Question

A $1.0MeV$ neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be $(Take,\frac{3}{2}{k}_{B}T=0.040eV)$

• A. $5$
• B. $25$
• C. $125$
• D. $625$

Answer 1

Answer: (B)

$25$

$\begin{array}{c}\text{Let "n" be required number}\\ \text{of collision}\\ \text{Given}\frac{3}{2}{k}_{B}T=0.04\\ \left(0.04\right)n=1\\ n=25\end{array}$