Every nuclear reaction and decay either releases energy or needs it. The binding energies tell you roughly which way it goes, but the exact bookkeeping is simpler: compare the total mass before with the total mass after.
Key idea
The Q-value of a reaction or decay is $$Q = \left(\sum m_{\text{before}} - \sum m_{\text{after}}\right)c^2,\qquad 1\ \text{u}\cdot c^2 = 931.5\ \text{MeV}.$$ $Q > 0$: energy is released (as kinetic energy of the products and gamma rays). $Q < 0$: energy must be supplied. Using atomic masses, the electrons cancel in alpha decay, beta-minus decay and most reactions; for beta-plus decay, subtract $2m_ec^2 = 1.022$ MeV.
Why it works
Total energy, including rest energy, is conserved. If the products have less rest mass, the difference has become kinetic energy. Tables list masses of neutral atoms, which include electrons. In alpha decay, $^{238}$U (92 electrons) becomes $^{234}$Th (90 electrons) plus $^4$He (2 electrons): the electrons balance. In beta-minus decay the nucleus gains a proton and emits an electron, and the daughter atom needs one more electron: again they balance. In beta-plus decay the nucleus loses a proton and emits a positron; the daughter atom has one electron too many, and a positron leaves as well, so the atomic masses miss two electron masses: $Q = [M_{\text{parent}} - M_{\text{daughter}} - 2m_e]c^2$.
Worked examples
Worked example: Alpha decay of uranium-238
Find the energy released when $^{238}$U decays to $^{234}$Th and an alpha particle. Atomic masses: 238.050788 u, 234.043601 u and 4.002603 u.
- Before: 238.050788 u. After: $234.043601 + 4.002603 = 238.046204$ u.
- $\Delta m = 0.004584$ u, so $Q = 0.004584 \times 931.5 = 4.27$ MeV.
Answer: $Q = 4.27$ MeV
Worked example: Fusion fuel
Find $Q$ for $^2$H + $^3$H → $^4$He + n. Masses: 2.014102, 3.016049, 4.002603 and 1.008665 u.
- Before: $5.030151$ u. After: $5.011268$ u.
- $\Delta m = 0.018883$ u, so $Q = 17.6$ MeV.
Answer: $17.6$ MeV per reaction, the fuel of fusion research
Worked example: A positron emitter
Fluorine-18 decays by beta-plus emission to oxygen-18. Atomic masses: $^{18}$F 18.000938 u, $^{18}$O 17.999160 u. Find $Q$.
- Atomic mass difference: $0.001778$ u, which is $1.656$ MeV.
- Beta-plus: subtract $2m_ec^2 = 1.022$ MeV.
Answer: $Q = 0.634$ MeV (the largest energy the positron can carry)
Who gets the energy? In alpha decay the products fly apart with equal and opposite momenta, so the light alpha particle carries most of it: $K_\alpha \approx Q\,\frac{A - 4}{A}$, about 4.20 MeV of uranium-238's 4.27 MeV.
Common mistake
Mixing nuclear masses with atomic masses and losing track of the electrons. If you subtract a bare-nucleus mass from an atomic mass, you are off by several electron masses (0.511 MeV each), which can be as large as the whole Q-value of a beta decay. Use atomic masses for every particle, and remember the $2m_ec^2$ for beta-plus decay. Check: hydrogen should appear as $^1$H (1.007825 u), not as a bare proton (1.007276 u), when the other masses are atomic.