INDEX
S.NO
CONTENT
1
Topic to be covered for supportive
learner
Important Formula for Numerical
2
PAGE NO
2
3
3
Short Answer Questions ( 1 Mark )
4-7
4
Short Answer Questions (2, 3 Mark)
8-20
5
Long Answer Questions
21-23
6
Hots (Higher Order thinking skills)
24-25
7
QUESTIONS BANK

26
Topic to be covered for supportive learner :-
(2)
ATOMS






Schematic arrangement of the Geiger-marasden experiment.
Alpha partical scattering and Rutherford model.
Bhor’s Postulate (statements only)
Energy levels in hydrogen atoms
Line spectra of hydrogen atoms.
Summary given in N.C.E.R.T text book.
NUCLEI






Isotopes, isobars, isotones.
Mass defect.
Mass energy relation.
Binding per nucleon /mass number graph.
Potential energy of nucleons / seprations of nucleon graph.
Radio active decay law.
N= N0e-λt
 Half life ,mean life and decay constant and their relation
T1/2 = log2 = log2
 Radio active decay
Alpha decay, beta decay, gamma decay.
 Nuclear fission and fusion Reaction
(3)
Important Formula for Numerical :1 Nucleus consist of protons and neutrons.
Nucleus of protons in a nucleus zXA is Z and number of neutrons ,N =A-Z
2 Radius of Nucleus :- R= R0A1/3 where R0 = 1.2 x 10-15m
3 Density of Nuclear Matter :-
Dn = 1017kg/m3
4 Einstein ‘s Mass Energy Equivalence Relation is E =mc2
1amu =1u =931 MeV
5.Mass Defect =mass of nucleons in given nucleus –mass of nucleus
=Zm p + (A - Z )mn– M nucleus
6.Rutherford –Soddy formula :(i) Number of atoms undecayed after time t
N=N0e-λt
(i)
N/N0 =[1/2]n
Where n = t\T is number of half lives.
7. Relation between half –life (T) mean life ( ) and disintegration constant (λ ) is
=1/λ and T = 0.693
= 0.693/λ
8. Displacement Laws:
(i)
For α -particle
zXA
(ii)
For β - particle
zXA
z-2YA-4 + 2He4
z+1Y
A
+
0
-1β
+ ν
(iii)
For gamma– ray
zXA
zXA + γ
9. In nuclear fission a heavy nucleus break into lighter nuclei .Nearly 0.1 % mass is converted
each fission of 92 U 235 with slow neutron 200 MeV energy is released .
into energy .In
10 In nuclear fusion two lighter nuclei combine to form a heavy and 0.7 % mass is converted into energy .
UNIT=VIII
ATOMS AND NUCLEI
 Question bank for the supporting learner
 Very Short answer Questions :-
(4)
(1 Mark)
Q:-1
Ans
Q:-2
Ans
Among alpha ,beta and gamma radiations ,which get affected by electric field ?
Alpha and beta radiations are charged ,so they are affected by electric field.
What will be ratio of two nuclei of mass numbers A1 and A2?
Radius of nucleus R=R0A1/3
1/3
Q:-3
Compare the radii of two nuclei numbers 1 and 27 respectively?
Ans
1/3
=
(1/ 27 )1/3 = 1/3
Q:-4
What is the nuclear radius of 125 Fe if that of 27 Al is 3.6 fermi?
Ans
Nuclear radius ,R= R0A1/3
R
(R Fe/RAl) = (A Fe/AAL)1/3
A1/3
= (125/ 27)1/3
R Fe = ( 5/3 ) RAl = (5/3) x 3.6 = 6.0 fermi
Q:-5
Assuming the nuclei to be spherical in shape ,how does the surface area of a nucleus of
mass number A1 compare with that of a nucleus of mass number A2?
Ans
Radius of nucleus of mass number A is
R= R0A1/3
Surface area of the nucleus , S= 4 πR2
R2
(S1/S2) = (R1/R2) 2 = (A1/A2)2/3
(5)
Q:-6
What is the ratio of nuclear densities of the two nuclei having mass number in the ratio 1:4?
Ans
Nuclear Density is independent of mass number so ratio of nuclear density is 1:1.
Q:-7
What happens to the neutron to proton ratio after the emission of α-particle?
Ans
Neutron to proton ratio increases after the emission of α-particle .
Q:-8
Two nuclei have mass number in the ratio of 2: 5 . What is the ratio of their nuclear densities ?
Ans
Nuclear density in independent of mass number ,so ratio of nuclear densities is 1:1
Q:-9
Which has greater ionizing particle : α-particle or β-particle .
Ans
α-particle has greater ionizing power than β-particle.
Q:-10 What is the difference between an electron and a β-particle?
Ans
β-particle are simply very fast moving electrons. The specific charge of electron is higher than that of βparticle.
Q:-11 Why do α-particle have high ionizing power?
Ans
α-particle are heavier ,they move slowly ; so posses large momentum .Due to this property they come in
contact with large number of particles ; so they possess high ionizing power .
