Atoms and Nuclei Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the primary reason nuclear reactions release more energy than chemical reactions?

Nuclear reactions involve changes in nuclear binding energy, which operates on a scale of MeV, vastly exceeding the eV scale of chemical bond energy, due to the strong nuclear force.

Higher nuclear density
Strong nuclear force
Increased electron interactions
Larger atomic radius
1

Which force is responsible for overcoming the repulsion between protons in the nucleus?

The nuclear force, being much stronger than the Coulomb force, binds protons and neutrons together, overcoming the electrostatic repulsion between positively charged protons.

Gravitational force
Coulomb force
Nuclear force
Weak force
3

What is the nuclear density of a nucleus with mass \( 2.33 \times 10^{-27} \, \text{kg} \) and radius \( 1.7 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

Density = \( \frac{\text{mass}}{\text{volume}} \), Volume = \( \frac{4}{3} \pi R^3 \).

\( R^3 = (1.7 \times 10^{-15})^3 = 4.913 \times 10^{-45} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 4.913 \times 10^{-45} \approx 2.06 \times 10^{-44} \, \text{m}^3 \).

Density = \( \frac{2.33 \times 10^{-27}}{2.06 \times 10^{-44}} \approx 1.13 \times 10^{17} \, \text{kg/m}^3 \).

1.13 × 10¹⁷ kg/m³
1.29 × 10¹⁷ kg/m³
1.50 × 10¹⁷ kg/m³
2.00 × 10¹⁷ kg/m³
1

What is the ratio of nuclear radii of two nuclei with mass numbers 64 and 8?

Radius ratio = \( \frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3} \).

\( A_1 = 64 \), \( A_2 = 8 \).

\( \frac{64}{8} = 8 \), \( (8)^{1/3} = 2 \).

1.5
2.0
2.5
3.0
2

Why does nuclear fission release more energy than chemical reactions?

Nuclear fission involves changes in nuclear binding energy (order of MeV), which is about a million times larger than the energy from chemical bond changes (order of eV).

Due to higher mass defect
Because of electron interactions
Larger energy scale of nuclear processes
Increased nuclear radius
3

What is the outcome when two light nuclei fuse into a heavier nucleus?

Fusion of light nuclei increases the binding energy per nucleon, making the resulting nucleus more tightly bound and releasing energy as a result of this increased stability.

Decrease in binding energy
Increase in nuclear radius
Reduction in nuclear force
Energy release
4

What does the mass defect of a nucleus represent?

The mass defect is the difference between the total mass of the individual protons and neutrons and the actual mass of the nucleus, representing the mass converted into binding energy that holds the nucleus together.

Mass converted into binding energy
Mass of the electrons in the atom
Excess mass due to nuclear repulsion
Mass lost during nuclear decay
1

Why does the binding energy per nucleon remain nearly constant for nuclei with mass numbers between 30 and 170?

The nuclear force is short-ranged, affecting only neighboring nucleons. In medium-sized nuclei (A = 30 to 170), most nucleons are inside the nucleus, surrounded by a constant number of neighbors, leading to a saturation effect and nearly constant binding energy per nucleon.

Due to the short-range nature of nuclear force
Due to the long-range nature of Coulomb force
Because of the increase in proton repulsion
Due to the variation in nuclear density
1

What is the ratio of nuclear radii of nuclei with mass numbers 125 and 27?

Radius ratio = \( \frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3} \).

\( A_1 = 125 \), \( A_2 = 27 \).

\( \frac{125}{27} \approx 4.63 \), \( (4.63)^{1/3} \approx 1.67 \).

1.5
1.67
2.0
2.5
2

What is the energy equivalent of \( 0.02 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.02 \, \text{kg} \), \( c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 0.02 \times 9 \times 10^{16} = 1.8 \times 10^{15} \, \text{J} \).

1.8 × 10¹⁵ J
1.6 × 10¹⁵ J
2.0 × 10¹⁵ J
2.2 × 10¹⁵ J
1

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