HC Verma Solutions Class 12 Chapter 24 The Nucleus

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 = Amp 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) = Amp (Given)

Mass of proton = mp = 1.007276 μ

μ = atomic mass unit = 1.6605 x 10-27 kg

Now,

Radius of nucleus = R = R0 A(1/3)

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

Volume = (4/3) π R3 = (4/3) π x R03 A(1/3)]3 = (4/3)πR03 A

(1)=> Density = [Ax1.007276 x1.6605 x 10-27]/[(4/3)πR03 A]

= 3.00006 x 1017 kg m-3

Specific gravity of nuclear matter = Density/1000 = = 3.00006 x 1014.

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 × 1030 kg (twice the mass of the sun).

Solution:

Mass of star(M) = 4 × 1030 kg (Given)

Density of nuclear matter(D) = 2.3 × 1017 kg/m3

Volume = Mass/Density = [4×1030]/[2.3×1017] m3

= [4×1013]/2.3 m3

Again, Volume = (4/3)πR3

Where, R = Radius of the neutron star

=> (4/3)πR3 = 4×1013]/2.3

=> R3 = [3×1013]/[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 = ΔMc2

=> ΔMc2 = 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:

7Li + p → α + α.

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

Solution:

Atomic mass of 7Li = 7.0160 u

Atomic mass of 4He = 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 = ΔMc2 = 0.018076 x 931.5 = 16.83 MeV

[Here c = 931.5 MeV/u]

Question 5: Find the binding energy per nucleon of 19779Au 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

HC Verma Class 12 Ch 24 Solution 5

Where, M = Atomic Mass

mn = mass of neutron

mp = 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 238U emits an α-particle.

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

Solution:

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

U238 -> Th234 + He42

So, energy released is the product of c2 and subtraction of mass of reactant and mass of products.

Here c2 = 931.5 MeV/u

Now,

=> E = Δm c2

=> 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 238U, 234Th and 4He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Reaction will be,

U233 -> Th234 + 2n + 2p

So, mass defect = Δm = m(U233 ) – m( Th234 ) + 2(mn) + 2(mp)

Where, mn = mass of neutron = 1.008665 u and mp = 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 c2 = 0.024712 x 931.5 = 23.02 MeV

Question 7: Find the energy liberated in the reaction

223Ra -> 209Pb + 14C.

The atomic masses needed are as follows.

223Ra 209Pb 14C
223.018u 208.981u 14.003u

Solution:

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

Energy = E = [223.018u – (208.981u + 14.003u)]c2

= 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 = (Mz–1, N + MH –MZ, N) c2

Where MZ, N = mass of an atom with Z protons and N neutrons in the nucleus and MH = 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) + p1

=> E(Z,N) -> E(Z-1,N) + 1H1

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

ΔE = (Mz–1, N + MH –MZ, N) c2

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 MZ, N, MZ, N–1 and the mass of the neutron.

Solution:

We know, energy released = Mass difference x c2

The reaction is,

EZ,N = EZ,N-1 + 1n0

The minimum energy needed to separate the neutron will be,

ΔE = (MZ,N-1 + MN – MZ,N) x c2

Question 10: 32P beta-decays to 32S. 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 32P = 31.974 u and that of 32S = 31.972 u.

Solution:

The reaction will be

P32 -> S32 + 1V0 + -1β0

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

E = m(P32 ) – m(S32) x c2

= (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 = [mn – (mp + mβ)] x c2

Where,

Mn = mass of Neutron = 1.008665 u

Mp = mass of proton = 1.007276 u

Mβ = 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.

HC Verma Class 12 Ch 24 Solution 12

Solution:

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

Resultant reaction:

HC Verma Class 12 Chapter 24 Solution 12

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]/3×108 x 103

= 2.67 x 1022 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 40Ar18, 40K19 and 40Ca20 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:

HC Verma Class 12 Ch 24 Solution 14

(b) We know, Q- value = [mass of reactants – Mass of products]c2

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 6Li and 7Li. 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 6Li and 7Li.

HC Verma Class 12 Ch 24 Solution 15

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

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: 228Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:

228Th -> 224Ra* + α

224Ra* -> 224Ra + γ (217 keV).

Atomic mass of 228Th is 228.028726u, that of 224Ra is 224.020196 u and that of 4H2 is 4.00260 u.

Solution:

Mass of 224Ra = 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:

12N -> 12C* + e+ + v

12C* -> 12C + γ (4.43 MeV).

The atomic mass of 12N is 12.018613 u

Solution:

Given reaction are:

12N -> 12C* + e+ + v

12C* -> 12C + γ (4.43 MeV).

Adding both the reactions, we get

12N -> 12C + e+ + v + γ (4.43 MeV).

Now, find K.E.max

Max K.E of β-particle = [(mass of 12N – mass of 12C) × c2] – 4.43Mev

= [12.018613 – 12] x 931 – 4.43

= 12.89 MeV

Question 19: The decay constant of 197Hg80 (electron capture to 197Au79 ) 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/A0 = (1/2)N

where N = Number of half lives

A= Activity of the substance

A0 = Initial activity

As per question,

Present activity = (1 – 0.25) A0 = 0.75 A0

=> (0.75A0)/A0 = (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 198Au is 2.7 days.

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

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

Solution:

The half-life of 198Au is 2.7 days.

(a) Disintegration constant = λ = 0.693/T1/2

= 0.693/[2.7x24x60x60]

= 2.97 x 10-6 s-1

Let N be the number of atoms left undecayed. then

N = [1×10-6x6.023×1023]/198

Activity, A = λN

= 2.97 x 10-6 x [1×10-6x6.023×1023]/198 Ci

= 0.244 Ci

(b) Using relation,

A/A0 = (1/2)N

where N = Number of half lives

A= Activity of the substance

A0 = Initial activity

Here, N = 2.592

A = (1/2)2.592 x 0.244 Ci

=>A = 0.0404 Ci