Q:-12 A nucleon of mass number A , has a defect Δm .Give the formula ,for the binding energy per nucleon , of
this nucleus .
Ans
B.E per nucleon ,
Bn =
Total binding energy
Number of nucleon
=
∆m.c2
A
Where ¢ is the speed of light in vacuum.
Q:-13 The binding energy per nucleon of the two nuclei A and B are 4 MeV and 8 -2 MeV .Which of accompanied
by release of energy ,Which of the two nucleus is more stable?
Ans
Q:-14
The nucleus (B) having larger binding energy is more stable.
Four nuclei of an element fuse together to form a heavier nucleus .If the process is accompanied by release
of energy ,which of the two the parent or the daughter nucleus would have a higher binding energy
/nucleon ?
Ans
The daughter nucleus will have higher binding energy per nucleon .
Q:-15 Name the reaction responsible for energy production in the sun
Ans
Fusion reaction.
(6)
Q:-16 Write a typical nuclear reaction in which a large amount of energy is released in the process of number
Fission?
Ans
Nuclear fission reaction is
92U
235
+ 0n1
141
56Ba
+ 36Kr92 + 3( 0n1) + 200MeV
Q:-17 Write any one equation representing nuclear fusion reaction?
Ans
Equation of fusion reaction
2
1H
+
2
1H
3
1H
+
1
1H
+4.0MeV
Q:-18 Give the mass number and atomic number of elements on the right hand side of the decay process?
220
86Ru
Ans
Po + He
The complete equation representing mass number and atomic number is give below
220
216
84Po
86Ru
+ 2 He4
For polonium Z=84,A =216
For helium ( α-particle )Z = 2 ,A =4 .
Q:- 19 State condition for controlled chain reaction in a nuclear reactor ?
Ans
In nuclear fission ,two or three neutron are released per fission .If on the average one neutron causes
further , the chain reaction is said to be controlled .
Q:-20 Name the absorbing material used to control the reaction rate of neutron in a nuclear reactor?
Ans
Cadmium is the absorbing material for neutrons produced in a nuclear reactor.
Q:-21 What a percentage of a given mass of a radioactive substance will be undecayed after four half periods?
Ans
Percentage of mass of radioactive substance undecayed after n=4 half –lives
= ( ½ )4x 100% = 100/16 = 6.25%
Q:-22 Write the nuclear decay process for nuclide β-decay of
Ans
15P
32
16S
32
32
15P
?
+ - 1e0 + v (Antineutrino )
(7)
Q:-23 Define the term activity of a radio nuclide .Write its S.I Unit .
Ans
Activity to the radionuclide is defined as the rate disintegration of given radioactive sample . It is
proportional to the number of undecayed radio nuclides in that sample.
. Its S.I unit is Becquarel
It is denoted R =
.
Q:-24 If the nucleon a nucleus are separated far apart from each other,the sum of masses of all these nucleons
is larger than the mass of the nucleus .where does this mass difference come from?
Ans
According to mass –energy equivalence relation E = mc2, this mass difference in nucleus remain in the
form binding energy . When nucleon are separated this binding energy of nucleus is converted into
mass.
Q:25
The radioactive isotope D –decays according to the sequence .
D
α
β
D1
D2
If the mass number and atomic number of D2 are 176 and 71 respectively ,what is the
I.
II.
Mass number
Atomic number ,of D?
Ans The sequence is represented as
I.
Given A-4 = 176
II.
Z -1 =71
z
α
DA
A-4
z-2D1
β
z-1D2
A-4
mass number of D, A =180
atomic number of D, Z =72
Q:-26 Write two characteristic features of nuclear force which distinguish it from Coulomb’s force ?
Ans
Characteristic features of Nuclear Force:
1. Nuclear Force are short range attractive force (range 2 to 3 fm ) while Coulomb’s force has range
upto infinity and may be attractive or repulsive.
2. Nuclear forces are charge independent forces ; while Coulomb force acts only between charged
particle .
(8)
 Short Answer Questions:Q:-1
( 2, 3 Marks)
The trajectories ,traced by different ά –particles, in Geiger –Marsden experiment were
shown in the
observed as
figure .
(a) What names are given to the symbols ‘ b’ and ‘θ’ shown here .
(b) What can we say about the value of b for (1) ‘θ’=00 (ii) ‘θ’=π radians.
θ
b
0 Target nucleus
Ans (a) The symbol ‘b’ represents impact parameter and ‘θ’ represents the scattering angle .
(b) When ‘θ’=00 ,the impact parameter will be maximum and represent the atomic size .
(c) When ‘θ’ = π radians ,the impact parameter ‘b’ will be minimum and represent the nuclear size .
Q:-2
Define ionization energy .What is its value for hydrogen atom?
Ans
The minimum energy required to removed an electron from atom to infinitely for away is called the
ionization energy.
The ionization energy for hydrogen atom is 13.6 eV.
Q:-3
Define half -life of a radioactive sample .which of the following radiations : α –rays ,β –rays and γ-rays
(i)
(ii)
(iii)
(iv)
Ans
Are similar to X-rays
Are easily absorbed by matter
Travel with the greatest speed
Are similar in nature to cathode rays ?
