Physics - Modern Physics Question with Solution | TestHub

$\underset{\text{Deuterium}}{{}_1^2\mathrm{H}}+\underset{\text{Tritium}}{{}_1^3\mathrm{H}}\to \underset{\text{Helium}}{{}_2^4\mathrm{He}}+\underset{\text{neutron}}{{}_0^1\mathrm{n}+Q}$ The energy released in the process is given by $\begin{array}{rl}& Q=\left[M_{{}_1^2\mathrm{H}}+M_{{}_1^3\mathrm{H}}-M_{{}_2^4\mathrm{He}}-M_{{}_0^1\mathrm{n}}\right]c^2\\ & =[2.014102+3.016050-4.002603-1.008665]u{c}^2\\ & =(0.018884\mathrm{u})\left[931.5\frac{\mathrm{MeV}}{\mathrm{u}}\right]=17.6\mathrm{MeV}.\end{array}$