Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(14)N` as represented by Eq.(iii). Equation (iii) refers to nuclei. Adding six electrons to both sides of Eq.(iii) gives
` .-6^(14)C "atom" rarr .-7^(14)N`.

Explanation

We know that `._(6)^(14)C` has a mass of `14.003 074u` . Here, the mass differrence between the initial and final states is
`Delta m=14.003 242 u -14.003 74 u = 0.000 168 u`
This corresponds to an energy release of
`E = (0.000 168 u)(931 . 494 MeV//u)=0.156 MeV`.