Half –life : The half –life of a radioactive sample is defined as the time in which the mass of sample is left
one half of the original mass.
(i)
(ii)
(iii)
(iv)
γ-rays are similar to X-rays
α –rays are easily absorbed by matter .
γ-rays travel with greatest speed
β –rays are similar to cathode rays .
(9)
Q:-4
Define the term ‘ Activity ‘ of a radioactive substance .State its SI unit .Give plot of activity of a radioactive
species versus time. Two different radioactive elements with half lives T1 and T2 have N1 and N2
(undecayed ) atoms respectively present at a given instant .Determine the ratio of their activities at this
instant.
Ans
The activity of a radioactive elements at any instant is equal to its rate of decay at that instant . S.I unit of
activity is Becquerel . (=1 disintegration /second ). The plot is shown is figure .
Activity R = dN
= λN
A0
dt
A
A/2
Decay constant λ = loge2/T
A/4
0
T
2T
t
Activity R = ( loge2) N/T
For two elements R1/R 2 = ( N1/N2 ) X ( T2 / T1 )
Q:-5
The half –life of 6C14 is 5700 years. What does it mean? Two radioactive nuclei X and Y initially contain
an equal number of atoms .Their half lives are 1 hour and 2 hours respectively .Calculate the ratio of their
rates of disintegration after two hours?
Ans
The half –life of 6C14 is 5700 years. It means that one half of the present number of radioactive nuclei of
14
6C will remain undecayed after 5700 years.
Number of nuclei Y after 2 hour . Nx = N0
1
2
Number of nuclei Y after 2 hour . Ny = N0
1
2
2/1
= N0
4
2/2
= N0
2
Radio of rates of disintegration
Rx
= No/4
Ry
Q:-6
=
N0/2
1
2
Draw the graph showing the variation of binding energy per nucleon with the mass number .What are the
main inferences from the graph? Explain with the help of this plot the release of energy in the processes
of nuclear fission and fusion .
(10)
Ans
The variation of binding energy per nucleon versus mass number is shown in figure:-
Binding
Fe56
Energy 9
Per
016
U238
8
N
Nucleon7
(MeV) 6
5
Li7
4
3
2
1
H2
0
20
40
60
80
120
140
160
180
200
220 240
Mass number
Inferences from graph
1.
The nuclei having mass number below 20 and above 180 have relatively small binding energy and hence
they are unstable .
2.
The nuclei having mass number 56b and about 56 have maximum binding energy -5.8 MeV and so they
are not stable.
3
Some nuclei have peaks ,e. g 2 He4 , 6C12 ,8O16 ; this indicates that theses are relatively more stable than
their neighbours .
Explanation : - When a heavy nucleus (A ≥ 235 say ) break into two lighter nuclei (nuclear fission ), the binding
energy per nucleon increase i.e nucleons get more tightly bound .This implies that energy would be
released in nuclear fission .
When Two very light nuclei (A ≤ 10) join to from a heavy nucleus ,the binding is energy per nucleon of
fused heavier nuclear more than the binding energy per nucleon of lighter nuclei ,so again energy would
be released in nuclear fusion .
(11)
Q:-7 Draw the graph showing the variation of binding energy per nucleon with mass numbers .Give the reason
for the decrease of binding energy per nucleon for nuclei with higher mass number?
Ans The graph of the binding energy per nucleon versus mass number a is shown in figure .The decrease of
binding energy per nucleon for nuclei with high mass number is due to increases coulombs repulsion between
protons inside the nucleus .
8.6
8
Bn
MeV
20
56
2
A
Q:-8 Draw a plot of binding energy per nucleon as a function of mass number of a larger number of nuclei .
Explain the energy release in the process of nuclear fission from the above plot ?
Or
Draw the graph showing the variation of binding energy per nucleon . Explain using this graph , why heavy nuclei
cab undergo fission?
Ans
The variation of binding energy per nucleon versus mass number is shown in figure.
Binding
Fe56
Energy 9
Per
016
U238
8
N
Necloun7
(MeV) 6
5
Li7
4
3
2
1
H2
0
20
40
60
80
120
140
(12)
160
180
200
220 240 (Mass number)
Reason for heavy nuclei undergoing fission :- The binding energy curve indicates that binding energy for
nucleon of heavy nuclei is less than that of middle nuclei . Clearly ,a heavy nucleus breaks into two lighter
nuclei ,then binding energy per nucleon will increased in the process .This process is called nuclear fission
.
Thus heavy nuclei undergo nuclear fission to achieve stability .
Q:-9
How does the size of nucleus depend on its mass number ? Hence explain why the density of nuclear
matter in independent of the size of nucleus?
Ans
The radius (size) R of nuclear is related to its mass number (A) as
R=R0 A1/3 where R0 =1.1 x 10- 15m
If m is the average mass of a nucleon ,then mass of nucleus =mA ,where A is mass number .
