HC Verma Solutions Class 12 Chapter 24 – The Nucleus, is one of the most useful study materials for students especially those who are in 12th standard studying physics or those who are making use of the HC Verma book part 2. The solutions contain answers that are described and formulated in the best way possible for students to clearly understand and learn the topics as well as the exercises discussed in the chapter. The solutions also cover questions that are frequent in exams like JEE. These solutions can be a great study material while preparing for competitive exams. HC Verma solutions are highly recommended and will enhance students’ abilities to tackle any type of questions that may be asked in the exams.

The solutions are also designed in such a manner in order to give students better learning experience and solve questions easily. Some of the problems given here include:

- Finding the binding energy of protons and the neutrons.
- There are questions related to half-life period and atoms of uranium and the counts radiated from the source, the number of atoms disintegrated and the number of decays.
- Students will also solve questions based on energy liberated and momentum of positron and neutrino particles and many more.

**Download This Solution As PDF: **HC Verma Solutions Chapter 24 PDF

## Key Topics Related To The Nucleus

- Properties of Nucleus
- Nuclear Forces
- Binding Energy
- Radioactive Decay
- Law of Radioactive Decay
- Properties and Uses of Nuclear Radiation
- Nuclear Fission
- Uranium Fission Reactor
- Nuclear Fusion

## Class 12 Important Questions In Chapter 24

**1.** When we separate nucleons of a nucleus from each other, there is an increase in the total mass. Can you explain where does this mass originate from?

**2.** During gamma decay, does the nucleus experience loss of mass?

**3.** An electron (or a positron) is generally released by a nucleus during beta decay. Do atoms that are usually left behind get oppositely charged when this occurs?

**4.** What happens to the binding energy of a nucleus when the mass number X is increased?

(a) the energy increases (b) It decreases (c) No change (d) It varies depending on the original value of X

**5.** What occurs in one average-life of nuclei?

(a) all the nuclei decay (b) half the active nuclei decay (c) less than half the active nuclei decay (d) more than half the active nuclei decay

## HC Verma Solutions Vol 2 The Nucleus Chapter 24

**Question 1:** Assume that the mass of a nucleus is approximately given by M = Am_{p} where A is the mass number. Estimate the density of matter in kgm^{–3} inside a nucleus. What is the specific gravity of nuclear matter?

**Solution: **

Density = Mass/Volume …(1)

Mass of nucleus(M) = Am_{p} (Given)

Mass of proton = m_{p} = 1.007276 μ

μ = atomic mass unit = 1.6605 x 10^{-27} kg

Now,

Radius of nucleus = R = R_{0} A^{(1/3)}

= 1.1 x 10^{-15 }A^{(1/3)} m

Volume = (4/3) π R^{3} = (4/3) π x R_{0}^{3} A^{(1/3)}]^{3} = (4/3)πR_{0}^{3} A

(1)=> Density = [Ax1.007276 x1.6605 x 10^{-27}]/[(4/3)πR_{0}^{3} A]

= 3.00006 x 10^{17 }kg m^{-3}

Specific gravity of nuclear matter = Density/1000 = = 3.00006 x 10^{14}_{.}

**Question 2:** A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0 × 10^{30} kg (twice the mass of the sun).

**Solution: **

Mass of star(M) = 4 × 10^{30} kg (Given)

Density of nuclear matter(D) = 2.3 × 10^{17} kg/m^{3}

Volume = Mass/Density = [4×10^{30}]/[2.3×10^{17}] m^{3}

= [4×10^{13}]/2.3 m^{3}

Again, Volume = (4/3)πR^{3}

Where, R = Radius of the neutron star

=> (4/3)πR^{3} = 4×10^{13}]/2.3

=> R^{3} = [3×10^{13}]/[2.3 π]

=> R = 16.07 km

**Question 3:** Calculate the mass of an α-particle. Its binding energy is 28.2 MeV.

**Solution: **

Binding energy = 28.2 MeV (given)

