It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is `p_d,` while for its similar collision with carbon nucleus at rest, fractional loss of energy is `p_c`. The values of `p_d and p_c` are respectively.

Options

A. (0,0)
B. (0,1)
C. (.89, .28)
D. (28, .89)

Explanation

To solve this problem, we need to analyze two elastic collisions: one between a neutron and deuterium, and the other between a neutron and a carbon nucleus. We will calculate the fractional loss of energy in both cases.

### Step 1: Understand the masses involved
- The mass of a neutron is denoted as \( m \).
- The mass of deuterium (which consists of one neutron and one proton) is \( 2m \).
- The mass of the carbon nucleus (which consists of 6 protons and 6 neutrons) is \( 12m \).

### Step 2: Analyze the collision with deuterium
1. **Initial Momentum**: The neutron is moving with velocity \( v \), and the deuterium is at rest.
\[ \text{Initial momentum} = mv + 0 = mv \]

2. **Final Momentum**: Let \( v_1 \) be the final velocity of the neutron and \( v_2 \) be the final velocity of the deuterium.
\[ \text{Final momentum} = mv_1 + 2m v_2 \]
Setting initial momentum equal to final momentum:
\[ mv = mv_1 + 2m v_2 \quad \Rightarrow \quad v = v_1 + 2v_2 \quad \text{(1)} \]

3. **Coefficient of Restitution**: For elastic collisions, the coefficient of restitution \( e = 1 \):
\[ e = \frac{v_2 - v_1}{v - 0} = 1 \quad \Rightarrow \quad v_2 - v_1 = v \quad \Rightarrow \quad v_2 = v + v_1 \quad \text{(2)} \]

4. **Substituting (2) into (1)**:
\[ v = v_1 + 2(v + v_1) \quad \Rightarrow \quad v = v_1 + 2v + 2v_1 \quad \Rightarrow \quad v = 3v_1 + 2v \quad \Rightarrow \quad -v = 3v_1 \quad \Rightarrow \quad v_1 = -\frac{v}{3} \]

5. **Finding \( v_2 \)**:
\[ v_2 = v + v_1 = v - \frac{v}{3} = \frac{2v}{3} \]

### Step 3: Calculate fractional loss of energy for deuterium
1. **Initial Kinetic Energy**:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]

2. **Final Kinetic Energy**:
\[ KE_{\text{final}} = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 = \frac{1}{2} m \left(-\frac{v}{3}\right)^2 + m \left(\frac{2v}{3}\right)^2 \]
\[ = \frac{1}{2} m \frac{v^2}{9} + m \frac{4v^2}{9} = \frac{1}{18} mv^2 + \frac{4}{9} mv^2 = \frac{1}{18} mv^2 + \frac{8}{18} mv^2 = \frac{9}{18} mv^2 = \frac{1}{2} mv^2 \]

3. **Fractional Loss of Energy**:
\[ \text{Fractional loss} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}} = \frac{\frac{1}{2} mv^2 - \frac{1}{2} mv^2}{\frac{1}{2} mv^2} = \frac{8}{9} \]
Thus, \( p_d = \frac{8}{9} \).

### Step 4: Analyze the collision with carbon nucleus
1. **Initial Momentum**: The neutron is moving with velocity \( v \), and the carbon nucleus is at rest.
\[ mv + 0 = mv_3 + 12m v_4 \quad \Rightarrow \quad v = v_3 + 12v_4 \quad \text{(3)} \]

2. **Coefficient of Restitution**:
\[ e = \frac{v_4 - v_3}{v} = 1 \quad \Rightarrow \quad v_4 - v_3 = v \quad \Rightarrow \quad v_4 = v + v_3 \quad \text{(4)} \]

3. **Substituting (4) into (3)**:
\[ v = v_3 + 12(v + v_3) \quad \Rightarrow \quad v = v_3 + 12v + 12v_3 \quad \Rightarrow \quad v = 13v_3 + 12v \quad \Rightarrow \quad -11v = 13v_3 \quad \Rightarrow \quad v_3 = -\frac{11}{13}v \]

4. **Finding \( v_4 \)**:
\[ v_4 = v + v_3 = v - \frac{11}{13}v = \frac{2}{13}v \]

### Step 5: Calculate fractional loss of energy for carbon nucleus
1. **Final Kinetic Energy**:
\[ KE_{\text{final}} = \frac{1}{2} m v_3^2 + \frac{1}{2} (12m) v_4^2 = \frac{1}{2} m \left(-\frac{11}{13}v\right)^2 + 6m \left(\frac{2}{13}v\right)^2 \]
\[ = \frac{1}{2} m \frac{121}{169} v^2 + 6m \frac{4}{169} v^2 = \frac{121}{338} mv^2 + \frac{24}{169} mv^2 = \frac{121}{338} mv^2 + \frac{48}{338} mv^2 = \frac{169}{338} mv^2 = \frac{1}{2} mv^2 \]

2. **Fractional Loss of Energy**:
\[ \text{Fractional loss} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}} = \frac{\frac{1}{2} mv^2 - \frac{169}{338} mv^2}{\frac{1}{2} mv^2} = \frac{169 - 121}{169} = \frac{48}{169} \]
Thus, \( p_c = \frac{48}{169} \).

### Final Result
The values of \( p_d \) and \( p_c \) are:
- \( p_d = \frac{8}{9} \)
- \( p_c = \frac{48}{169} \)