A neutron travelling with a velocity `v` and kinetic energy `E` collides perfectly elastically head on with the nucleus of an atom of mass number `A` at rest. The fraction of the total kinetic energy retained by the neutron is

Options

A. `((1-A)/(A+1))^(2)`
B. `((A+1)/(A-1))^(2)`
C. `((A-1)/(A))^(2)`
D. `((A+1)/(A))^(2)`

Answer

To solve the problem, we need to analyze the elastic collision between a neutron and a nucleus of an atom. Let's break down the solution step by step.

### Step 1: Understand the scenario
We have a neutron of mass \( m_n \) (approximately 1 unit) moving with velocity \( v \) and kinetic energy \( E \). The nucleus of an atom with mass number \( A \) (mass \( m_A \approx A \) units) is initially at rest.

### Step 2: Apply conservation of momentum
In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum of the system is given by:
\[ p_{\text{initial}} = m_n v + m_A \cdot 0 = m_n v \]

After the collision, let the final velocity of the neutron be \( v' \) and the final velocity of the nucleus be \( V' \). The final momentum is:
\[ p_{\text{final}} = m_n v' + m_A V' \]

Setting the initial momentum equal to the final momentum:
\[ m_n v = m_n v' + m_A V' \]

### Step 3: Apply conservation of kinetic energy
The initial kinetic energy of the neutron is:
\[ KE_{\text{initial}} = \frac{1}{2} m_n v^2 \]

The final kinetic energy after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} m_n (v')^2 + \frac{1}{2} m_A (V')^2 \]

Setting the initial kinetic energy equal to the final kinetic energy:
\[ \frac{1}{2} m_n v^2 = \frac{1}{2} m_n (v')^2 + \frac{1}{2} m_A (V')^2 \]

### Step 4: Solve for final velocities
From the conservation of momentum:
\[ m_n v = m_n v' + m_A V' \]
Rearranging gives:
\[ V' = \frac{m_n v - m_n v'}{m_A} \]

Substituting \( V' \) into the kinetic energy equation leads to a system of equations that can be solved simultaneously. However, we can use the formulas for elastic collisions directly.

For an elastic collision between two bodies, the velocities after the collision can be expressed as:
\[ v' = \frac{m_n - m_A}{m_n + m_A} v \]
\[ V' = \frac{2 m_n}{m_n + m_A} v \]

### Step 5: Calculate the fraction of kinetic energy retained by the neutron
The fraction of kinetic energy retained by the neutron is given by:
\[ \text{Fraction} = \frac{KE_{\text{final, neutron}}}{KE_{\text{initial}}} = \frac{\frac{1}{2} m_n (v')^2}{\frac{1}{2} m_n v^2} = \frac{(v')^2}{v^2} \]

Substituting \( v' \):
\[ \frac{(v')^2}{v^2} = \left(\frac{m_n - m_A}{m_n + m_A}\right)^2 \]

Since \( m_n \approx 1 \) and \( m_A \approx A \), we can simplify:
\[ \text{Fraction} = \left(\frac{1 - A}{1 + A}\right)^2 \]

### Final Answer
Thus, the fraction of the total kinetic energy retained by the neutron after the collision is:
\[ \frac{(1 - A)^2}{(1 + A)^2} \]