Chapter 42 - Nuclear Physics - Problems - Page 1303: 22b

$$6.26 \mathrm{\ MeV}$$

Now, we add energy to produce as second step $^{3} \mathrm{H} \rightarrow n+^{2} \mathrm{H}$. The energy needed is $$\Delta E_{2}=\left(m_{\mathrm{th}}+m_{n}-m_{\mathrm{i} \mathrm{h}}\right) c^{2}$$ $$=(2.01410 \mathrm{u}+1.00867 \mathrm{u}-3.01605 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})$$ $$ =6.26 \mathrm{\ MeV}$$