Proton with kinetic energy `T=1.0MeV` striking a lithium target induce a nuclear reaction `p+Li^(7)rarr 2He^(4)`. Find the kinetic energy of each alpha-particles and the angle of their divergence provided their motion directions are symmertical with respect to that of incoming protons.

Explanation

The `Q`value of the reaction `Li^(7)(p,alpha)He^(4)` is
`Q=(Delta_(Li)^(7)+DeltaH-2Delta_(He)^(4))c^(2)`
`=(0.01601+0.00783-0.00520)am uxxc^(2)`
`=0.01864 am uxxc^(2)= 17.35MeV`
Since the direction of `He^(4)` nuclei is symmertical, their momenta must also be equal. Let `T` to be `K.E` of each `He^(4)`. Then
`P_(p)=2sqrt(2m_(He)T) `cos`(theta)/(2)`
`(P_(p)` is the number of proton). Also
`(p_(p)^(2))/(2m_(p))+Q= 2T=T_(P)+Q`
Hence `T_(p)+Q=2(P_(p)^(2)"sec"^(2)(theta)/(2))/(8m_(He))`
`T_(p)(m_(p))/(2m_(He))"sec"^(2)(theta)/(2)`
Hence `"cos"(theta)/(2)=sqrt((m_(p))/(2m_(He))(T_(p))/(T_(p)+Q))`
Substitution gives
`theta= 170.53^(@)`
Also `T=(1)/(2)(T_(p)+Q)= 9.18MeV`