Volume of nucleus = (4/3) π R3 = (4/3) π R 0 3 A
Density of nucleus = Mass / Volume = 3m / 4 π R 0 3
Nuclear density is independent of mass number .
Q:-10 A radio nuclide sample has N0 nuclei at t=0 .Its number of undecayed nuclei get reduced to N 0 /e at t= .
What does the term stand for ? write the term of , the time interval ‘T’ in which half of the original
number of nuclei of this radio nuclei would have got decayed ?
Ans
is the mean life time of radio nuclei T is the half –life period of ratio nuclide ,the relation is
=1.44 T
i.e Mean life period =1.44 x half life period .
Q:-11 Explain with example ,whether the neutron –proton ratio in a nucleus in increases or decreases due to
β-decay?
Ans
In β-decay a neutron is converted into a proton ,so the neutron – proton ratio decreases .
Equation of β-decay is
zX
A
90Th
234
z+AY
+A
91Pa
234
+ -1β0 + ν
+ -1β0 +
ν
Neutron to proton ratio before β-decay
= (234 -90 ) / 90 =1.60
Neutron to proton ratio after β-decay
= (234-91 ) / 91 = 1.57
So Neutron to proton ratio in β decay decreases
(13)
Q:-12 Whit the help of an example how the neutron to proton ratio changes during α –decay of a nucleus ?
38
92U .
Ans Let us take the example of α –decay of
38
92U
90Th
234
The decay scheme is
+ 2α 4
Neutron to proton ratio before α –decay = (238 -92 ) / 92 = 146 / 92 = 1.59
Neutron to proton ratio after α –decay =
(234 – 90 ) / 90 = 144 /90 = 1.60
This shows that the neutron to proton ratio increases during α –decay of a nucleus .
Q:-13 Distinguish between isotops and isobars. Give one example for each of the species?
Ans
Isotopes
Isobars
The nuclides having the same The nuclides having the same atomic
atomic number Z but different mass (A) but different atomic numbers (Z)
atomic mass (A) are called isotopes are called isobars .
Examples :
Example :1
2
3
3
1H ,1H
1H , 2He
Q:-14 Group the following six nuclides into three pairs (i) isotones (ii) isotopes(iii) isobars ?
12
3
6C , 2He
Ans Isotones:-
, 80Hg198 , 1H3 ,79Au197 , 6C14
198
80Hg
Isotopes :
12
6C
Isobars :-
2He
3
and
and
and
197
79Au
14
6C
3
1H
(same number of neutrons A-Z )
(same atomic number )
( same mass number )
Q:-15 In a radioactive decay as follows
A
0
+1e
A1
α
A2
The mass number and atomic number of A2 are 176 and 71 respectively ,what are the mass number and atomic
numbers of A1 and A .Which of these elements are isobars?
Ans The reaction may be expressed as
zX
A
Z-1Y
(A)
A
+
+1e
0+
ν
Z-3Y1
(A1)
Given Z-3 =71
Z =74 and Z -1 =73
Also A -4 =176
A =180
A-4
( A2 )
Thus ,mass numbers of A1 and A are 180 each
(14)
+ 2He4
The atomic numbers of A1 and A are 73 and 74 respectively
The elements A and A1 are isobars.
Q:-16 A radioactive nucleus A undergoes a series of decay according to following scheme:
A
α
β-1
A1
A2
α
A3
γ
A4
The mass number and atomic number of A are 180 and 72 respectively .What are these numbers for A 4?
Ans The decay scheme may completely be represented as
72A
180
76
70A
α
β-1
71A
176
α
69A3
172
γ
69A4
172
Clearly ,mass number of A4 is 172 and atomic number is 69.
Q:-17 A radioactive isotope D decays according to sequence
D
0n
1
D1
α
D2
If the mass number of atomic number of D2 are 176 and 71 respectively ,find the mass number and atomic
number of D .Amongst D,D1 and D2 do we have any isobars isotopes?
Ans The scheme may be represented as
ZD
A
0n
1
A-1
ZD 1
α
Z-2D 2
A-5
Given Z-2 =71 and A -5 =176
=
Z=73 and A =181
The mass number of D is 181 and atomic number 73 .clearly D and D 1 are isotopes.
Q:-18 You are given two nuclides 3X7 and 3Y4
(ii)
(iii)
Are they isotopes of the same element ? Why ?
Which one of the two is likely to be more stable?
Ans (i) The two nuclides are isotopes of the same elements because they have the same Z.
(2) The nuclide 3Y4 is more stable because is has less neutron to proton ratio .
Q:- 19 Derive the relation Nt = N0e-λt
Or
Use basic law of radioactive decay to show that radioactive nuclei follow an exponential decay law?
Or
State the law of radioactive decay .If No is the number of radioactive nuclei at some initial time t 0 ,find out the
relation to determine the number N present at a subsequent time .Draw a plot of N as a function of time ?
(15)
Ans Radioactive decay Law:The rate of decay of radioactive nuclei is directly proportional to the number of undecayed nuclei at that
time .
