Nuclear Physics Numerical Problems - Chapter 21 of FSc/ICS Physics

Nuclear Physics Numerical Problems - Chapter 21

Problem 21.1: Find the mass defect and the binding energy for tritium, if the atomic mass of tritium is 3.016049 u.

Given Data:

• Mass of Tritium, \( m_T = 3.016049 \text{u} \)
• Mass of Proton, \( m_P = 1.007276 \text{u} \)
• Mass of Neutron, \( m_N = 1.008665 \text{u} \)
• Atomic Number, \( Z = 1 \)
• Mass Number, \( A = 3 \)

To Determine:

• (a) Mass Defect, \( \Delta m = ? \)
• (b) Binding Energy, \( E_B = ? \)

Calculations:

\[ \Delta m = [Z m_P + (A - Z) m_N] - m_T \] \[ \Delta m = [1 \times 1.007276 + 2 \times 1.008665] - 3.016049 = 0.008557 \text{u} \]

\[ E_B = \Delta m c^2 = 0.008557 \times 931 = 7.97 \text{MeV} \]

Problem 21.2: The half-life of \( ^{91}_{38}\text{Sr} \) is 9.70 hours. Find its decay constant.

Given Data:

• Half-life, \( T_{1/2} = 9.70 \text{hours} = 34920 \text{s} \)

To Determine:

• Decay Constant, \( \lambda = ? \)

Calculations:

\[ T_{1/2} = \frac{0.693}{\lambda} \Rightarrow \lambda = \frac{0.693}{T_{1/2}} \] \[ \lambda = \frac{0.693}{34920} = 1.98 \times 10^{-5} \text{s}^{-1} \]

Problem 21.3: The element \( ^{234}_{91}\text{Pa} \) is unstable and decays by β-emission with a half-life 6.66 s. State the nuclear reaction and the daughter nuclei.

Given Data:

• Parent Nucleus: \( ^{234}_{91}\text{Pa} \)
• Half-life: \( T_{1/2} = 6.66 \text{s} \)

To Determine:

• Nuclear Reaction and Daughter Nucleus

Calculations:

\[ ^{234}_{91}\text{Pa} \rightarrow ^{234}_{92}\text{U} + ^0_{-1}\beta \]

Daughter nucleus is Uranium-234.

Problem 21.4: Find the energy associated with the following reaction. What does negative sign indicate?

\[ ^{14}_{7}\text{N} + ^4_2\text{He} \rightarrow ^{17}_{8}\text{O} + ^1_1\text{H} \]

Given Data:

• Mass of \( ^{14}_{7}\text{N} = 14.0031 \text{u} \)
• Mass of \( ^4_2\text{He} = 4.00264 \text{u} \)
• Mass of \( ^{17}_{8}\text{O} = 16.991 \text{u} \)
• Mass of \( ^1_1\text{H} = 1.00784 \text{u} \)

To Determine:

• Associated Energy, \( Q = ? \)

Calculations:

\[ \Delta m = (m(^{14}_{7}\text{N}) + m(^4_2\text{He})) - (m(^{17}_{8}\text{O}) + m(^1_1\text{H})) \] \[ \Delta m = (14.0031 + 4.00264) - (16.991 + 1.00784) = -0.0012 \text{u} \]

\[ Q = \Delta m c^2 = -0.0012 \times 931 = -1.12 \text{MeV} \]

The negative sign indicates that energy is needed to start this nuclear reaction.

Problem 21.5: Determine the energy associated with the following reaction:

\[ ^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^0_{-1}e \]

Given Data:

• Mass of \( ^{14}_{6}\text{C} = 14.003918 \text{u} \)
• Mass of \( ^{14}_{7}\text{N} = 14.00067 \text{u} \)
• Mass of \( ^0_{-1}e = 0.0005484 \text{u} \)

To Determine:

• Associated Energy, \( Q = ? \)

Calculations:

\[ \Delta m = m(^{14}_{6}\text{C}) - [m(^{14}_{7}\text{N}) + m(^0_{-1}e)] \] \[ \Delta m = 14.003918 - (14.00067 + 0.0005484) = 0.0026996 \text{u} \]

\[ Q = \Delta m c^2 = 0.0026996 \times 931 = 2.51 \text{MeV} \]

Problem 21.6: If \( ^{233}_{92}\text{U} \) decays twice by α-emission, what is the resulting isotope?

