A neutron with velocity v suffers head on elastic collision with the nucleus of an atom of mass number A at rest. The fraction of energy retained by neutron is

Options

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

Answer

c

Explanation

Here `m_(1)=m_(1) m_(2)=Am,u_(1)=u,u_(2)=0`
`:.` velocity of neutron after collison is
`v=((m_(1)-m_(2))u_(1)+2m_(2)u_(2))/(m_(1)+m_(2))=((m-Am)u)/(m+Am)=((1-A)u)/((1+A))`
`:.` K.E. of neutron after collison, `1/2 m_(1)v_(2)`
or `E_(1)=1/2m_(1)((1-A)/(1+A))^(2)u^(2)` and K.E. before collison,
`E_(1)=1/2"mu"^(2)`
`:. (E_(1))/(E_(1))=(1/2m((1-A)/(1+A))^(2)u^(2))/(1/2mu^(2))=((1-A)/(1+A))^(2)`