Passing down to the ground state, excited `Ag^(109)` nuclei emit either gamma quanta with energy `87 keV or K` conversion electrons whose binding enrgy is `26keV`. Find the velocity of these electrons.

Explanation

In internal conversion, the total enrgy is used to Knock out `K` electrons. The `K.E` of these electrons is enrgy availiable `B.E` of `K` electrons
`=(87-26)= 61keV`
The total energy including rest mass of electrons is `0.511+0.061= 0.572MeV`
The momentum corresponding to this total enrgy is
`sqrt((0.572)^(2)-(0.511)^(2))//c=0257MeV//c`
The velocity is then `(c^(2)P)/(E )= cxx(0.257)/(0.572)= 0.449c`