A neutron collides, head-on with a deuterium at rest. What fraction of the neutron's energy would be transferred to the deuterium?

Options

A. `89%`
B. `11%`
C. `79%`
D. `21%`

Explanation

To solve the problem of determining the fraction of the neutron's energy that is transferred to the deuterium during a head-on elastic collision, we can follow these steps:

### Step 1: Define the masses and initial conditions
Let:
- Mass of the neutron = \( m \)
- Mass of the deuterium (which consists of one proton and one neutron) = \( 2m \)
- Initial velocity of the neutron = \( u \)
- Initial velocity of the deuterium = \( 0 \) (at rest)

### Step 2: Apply conservation of momentum
According to the law of conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
This gives us:
\[ mu = mv_1 + 2mv_2 \]
where \( v_1 \) is the final velocity of the neutron and \( v_2 \) is the final velocity of the deuterium. We can simplify this equation by dividing through by \( m \):
\[ u = v_1 + 2v_2 \tag{1} \]

### Step 3: Apply the coefficient of restitution
For an elastic collision, the coefficient of restitution \( e \) is equal to 1. This means:
\[ \frac{v_2 - v_1}{u - 0} = 1 \]
From this, we can derive:
\[ v_2 - v_1 = u \tag{2} \]

### Step 4: Solve equations (1) and (2)
Now we have two equations:
1. \( u = v_1 + 2v_2 \)
2. \( v_2 - v_1 = u \)

From equation (2), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = u + v_1 \tag{3} \]

Substituting equation (3) into equation (1):
\[ u = v_1 + 2(u + v_1) \]
Expanding this gives:
\[ u = v_1 + 2u + 2v_1 \]
Combining like terms:
\[ u = 3v_1 + 2u \]
Rearranging gives:
\[ 3v_1 = u - 2u = -u \]
Thus:
\[ v_1 = -\frac{u}{3} \tag{4} \]

Now substituting \( v_1 \) back into equation (3) to find \( v_2 \):
\[ v_2 = u + \left(-\frac{u}{3}\right) = \frac{3u}{3} - \frac{u}{3} = \frac{2u}{3} \tag{5} \]

### Step 5: Calculate the kinetic energy transferred to deuterium
The kinetic energy of the deuterium after the collision is given by:
\[ KE_{D} = \frac{1}{2} \times 2m \times v_2^2 = m \left(\frac{2u}{3}\right)^2 = m \times \frac{4u^2}{9} = \frac{4}{9} mu^2 \]

The initial kinetic energy of the neutron is:
\[ KE_{n} = \frac{1}{2} mu^2 \]

### Step 6: Determine the fraction of energy transferred
The fraction of the neutron's energy transferred to the deuterium is:
\[ \text{Fraction} = \frac{KE_{D}}{KE_{n}} = \frac{\frac{4}{9} mu^2}{\frac{1}{2} mu^2} = \frac{\frac{4}{9}}{\frac{1}{2}} = \frac{4}{9} \times \frac{2}{1} = \frac{8}{9} \]

### Step 7: Convert to percentage (if needed)
To convert this fraction to a percentage:
\[ \frac{8}{9} \times 100 \approx 88.89\% \]

### Final Answer
The fraction of the neutron's energy that is transferred to the deuterium is approximately \( \frac{8}{9} \) or about 89%.

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