Discussion
Binding energy and rest mass energy of a two-nucleon bound state are denoted by \(B\) and \(mc^{2}\), respectively, where \(c\) is the speed of light. The minimum energy of a photon required to dissociate the bound state is
Answer: B
The minimum photon energy is \(B\left(1 + \dfrac{B}{2mc^{2}}\right)\).
The photon must supply the binding energy B and also the recoil kinetic energy of the products, because its momentum E/c must be conserved. At threshold the two nucleons move off together with total momentum E/c and total rest energy mc2 + B. Energy conservation with the bound state at rest:
\[ \begin{aligned} E + mc^{2} &= \sqrt{(mc^{2} + B)^{2} + E^{2}} \\ E &= B + \frac{B^{2}}{2mc^{2}} = B\left(1 + \frac{B}{2mc^{2}}\right) \end{aligned} \]
B alone ignores the recoil. The minus sign would make the threshold smaller than the binding energy, which energy conservation forbids, and B - mc2 is negative for any real nucleus.
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