A neutron in a nuclear reactor collides head on elastically with the nucleus of a carbon atom initially at rest. The fraction of kinetic energy transferred from the neutron to the carbon atom is

Options

A. `11/12`
B. `2/11`
C. `48/121`
D. `48/169`

Answer

D

Explanation

Let m and M be the masses of neutron and carbon nucleus (at rest ) respectively.
If u and `v_1` are the velocity of neutron before and after collision, then
`K_i=1/2m u^(2)` and `K_f=1/2mv^(2)`, But `v_1=((m-M)u)/(m+M)`
`K_f=1/2m ((m-M)/(m+M))^2 u^2 therefore K_f/K_i=((m-M)/(m+M))^2`
The fraction of kinetic energy transferred from the neutron to the carbon atom is `f=(4mM)/((m+M)^2)`
But the carbon M = 12 m
`f=(4m(12m))/((m+12m)^2)=48/169`