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`

Answer

c

Explanation

Let n collisions are required for the given conditiaon. Then,
`((1)/(2))^(n) xx 2 MeV=0.04 xx 10^(-6) MeV`
`2^(n) =(2)/(0.04 xx 10^(6)) =50xx10^(6)`
After solving above equation, `n=26.