum of an e- is
nh
an integral multiple of h/2 i.e. L = mr = .
2
3. Stationary orbits : While revolving in the permissible orbits called stationary orbits, an e- does not radiate
energy.
4. Frequency condition ; An atom can emit or absorb radiation in the form of discrete energy photons only
when an e- jumps from a higher to a lower orbit or from a lower to higher orbit.
i.e. hf = E 2 − E1
(1) Radii of permitted orbits
The electrostatic force of attraction between nucleus and the e-s is
kZe 2
F= = Centripetal force
r2
mv 2
=
r
kZe 2
 r= (1)
m 2
nh nh
Since L = mvr = r= (2)
2 2m
nh kZe 2 2kZe 2
Equating (1) & (2) =   = (3)
2mv m 2 nh
Substitution of (3) in (2)
nh nh n2h2
r= .   r  n2
2m 2kZe 2 4 2 mkZe 2
r0 : r1:r2:r3:--- = 1:4:9:16---
Bohr’s Radius : Radius of the innermost orbit

r0 =
(6.63  10 −34 )2 o
= 0.53 A
4 2  9.1  10 −31  9  10 9  (1.6  10 −19 )
2

2kZe 2
Speed of e- = v = ;
nh
For Z = 1 and H atom
 2ke 2   c  c
 =       = 
 ch   n  n

 = fine structure constt. =

(
2  9  10 9  1.6  10 −19 )
2

=
1
3  10 8  6.63  10 −34 137
1 c c
= ; for first orbit n=1,  = .
137 n 137
Thus the speed of e- in the innermost, orbit is 1/137 of the speed of light in vacuum, while the speed in the outer
orbits varies inversely with n.
Energy of e- = K.E.+ P.E.
1 kZe 2 kZe 2
 K.E = m 2 = (using v 2 = )
2 2r mr
P.E. of e- in nth orbit
kq1q2 k Ze(e ) − kZe 2
P.E. = = =
r r r
kZe 2 kZe 2 − kZe 2
E n = K.E. + P.E. = − =
2r r 2r
− kZe 2 4mkZe 2  n2h2 
En = .  r = 
2 n2h2  4mkZe 2 
− 2 2 mk 2 Z 2 e 4
 En =
n 2h 2
The negative value of total energy indicates that the e- is bound to the nucleus by means of electrostatic attraction
and work is required to be done to pull it away from the nucleus.
SPECTRAL ISERIES OF HYDROGEN ATOM
− 2 2 mk 2 Z 2 e 4 1
En = . 2 ; h = En − En
h2 n 2 1

2 2 mk 2 e 4  1 1 
 h =  2 − 2 
h2  n1 n 2 

2 2 mk 2 e 4  1 1 
 =  2 − 2 
h3  n1 n 2 
as c =  ,
1
 = wave no.


 2 2 mk 2 e 4  1 1 
= =  2 − 2 
c ch 3  n1 n 2 

1  1 1 
= = R  2 − 2 
  n1 n 2 
 Where
2 2 mk 2 e 4
R=
ch 3
called Rydberg constt.
R = 1.0973 x 107 m-1
1. Lyman series : If an e-jump any higher energy level n 2 = 2,3,4,... to a lower energy level n1 = 1, we get
spectral lines called lyman series which belong to the U.V. region of the e.m. spectrum.

1 1 1 
 = = R  2 − 2  n2 = 2,3,4,...
 1 n2 
2. Balmer Series : n 1 = 2; n 2 = 3,4,5,... , spectral lines in the visible region

1  1 1 
= = R  2 − 2  n2 = 3, 4, 5, …
 2 n2 
the visible region
3. Paschen Series : Transition from n 2 = 5,6,7 … and n = 3. We get a spectral series in the infrared region

1 1 1
= = R 2 − 2  ; n 2 = 4,5,6 …
 3 n2 

4. Brackett Series : If n2 = 5, 6, 7 …… & n1 = 4 we get a spectral series in the infrared region which is called.
Brakett series
5. P-fund series : If n2 = 6, 7, 8 …… and n1 = 5 we get spectral lines in the infrared region.
1= R  1 1
=  2 − 2 , n 2 = 6,7,8
 5 n2 