Mass of proton = 1.007276 u

Mass of neutron = 1.008665 u

Where, Atomic mass unit = u = 1.6605402 × 10^{-27} kg

Now,

ΔM = (number of proton x mass of proton + number of neutron x mass of neutron ) – M

Where, M = mass of an alpha particle

ΔM = (2 x 1.007276 u + 2 x 1.008665 u) – M …(1)

Also, Binding energy = ΔMc^{2}

=> ΔMc^{2} = 28.2

=> ΔM = 28.2/[931.5] [Here c = 931.5 MeV/u]

=> ΔM = 0.030273 u …(2)

From (1) and (2)

(2 x 1.007276 u + 2 x 1.008665 u) – M = 0.030273 u

M = 4.0016 u

**Question 4:** How much energy is released in the following reaction:

^{7}Li + p → α + α.

Atomic mass of ^{7}Li = 7.0160 u and that of ^{4}He = 4.0026 u.

**Solution: **

Atomic mass of ^{7}Li = 7.0160 u

Atomic mass of ^{4}He = Mass of alpha particle = 4.0026 u

Mass of proton = 1.007276 u

Mass defect (ΔM) = Mass of reactants – Mass of products

= [7.0160 + 1.007276] – 2 x 4.0026

=> ΔM = 0.018076 u

Now, energy release, E = ΔMc^{2} = 0.018076 x 931.5 = 16.83 MeV

**Question 5:** Find the binding energy per nucleon of ^{197}_{79}Au if its atomic mass is 196.96 u.

**Solution: **

Atomic mass of Au = 196.96 u

Atomic Number of Au = Z = 79

Number of nucleons = A = 197

Number of neutrons = N = A – Z = 118

Now, Binding energy is

Where, M = Atomic Mass

m_{n} = mass of neutron

m_{p} = mass of proton

c = Speed of light = 931.5 MeV/u

=> B = [(79 x 1.007276+118×1.008665)-196.96] x 931.5 = 1525.12 MeV

Therefore, Binding energy per nucleon = 1525.12/197 = 7.741

**Question 6:** (a) Calculate the energy released if ^{238}U emits an α-particle.

(b) Calculate the energy to be supplied to ^{238}U if two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U, ^{234}Th and ^{4}He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

**Solution:**

When ^{238}U emits an α-particle, the reaction is as follow,

U^{238} -> Th^{234} + He^{4}_{2}

So, energy released is the product of c^{2} and subtraction of mass of reactant and mass of products.

Here c^{2} = 931.5 MeV/u

Now,

=> E = Δm c^{2}

=> E = [238.0508 – (234.04363+4.00260)] x 931.5 = 4.2569 MeV

(b) When two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U, ^{234}Th and ^{4}He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Reaction will be,

U^{233} -> Th^{234} + 2n + 2p

So, mass defect = Δm = m(U^{233} ) – m( Th^{234} ) + 2(m_{n}) + 2(m_{p})

Where, m_{n} = mass of neutron = 1.008665 u and m_{p} = mass of proton = 1.007276 u

Δm = 238.0508 – [234.04363 + 2×1.008665 + 2×1.007276]

Δm = 0.024712 u

So, energy released = E = Δm c^{2} = 0.024712 x 931.5 = 23.02 MeV

**Question 7:** Find the energy liberated in the reaction

^{223}Ra -> ^{209}Pb +^{ 14}C.

The atomic masses needed are as follows.

^{223}Ra |
^{209}Pb |
^{14}C |

223.018u | 208.981u | 14.003u |

**Solution:**

Given reaction is: ^{223}Ra -> ^{209}Pb +^{ 14}C.

Energy = E = [223.018u – (208.981u + 14.003u)]c^{2}

= 0.034 x 931

= 31.65 MeV

**Question 8:** Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is

ΔE = (M_{z–1}, N + M_{H} –M_{Z}, N) c^{2}

Where M_{Z}, N = mass of an atom with Z protons and N neutrons in the nucleus and M_{H} = mass of a hydrogen atom. This energy is known as proton-separation energy.

