A neutron collides head-on and elasticity with an atom of mass number `A`, which is initially at rest. The fraction of kinetic energy retained by neutron is

Options

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

Answer

To solve the problem of finding the fraction of kinetic energy retained by a neutron after it collides elastically with an atom of mass number \( A \), we can follow these steps:

### Step-by-Step Solution:

1. **Define Initial Conditions**:
- Let the mass of the neutron be \( m \).
- The mass of the atom (with mass number \( A \)) is \( Am \).
- The initial velocity of the neutron is \( v_0 \).
- The initial velocity of the atom is \( 0 \) (it is at rest).

2. **Conservation of Momentum**:
- According to the conservation of momentum:
\[ mv_0 = mv_1 + Amv_2 \]
- Here, \( v_1 \) is the final velocity of the neutron, and \( v_2 \) is the final velocity of the atom.

3. **Elastic Collision Condition**:
- For an elastic collision, the relative velocities before and after the collision are related by:
\[ v_2 - v_1 = v_0 \]

4. **Rearranging the Equations**:
- From the momentum equation, we can simplify:
\[ v_0 = v_1 + Av_2 \quad \text{(1)} \]
- From the elastic collision condition, we can express \( v_2 \):
\[ v_2 = v_1 + v_0 \quad \text{(2)} \]

5. **Substituting Equation (2) into Equation (1)**:
- Substitute \( v_2 \) from equation (2) into equation (1):
\[ v_0 = v_1 + A(v_1 + v_0) \]
- This simplifies to:
\[ v_0 = v_1 + Av_1 + Av_0 \]
- Rearranging gives:
\[ v_0 - Av_0 = (1 + A)v_1 \]
- Factoring out \( v_0 \):
\[ v_0(1 - A) = (1 + A)v_1 \]
- Thus, we find:
\[ \frac{v_1}{v_0} = \frac{1 - A}{1 + A} \]

6. **Finding the Fraction of Kinetic Energy Retained**:
- The fraction of kinetic energy retained by the neutron is given by the square of the ratio of velocities:
\[ \frac{E'}{E} = \left(\frac{v_1}{v_0}\right)^2 = \left(\frac{1 - A}{1 + A}\right)^2 \]

7. **Final Expression**:
- Therefore, the fraction of kinetic energy retained by the neutron is:
\[ \frac{E'}{E} = \frac{(1 - A)^2}{(1 + A)^2} \]

### Conclusion:
The fraction of kinetic energy retained by the neutron after the elastic collision is:
\[ \frac{(1 - A)^2}{(1 + A)^2} \]