An alpha-particle with kinetic energy `T_(alpha)= 7.0MeV` is scattered elastically by an initially stationary `Li^(6)` nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particle is `Theta= 60^(@)`.

Explanation

Initial momentum of the `alpha` particle is `sqrt(2mT_(alpha)) hat(i)` (where `hat(i)` is a unit vector in the incident direction.) Final momenta are respectively `vec(p)_(alpha)` and `vec(p_(Li)`. Conservation of momentum reads
`vec(P)_(a)+vecP_(Li)= sqrt(2mT_(alpha))hat(i)`
Squarting `p_(alpha)^(2)+p_(Li)+2p_(alpha)p_(Li)cos Theta= 2mT_(alpha)` (1)
where `Theta` is the angle between `vec(p)_(alpha)` and `vec(p)_(Li)`.
Also by energy conservation `(p_(alpha)^(2))/(2m)+(p_(L)^(2))/(2M)=T_(alpha)`
(`m & M` are respectively the masses of `alpha` particle and `Li^(6)`.)So
`p_(alpha)^(2)+(m)/(M)p_(Li)^(2)= 2mT_(alpha)` (2)
Substituting (2) from (1) we see that
`P_(Li)[(1-(m)/(M))p_(Li)+2p_(alpha)cos Theta]=0`
Thus if `p_(Li) ~~ 0`
`p_(alpha)= -(1)/(2)(1-(m)/(M))p_(Li)sec Theta`.
Since `p_(alpha), p_(Li)` are both positive number (being magnitudes of vectors) we must have
`-le cos Theta lt 0 if m lt M`
This being understood, we write
`(P_(Li)^(2))/(2M)[1+(M)/(4m)(1-(m)/(M))^(2) sec^(2)Theta]=T_(alpha)`
Hence the recoil energy of the `L_(i)` nucleus is
`(P_(Li)^(2))/(2M)=(T_(alpha))/(1+((M-m)^(2))/(4mM)sec^(2)Theta)`
As we pointed out above `Theta ~~ 60^(@)`. If we take `Theta= 120^(@)`, we get recoil energy of `Li= 6MeV`.