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. `2//3`
C. `8//27`
D. `8//9`

Answer

D

Explanation

Let the two balls of mass `m_(1)` and `m_(2)` collide each other elastically with velocities `u_(1)` and `u_(2)`. Their velocities become `v_(1)` and `v_(2)` after the collision.
Applying conservation of linear momentum, we get
`m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) + m_(2)v_(2) .........(1)`
Also from conservation of kinetic enegry
`(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) = (1)/(2)m_(1)v_(1)^(2) + (1)/(2)m_(2)v_(1)^(2) .....(2)`
Solving Eqs.(1) and (2), we get
`v_(1) = ((m_(1)-m_(2))/(m_(1) + m_(2)))u_(1) + ((2m_(2))/(m_(1)+m_(2)))u_(2) ....(3)`
and `v_(2) = ((m_(2)-m_(1))/(m_(1) + m_(2)))u_(2) + ((2m_(1))/(m_(1)+m_(2)))u_(1) ....(4)`
On taking approximate value the mass of deuteron is twice the mass of neutron.
Given, `u_(1) = u, u_(2) = 0, m_(1) = m, m_(2) = 2m`
Velocity of neutron, `v_(1) = ((m-2m)/(m+2m))u =- (u)/(3)`
Velocity of deuteron, `v_(2) = (2mu)/(m+2m) = (2)/(3)u`
Fractional enegry loss `= ((1)/(2)mu^(2) - (1)/(2)m(-(u)/(3))^(2))/((1)/(2)mu^(2))`
`=1 - (1)/(9)`
`= (8)/(9)`