λN
Where λ .is the decay constant . Suppose initially the number of atoms in radioactive elements is N 0 and the
number of atoms after time t . According to Rutherford and Soddy law .
λ N where λ disintegration constant .
λ dt
Integration loge N =
λ t +C
…(1)
Where c is a constant of integration . If N0 is initial number of radioactive nuclei ,then at t = 0,N= N0 ; so
Loge N0 =0 + C C = loge N0
N0
Substituting this equation in (1) ,we get
Loge N – loge N0 =
λt
Loge N / loge N0 =
λt
N
N=N0e—λt
t
The graph is shown in fig.
Q:-20 Derive expression for average life of a radio nuclei .Give its relationship with half life ?
Ans All the nuclei of a radio active do not decay simultaneously ; but nature of decay process is statistical , i.e it
can not be stated with certainly which nucleus will decay when .The time of decay of a nucleus may be between 0
and infinity . The mean of lifetimes of all nuclei of a radioactive elements is called its mean life .It is denoted by .
Expression for Mean Life :According to Rutherford –Soddy Law .rate of decay of a radioactive elements
R( t ) =
=λN
(16)
Therefore ,the number of nuclei decaying in between time t and t + dt is
dN = λ Ndt
If N0 is the total number of nuclei at t= 0 , then mean lifetime
=
Total life time of all the nuclei
Total number of nuclei
N0
N=N0e—λt
Also we have
=
= ∑ t. dN
∑t λ (N0e—λt ) dt
= λ ∑t e -λt dt
N0
As nuclei decay indefinitely , we may replace the summation into integration with limits from t=0 to t=
i.e.
te - λt dt.
= λ
Integrating by parts ,we get = 1
λ
Thus ,
= 1
λ
i.e the mean
lifetime of a radioactive
Relation Between Mean Life and Half Life .
Half life
T=
elements
is
reciprocal
of
its
decay
constant
.
0.6931
λ
Mean life
= 1
λ
Substituting value of λ form (2) in (1) ,we get
T = 0.6931
Q:21 Define half life of radioactive substance .Establish its relation with the decay constant ?
Or
Define – life of a radioactive sample .Using exponential decay law obtain the formula for the half –life of a
radioactive in terms of its disintegration constant/
(17)
Ans Half-life of a radioactive elements is define as the time in which number of radioactive nuclei becomes half
of its initial value
Expression for half time :-time radioactive decay equation is
N=
When
N0e—λt
t = T , N = N0
…(1)
2
N0e—λt
N0 =
2
e-λt
Or
= 1/2
Taking log of both sides
….(2)
- λT loge e = loge 1- loge 2
Or
λT = log e 2
T = loge 2
λ
….(3)
= 2.3026 log10 2 = 2.3026 X 0.3010
λ
or
λ
T = 0.6931
….(4)
λ
Q:-22 What are α –particles ? In the reaction
ZX
A
α+γ give the atomic number and mass numbers of Y?
Ans α –particles are doubly ionized helium atoms (or nuclei of helium ) .When a radioactive nuclide emits an are
α –particles ,its mass number is reduced by 4 and atomic number by 2 ; so the atomic number of Y is Z-2 and
mass number (A- 4).
Q:-23 Explain why is the energy distribution of β –rays continuous/
Ans During β-decay ,a neutron is converted into a proton with the emission of a β- particle with an antineutrino
1
0n
1P
1
+ 1β0 + ν
The energy produced in this decay is shared by β-particle and antineutrino ; therefore β –particle may have
varying amount of energy starting from zero to a certain maximum value.
(18)
Q:-24 In a nuclear reactor give the function of the (i) moderator (ii) control rods (iii) coolant (iv) heavy water .
In the reaction
92P
236
141
aQ
+
b
36R
+ 3( 0n1)
What are the values of a and b .
Ans (i) A moderator slows down the fast moving neutrons to convert them to thermal neutrons.
(2) Control rods absorb excess neutrons to control the chain reaction .
(3) Coolant transfers heat from the core of reactor to surroundings.
(4) Heavy water in a reactor is used as a moderator and coolant.
By conservation of mass number
236 = 141 +b +3x1
b = 92
By conservation of charge
92 = a + 36
a = 56
Q:-25 Why is the heavy water used as a moderator in a nuclear reactor ?
Ans The basic principal of mechanics is that momentum transfer is maximum when the mass of colliding particle
and target are equal . Heavy water has negligible absorption cross –section for neutron and its mass is small ; so
heavy molecules do not absorb fast neutrons ;but simply slow them.
Q:-26 Draw a graph showing the variation of potential energy between a pair of nucleon as a function of their
separation .Indicate the regions in which the nucleus force is (i) attractive (ii) repulsive?
Ans Part AB represent repulsive force and part BCD represents attractive force .
+ 100 MeV
Repulsive
0
-100MeV
AttrA
Attractive
r (fm)
Q:-27 Why is the mass of a nucleus always less than the sum of the masses of constituents , neutrons and
protons?
If the total number of neutrons and protons in a nuclear reaction is conserved how then is the energy absorbed
or evolved in the reaction ? Explain ?