ENERGY-LEVEL DIAGRAM FOR HYDROGEN

It is a diagram in which the energies of the different

stationary states of an atom are represented by Parallel
horizontal lines, drawn according to some suitable scale.
ENERGY LEVELS OF HYDROGEN
− 2 2 mk 2 z 2 e 4
En =
n2h2
For hydrogen z = 1, therefore En =
− 2 2 mk 2 e 4 1
. 2 = −13.6ev
h2 n
E1 − 13.6
 En = = ev
n2 n2
Subs.: m = 9 x 10-31 kg
For n = 2 E2 = -3.4eV k = 9 x 109 N-m2/c2
=3 E3 = -1.51 eV e = 1.6 x 10-19 c
 =4 E4 = -0.85eV h = 6.63 x 10-34Js
=5 E5 = -0.54 eV
=6 E6 = -0.37 eV an e- can have only certain definite values of energy while revolving in
different orbits.
This is called ENERGY QUANTISATION
n = 1 ; E1 = -13.6 eV Ground State
n = 2 ; E2, E3 , E4 … Excited States
n = 3; ----
n = 4; ----
LIMITATIONS OF BOHR’S THEORY
This theory is applicable to only hydrogen like atoms (single e-) and fails in the case of 2 or more e-s. In the
spectrum of hydrogen, certain spectral lines are not single lines but a group of closed lines with different
frequencies. This could not be explained. It could not explain why circular orbits are chosen when elliptical can
be used.
1. As e-s exhibit wave properties also, so orbits of e-s cannot be exactly defined as in Bohr’s theory.
2. This theory doesn’t tell anything about relative intensities of the various spectral lines.
3. It doesnot explain the further splitting of spectral lines in a magnetic effect (Zeeman effect) and electric
fied (Stark effect).
Excitation Energy : It is defined as energy required by e- to jump from ground state to anyone of excited state
e.g. E2 – E1 = -3.4 - (-13.6) = = 10.2 eV
E3 – E1 = -1.51 – (-13.6) = 12.09 eV
Ionisation Energy : Energy required to knock out an e- completely from an atom i.e. to take e- from ground state
(n = 1) to outermost (n = 8) orbit
Ionisation energy and hydrogen E - E1 = 0 – (13.6) = 13.6 eV
Excitation Potential : It is the acceleration potential which gives to a bombarding e-, sufficient energy to excite
the target atom by raising one of its e- from an inner to outer orbit.
e.g. First excitation potential = 10.2V
Second excitation potential = 12.09 V
Ionisation Potential : It is the acceleration potential which gives to a bombarding e- sufficient energy to ionise
the target atom by knocking one of its e- completely out of atom.
Ionisation potential of Hydrogen = 0 – (-13.6) = +13.6V
 NUCLEI

Isotopes : The atoms of an element which have the some atomic number but different mass number are
called isotopes of an element exhibit similar chemical properties and occupy the same position in the
periodic table
e.g. Hydrogen → 1 H1 , 1H 2 , 1H 3
Isobars – The atoms having same mass number but different atomic number are called isobars.
Only differ is chemical properties & occupy different positions in periodic table
Isotones : The nucleus having the same no. of neutrons are called isotones

Discovery of Neutron : It was observed by James Chadwick in 1932.

EXPERIMENT SETUP USED BY CHADWICK TO

DISCOVERY NEUTRONS
Polonium source was used to bombard nuclei of Berylium.
Highly penetrating rays were found to come out of the
Berylium. nuclei which could not be deflected by electric and magnetic fields. These were made to
bomard hydrocarbons like Paraffin wax – high energy protons were knocked out from it. Chadwick
concluded – that the penetrating radiation consisted of neutral particles each having mass nearly that of
a proton called neutrons.
4
2 He+ 49Be→ 01 n+126C
Properties of Neutrons
1) It has no charge and its mass is slightly more that of a proton.
2) Inside a nucleus, a neutron is stable but outside it is unstable → a free neutron decays into a
proton, electron and antineutrino (an elementary particle with zero charge and zero rest mass)
with a mean-life of 1000s
1
0 n→11H + −10 e + 
3) Neutrons are neutral they do not interact with e-1, so neutrons have low ionising power.
4) They induce radioactivity.
5) They can penetrate heavy nuclei being neutral.
6) In heavier nuclei, the number of neutrons is more than that of protons. (Protons being +vely
charged, repel each other and in order to maintain stability of nucleus.)
1
a.m.u. It is th of actual mass of C – 12 atom
12
 1 12
a.m.u. =  = 1.6602 x 10-27 kg
12 6.02  10 23

