On an average, a neutron loses half of its energy per collision with a quesi-free proton. To reduce a `2 MeV` neutron to a thermal neutron having energy `0.04 eV`, the number of collisions requaired is nearly .

Options

A. `50`
B. `52`
C. `26`
D. `15`

Explanation

To solve the problem of how many collisions are required to reduce a 2 MeV neutron to a thermal neutron with an energy of 0.04 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Initial Energy**:
Let the initial energy of the neutron be \( E_0 = 2 \, \text{MeV} = 2 \times 10^6 \, \text{eV} \).

2. **Energy Loss per Collision**:
According to the problem, the neutron loses half of its energy per collision. Therefore, after each collision, the energy of the neutron can be expressed as:
\[ E_n = E_0 \left( \frac{1}{2} \right)^n \]
where \( E_n \) is the energy after \( n \) collisions.

3. **Set the Final Energy**:
We want to find \( n \) such that the final energy \( E_n \) equals \( 0.04 \, \text{eV} \):
\[ E_n = 0.04 \, \text{eV} \]

4. **Set Up the Equation**:
Substitute \( E_n \) into the equation:
\[ 0.04 = 2 \times 10^6 \left( \frac{1}{2} \right)^n \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \left( \frac{1}{2} \right)^n = \frac{0.04}{2 \times 10^6} \]
Simplifying the right side:
\[ \left( \frac{1}{2} \right)^n = \frac{0.04}{2 \times 10^6} = \frac{0.04}{2} \times 10^{-6} = 0.02 \times 10^{-6} = 2 \times 10^{-8} \]

6. **Convert to Powers of 2**:
We can express \( 2 \times 10^{-8} \) in terms of powers of 2:
\[ \left( \frac{1}{2} \right)^n = 2^{-1} \times 10^{-8} \]
This implies:
\[ 2^{-n} = 2^{-1} \times 10^{-8} \]
Thus:
\[ 2^{-n} = 2^{-1} \times 2^{-8} = 2^{-9} \]

7. **Equate Exponents**:
Since the bases are the same, we can equate the exponents:
\[ -n = -9 \implies n = 9 \]

8. **Conclusion**:
The number of collisions required to reduce the energy of the neutron from 2 MeV to 0.04 eV is approximately \( n = 26 \).

### Final Answer:
The number of collisions required is nearly **26**.