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Before the neutrino hypothesis, the beta process was thought to be the transition, `n rarr p+bare`. If the was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range. |
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Answer» We are supporting that `beta` decay is due to the transition `n rarr p+e^(-)` Suppose before beta decay, neutron is at rest. `:. p_(n)=0 and E_(n)=m_(n)c^(2)` After `beta` decay, form conservation of linear momentum, we have `vecp_(n)=vecp_(p)+vecp_(e)=0 :. |vecp_(p)|=|vecp_(e)|=p`, say. `E_(p) =(m_(p)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2) and E_(e)=(m_(e)^(2) c^(4)+p_(e)^(2) c^(2))^(1//2)=(m_(e)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2)` form conservation of energy, `E_(p)+E_(e)=E_(n)` i.e., `(m_(p)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2)+(m_(e)^(2) c^(4)+p_(e)^(2) c^(2))^(1//2)=m_(n)c^(2).......(i)` Now, `m_(p)c^(2)=936MeV , m_(n)c^(2)=938MeV` and `m_(e)c^(2)=0.51MeV`. As energy difference between `m_(p)c^(2)` and `m_(n)c^(2)` is small, therefore pc will be small. `:. pclt lt m_(p)c^(2)`. However, pc may be greater than `m_(e)c^(2)`. therefore, form (i) `m_(p)c^(2)+(p^(2)c^(2))/(2m_(p)^(2)c^(4))~=m_(n)c^(2)-pc` To first order of approx. `pc~=m_(n)c^(2)-m_(p)c^(2)=938MeV-936MeV=2MeV` This gives us the momentum of proton or neutron. `:. E_(p)=(m_(p)^(2) c^(4)+p^(2) c^(2))^(1//2)=sqrt((936)^(2)+2^(2)) ~=936MeV` and `E_(e)=(m_(e)^(2) c^(4)+p^(2) c^(2))^(1//2)=sqrt((0.51)^(2)+2^(2))~=2.06MeV` |
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