An accelerated electron is absorbed by a proton at rest and a neutron is formed. Write the reaction equation. Assuming that the resulting neutron remains at rest, calculate the minimum kinetic energy of the electron at which the reaction is possible.

Answer

`-1^(e^(0))+1^(p^(1))to0^(n^(1))+0^(v^(0));0.24MeV`.

Explanation

The equation of the reaction is of the form `""_(-1)e^(0)+""_(1)p^(1)to""_(0)n^(1)+""_(0)v^(0)`. Since we assume the neutron produced in the reaction to be at rest, the electron momentum is completely transmitted to the neutrino. The laws of conservation of energy and the momentum will be written in the form
`p_(0)=p_(v),epsi_(e)+epsi_(0p)=epsi_(on)+epsi_(v)`
Since the rest mass of a neutrino is zero, its energy is
`epsi_(v)=p_(v)c=p_(e)c=sqrt(epsi_(e)^(2)-epsi_(0e)^(2))`
Hence
`epsi_(e)=epsi_(0p)=epsi_(0n)+sqrt(epsi_(e)^(2)-epsi_(0e)^(2))`
Solving this equation, we obtain for the total energy of an electron
`epsi_(e)=((epsi_(0n)-epsi_(0p))^(2)+epsi_(0e)^(2))/(2(epsi_(0n)-epsi_(0p)))`
The kinetic energy of the electron is
`K_(e)=epsi_(e)-epsi_(0e)=((epsi_(0n)-epsi_(0p)-epsi_(0e))^(2))/(2(epsi_(0n)-epsi_(0p)))`
Substituting the respective values, we obtain the desired result.