Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta-particles emitted by `Be^(10)` nuclei formed directly in the ground state.

Explanation

The reaction is `Be^(10) rarr B^(10)+e^(-)+overset(-)v_(e )`
For maximum `KE` of electrons we can put the energy of `vec(v)_(e )` to be zero. The atomic massed are `Be^(10)=10.016711 amu`
`B^(10)= 10.016114 amu`
So the `K.E` of electrons is (see previous problem)
`597xx10^(-6) amuxxc^(2)= 0.56MeV`
The momentum of electrons with this `K.E` is `0.941(MeV)/(c )`
and the recoil energy of the daughter is
`((0.941)^(2))/(2xxm_(d)c^(2))=((0.941)^(2))/(2xx10xx938)MeV= 47.2eV`