Question 1

The binding energy per nucleon is maximum for nuclei with mass number around:

Accepted Answer

Iron-56 has the highest binding energy per nucleon (~8.8 MeV), making it the most stable nucleus.

Question 2

A nucleus undergoes beta-minus decay. What happens to the atomic number (Z) and mass number (A)?

Accepted Answer

In β⁻ decay, a neutron converts to a proton, emitting an electron; Z increases by 1 while A stays constant.

Question 3

The radius of a nucleus is given by R = R₀A^(1/3), where R₀ ≈ 1.2 fm. If nucleus A has mass number 8 and nucleus B has mass number 27, what is the ratio of their radii R_A : R_B?

Accepted Answer

R_A/R_B = (8/27)^(1/3) = 2/3, so R_A : R_B = 2 : 3.

Question 4

An alpha particle (He-4 nucleus) has a mass defect of 0.0305 u. What is its binding energy? (1 u = 931 MeV/c²)

Accepted Answer

Binding Energy = mass defect × c² = 0.0305 × 931 ≈ 28.4 MeV.

Question 5

In nuclear fission, a heavy nucleus splits into two lighter nuclei. Which statement is correct?

Accepted Answer

Fission converts mass defect into kinetic energy of products (E=mc²); lighter nuclei have higher binding energy per nucleon.

Question 6

A nucleus has mass number A = 64 and atomic number Z = 29. The number of neutrons in this nucleus is:

Accepted Answer

Number of neutrons = A - Z = 64 - 29 = 35.

Question 7

The binding energy per nucleon for a nucleus is maximum for which mass number range?

Accepted Answer

Binding energy per nucleon is maximum around A = 50-60 (Iron-56), indicating maximum nuclear stability.

Question 8

A radioactive nucleus undergoes beta-minus decay. Which of the following changes occur?

Accepted Answer

In β⁻ decay, a neutron converts to a proton, so Z increases by 1 while A remains unchanged.

Question 9

The radius of a nucleus is approximately given by R = R₀A^(1/3), where R₀ = 1.2 × 10⁻¹⁵ m. The radius of a nucleus with A = 216 is approximately:

Accepted Answer

R = 1.2 × 10⁻¹⁵ × (216)^(1/3) = 1.2 × 10⁻¹⁵ × 6 = 7.2 × 10⁻¹⁵ m.

Question 10

A nucleus emits an alpha particle (Helium-4 nucleus). The mass number and atomic number of the parent nucleus change by:

Accepted Answer

Alpha decay emits ⁴₂He, so mass number decreases by 4 and atomic number decreases by 2.

Question 11

The binding energy of a nucleus with Z = 8, A = 16 is 127.6 MeV. The binding energy per nucleon is:

Accepted Answer

Binding energy per nucleon = Total BE / A = 127.6 / 16 = 7.975 ≈ 7.98 MeV.

Question 12

Two nuclei have mass numbers A₁ = 100 and A₂ = 200. The ratio of their nuclear densities (ρ₁/ρ₂) is:

Accepted Answer

Nuclear density ρ = (mass)/(volume) ∝ A/(A^(1/3))³ is approximately constant for all nuclei, so ρ₁/ρ₂ ≈ 1.

Question 13

A nucleus undergoes two successive alpha decays starting from mass number A = 232. The final mass number is:

Accepted Answer

Each alpha decay reduces A by 4; after two decays: A = 232 - 4 - 4 = 224.

Question 14

The packing fraction of a nucleus is defined as (M - A)/A × 10³, where M is the mass defect in atomic mass units. If a nucleus has mass defect M = 0.5 u and A = 50, the packing fraction is:

Accepted Answer

Packing fraction = (0.5 - 50)/50 × 10³ = (-49.5/50) × 10³ = -990 ≈ -10 (or more precisely -9900); rechecking: (M-A) gives negative value, so packing fraction = (0.5-50)/50 × 10 = -9.9 ≈ -10.

Question 15

A nucleus of mass number A = 200 undergoes alpha decay. What is the mass number of the daughter nucleus?

Accepted Answer

Alpha decay emits a helium nucleus (⁴₂He), so mass number decreases by 4: 200 - 4 = 196.

Question 16

The binding energy per nucleon for a nucleus is 8.5 MeV. If the nucleus has 60 nucleons, what is its total binding energy?

Accepted Answer

Total binding energy = (binding energy per nucleon) × (number of nucleons) = 8.5 × 60 = 510 MeV.

Question 17

A radioactive nucleus has a half-life of 10 years. What fraction of the original sample will remain after 30 years?

Accepted Answer

After 30 years, three half-lives pass (30/10 = 3), so remaining fraction = (1/2)³ = 1/8.

Question 18

The radius of a nucleus with mass number A = 125 is approximately 6 fm. Using the relation R = R₀A^(1/3), find the nuclear radius constant R₀ in fermis.

Accepted Answer

Using R = R₀A^(1/3): 6 = R₀ × 125^(1/3) = R₀ × 5, therefore R₀ = 6/5 = 1.2 fm.

Agniveer Army Technical Nuclei

Nuclei — Practice Questions