(19)
Ans When nucleon combine to form nucleus , some mass is converted into binding energy in accordance with
mass energy equivalence relation ∆E=∆m.c2 .Hence ,the mass of a nucleus is less than that of nucleons.
2nd Part:-
In fact the number of protons and number of neutron are the same before and after a nuclear
reaction ,but the binding energies of nuclei present before and after a nuclear reaction are different .This
different is called the mass defect (∆M).
∆M =[Zmp+(A-Z)mn] –M
Einstein’s mass energy relation ,E =mc2
Can express this mass different in term energy as ∆Eb = ∆Mc2.
This show that if a certain number of neutron and protons are brought together to from a nucleus of a certain
charge and mass , an energy ∆Eb will be released in the process .The energy ∆Eb is called the binding energy of
the nucleus .If we separate a nucleus into its nucleons we would have to transfer a total energy equal to ∆Eb ,to
the particles .
Q:-28 Draw a graph showing the variation of binding energy per nucleon of lighter nuclei is small .In different
nuclei .Explain ,with the help of this graph ,the release of energy by the process of nuclear fusion ?
Ans The variation of binding energy per nucleon verus mass number is shown in figure.
Binding
Fe56
Energy 9
Per
016
U238
8
N
Necloun7
(MeV) 6
5
Li7
4
3
2
1
H2
0
20
40
60
80
120
140
(Mass number)
160
180
200
220 240
(20)
Energy released during nuclear fusion :- Binding energy per nucleon of lighter is small .In order to attain high
binding energy per nucleon , lighter nuclei fuse together to form a heavier nucleus releasing an enormous
amount of energy .This is converted from mass defected in the process in according with Einstein ‘ s mass energy
equivalence relation (E=∆Mc2).
Q:- 29 Show that the decay rate ‘R’ of a sample of a radionuclide is related to the number of radioactive nuclei
‘N’ at the same instant by the expression R =N
Ans From Rutherford and soddy law
N=N0e-λt
Decay rate R = -dN
=-
N0(- λ)e -λt
dt
using (1) ,we get
R= λN
Q:-30 (a)Draw the energy level diagram showing the emission of β-particle followed by γ-rays by a
nucleus?
27
60
Co
(b)Plot the distribution of kinetic energy of β-particle and state why the energy spectrum is continuous ?
Ans (a) The energy level diagram is shown in figure. (a)
(b) Plot of distribution is shown in fig. (b).
60
27Co
b-
Number
Of β-particles
per unit energy
Er = 1.17 MeV
Er = 1.33 Mev
28
(a)Energy level diagram
Kinetic energy
Of β-particles
60Ni
(b) Energy distribution of β –particles
The energy spectrum of β-particle is continuous because an antineutrino is simultaneously emitted in β- decays ;
the total released in β- decays is shared β- particle and the antineutrino so that moment of system may remain
conserved .
(21)
 Long Answer Questions:Q:-1 . Draw s schematic arrangement of Geiger –Marsden experiment for studying α-particle scatting by a thin
foil of gold . Describe briefly ,by drawing trajectories of the scattered α-particle ,how this study can be
used to estimate the size of the nucleus ?
Or
Describe –Geiger Marsden experiment . What are its observation and conclusions?
Ans
At the suggestion of Rutherford ,in 1911 ,H. Geiger and E. Marsden performed an important experiment
called Geiger –Marsden experiment (or Rutherford ‘s scattering ) .It consists of .
1. Sources of α-particle:- The radioactive source polonium emits high energetic alpha (α-)particles .There for
,polonium is used as a source of α-particle .This source is placed in an enclosure containing a hole and a
few slits A1,A2….. etc. are placed in front of the hole .This arrangement provides a fine of α-particles.
2. Thin gold foil :- It is a gold foil * thickness nearly 10-6m , α-particle are scattered by this foil .The foil taken
is thin to avoid multiple scatting of α-particle ,i.e to ensure that α-particle be deflected by a single collision
with a gold atoms.
3. Scintillation counter :- By this the number of α-particle scatting in a given direction may be counted .The
entire apparatus is placed in a vacuum chamber to prevent any energy loss of α-particle due to their
collisions with air molecules .
Method : - When α-particle beam falls on gold foil ,the α-particle are scattered due to collision with gold
atoms .This scatting takes place in all possible direction .The number of a particles scatted in any direction
is counted by scintillation counter.
Observation and Conclusions:(1) Most of α-particle pass through the gold foil undeflected .This implies that “ most part of the atom is
hollow.
(2) α-particle are scattered through all angles .Some α-particles (nearly 1 in 2000) , suffer scatting though
angles more than 900,while a still smaller number (nearly 1 in 8000) retrace their path .This implies
that when fast moving positively charged α-particle come near gold . come near gold atom, then a few
of them experience such a strong repulsive force that they turn back .On this basis Rutherford
concluded that whole of positive charger of atoms is concentrated in a small central core ,called the
nucleus .
