Question

`.^(40)K` is an unusual isotope, in that it decays by negative beta emission, positive beta emission, and electron capture. Find the Q values for these decays.

Answer 1

The process for negative beta decay is given by equation
`._(19)^(40)K_(21) rarr ._(20)^(40)Ca_(20) +e^(-) + barv`
and the Q value is found from equation using atomic masses :
`Q_(beta) [m(.^(40)K) - m(.^(40)Ca)]c^(2)`
`= (39.963999u - 39.962591u)(931.5 MeV//u)`
`= 1.312 MeV`
Equation gives the decay process for positive beta emission :
`._(19)^(40)K_(21) rarr ._(18)^(40)Ar_(22) + e^(+) + v`
and the Q value is given by equation
`Q_(beta) = [m(.^(40)K) - m(.^(40)Ar) - 2m_(e)]c^(2)`
`= 0.482 MeV`
For electron capure,
`._(19)^(40)K_(21) +e^(+) rarr ._(18)^(40)Ar_(22) + v`
and from equation
`Q_(ec) = [m(.^(40)Ar)]c^(2)`
`= (36.963999 u - 39.962384 u)(931.5 MeV//u)`
`= 1.504 MeV`