Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1152: 20

$H^3_1\rightarrow He^3_2+e^{-}+ \nu^{-}$ $19KeV$

We know that $H^3_1\rightarrow He^3_2+e^{-}+ \nu^{-}$ Now $\Delta m=m_f-m_i$ $\Delta m=3.016029u-3.016049u$ $\Delta m=-0.00002u$ Now the amount of energy released is $E=|\Delta m|c^2$ We plug in the known values to obtain: $E=|-(0.00002u)(\frac{931.494\frac{MeV}{c^2}}{1u})|c^2$ $E=0.0186MeV=19KeV$