A neutron in a nuclear reactor collides head on elastically with the nucleus of a carbon atom initially at rest. The fraction of kinetic energy transferred from the neutron to the carbon atom is

Options

A. `11/12`
B. `2/11`
C. `48/121`
D. `48/169`

Explanation

To solve the problem of finding the fraction of kinetic energy transferred from a neutron to a carbon nucleus during an elastic collision, we will follow these steps:

### Step 1: Understand the System
We have a neutron with mass \( m \) and a carbon nucleus with mass \( 12m \) (since the mass of carbon is approximately 12 times that of a neutron). The neutron is moving with an initial velocity \( u \), while the carbon nucleus is initially at rest.

### Step 2: Apply Conservation of Momentum
In an elastic collision, the total momentum before the collision equals the total momentum after the collision. We can write this as:
\[ mu = mv_1 + 12mv_2 \]
where \( v_1 \) is the final velocity of the neutron and \( v_2 \) is the final velocity of the carbon nucleus. Simplifying, we get:
\[ u = v_1 + 12v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision. Thus, we have:
\[ \frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(12m)v_2^2 \]
Cancelling \( \frac{1}{2}m \) from all terms gives:
\[ u^2 = v_1^2 + 12v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Relate Velocities Using the Elastic Collision Condition
For elastic collisions, the relative velocity of approach equals the relative velocity of separation:
\[ v_2 - v_1 = u \quad \text{(Equation 3)} \]

### Step 5: Solve the Equations
From Equation 3, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = u + v_1 \]

Substituting \( v_2 \) from Equation 3 into Equation 1:
\[ u = v_1 + 12(u + v_1) \]
This simplifies to:
\[ u = v_1 + 12u + 12v_1 \]
\[ u = 13v_1 + 12u \]
Rearranging gives:
\[ -11u = 13v_1 \implies v_1 = -\frac{11}{13}u \]

Now substituting \( v_1 \) back into the expression for \( v_2 \):
\[ v_2 = u + \left(-\frac{11}{13}u\right) = \frac{2}{13}u \]

### Step 6: Calculate the Kinetic Energies
The initial kinetic energy of the neutron is:
\[ KE_{initial} = \frac{1}{2}mu^2 \]

The final kinetic energy of the carbon nucleus is:
\[ KE_{final} = \frac{1}{2}(12m)v_2^2 = \frac{1}{2}(12m)\left(\frac{2}{13}u\right)^2 = \frac{1}{2}(12m)\left(\frac{4}{169}u^2\right) = \frac{24m}{169}u^2 \]

### Step 7: Find the Fraction of Kinetic Energy Transferred
The fraction of kinetic energy transferred from the neutron to the carbon nucleus is given by:
\[ \text{Fraction} = \frac{KE_{final}}{KE_{initial}} = \frac{\frac{24m}{169}u^2}{\frac{1}{2}mu^2} = \frac{24}{169} \cdot \frac{2}{1} = \frac{48}{169} \]

### Final Answer
Thus, the fraction of kinetic energy transferred from the neutron to the carbon atom is:
\[ \frac{48}{169} \]