Question

`.^(12)N` beta decays to an ecited state of `.^(12)C`, which subsequenctly decays to the ground state with the emission of a `4.43 -MeV` gamma ray. What is the maxium kinetic energy of the emitted beta particle?

Answer 1

To determine the Q value for this decay, we first need to find the mass of the product nucleus `.^(12)C` in its excited state. In the ground state, `.^(12)C` has a mass of `12.000000u`, so its mass in the excited state is
`12.000000u + (4.43 MeV)/(931.5 MeV//u) = 12.004756 u`
In this decay, a proton is converted to a neutron, so it must be an example of position decay. The Q value is, accoding to equation
`Q = (12.018613 u - 12.004756 u - 2 xx 0.000549 u) (931.5 MeV//u)`
`= 11.89 MeV`
(We could have just as easily found the Q value by first finding the Q value for decay to the ground state, `16.32 MeV`, and the subtracting the excitation energy of `4.43 MeV` since the decay to the excited state has that much less availbale energy.)
Neglecting the small correction for the recoil kinetic energy of the `.^(12)C` nucleus, the maximum electron kinetic energy is `11.89 MeV`.