A neutron makes is 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 the neutron making a head-on elastic collision with a stationary deuteron, we will follow these steps:

### Step 1: Understand the Problem
We have a neutron (mass \( m_1 \)) colliding elastically with a stationary deuteron (mass \( m_2 = 2m_1 \)). We need to find the fraction of energy lost by the neutron after the collision.

### Step 2: Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum before and after the collision can be expressed as:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Here, \( u_1 \) is the initial velocity of the neutron, \( u_2 = 0 \) (the deuteron is stationary), \( v_1 \) is the final velocity of the neutron, and \( v_2 \) is the final velocity of the deuteron.

Substituting the values, we get:
\[ m_1 u_1 + 0 = m_1 v_1 + 2m_1 v_2 \]
This simplifies to:
\[ u_1 = v_1 + 2v_2 \quad \text{(Equation 1)} \]

### Step 3: Conservation of Kinetic Energy
The conservation of kinetic energy before and after the collision can be expressed as:
\[ \frac{1}{2} m_1 u_1^2 + 0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting \( m_2 = 2m_1 \):
\[ \frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} (2m_1) v_2^2 \]
This simplifies to:
\[ u_1^2 = v_1^2 + 2v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations (Equation 1 and Equation 2) to solve for \( v_1 \) and \( v_2 \).

From Equation 1:
\[ v_1 = u_1 - 2v_2 \]
Substituting \( v_1 \) in Equation 2:
\[ u_1^2 = (u_1 - 2v_2)^2 + 2v_2^2 \]
Expanding and simplifying:
\[ u_1^2 = u_1^2 - 4u_1v_2 + 4v_2^2 + 2v_2^2 \]
\[ 0 = -4u_1v_2 + 6v_2^2 \]
Factoring out \( v_2 \):
\[ v_2(6v_2 - 4u_1) = 0 \]
Thus, \( v_2 = \frac{2}{3} u_1 \).

Substituting \( v_2 \) back into Equation 1:
\[ v_1 = u_1 - 2\left(\frac{2}{3} u_1\right) = u_1 - \frac{4}{3} u_1 = -\frac{1}{3} u_1 \]

### Step 5: Calculate Initial and Final Kinetic Energies
Initial kinetic energy of the neutron:
\[ KE_{initial} = \frac{1}{2} m_1 u_1^2 \]

Final kinetic energy of the neutron:
\[ KE_{final} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 \left(-\frac{1}{3} u_1\right)^2 = \frac{1}{2} m_1 \frac{1}{9} u_1^2 = \frac{1}{18} m_1 u_1^2 \]

### Step 6: Calculate Energy Loss
Energy lost by the neutron:
\[ \Delta KE = KE_{initial} - KE_{final} = \frac{1}{2} m_1 u_1^2 - \frac{1}{18} m_1 u_1^2 = \left(\frac{9}{18} - \frac{1}{18}\right) m_1 u_1^2 = \frac{8}{18} m_1 u_1^2 = \frac{4}{9} m_1 u_1^2 \]

### Step 7: Fraction of Energy Loss
The fraction of energy lost by the neutron is given by:
\[ \text{Fraction of energy loss} = \frac{\Delta KE}{KE_{initial}} = \frac{\frac{4}{9} m_1 u_1^2}{\frac{1}{2} m_1 u_1^2} = \frac{4/9}{1/2} = \frac{4}{9} \times \frac{2}{1} = \frac{8}{9} \]

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