A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron
(a)in a head-on collision
(b) in scatterting at right angles.

Explanation

In a head on collision
`sqrt(2mT)=p_(d)+p_(n)`
`sqrt(2mT)=p_(d)+p_(n)`
`T=(p_(d)^(2))/(2M)+p_(n^(2))/(2m)`
where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring
`p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT`
`p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT`
or since `p_(d)=0` in a head on collisions
`P_(n)= -(1)/(2)(1-(m)/(M))p_(d)`
Going back to energy con
where `p_(d)` and `p_(n)` are the momenta of deuteron and neutron after the collision. Squaring
`p_(d)^(2)+p_(n)^(2)+2p_(d)p_(n)=2mT`
`p_(n)^(2)+(m)/(M)p_(d)^(2)= 2mT`
or since `p_(d)=0` in a head on collisions
`P_(n)= -(1)/(2)(1-(m)/(M))p_(d)`
Going back to energy conservation
`(p_(d)^(2))/(2M)[1+(M)/(4m)(1-(m)/(M))^(2)]=T`
so `(p_(d)^(2))/(2M)=(4mM)/((m+M)^(2))=(8)/(9)`
(b) In this case neutron is scattered by `90^(@)`. Then
we have from the diagram
`vec(p)_(d)= p_(n)hat(j)+sqrt(2mT)hat(i)`
Then by energy conservation
`(p_(n)^(2)+2mT)/(2M)+(p_(n)^(2))/(2m)=T`
or `(p_(n)^(2))/(2m)(1+(m)/(M))=T(1-(m)/(M))`
or `(p_(n)^(2))/(2m)=(M-m)/(M+m).T`
The energy lost by neutron in then
`T-(p_(n)^(2))/(2m)=(2m)/(M+m)T`
or fraction of energy lost is `eta=(2m)/(M+m)=(2)/(3)`