A neutron makes a head-on elastic collision with a stationary deuteron. The fraction energy loss of the neutron in the collision is

Options

A. 16/81
B. `8//9`
C. `8//27`
D. `2//3`

Explanation

To solve the problem of finding the fraction of energy loss of a neutron in a head-on elastic collision with a stationary deuteron, we can follow these steps:

### Step 1: Understand the System
- The neutron (mass = \( m \)) is moving with an initial velocity \( u \).
- The deuteron (mass = \( 2m \)) is stationary, so its initial velocity \( u_2 = 0 \).

### Step 2: Apply Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum can be expressed as:
\[ m u + 2m \cdot 0 = m v_1 + 2m v_2 \]
where \( v_1 \) is the final velocity of the neutron and \( v_2 \) is the final velocity of the deuteron.

### Step 3: Apply Conservation of Kinetic Energy
The conservation of kinetic energy can be expressed as:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 \]
This simplifies to:
\[ u^2 = v_1^2 + 2 v_2^2 \]

### Step 4: Solve for Final Velocities
Using the equations from conservation of momentum and kinetic energy, we can solve for \( v_1 \) and \( v_2 \).

1. From the momentum equation:
\[ u = v_1 + 2 v_2 \quad \text{(1)} \]

2. From the kinetic energy equation:
\[ u^2 = v_1^2 + 2 v_2^2 \quad \text{(2)} \]

Substituting \( v_1 \) from equation (1) into equation (2):
\[ u^2 = (u - 2v_2)^2 + 2v_2^2 \]
Expanding and simplifying:
\[ u^2 = u^2 - 4uv_2 + 4v_2^2 + 2v_2^2 \]
\[ 0 = -4uv_2 + 6v_2^2 \]
Factoring out \( v_2 \):
\[ v_2(6v_2 - 4u) = 0 \]
Thus, \( v_2 = \frac{2u}{3} \).

Substituting \( v_2 \) back into equation (1):
\[ u = v_1 + 2 \left(\frac{2u}{3}\right) \]
\[ u = v_1 + \frac{4u}{3} \]
Rearranging gives:
\[ v_1 = u - \frac{4u}{3} = -\frac{u}{3} \]

### Step 5: Calculate the Energy Loss
The initial kinetic energy of the neutron is:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
The final kinetic energy of the neutron is:
\[ KE_{\text{final}} = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(-\frac{u}{3}\right)^2 = \frac{1}{2} m \frac{u^2}{9} = \frac{1}{18} m u^2 \]
The energy lost by the neutron is:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m u^2 - \frac{1}{18} m u^2 = \left(\frac{9}{18} - \frac{1}{18}\right) m u^2 = \frac{8}{18} m u^2 = \frac{4}{9} m u^2 \]

### Step 6: Calculate the Fractional Energy Loss
The fractional energy loss is given by:
\[ \text{Fractional Energy Loss} = \frac{\Delta KE}{KE_{\text{initial}}} = \frac{\frac{4}{9} m u^2}{\frac{1}{2} m u^2} = \frac{4/9}{1/2} = \frac{4}{9} \cdot \frac{2}{1} = \frac{8}{9} \]

### Final Answer
The fraction of energy loss of the neutron in the collision is:
\[ \frac{8}{9} \]