Atoms and Nuclei Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the nuclear density of a nucleus with mass \( 6.68 \times 10^{-27} \, \text{kg} \) and radius \( 2.4 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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

\( R^3 = (2.4 \times 10^{-15})^3 = 1.382 \times 10^{-44} \, \text{m}^3 \).

Volume = \( \frac{4}{3} \times 3.14 \times 1.382 \times 10^{-44} \approx 5.79 \times 10^{-44} \, \text{m}^3 \).

Density = \( \frac{6.68 \times 10^{-27}}{5.79 \times 10^{-44}} \approx 1.15 \times 10^{17} \, \text{kg/m}^3 \).

1.0 × 10¹⁷ kg/m³
1.1 × 10¹⁷ kg/m³
1.15 × 10¹⁷ kg/m³
1.2 × 10¹⁷ kg/m³
3

What is the binding energy per nucleon if a nucleus with mass number 56 has a total binding energy of \( 490 \, \text{MeV} \)?

\( E_{bn} = \frac{E_b}{A} \).

\( E_b = 490 \, \text{MeV} \), \( A = 56 \).

\( E_{bn} = \frac{490}{56} \approx 8.75 \, \text{MeV} \).

8.75 MeV
8.50 MeV
9.00 MeV
8.25 MeV
1

What is the energy equivalent of \( 0.002 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.002 \, \text{kg} \), \( c^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 0.002 \times 9 \times 10^{16} = 1.8 \times 10^{14} \, \text{J} \).

1.6 × 10¹⁴ J
1.7 × 10¹⁴ J
1.8 × 10¹⁴ J
1.9 × 10¹⁴ J
3

A nucleus has a radius of \( 3.6 \times 10^{-15} \, \text{m} \). What is its approximate mass number? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

\( R = R_0 A^{1/3} \).

\( 3.6 \times 10^{-15} = 1.2 \times 10^{-15} \times A^{1/3} \).

\( A^{1/3} = \frac{3.6}{1.2} = 3 \).

\( A = 3^3 = 27 \).

8
16
27
64
3

What is the energy equivalent of \( 0.1 \, \text{g} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

\( m = 0.1 \times 10^{-3} \, \text{kg} = 10^{-4} \, \text{kg} \), \( c^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \).

\( E = 10^{-4} \times 9 \times 10^{16} = 9 \times 10^{12} \, \text{J} \).

8.5 × 10¹² J
9.0 × 10¹² J
9.5 × 10¹² J
10.0 × 10¹² J
2

Which process is responsible for the energy release in an atomic bomb?

The energy in an atomic bomb comes from uncontrolled nuclear fission, where a heavy nucleus splits into lighter fragments, releasing energy due to increased binding energy per nucleon.

Nuclear fission
Nuclear fusion
Chemical explosion
Thermonuclear fusion
1

What is the radius of a nucleus with mass number 200? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

\( R = R_0 A^{1/3} \).

\( A = 200 \), \( A^{1/3} = 200^{1/3} \approx 5.85 \).

\( R = 1.2 \times 10^{-15} \times 5.85 \approx 7.0 \times 10^{-15} \, \text{m} \).

6.8 × 10⁻¹⁵ m
6.9 × 10⁻¹⁵ m
7.1 × 10⁻¹⁵ m
7.0 × 10⁻¹⁵ m
4

What is the energy equivalent of \( 2 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = m c^2 \).

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

\( E = 2 \times 9 \times 10^{16} = 1.8 \times 10^{17} \, \text{J} \).

1.6 × 10¹⁷ J
1.7 × 10¹⁷ J
1.8 × 10¹⁷ J
1.9 × 10¹⁷ J
3

Which factor primarily determines the nuclear radius?

The nuclear radius is given by \( R = R_0 A^{1/3} \), where \( A \) (mass number) is the key factor determining the size, as the radius scales with the cube root of the number of nucleons.

Atomic number
Mass number
Binding energy
Nuclear density
2

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

Radius ratio = \( \frac{R_1}{R_2} = \frac{R_0 A_1^{1/3}}{R_0 A_2^{1/3}} = \left( \frac{A_1}{A_2} \right)^{1/3} \).

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

\( \frac{27}{125} = 0.216 \), \( (0.216)^{1/3} \approx 0.6 \).

0.5
0.6
0.8
1.0
2

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