**Solution:**

As hydrogen contains only protons, so we can write energy equation as follows:

E_{(Z,N)} -> E_{(Z-1,N)} + p_{1}

=> E_{(Z,N)} -> E_{(Z-1,N)} + ^{1}H_{1}

So, minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is

ΔE = (M_{z–1}, N + M_{H} –M_{Z}, N) c^{2}

**Question 9:** Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons in terms of the masses M_{Z, N}, M_{Z, N–1} and the mass of the neutron.

**Solution:**

We know, energy released = Mass difference x c^{2}

The reaction is,

E_{Z,N} = E_{Z,N-1} + ^{1}n_{0}

The minimum energy needed to separate the neutron will be,

ΔE = (M_{Z,N-1} + M_{N} – M_{Z,N}) x c^{2}

**Question 10:** ^{32}P beta-decays to ^{32}S. Find the sum of the energy of the antineutrino and the kinetic energy of the β-particle. Neglect the recoil of the daughter nucleus. Atomic mass of ^{32}P = 31.974 u and that of ^{32}S = 31.972 u.

**Solution:**

The reaction will be

P^{32} -> S^{32} + _{1}V^{0} + _{-1}β^{0}

The sum of the energy of the antineutrino and the kinetic energy of the β-particle:

E = m(P^{32} ) – m(S^{32}) x c^{2}

= (31.974 – 31.972) x 931.5

= 1.863 MeV

**Question 11:** A free neutron beta-decays to a proton with a half-life of 14 minutes.

(a) What is the decay constant?

(b) Find the energy liberated in the process.

**Solution: **

(a) Decay constant:

Half-life of 14 minutes = 840 sec (Given)

We know, half life = ln(2)/λ

Where λ = decay constant

=> ln(2)/λ = 840

= 8.25 x 10^{-4} s^{-1}

(b) Find the energy liberated in the process:

E = [m_{n} – (m_{p} + m_{β})] x c^{2}

Where,

M_{n} = mass of Neutron = 1.008665 u

M_{p} = mass of proton = 1.007276 u

M_{β} = mass of β-particle = 0.0005486 u

c = Speed of light = 931.5 MeV/u

=> E = [1.008665 – (1.007276 + 0.0005486 )] x 931.5

= 0.78283 MeV

= 782.83 KeV

**Question 12:** Complete the following decay schemes.

**Solution: **

(a) One α particle is produced so atomic number will decrease by 2 and the mass by 2,

Resultant reaction:

**Question 13:** In the decay 64Cu -> 64Ni + e^{+} + v, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.

(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?

(b) What is the momentum of this neutrino in kg ms^{–1}? Use the formula applicable to a photon.

**Solution: **

(a) Energy the neutrino = 0.650 – K.E. of given positron

Maximum kinetic energy carried by the positron is found to be 0.650 MeV. (Given)

= 0.650 – 0.150

= 0.5 MeV

= 500 keV

(b) Momentum of this neutrino is

P = E/c = [500 x 1.6×10^{-19}]/3×10^{8} x 10^{3}

= 2.67 x 10^{22} Kg m/s

**Question 14:** Potassium -40 can decay in three modes. It can decay by β^{–}-emission, β^{+}-emission or electron capture.

(a) Write the equations showing the end products.

(b) Find the Q-value in each of the three cases. Atomic masses of ^{40}Ar_{18}, ^{40}K_{19} and ^{40}Ca_{20} are 39.9624 u, 39.9640 u and 39.9626 u respectively.

**Solution:**

(a) Decay of Potassium -40 by β^{–}-emission, β^{+}-emission or electron capture:

(b) We know, Q- value = [mass of reactants – Mass of products]c^{2}

Where c = 931.5 MeV/u

Now,

Q-value in β^{– } decay = [39.9640 – 39.9626] x 931.5 = 1.30141 MeV

Q-value in β^{+} decay = [39.9640 – (39.9626+2×0.005486)] x 931.5 = 0.4683 MeV

Q-value in electron capture = [39.9640 – 39.9624] x 931.5 = 1.490 MeV

**Question 15:** Lithium (Z = 3) has two stable isotopes ^{6}Li and ^{7}Li. When neutrons are bombarded on lithium sample, electrons and α-particles are ejected. Write down the nuclear processes taking place.

