Question

A 2 MeV neutron is emitted in a fission reactor. If it loses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of 0.039 eV?

• A. 20
• B. 26
• C. 30
• D. 42
• E. 48

Hint:

To drop energy by a factor of 10, it takes about 3.32 collisions (since $2^{3.32} \approx 10$). For a drop from $10^6$ to $10^{-2}$ (8 orders of magnitude), $8 \times 3.32 \approx 26.5$.

Answer 1

The Correct Option is B

Concept:
The energy after $n$ collisions is given by $E_n = E_0 (1/2)^n$, where $E_0$ is the initial energy.

Step 1: Setting up the equation.
$E_0 = 2 \text{ MeV} = 2 \times 10^6 \text{ eV}$
$E_n = 0.039 \text{ eV}$ \[ 0.039 = (2 \times 10^6) \times \left(\frac{1}{2}\right)^n \]

Step 2: Solving for $n$.
\[ 2^n = \frac{2 \times 10^6}{0.039} \approx 5.128 \times 10^7 \] Taking logs on both sides: \[ n \log 2 = \log(5.128 \times 10^7) \] \[ n \approx \frac{7.71}{0.301} \approx 25.6 \] The neutron must undergo approximately 26 collisions.