A neutron makes a head-on elastic collision with a stationary deuteron. The fraction energy loss of the neutron in the collision is

Options

A. 16/81
B. `8//9`
C. `8//27`
D. `2//3`

Answer

B

Explanation

One deuteron consists of one proton and one neutron. As the mass of a proton and a neutron is approximately same `(~~1.67xx10^(-27) kg)`, we assume that the mass of the neutron be m and the mass of the deuteron be 2m as the electron mass is negligible small compared to that of proton and neutron.
Let the initial velocity of the neutron be u. As the deuteron is initially at rest, the final velocity of the neutron is
`v_1=((m-2m)/(m+2m))u = - (mu)/(3m) = - u/3`.
and the velocity of the deuteron is
`v_2=(2m)/(m+2m)u=2/3u`.
Total energy before collosion is `E_1=1/2mu^2`
After collision, the energy gained by the deuteron is
`E_d=1/2xx(2m)v_2^2=mxx(2/3u)^2=4/9mu^2`.
`E_d` is the amount of energy lost by neutron. therefore the fractional loss of energy of the neutron is
`E_d/E_1=((4//9)mu^2)/((1//2)mu^2)=8/9`.