E = mc 2 = 1.66  10 −27  (3  108 )

2

a.m.u. = 931 MeV as eV

1.6  10 −19
It is found that volume of nucleus is proportional to Mass No. A
4 3
r  A  RA1 / 3  R = R0 A1 / 3 , RO = 1.2 x 10-15 m
3
Ro – average nucleon size and called nuclear radius
Nuclear Density- It is independent of size of nucleus
M mA mA 3m
= = = = = 2.3  1017 kg / m 3
V V ( 4 )R0 A 4R0
3 3
3
Nuclear Forces : These are the strong for a of attraction which hold together the nucleons.
Properties
1) Nuclear force is charge independent. (It
is same between proton-proton, neutron-
neutron and proton-neutron.)
2) It is strongest force in nature ~ around
1038 times the gravitational force.
3) It is short range force with range ~ 1.5
fermi
4) Nuclear force between two nucleons becomes repulsive if distance between them reduces to 0.5
fermi & less.
5) It is an exchange force – two nucleons exchange  mesons between them.
6) It exhibits saturation properties – i.e. nucleons interact, with immediate neighbours only.
7) It is spin dependent.
8) It is a non central force.
Nuclear Binding Energy
The energy required to break up the nucleus into its constituent nucleons upto a distance so that they do
not interact with each other. It is called Binding energy or it is the energy with which the nucleons are
bound to the nucleus.
Cause : Mass of nucleus is always little smaller than the sum of mass of its constituent nucleons. A small
fraction of mass disappears in the formation of nucleus. The difference in mass of nucleus and sum of
mass of its constituting nucleons is called Mass Defect. Mass defect appears in the form of binding
energy.
Expression for Binding Energy : zXA – Nucleus; A – Mass No.
No. of Neutrons = A – Z = N, Z – At No.
Let mp – Mass of proton, mn – mass of neutron; mN – mass of nucleus.
 Mass Defect = Mass of Z protons + Mass of (A – Z) neutrons - Mass of nucleus (zXA)
M = Zm p + ( A − Z )mn − m N 

 
 B.E. = Mc 2 =  Zm + ( A − Z )m − m  c 2
 p n N

B.E. [ Zm p + ( A − Z )mn − m N ]c 2
 B.E. per nucleon = =
A A
NUCLEAR STABILITY
Nuclei are able to stay together despite the repulsive force between them because of much stronger force,
called nuclear force.
There are 103 different elements including Z = 1 (H) & Z = 103 is Lawrencilum. All these elements have
isotopes and as such there are 1000 different nucleides – out of these 280 are stable and the rest are
unstable.
Factors on Which Nuclear Stability depends
1. The stability of a nucleus is determined by the value of binding
energy per nucleon. Higher the B.E. per nucleon, more stable is the
nucleus.
2. The stability is also determined by neutron to proton ratio. The solid
line in the plot shows the locations of nuclei that have an equal number
of (N = Z) is protons and neutrons) only light nuclei are stable only if
they contain about same no. of protons and neutrons. Heavy nuclei, on
the other hand are stable one when they have more neutrons than
protons. So heavy nuclei are neutron rich compared to lighter nuclei
which makes it stable. More neutrons provide the strong attractive force, against the repulsive force due
to more no. of protons. The long narrow region in
figure which contains, the cluster of dots
representing stable nuclei is referred to as the valley
of stability.
BINDING ENERGY CURVE
Main features of the Plot
1) B.E./Nucleon is practically constant is
independent of the atomic no. for nuclei of
middle mass no. (30 < A < 170).
2) Curve has maximum of about 8.75 MeV for A = 56 and value 7.6 MeV for A = 238.
3) Ebn is lower for both light nuclei (A < 30) and for heavy nuclei (A > 170). The maximum like 3,
6, 7, 8 higher stability of these nuclei than the neighbouring ones and minima shows lower.
Some Conclusions from Above Features (Observations of Curve)
1) The force is attractive and sufficiently strong to produce a B.E. of few MeV/nucleon (order of 1
to 10)
2) The constancy of B.E. in the range of 30 < A < 170 is a consequence of the fact that nuclear force
is a short ranged force. If a nucleon can have a maximum a P neighbours within the range of
nuclear force, it’s B.E. found be proportional to P. Let the B.E. of nucleus be Pk, k is was constt.
If we increase A by increasing no. of neutrons, they will not change B.E. of nucleon
3. For heavy nucleus A = 240 has lower B.E./A compared both of a nucleus with A = 120 i.e. if
nucleus A = 240 breaks into the A = 120 nuclei, nuclei gets more tightly bound. This implies
energy should be released in the process (Fission)
4. For nuclei A  10 joining to form a heavier nucleus. The B.E./A of fused, heavier nuclei is more
than the B.E./nucleon of lighter nuclei. This means finally system is more tightly bound than the
initial system. Energy will be released in such a process of fusion.
RADIOACTIVITY
The natural phenomenon of spontaneous emission of radioactive i.e disintegration of nucleus is called
radioactivity. The elements showing radioactivity are called radio-active elements.
TYPES OF NUCLEAR RADIATIONS
Radiation emitted by nuclei of three types : (i) Alpha () Rays (ii) Beta rays () (iii) gamma ( rays). In
a small cavity, a sample of a radioactive element is placed.
A narrow beam of radiation is subjected to an electric field and magnetic field.

