Part 1: A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.

Question

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