Given Data:

• Parent Nucleus: \( ^{233}_{92}\text{U} \)

To Determine:

• Resulting Isotope

Calculations:

First decay: \( ^{233}_{92}\text{U} \rightarrow ^{229}_{90}\text{Th} + ^4_2\text{He} \)

Second decay: \( ^{229}_{90}\text{Th} \rightarrow ^{225}_{88}\text{Ra} + ^4_2\text{He} \)

The resulting isotope is Radium-225.

Problem 21.7: Calculate the energy (in MeV) released in the following fusion reaction:

\[ ^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0n \]

Given Data:

• Mass of \( ^2_1\text{H} = 2.014102 \text{u} \)
• Mass of \( ^3_1\text{H} = 3.01605 \text{u} \)
• Mass of \( ^4_2\text{He} = 4.00263 \text{u} \)
• Mass of \( ^1_0n = 1.008665 \text{u} \)

To Determine:

• Energy Released, \( Q = ? \)

Calculations:

\[ \Delta m = (m(^2_1\text{H}) + m(^3_1\text{H})) - (m(^4_2\text{He}) + m(^1_0n)) \] \[ \Delta m = (2.014102 + 3.01605) - (4.00263 + 1.008665) = 0.018857 \text{u} \]

\[ Q = \Delta m c^2 = 0.018857 \times 931 = 17.56 \text{MeV} \]

Problem 21.8: A sheet of lead 5.0 mm thick reduces the intensity of a beam of γ-rays by a factor 0.4. Find half value thickness of lead sheet which will reduce the intensity to half of its initial value.

Given Data:

• Thickness, \( x_1 = 5.0 \text{mm} = 5 \times 10^{-3} \text{m} \)
• Intensity Reduction: \( I_1 = 0.4 I_0 \)
• Desired Intensity: \( I_2 = 0.5 I_0 \)

To Determine:

• Half-value thickness, \( x_2 = ? \)

Calculations:

Using the attenuation formula \( I = I_0 e^{-\mu x} \):

For first case: \( 0.4 = e^{-\mu x_1} \Rightarrow \ln(0.4) = -\mu x_1 \quad (1) \)

For second case: \( 0.5 = e^{-\mu x_2} \Rightarrow \ln(0.5) = -\mu x_2 \quad (2) \)

Dividing (2) by (1): \( \frac{\ln(0.5)}{\ln(0.4)} = \frac{-\mu x_2}{-\mu x_1} = \frac{x_2}{x_1} \)

\[ x_2 = x_1 \cdot \frac{\ln(0.5)}{\ln(0.4)} = 5 \times \frac{-0.6931}{-0.9163} = 3.78 \text{mm} \]

Problem 21.9: Radiation from a point source obeys the inverse square law. If the count rate at a distance of 1.0 m from Geiger counter is 360 counts per minute, what will be its count rate at 3.0 m from the source?

Given Data:

• Count rate at \( r_1 = 1.0 \text{m} \): \( I_1 = 360 \text{counts/min} \)
• Distance \( r_2 = 3.0 \text{m} \)

To Determine:

• Count rate at \( r_2 \), \( I_2 = ? \)

Calculations:

Using inverse square law \( I \propto \frac{1}{r^2} \):

\[ \frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} \Rightarrow I_2 = I_1 \cdot \frac{r_1^2}{r_2^2} \] \[ I_2 = 360 \times \frac{1^2}{3^2} = 360 \times \frac{1}{9} = 40 \text{counts/min} \]

Problem 21.10: A 75 kg person receives a whole body radiation dose of 24 m-rad, delivered by α-particles for which RBE factor is 12. Calculate (a) the absorbed energy in joules, and (b) the equivalent dose in rem.

Given Data:

• Mass of Person \( m = 75 \text{kg} \)
• Radiation Dose \( D = 24 \text{mrad} = 24 \times 10^{-3} \text{rad} \)
• RBE factor = 12

To Determine:

• (a) Absorbed Energy in joules \( E = ? \)
• (b) Equivalent Dose in rem \( D_e = ? \)

Calculations:

(a) Absorbed energy:

\[ E = D \times m = (24 \times 10^{-3} \text{rad}) \times (0.01 \text{J/kg/rad}) \times 75 \text{kg} = 0.018 \text{J} \]

(b) Equivalent dose:

\[ D_e = D \times \text{RBE} = (24 \times 10^{-3} \text{rad}) \times 12 = 0.288 \text{rem} \]