**Solution:**

Given: Lithium (Z = 3) has two stable isotopes ^{6}Li and ^{7}Li.

**Question 16: **The masses of ^{11}C and ^{11}B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β^{+}-decay of ^{11}C to ^{11}B.

**Solution: **

The maximum energy for the positron in the β^{+}-decay = energy due to the mass defect (ΔM)

=> ΔM = [11.0114 – 11.0093] x 931 = 1.955 MeV

**Question 17:**^{ 228}Th emits an alpha particle to reduce to ^{224}Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:

^{228}Th -> ^{224}Ra* + α

^{224}Ra* -> ^{224}Ra + γ (217 keV).

Atomic mass of ^{228}Th is 228.028726u, that of ^{224}Ra is 224.020196 u and that of ^{4}H_{2 }is 4.00260 u.

**Solution:**

Mass of ^{224}Ra = 224.020196 x 931 + 0.217 = 208563.0195 MeV

K.E. of α particle = (mass of Th – mass of Ra – mass of alpha particle) x (speed of light)^{2}

= (228.028726 x 931) – (208563.0195 + 4.00260 x 931)

= 5.304 MeV

**Question 18:** Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:

^{12}N -> ^{12}C* + e+ + v

^{12}C* -> ^{12}C + γ (4.43 MeV).

The atomic mass of ^{12}N is 12.018613 u

**Solution:**

Given reaction are:

^{12}N -> ^{12}C* + e^{+} + v

^{12}C* -> ^{12}C + γ (4.43 MeV).

Adding both the reactions, we get

^{12}N -> ^{12}C + e^{+} + v + γ (4.43 MeV).

Now, find K.E._{max}

Max K.E of β-particle = [(mass of ^{12}N – mass of ^{12}C) × c^{2}] – 4.43Mev

= [12.018613 – 12] x 931 – 4.43

= 12.89 MeV

**Question 19:** The decay constant of ^{197}Hg_{80} (electron capture to ^{197}Au_{79} ) is 1.8 × 10^{–4} s^{–1}.

(a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

**Solution:**

(a) Half life = ln(2)/[decay constant]

= ln(2)/[1.8 x 10^{-4}]

= 3850.81 sec

= 64 minutes (approx)

(b) Average life = 1/[decay constant]

=1/[1.8 x 10^{-4}]

= 5555.56 sec

= 92 minutes (approx.)

(c) Using relation,

A/A_{0} = (1/2)^{N}

where N = Number of half lives

A= Activity of the substance

A_{0} = Initial activity

As per question,

Present activity = (1 – 0.25) A_{0} = 0.75 A_{0}

=> (0.75A_{0})/A_{0} = (1/2)^{N}

Taking “*ln*” both the sides,

*ln* 0.75 = N x *ln*(1/2)

=> N = 1598.23 sec

**Question 20:** The half-life of ^{198}Au is 2.7 days.

(a) Find the activity of a sample containing 1.00 μg of ^{198}Au.

(b) What will be the activity after 7 days? Take the atomic weight of ^{198}Au to 198 g mol^{-1}.

**Solution: **

The half-life of ^{198}Au is 2.7 days.

(a) Disintegration constant = λ = 0.693/T_{1/2}

= 0.693/[2.7x24x60x60]

= 2.97 x 10^{-6} s^{-1}

Let N be the number of atoms left undecayed. then

N = [1×10^{-6}x6.023×10^{23}]/198

Activity, A = λN

= 2.97 x 10^{-6} x [1×10^{-6}x6.023×10^{23}]/198 Ci

= 0.244 Ci

(b) Using relation,

A/A_{0} = (1/2)^{N}

where N = Number of half lives

A= Activity of the substance

A_{0} = Initial activity

Here, N = 2.592

A = (1/2)^{2.592 }x 0.244 Ci

=>A = 0.0404 Ci