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A neutron with kinetic energy T= 10MeV activates a nuclear reaction C^(12)(n,alpha)Be^(9) whose threshold is T_(th)= 6.17MeV. Find the kinetic energyof the alpha-particles outgoing at right angles to the incoming neutrons' direction. |
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Answer» Solution :The momentum of incident neutron is `sqrt(2m_(N)T)hat(i)`, that of `ALPHA` particle is `sqrt(2m_(alpha)T_(alpha))hat(j)` and of `Be^(9)` is `-sqrt(2m_(alpha)T_(alpha))hat(j)+sqrt(2m_(n)T)hat(i)` By conservation of energy `T_(alpha)=[T(1-(m_(n))/(M))-|Q|](M)/(M+m_(alpha))`. Using `T_(th)=(1+(m_(n))/(M))|Q|` we GET `T_(alpha)=(M)/(M+m_(alpha))[(1-(m_(n))/(M))T-(T_(th))/(1+(m_(n))/(M)]]` `M'` is the mass of `C^(12)` nucleus. LTBRGT or `T_(alpha)=(1)/(M+m_(alpha))[(M-m_(n))T-(MM')/(M'+m_(n))T_(th)]=2.21MeV` 
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