Question

Part 1: A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is about 11.7 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?

Part 2: If the initial kinetic energy of the neutron is
9.55 × 10−13 J, find its final kinetic energy.
Answer in units of J.

Answer 1

To solve this problem, we can use the principles of conservation of momentum and kinetic energy in elastic collisions.

### Part 1: Fraction of Kinetic Energy Transferred

Let:
- \( m_n \) = mass of the neutron
- \( m_a \) = mass of the atomic nucleus = 11.7 \( m_n \)

Let \( v_{n_i} \) be the initial velocity of the neutron, and \( v_{a_i} = 0 \) be the initial velocity of the atomic nucleus.

After the elastic collision:
- \( v_{n_f} \) = final velocity of the neutron
- \( v_{a_f} \) = final velocity of the atomic nucleus

Using the equations for velocity after an elastic collision, we can derive:
1. **Conservation of Momentum**:
\[
m_n v_{n_i} + m_a v_{a_i} = m_n v_{n_f} + m_a v_{a_f}
\]
Since \( v_{a_i} = 0 \):
\[
m_n v_{n_i} = m_n v_{n_f} + m_a v_{a_f}
\]

2. **Conservation of Kinetic Energy**:
\[
\frac{1}{2} m_n v_{n_i}^2 = \frac{1}{2} m_n v_{n_f}^2 + \frac{1}{2} m_a v_{a_f}^2
\]

Using the formulas for final velocities after an elastic collision, we find:
\[
v_{n_f} = \frac{m_n - m_a}{m_n + m_a} v_{n_i}
\]
\[
v_{a_f} = \frac{2 m_n}{m_n + m_a} v_{n_i}
\]

Next, substituting \( m_a = 11.7 m_n \):

1. **Final velocity of the neutron**:
\[
v_{n_f} = \frac{m_n - 11.7 m_n}{m_n + 11.7 m_n} v_{n_i} = \frac{-10.7 m_n}{12.7 m_n} v_{n_i} = -\frac{10.7}{12.7} v_{n_i}
\]

2. **Final velocity of the atomic nucleus**:
\[
v_{a_f} = \frac{2 m_n}{m_n + 11.7 m_n} v_{n_i} = \frac{2}{12.7} v_{n_i}
\]

Now we can find the fraction of the kinetic energy transferred to the atomic nucleus:

- The initial kinetic energy of the neutron is:
\[
KE_{n_i} = \frac{1}{2} m_n v_{n_i}^2
\]

- The final kinetic energy of the atomic nucleus:
\[
KE_{a_f} = \frac{1}{2} m_a v_{a_f}^2 = \frac{1}{2} (11.7 m_n) \left(\frac{2}{12.7} v_{n_i}\right)^2
\]
\[
= \frac{1}{2} (11.7 m_n) \cdot \frac{4}{161.29} v_{n_i}^2 = \frac{23.4 m_n v_{n_i}^2}{161.29}
\]

The fraction of kinetic energy transferred is:
\[
\text{Fraction} = \frac{KE_{a_f}}{KE_{n_i}} = \frac{\frac{23.4 m_n v_{n_i}^2}{161.29}}{\frac{1}{2} m_n v_{n_i}^2} = \frac{46.8}{161.29}
\]
Calculating this gives:
\[
\text{Fraction} \approx \frac{46.8}{161.29} \approx 0.290
\]

### Part 2: Final Kinetic Energy of the Neutron

Initial kinetic energy of the neutron is:
\[
KE_{n_i} = 9.55 \times 10^{-13} \, J
\]

The fraction of kinetic energy retained by the neutron is:
\[
1 - \text{Fraction} \approx 1 - 0.290 = 0.710
\]

Now we can find the final kinetic energy of the neutron:
\[
KE_{n_f} = 0.710 \cdot KE_{n_i} = 0.710 \cdot 9.55 \times 10^{-13} \, J
\]
Calculating this gives:
\[
KE_{n_f} \approx 6.78 \times 10^{-13} \, J
\]

### Summary of answers:
1. Fraction of kinetic energy transferred to the atomic nucleus: **~0.290**
2. Final kinetic energy of the neutron: **~6.78 x 10^{-13} J**