1. The component which bends towards left (-ve plate) consists of positively charged particles called
 particles, known as -rays.
2. The component which bends towards right (+ve plate) consist of –vely charged particles called 
particles, known as  rays.
3. The component which goes undeflected consists of uncharged particles photons this component
as r rays.
PROPERTIES OF -RAYS
1. An -rays are +vely charged. He nucleus, carrying a charge +2e.
2. It consists of 2 protons and two neutrons, therefore its mass is 4 times mass of hydrogen atom.
3. They are defected by electric and magnetic fields.
4. They affect photographic plate.
5. The velocity of  particles very from 1.4 x 107 m/s to 2.1 x 107 m/s depend upon source emitting
them.
6. Due to large mass, they have small penetrating power.
7. Due to large mass and velocity, they produce strong ionisation.
8. They produce fluorescence when they fall on zinc sulphide.
Properties of -rays
(i) -rays are fast moving electrons. The mass and charge of each -particle is equal to that of
electron.
(ii) -rays are deflected by electric and magnetic fields. The direction of deflection indicates to they
are negatively charged particles.
(iii) -rays affect photographic plates.
(iv) The velocity of -particles can be upto 90% of the velocity of light.
(v) The penetrating power of -rays is much higher than that of -rays due to much lower of -
particles. They can penetrate aluminium sheets of thickness upto 1 cm.
(vi) Due to lower mass of -particles, they produce less ionisation.
(vii) -particles produce fluorescence when they strike certain substances like zinc-sulphide barium-
platinocyanide.
Properties of -rays
(i) -rays are electromagnetic waves of very high frequencies consisting of photons.
(ii) -rays are not deflected by electric and magnetic fields indicating that they do not any change.
(iii) They affect photographic plates.
(iv) They travel with the speed of light.
(v) The penetrating power of -rays is highest among the nuclear radiations. They can pass several
centimetres of iron and lead.
(vi) These rays have very weak ionising power.
(vii) These rays produce fluorescence when they strike certain substances like barium-plato etc.
(viii) They are harmful if incident on human tissues.
(ix) They cause photoelectric effect when incident on metallic surfaces.
RADIOACTIVE DECAY LAW
Let us consider a radioactive elements containing No atoms to t = 0.
Let at time t, no reduces to N, due to disintegrations of atom, let dN atoms disintegrate in time interval
between t and t + dt
 rate of disintegration during this time interval is dN/dt.
dN
i.e. N
dt
dN
 = −.N [-ve sign shows disintegration of atoms)
dt
where  - disintegration constt. or decay constt.

= dt (−  ) i.e.
dN dN
i.e. = − dt
N N
Integrating both sides
dN
 N
= −  dt

 log e N = −t + C (1)

where C – constt. of integration

at t = 0; N = N0 loge N0 = C (2)
Putting (2) in (1), we get
log e N = −t + log e N 0

 N 
 log e   = −t
 N0 
N
 = e − t
N0

 N = N 0 e − t

This indicates that the radioactive decay is exponential

 n t /T
N 1 1
 =  = 
N0  2   2 
at t = 0, N = N0
Disintegration constants : N = N 0 e − t Let t = 1/

N = N0 e-/λ
N = N0/e
Disintegration constt. can be defined as the reciprocal of time during which the original no. of atoms
(N0) in a radioactive sample reduces to N0/e.
Radioactivity: Three types of radioactive decay occur in nature :
4
1)  decay in which a He nucleus He is emitted
2

2) β decay in which electrons or positions are emitted.

