nuclear physics - Binding energies of $Be$

Just a quick question.

How does Beryllium 8 decay into 2 alpha particles?

Beryllium 8 has a binding energy of 56.499508 Mev An alpha particle has a binding energy of 28.3 so two of these would have 56.6 Mev

So how does this reaction happen?

Beryllium doesn't have enough energy.

Answer

I always get confused when I try to use binding energies directly, partly because of the sign issue, and partly because differences in binding energies show up in the third or fourth significant figure. My favorite reference instead lists for each isotope a "mass excess," which is the difference (in energy units) between an observed mass and the naïve mass of $A$ atomic mass units. For beryllium and helium we have \begin{align} \rm\Delta({}^8Be) &= \rm 4.9416\,MeV \\ \rm\Delta({}^4He) &= \rm 2.4249\,MeV \\ 2\cdot\rm\Delta({}^4He) &= \rm 4.8498\,MeV \end{align} This makes it clear that a $^8\rm Be$ is heavier than two $^4\rm He$ by about $0.1\,\rm MeV$ — it has more excess mass.

So what was wrong with your binding energy approach? You have \begin{align} \rm BE({}^8Be) &= 56.49955\,\rm MeV \\ \rm BE({}^4He) &= 28.3\rm\,MeV &\text{(careful with precision here)}\\ 2\cdot\rm BE({}^4He) &= 56.6\rm\,MeV \end{align} Here you have the same energy difference, about $0.1\rm\,MeV$, but it appears to go the other way, as you point out in your question. That's because helium is more tightly bound than beryllium; when beryllium-8 fissions, the extra binding energy is released.