# 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.

• What is the sign of a binding energy, again? Jan 20, 2015 at 4:12
• The sign? Positive....? Jan 20, 2015 at 4:21
• Hmmm ... binging energies measure the extent to which the system is bound. That is how much less energy they have than the bits would have if free (or equivalently how much energy you would have to supply to free all the bits). In other words a higher value is more stable; so when using them in a conservation of energy context they come in with a negative sign. Jan 20, 2015 at 4:27
• Hmm.. I guess we can disregard the signs for a moment. I have 56.499508 Mev on one side, and then it decays to two 28.3 MeV particles? Where does this extra energy come from. Even if I do it with masses, there still isnt enough energy for this decay to happen. So why does it? Jan 20, 2015 at 4:32

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.
As the total binding energy of $^8Be$ is less than of two $\alpha$, it means that in creating $^8Be$ there remained inside it more energy than remains in the $\alpha$ particles if we would create them. Thus, inside $^8Be$ there is some additional energy than in the two $\alpha$, and it can decay into 2 $\alpha$. The excess of energy in the $^8Be$ will transform into kinetic energy of the $\alpha$ nuclei.