3)  decay in which high energy (hundred & kev or n photons are emitted.
HALF LIFE:- t= T; N = N0,
N0/2 = N0e-λT ; loge (1/2) = -λT ; T = 0.693/λ
AVERAGE LIFE:-
1 N 0

 av. life  =  tdN [(dN = − N 0 e − t dt )

N0 0
When N = N0 t = 0
N = 0, t = α]

( )
0
1
= 
N0 
t − N 0 e −t dt


1
= 
N0 0
t 0 e −t dt

  te − t



1.e −t 
  =   te dt =  
− t
 − dt 
0  −   0 0 −  

  − t  
 e − t 
= 0 +  e dt =  e dt =  e dt = 
− t − t

0 0 0  −   0

=−
1
e −
− e0  = −
1
(0 − 1)
 
1 0.693
 = also Ty = = 0.6933
 2

T1 / 2
 = = 1.44 T1 / 2
0.693
Activity of a Radioactive Substance : The activity is defined as
dN
i.e. R=− -ve sign shows activity of the sample decreases with the passage
dt
of time.
− dN
i.e. According to decay law = N
dt
 R = N
 R =  N0 e-t
R = Ro e-t
Decay rate at time t = 0; R0 = N0
R – decay rate at time t.
UNIT OF RADIOACTIVITY
1. Bacquerel (Bq) – It is defined as decay rate of one disintegration. 1 Bq = 1 decay/sec.
2. Curie (Ci) – It is the decay rate of 3.7 x 1010 disintegration per second . 3.7 X 1010 Bq = 1 Ci
3. Rutherford (rd) = Decay rate of 106 disintegration/sec
1rd = 106 decay/sec. = 106 Bq.
ALPHA DECAY
A A−4 4
X → Y + He
Z Z−2 2

In inn process mass number and atomic no. of daughter nucleus decreases by four and two respectively.
Q-Value or disintegration energy = Difference between initial mass energy and final mass energy of
decay products.
Q = (m x − m y − m He )c 2 - This energy is shared by daughter nucleus and ‘’ particles in the form of K.E.

Beta decay : A nucleus that decays spontaneously by emitting an e - or a position is said to undergo beta
decay.

A A 0
e.g. X → Y+ e
Z Z +1 −1

A A 0
X → Y+ e
Z Z −1 +1

→ S + e +  (1 / 2 = 14.3d )
32 32 0
P
15 16 −1

Na → Ne + e +  (T1 / 2 = 2.6 y )
22 22 0

11 10 +1

0
Neutron Transforms into a Proton within a nucleus n → p + e + 
−1

0
Proton transforms into neutron inside the nucleus. p → n + e + 
+1
Energy released in the process of  decay is shared between  particle and  or .
Gamma Decay : ZXA → ZYA +  when a nucleus is in excited state, it makes ** to a lower energy state
by emission of electromagnetic radiation. The photons emitted by the nuclei have Mey energies and are
called Gamma rays. Most radionuclides after  or  decay leave, the daughter nucleus in an excited state
and daughter nucleus reaches the ground state by a single transition or by successive, transitions by
60 60
emitting one or more gamma rays. E.g. CO transforms into Ni in excited state which deexcites to its
27 28

ground state by successive emission of 1.17 MeV, 1.33 MeV  rays.

