Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 53

$4.033MeV$

We can find the required energy as follows: First we find the mass of $H^2_1+H^2_1$ $m_i=2(2.014102u)=4.028204u$ and $m_f=3.016049u+1.007825u=4.023874u$ $\Delta m=4.023874u-4.028204u=-0.004330u$ Now $E=\Delta mc^2$ We plug in the known values to obtain: $E=0.004330u(\frac{931.5MeV/c^2}{1u})c^2$ $E=4.033MeV$