The distance of closest approach of α-particle gives the estimate of nuclear size .If Ze is charge of
nucleus Ek kinetic energy of α-particle 2e charge on a particle the size of nucleus r0 is give n by
Note ( For Fig. please refer N.C.E.R.T Text book)
Ek = 1
4π
( Ze ) (2e)
0
r0
1
2Ze2
4π 0
Ek
Calculation show that the size of nucleus is of the order of 10-14 m, while size of atoms is of the order
of 10-10 m; therefore the size of nucleus is about 10-14/10-10
= 1/10,000
times
the size of
atom
(3) The negative chares (electrons) do not influence the scattering process. This implies that nearly whole
mass of atom is concentrated in nucleus.
r0 =
(22)
Q:-2
State the laws of a radioactive decay .Deduce the relation N=N 0e-λt . sketch a graph illustrate radioactive
decay .define half –life ?
Ans
Rutherford and Soddy ‘s Laws of Radioactivity :Rutherford and Soddy in 1902 ,studied spontaneous disintegration of radioactive nuclei , and on the basis
of experimental observation they came to the following conclusions :
(i)
(ii)
(iii)
Radioactive is a nuclear phenomenon and cannot be accelerated or related by any physical or
chemical process.
The nuclei of radioactive substances continue to decay spontaneously and randomly .it is not
certain which nucleus will decay at what time . In the process of disintegration α - β particles are
emitted along with γ-rays .thus new radioactive elements are formed which have entirely new
chemical and radioactive properties .
The rate of disintegration depends upon the elements and for a particular radioactive elements
the rate of disintegration is directly proportional to the number of undecayed nuclei (or mass ) of
the sample present .
Derivation of Formula :Suppose in initially the number of atoms in radioactive elements is N 0 and n the number of atoms
after time t.
After time t, let dN be the number of atoms which disintegrate in a short interval dt , then rate of
disintegration will be dN ,this is also called the activity of the substance /element.
dt
According to Rutherford –Soddy law
Or
Where λ is a constant ,called decay constant or disintegration constant of the elements .its unit is s -1 . Negative
sign shows that the rate of disintegration decreases with increase of time .For a given elements / substance λ is a
constant and is different elements .equation (1) may be rewritten as
Integration
loge N =-λt +C
Where C is a constant of integration
At
t = 0 , N = N0
loge N0 = + C
C = loge N0
Equation (2) gives
loge N = - λt + loge N0
Or
loge N – loge N0 = - λt
Or
loge N/N0 = - λt
(20)
Or
N/ N0e-λt
Or
N= N0e-λt
According to this equation , the number of undecayed atoms /nuclei of a given radioactive element decrease
exponentially with time ( i.e more rapidly at first and slowly afterwards .)
Definition of Half –Life
Thus half –life of a radioactive elements may be define as the time in which number of undecayed atoms (nuclei )
of that radioactive elements falls to half of its initial value .
Q:-3 Define the term : half –life period and decay constant of a radioactive sample .Derive of relation between
these terms?
Ans Half - life Period : The half –life period of an elements is defined as the which the number of radioactive
nuclei decay to half of this its initial value.
Decay Constant :- The decay constant of a radioactive elements is define as the reciprocal of time in which the
number of undecayed nuclei of that radioactive elements falls to 1/e times of is of its initial value.
Relation between half – life and decay Constant :- The radioactive decay equation is
N= N0e- λt
When
t = T, N = N0/2
N0e -λt
2
e-λT = ½
Or
Taking log of both sides
λT loge e = loge 2
or
λT = loge 2
λ
T = loge 2
λ
= 2.3026log102 = 2.3026 X 0.3010
λ
or
λ
T = 0.6931
λ
(23)
Q:-4 State three properties of nuclear forces .Show that the density of nuclear matter is independent of mass number A.
Ans
(i)
Nuclear forces are the strongest attractive forces.
(ii)
Nuclear forces are short ranged upto 10-15 m.
(iii)
Nuclear forces are charge independent .
Density of nuclear matter : Let mass number of an elements be A amu.
Density ,pn = mass
=
Volume
A X 1.66 X 10-27 kg
4
πR3
3
Radius of nucleus R=R0A1/3 = 1.2 X 10-15 A1/3 meter
Pn = A X 1.66 X 10-27 kg
≈ 1017kg /m3
( Order of magnitudes )
4/3 π X (1.2 x 10-15 A 1/3)
This is independent of number A.
(24)

HOTS (Higher Order Thinking Skills )
Q:-1 Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atoms from
its
(i)
(ii)
Sol
Seconds permitted energy level to the first level ,and
Highest permitted energy level to the second permitted level .
Energy of electron in permitted level
En = - Rhc
N2
(i)
When an electron jump from second to the permitted energy level ,
energy of photons = E2 -1 = Rhc ( 1/12 – ½2) = 3/4 Rhc
(2)When an electron jumps from the permitted level (n=) to the second permitted level (n=2)
( 1/22 - 1/
E
) =Rhc
4
Ratio
E2-1
E
= 3Rhc /4 = 3
Rhc /4
; Ratio = 3:1
1
Q:-2 Some scientists have predicted that a global nuclear war on earth would be followed by ‘nuclear winter.’
