How many head-on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from `6.561 MeV` to `1 keV`?

Options

A. 4
B. 5
C. 8
D. 3

Explanation

To solve the problem of how many head-on elastic collisions a neutron must have with deuterium nuclei to reduce its energy from 6.561 MeV to 1 keV, we can follow these steps:

### Step 1: Understand the Energy Loss in a Collision
In a head-on elastic collision, the energy loss can be calculated using the formula:
\[ \Delta E = E_{\text{initial}} \cdot \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]
where \( m_1 \) is the mass of the neutron and \( m_2 \) is the mass of the deuterium nucleus.

### Step 2: Assign Mass Values
For our case:
- Mass of neutron, \( m_1 = 1 \) (in atomic mass units)
- Mass of deuterium, \( m_2 = 2 \) (in atomic mass units)

### Step 3: Calculate the Energy Loss Factor
Substituting the values into the formula:
\[ \Delta E = E_{\text{initial}} \cdot \frac{4 \cdot 1 \cdot 2}{(1 + 2)^2} = E_{\text{initial}} \cdot \frac{8}{9} \]
This means that after each collision, the energy is reduced to:
\[ E_{\text{final}} = E_{\text{initial}} - \Delta E = E_{\text{initial}} - E_{\text{initial}} \cdot \frac{8}{9} = E_{\text{initial}} \cdot \frac{1}{9} \]

### Step 4: Establish the Pattern
After the first collision:
\[ E_2 = \frac{E_1}{9} \]
After the second collision:
\[ E_3 = \frac{E_2}{9} = \frac{E_1}{9^2} \]
After the third collision:
\[ E_4 = \frac{E_3}{9} = \frac{E_1}{9^3} \]
Thus, after \( n \) collisions:
\[ E_{n+1} = \frac{E_1}{9^n} \]

### Step 5: Set Up the Equation
We need to find \( n \) such that:
\[ E_{n+1} = 1 \text{ keV} = 1 \times 10^3 \text{ eV} \]
Setting up the equation:
\[ \frac{6.561 \times 10^6 \text{ eV}}{9^n} = 1 \times 10^3 \text{ eV} \]

### Step 6: Solve for \( n \)
Rearranging gives:
\[ 9^n = \frac{6.561 \times 10^6}{1 \times 10^3} = 6.561 \times 10^3 \]
This simplifies to:
\[ 9^n = 6561 \]
Recognizing that \( 6561 = 9^4 \), we find:
\[ n = 4 \]

### Conclusion
Thus, the neutron must undergo **4 head-on elastic collisions** with deuterium nuclei to reduce its energy from 6.561 MeV to 1 keV.

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