Question

Before the neutrino hypothesis the beta decay process was throught to be the transition.
`n to p+bar(e)`
If this 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.

Answer 1

Before `beta-`decay neutron is at rest Hence `E_(n) =m_(n)c^(2), p_(n)=0`
`p_(n) = p_(p) +p_(e)`
`" or " " " p_(p) +p_(e) =0 rArr |p_(p)| =|p_(e)| = p`
`"Also"" " E_(p) =(m_(p)^(2) c^(4) +p_(p)^(2) c^(2))^(1/2)`
`E_(p) =(m_(p)^(2) c^(4) +p_(p)^(2) c^(2))^(1/2)`
`=(m_(e)^(2)c^(4) +p_(e)^(2)c^(2))^(1/2)`
From conservation of energy
`(m_(p)^(2)c^(2)+p^(2)c^(2))^(1/2) + =(m_(e)^(2) c^(4) +p^(2)c^(2))^(1/2) =m_(n)c^(2)`
`m_(p)c^(2) ~~936MeV , m_(n)c^(2) ~~ 938MeV , m_(e)c^(2) =0.51MeV`
since the energy difference between n and p is small pc will be small `pc lt lt lt m_(p)c^(2)` while pc may be breater than `m_(e)c^(2)`
`rArr" "m_(p)c^(2) + (p^(2)c^(2))/(2m_(p)^(2)c^(4)) ~~ m_(n)c^(2) =pc`
To first order `pc~~ m_(n)c^(2) -m_(p)c^(2) =938MeV -936MeV = 2MeV`
this gives the momentum of proton or neutron. then
`E_(p) =(m_(p)^(2) c^(4) +p^(2)c^(2))^(1/2) =sqrt(936^(2)+2^(2))`
`~~936 MeV`
`E_(e) =(m_(e)^(2)c^(4)=p^(2)c^(2))^(1/2) =sqrt((0.51)^(2)+2^(2))`
`=2.06MeV`