What would cause nuclear winter ?
Ans The radioactive emitted from radioactive substances after the nuclear war will hang in the atmosphere
and obstruct the solar radiations to come on earth .this will cause the nuclear winter.
Q:-3 The mass number of β-particles is higher than the mass of electrons obtained by other means where as
β-particles are also electrons ,why ?
Ans The speed of β-particles is very high while speed of electrons obtained by other means is small hence the
mass of β-particles according to Einstein ‘s equation .
m=
m0
Is higher than ordinary electron mass m0.
(25)
Q:- 4 State whether the following statement is true or false :
“ β-particles are those electrons which are orbiting around the nucleus .”
Ans The statements is false ; because β-particles are emitted from nucleus due to disintegration of a
neutron into a protons according to equation
0n
Q:-5
1
1
1H
+
0
1β
+ν
Which is easier to remove : orbital electron from an atom or a nucleon from a nucleus ?
Ans
It is easier to remove an orbital electron from an atom .the reason is the binding energy of orbital electron
is a few electron –volts while that of nucleon in a nucleus is quite large (nearly 8 MeV ). This mean that
the removal of an orbital electron requires few electron volt energy while the removal of a nucleon from
a nucleus require nearly 8 MeV energy.
Q:-6
A general impression exist that mass –energy interconversion takes place in nuclear reaction and never
in chemical reactions .this is strictly speaking , incorrect .Explain?
Ans
Mass –energy interconversion actually takes place in all reaction –chemical and nuclear .But in a chemical
reaction the effect is about 10-6 times less than in a nuclear reaction .hence ,in a chemical reaction the
changes in masses of initial reactants and final products are small that they cannot be observed .
Q:-7 State the process taking place in the following situations:
(i) When a slow neutron goes sufficiently close to a U235 nucleus .
(ii) When a neutron goes sufficiently close to a proton.
Ans
(i) When a slow neutron approaches a U235 nucleus ,the fission of U235 into lighter nuclei takes place with
the emission of 2 or 3 neutron and enormous energy .
(ii) Fusion of neutron and proton to form deuteron .
Thus ,the first one is nuclear fission process and the second one is nuclear fusion process.
Q:-8
Neutrons can be slowed down even by ordinary water has hydrogen nuclei ( 1H1) having mass equal to
almost that of neutron .Then why heavy hydrogen is used for this purpose in a reactor?
Ans
Hydrogen nuclei in ordinary water has greater absorption capture for neutrons i.e hydrogen nuclei ( 1H1)
2
have the tendency to absorb neutron rather than to slow them. The equation is 1H1 + 0n1
1H + γ .
On the other hand heavy hydrogen nuclei have negligible absorption capture for neutrons ; they slow
them sharing energy from them .
Q:- 9
Explain how radioactive nuclei can emit β-particles even through atomic nuclei do not contain these
particles . Hence explain why the mass number of radioactive nuclide does not change during β-decay .
Ans
Radioactive nuclei do not contain electron ( β-particles ), but β-particles are formed due to conversion
of a neutron into proton according to equation
1
1
0
0n
1P + - 1 β + ν
The β-particles so formed is emitted at once .In this process one neutron is covered into one proton ; so
that the number of nucleon of nucleons in the nucleus remain unchanged ;hence mass number of the
nucleus does not change during β-deacy.
(26)

QUESTIONS BANK :-
 1. Draw a labelled diagram of Geiger – Marsden experiment on scattering of α particles.
How is the size of the nucleus estimated in this experiment.
 2. Which particles constitute the nucleus.
 3. Distinguish between mass number and atomic number.
 4. What are isotopes? Give one example.
 5. What are isotopes? Give one example.
 6. What are isobars? Give example.
 7. How is the radius of a nucleus related to its mass number?
 8. What is the ratio of the radii of two nuclei of mass numbers A1 and A2 ?
 9. How does the nuclear density depend on the size of the nucleus?
 10. Who discovered neutrons? Write the nuclear reaction.
 11. Is free neutron stable’?
 12. What is natural radioactivity?
 13. Arrange α, β and γ rays in order of increasing ionizing power.
 14. What is the practical unit of rate of decay (activity)?
 15. What are radioisotopes?
 16. A radioactive nucleus undergoes a series of decays according to the sequence
 If the mass number and atomic number of A3 are 172 and 69, respectively, what are the
mass number and atomic number of A?
 17. Write the important properties of neutrons.
 18. What do you mean by alpha decay?
 19. What is beta decay?
 20. Write the equation of beta decay of a neutron.
 21. What are gamma rays?
 22. Define one roentgen.
 23. Which part of the atom is responsible for radioactivity?
 24. Arrange alpha. beta and gamma rays in the decreasing order of their penetrating power
 25. For which of the nuclear radiations a, 13 or a, the path is not affected by an electric
field and why?
 26. In a magnetic field, the radius of the circular path of β particles is more than the that
of the path of α particles. Why?
Guided by = Mr.P.S.Rawat (PGT Physics)