Page No 463: - Chapter 13 - Nuclei - Exercise Solutions - Ncert Solutions class 12 - Physics
Updated On 26-08-2025 By Lithanya
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Question 13.9:
Obtain the amount of 
necessary to provide a radioactive source of 8.0 mCi strength. The half-life of is 5.3 years.
Answer:
The strength of the radioactive source is given as:
Where,
N = Required number of atoms
Half-life of , 
= 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 108 s
For decay constant λ, we have the rate of decay as:
Where, λ 
For
:
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
∴Mass of 
atoms 
Hence, the amount of necessary for the purpose is 7.106 × 10−6 g.
Question 13.10:
The half-life of 
is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:
Half life of , 
= 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m = 15 mg
90 g of atom contains 6.023 × 1023 (Avogadro’s number) atoms.
Therefore, 15 mg of contains:
Rate of disintegration, 
Where,
λ = Decay constant 
Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 1010 atoms/s.
Question 13.11:
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.
Question 13.12:
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) 
and (b)
.
Given
= 226.02540 u, 
= 222.01750 u,
= 220.01137 u, 
= 216.00189 u.
Answer:
(a) Alpha particle decay of 
emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.
Q-value of
emitted α-particle = (Sum of initial mass − Sum of final mass) c2
Where,
c = Speed of light
It is given that:
Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u c2
But 1 u = 931.5 MeV/c2
∴Q = 0.005297 × 931.5 ≈ 4.94 MeV
Kinetic energy of the α-particle 
(b) Alpha particle decay of 
is shown by the following nuclear reaction.
It is given that:
Mass of = 220.01137 u
Mass of 
= 216.00189 u
∴Q-value = 
≈ 641 MeV
Kinetic energy of the α-particle 
= 6.29 MeV
Question 13.13:
The radionuclide 11C decays according to
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
calculate Q and compare it with the maximum energy of the positron emitted
Answer:
The given nuclear reaction is:
Atomic mass of 
= 11.011434 u
Atomic mass of 
Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the 
nucleus is given as:
Where,
me = Mass of an electron or positron = 0.000548 u
c = Speed of light
m’ = Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of
and 5 me in the case of
.
Hence, equation (1) reduces to:
∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2
= (0.001033 c2) u
But 1 u = 931.5 Mev/c2
∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV
The value of Q is almost comparable to the maximum energy of the emitted positron.
Question 13.14:
The nucleus 
decays by
emission. Write down the
decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
= 22.994466 u
= 22.989770 u.
Answer:
In emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
emission of the nucleus is given as:
It is given that:
Atomic mass of = 22.994466 u
Atomic mass of = 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
There are 10 electrons in and 11 electrons in
. Hence, the mass of the electron is cancelled in the Q-value equation.
The daughter nucleus is too heavy as compared to 
and 
. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.
Question 13.15:
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.
Question 13.16:
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.
Question 13.17:
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 NA atoms.
∴m g of contains
∴Total energy released during the fission of 1 kg of is calculated as:
Hence, 
is released if all the atoms in 1 kg of pure undergo fission.
