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

Options

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

Answer

To solve the problem of a neutron colliding elastically with a nucleus, we will follow these steps:

### Step 1: Understand the Problem
We have a neutron of mass \( m \) traveling with velocity \( v \) and kinetic energy \( E \). It collides elastically with a nucleus of mass \( A \cdot m \) (where \( A \) is the mass number of the nucleus) that is initially at rest. We need to find the fraction of total energy retained by the neutron after the collision.

### Step 2: Initial Kinetic Energy of the Neutron
The initial kinetic energy \( E \) of the neutron is given by:
\[ E = \frac{1}{2} m v^2 \]

### Step 3: Conservation of Momentum
Since the collision is elastic, we can use the conservation of momentum. The initial momentum of the system is:
\[ p_{initial} = m v + 0 = m v \]
After the collision, let the neutron's velocity be \( v_1 \) and the nucleus's velocity be \( v_2 \). The conservation of momentum gives us:
\[ m v = m v_1 + A m v_2 \]
Dividing through by \( m \):
\[ v = v_1 + A v_2 \quad \text{(1)} \]

### Step 4: Conservation of Kinetic Energy
For elastic collisions, the kinetic energy is also conserved. The initial kinetic energy is:
\[ E_{initial} = \frac{1}{2} m v^2 \]
The final kinetic energy is:
\[ E_{final} = \frac{1}{2} m v_1^2 + \frac{1}{2} A m v_2^2 \]
Setting these equal gives us:
\[ \frac{1}{2} m v^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} A m v_2^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ v^2 = v_1^2 + A v_2^2 \quad \text{(2)} \]

### Step 5: Solve the Equations
From equation (1), we can express \( v_1 \) in terms of \( v \) and \( v_2 \):
\[ v_1 = v - A v_2 \]
Substituting \( v_1 \) into equation (2):
\[ v^2 = (v - A v_2)^2 + A v_2^2 \]
Expanding the left side:
\[ v^2 = v^2 - 2A v v_2 + A^2 v_2^2 + A v_2^2 \]
This simplifies to:
\[ 0 = -2A v v_2 + (A^2 + A) v_2^2 \]
Factoring out \( v_2 \):
\[ v_2 (A^2 + A - 2A v) = 0 \]
Since \( v_2 \neq 0 \), we have:
\[ A^2 + A - 2A v = 0 \]
Solving for \( v_2 \):
\[ v_2 = \frac{v}{A + 1} \]

### Step 6: Find \( v_1 \)
Substituting \( v_2 \) back into equation (1):
\[ v_1 = v - A \left(\frac{v}{A + 1}\right) = v \left(1 - \frac{A}{A + 1}\right) = v \left(\frac{1}{A + 1}\right) \]

### Step 7: Calculate the Final Kinetic Energy of the Neutron
The final kinetic energy of the neutron is:
\[ E_{final} = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(\frac{v}{A + 1}\right)^2 = \frac{m v^2}{2(A + 1)^2} \]

### Step 8: Find the Fraction of Energy Retained
The fraction of the total energy retained by the neutron is:
\[ \text{Fraction} = \frac{E_{final}}{E_{initial}} = \frac{\frac{m v^2}{2(A + 1)^2}}{\frac{1}{2} m v^2} = \frac{1}{(A + 1)^2} \]

### Final Answer
The fraction of total energy retained by the neutron is:
\[ \frac{1}{(A + 1)^2} \]