Free NCERT solutions for Chapter 13 Nuclei are available here, prepared by expert teachers as per the CBSE guidelines. This chapter explains many important concepts such as the atomic nucleus, nuclear forces, nuclear stability, and radioactivity. These solutions include clear, step-by-step answers to all NCERT questions, making it easier for students to understand both theory and numerical problems.
The chapter is very important for students who want to study physics or nuclear physics further, because it builds a strong base. By going through these solutions, you will learn about the structure of the nucleus, binding energy, mass defect, radioactive decay, half-life, fission, and fusion. All these topics are explained in a simple way so that you can revise quickly and score well in exams.
NCERT Solutions for Class 12 Physics Chapter 13 Nuclei
Q1 : Obtain the binding energy (in MeV) of a nitrogen nucleus
, given
=14.00307 u
Answer :
Atomic mass of nitrogen
, m = 14.00307 u
A nucleus of nitrogen
contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7mH+ 7mn – m
Where,
Mass of a proton, mH= 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm= 7 × 1.007825 + 7 × 1.008665 – 14.00307
= 7.054775 + 7.06055 – 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.11236 × 931.5 MeV/c2
Hence, the binding energy of the nucleus is given as:
Eb= Δmc2
Where,
c= Speed of light
∴Eb = 0.11236 × 931.5
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.
Q2 : Obtain the binding energy of the nuclei
and
in units of MeV from the following data:
= 55.934939 u 
= 208.980388 u
Answer :
Atomicmass of, m1= 55.934939 u
nucleus has 26 protons and (56 – 26) = 30 neutrons
Hence, the mass defect of the nucleus, Δm = 26 × mH+ 30 × mn – m1
Where,
Mass of a proton, mH= 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 26.20345 + 30.25995 – 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.528461 × 931.5 MeV/c2
The binding energy of this nucleus is given as:
Eb1= Δmc2
Where,
c= Speed of light
∴Eb1= 0.528461 × 931.5
= 492.26 MeV
Average binding energy per nucleon
Atomic mass of
, m2= 208.980388 u
nucleus has 83 protons and (209 – 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm’ = 83 × mH+ 126 × mn – m2
Where,
Mass of a proton, mH= 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm’ = 83 × 1.007825 + 126 × 1.008665 – 208.980388
= 83.649475 + 127.091790 – 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∴Δm’ = 1.760877 × 931.5 MeV/c2
Hence, the binding energy of this nucleus is given as:
Eb2= Δm’c2
= 1.760877 × 931.5
= 1640.26 MeV
Average bindingenergy per nucleon =
Q3 : A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of
atoms (of mass 62.92960 u).
Answer :
Mass of a copper coin, m’ = 3 g
Atomic mass of 
atom, m = 62.92960 u
The total number of atoms in the coin
Where,
NA= Avogadro’s number = 6.023 × 1023atoms /g
Mass number = 63 g
29Cu63 nucleus has 29 protons and (63 – 29) 34 neutrons
∴Mass defect of this nucleus, Δm’ = 29 × mH+ 34 × mn – m
Where,
Mass of a proton, mH= 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm’ = 29 × 1.007825 + 34 × 1.008665 – 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, Δm= 0.591935 × 2.868 × 1022
= 1.69766958 × 1022u
But 1 u = 931.5 MeV/c2
∴Δm = 1.69766958 × 1022× 931.5 MeV/c2
Hence, the binding energy of the nuclei of the coin is given as:
Eb= Δmc2
= 1.69766958 × 1022× 931.5
= 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10-13J
Eb= 1.581 × 1025 × 1.6 × 10-13
= 2.5296 × 1012J
This much energy is required to separate all the neutrons and protons from the given coin.
Q4 : Obtain approximately the ratio of the nuclear radii of the gold isotope
and the silver isotope
.
Answer :
Nuclear radius of the gold isotope
= RAu
Nuclear radius of the silver isotope 
= RAg
Mass number of gold, AAu= 197
Mass number of silver, AAg= 107
The ratio of the radii of the two nuclei is related with their mass numbers as:
Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.
Q 5 : The Q value of a nuclear reaction A + b→C + d is defined by
Q = [ mA+ mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i)
(ii)
Atomic masses are given to be
Answer :
(i) The given nuclear reaction is:
It is given that:
Atomic mass
Atomic mass
Atomic mass
According to the question, the Q-value of the reaction can be written as:
The negativeQ-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
It is given that:
Atomic mass of
Atomic mass of
Atomic mass of
The Q-value of this reaction is given as:
The positive Q-value of the reaction shows that the reaction is exothermic.
Q 6 : Suppose, we think of fission of a 
nucleus into two equal fragments,
. Is the fission energetically possible? Argue by working out Q of the process. Given
and
.
Answer :
The fission of can be given as:
It is given that:
Atomic mass of 
= 55.93494 u
Atomic mass of
The Q-value of this nuclear reaction is given as:
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.
Q 7 : The fission properties of 
are very similar to those of
.
The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure undergo fission?
Answer :
Average energy released per fission of,
Amount of pure
,m = 1 kg = 1000 g
NA= Avogadro number = 6.023 ×1023
Mass number of= 239 g
1 mole of contains NAatoms.
∴mg of contains
∴Total energy released during the fission of 1 kg ofis calculated as:
Hence, 
is released if all the atoms in 1 kg of pure undergo fission.
Q8 : How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
Answer :
The given fusion reaction is:
Amount of deuterium, m= 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 ×1023atoms.
∴2.0 kg of deuterium contains
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴Total energy per nucleus released in the fusion reaction:
Power of the electric lamp, P= 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:
Q9 : Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Answer :
When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm = 2 ×10-15m
∴d= 2 ×10-15+ 2 ×10-15 = 4 ×10-15m
Charge on a deuteron nucleus = Charge on an electron = e= 1.6 ×10-19C
Potential energy of the two-deuteron system:
Where,
= Permittivity of free space
Hence, the height of the potential barrier of the two-deuteron system is
360 keV.
Q10 : From the relation R = R0A1/3, where R0is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
Answer :
We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0= Constant.
A= Mass number of the nucleus
Nuclear matter density,
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
Hence, the nuclear matter density is independent of A. It is nearly constant.