FISSION : The phenomena in which a heavy nucleus (A > 230) when excited spirits into two smaller
nuclei of nearly comparable masses is called nuclear fission. e.g.
235 1 141 92 1
 + n → Ba + Kr + 3 n + Q
0 56 36 0
92

235
 value is about 200 MeV. The fission of  yields broadly two groups of nuclei one of the groups is a

light group with mass no. range from 85 to 104. The other group is a heavy group with mass no. range
from 130 to 149.
Fission can produced other fragments also.
For a = 240, breaking into two fragments each of A = 120. Then
Ebn for A = 240 nucleus is about 7-6 MeV
Ebn for two A = 120 fragments nuclei is about 8.5 MeV
 Gain in B.E. for nucleon is about 0.9 MeV
Hence total gain in B.E. is 240 x 0.9 = 216 MeV
The disintegration energy is fission events first appears as the K.E of fragments and neutrons. Eventually
it is transferred to the surrounding matter appear as hear.
Nuclear Fusion : The process in which two light nuclei combine to form a single heavier nucleus is
called. Nuclear fusion
1 1 2 f1
e.g. H + H → H + e +  + 0.421xe v
1 1 1 1

2 2 3 1
H + H → He+ n + 3.27 Me v
1 1 2 0

2 2 3 1
H + H → H + H + 4.03 Me v
1 1 1 1

The fusing nuclei have to overcome very high electrostatic repulsion between them at extremely small
distance. This repulsion prevents the two nuclei from getting dose enough to be within range of their
attractive nuclear forces and thus fusing. The height of the Coulomb barrier depends on the charges and
radii of two colliding nuclei. The height of potential barrier is higher for more highly charged nuclei. To
carry nuclear fusion in a bulk material, the temperature of the material has to be raised to 10 6K, so that
colliding nuclei have enough energy due to their thermal motion and they can penetrate the Coulomb
barrier. This process is called thermonuclear fusion.
Necessary Conditions
i) High temperature
ii) High density or pressure increases the frequency of collision of light nuclei and hence increases
the rate of fission.
These conditions exist in the interior of sun where temperature is 2x106 K.
Within the sun, Hydrogen is burned into Helium, Hydrogen being Fuel and Helium being ‘ashes’.
Two cycles taking place inside sun
B.P. Cycle (Proton-Proton) Cycle
1 1 2 +1
H1 + H1 → H1 + e +  + .042 MeV (i)
0

+1 −1 0 0
e + e →  +  + 1.02 MeV (ii)
0 0 0 0

2 1 3
H1 + H1 → He+  + 5.49 MeV (iii)
2

3 3 4 1 1
H2 + H2 → He+ H1 + H1 + 12.86 MeV (iv)
2

For the fourth reaction to occur, the first three reactions must occur twice.
The net effect of 2(i) + 2(ii) + 2(iii) + (iv) →
1 −1 4
4 H1 + 2e → He+ 2 + 6 + 26.7 MeV
0 2

Burning of hydrogen in sun is going for about 5 x 109 Y and there is enough H to keep the sun going for
about same time.
Controlled Thermonuclear Reaction
If the energy released in a thermonuclear reaction is controlled in such a manner that a limited amount
of energy is produced continuously, it can be used for many useful purposes.
The easiest thermonuclear reaction that can be carried on earth is the fusion of two deuteron (d – d)
reactions or fusion of a deuteron with a triton (d – t) reaction.
2 2 3 1
H1 + H1 → He+ n + 3.2 MeV (d – d)
2 0

2 2 3 1
H1 + H1 → H1 + H1 + 4.0 MeV (d – d)
2 3 4 1
H1 + H1 → He+ n = 17.6 MeV
2 0

 Decay -Speed of emitted  particle

A A−4 4
X → Y + He+ Q (4)
Z Z− 2 2

Speed can be calculated using law of conservation of energy and momentum.

A
Let X be at rest before decay.
Z

According to conservation of momentum

m y v y = m He  He

As Q released in the decay appears as K.E of  ptdl. and daughter nucleus.

According to cons. Of K.E.
1 1
m He v 2He + m y − v 2y + Q (2)
2 2
Sub,. (1) in (2)
2
1 1 m  
2
m He VHe + m y  He He  =Q
2 2  m 
 y 
1 1 m 2He 2
 m He  2He + VHe = Q
2 2 my

1  m He 
 2
m He VHe 1 + =Q
2  m y 

1 Q
 2
m He VHe =
2  m y + m He 
 
 my 
Q(A − 4 ) Q(A − 4 )
 K.E| = =
(A − 4) + 4 A

2(A − 4 )Q
VHe
A.m He
222
e.g. Q = 5.587 MeV for radon R n
86

A−4  222 − 4 
K He = Q=   5.587 Me v
A  222 
= 5.486 MeV
2  5.486  1.6  10 −19
VHe =
4  1.667  10 27
= 1.62 x 107 MeV