A neutron travelling with a velocity `v` and kinetic energy `E` collides perfectly elastically head on with the nucleus of an atom of mass number `A` at rest. The fraction of the total kinetic energy retained by the neutron is

Options

A. `[(A-1)// (A + 1)]^(2)`
B. `[(A + 1)// (A - 1)]^(2)`
C. `[(A - 1) // A]^(2)`
D. `[(A + 1)// A]^(2)`

Answer

A

Explanation

`v_(1) = ((m_(1) - m_(2) ) v_(1) + 2m_(2) v_(2))/(m_(1) + m_(2))`
As `v_(2)` is zero , `m_(2) gt m_(1) , v_(1)` is in the opposite direction .
`m_(1) = 1 , m_(2) = A` .
`|v_(1)| = ((A - 1)/(A + 1)) v_(1)`
The fraction of total energy retained is
`(1//2 mv_(1)^(2))/(1//2 mv_(1)^(2)) = ((A - 1)^(2))/((A + 1)^(2))`