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

Linear momentum conserved
`m_(1)u_(n)=m_(d)v_(d)+m_(n)v_(n)`
`u_(n) =2v_(d)+v_(n)`
`u_(n)-v_(n)=2v_(d)" "…(i)`
`1/2m_(n)u_(n)^(2)=1/2m_(d)v_(d)^(2)+1/2m_(n)v_(n)^(2)`
`u_(n)^(2)=2v_(d)^(2)+v_(n)^(2)`
From equation (i)
`2vd(u_(n)+v_(n))=2v_(d)^(2)`
`u_(n)+v_(n)=v_(d)`
`u_(n)=v_(d)-v_(n)`
Put in (i) `v_(d)-2v_(n)=2v_(d)`
`v_(d)=-2v_(n)`
`u_(n)=-3v_(n)`
Intial K.E of neutron `E_(n) =1/2m_(n)u_(n)^(2)=9/2m_(n)v_(n)^(2)`
`= 9 (1/2m_(n)v_(n)^(2))`
`E_(n) = 9 (E_(n)^(.))`
Fractions loss of K.E of neutron = `(E_(n)-E_(n))/(E_(n))`
`=(9E_(n)-E_(n))/(9E_(n)) = 8/9 `