Chapter 30 - Nuclear Reactions and Elementary Particles - Learning Path Questions and Exercises - Exercises - Page 1031: 20

We have $Q=\Delta mc^{2}$ For a beta decay, $A_{p}=A_{d}+e_{-1}$, $M_{D}=m_{d}+m_{e}$ & $M_{p}=m_{p}$ Here, $\Delta m=m_{p}-m_{d}-m_{e}$ So, $Q=\Delta mc^{2}=(m_{p}-m_{d}-m_{e})c^{2}=(m_{p}-(m_{d}+m_{e}))c^{2}=(M_{p}-M_{D})c^{2}$ Hence, proved.

We have $Q=\Delta mc^{2}$ For a beta decay, $A_{p}=A_{d}+e_{-1}$, $M_{D}=m_{d}+m_{e}$ & $M_{p}=m_{p}$ Here, $\Delta m=m_{p}-m_{d}-m_{e}$ So, $Q=\Delta mc^{2}=(m_{p}-m_{d}-m_{e})c^{2}=(m_{p}-(m_{d}+m_{e}))c^{2}=(M_{p}-M_{D})c^{2}$ Hence, proved.