Question

Deuteron is a bound state of a neutron and a proton with a binding energy `B=2.2MeV. A gamma`-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron +proton) such that the n and p move in the direction of the incident `gamma`-rays. If E=B, show that this can not happen. Hence, calculate how much bigger than B must E be for such a process to happen.

Answer 1

Given binding energy B =2.2 MeV
From the energy conservation law
`E -B =K_(n) +K_(p) =(p_(n)^(2))/(2m) + (p_(p)^(2))/(2m) " ".........(i)`
From conservation of momentum,
`p_(n) +p_(p) =E/C" ".........(ii)` ltbr. `"As"" "E=B, Eq.(i) p_(n)^(2) =p_(p)^(2)=0`
In only happen if `p_(n) =p_(p) =0`
So the Eq (ii) cannot satisfied and the process cannot take place Let E=B +X, where `X lt lt B` for the process to take place. Put value of `p_(n)` from Eq.(ii) in Eq. (i) , we get
`X=(((E)/(C) -Pe))/(2m) +(p_(0)^(2))/(2m)`
`"or "" " 2p_(p)^(2) =(2Ep_(p))/(c) +(E^(2))/(c^(2)) -2mX=0`
Using the formula of quadratic equation we get
`p_(p)=((2E)/(c)+- sqrt((4E^(2))/(c^(2))-8((E^(2))/(c^(2)) -2 mX)))/(4)` For the real value `p_(p)` the discriminant is positive
`(4E^(2))/(c^(2)) =8 ((E^(2))/(c^(2))-2mX)`
`16mX =(4E^(2))/(c^(2))` ,
`X=(E^(2))/(4mc)~~ (B^(2))/(4mc^(2))" "[:. X lt B rArr E~~ B]`