Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :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` 
|
|