Binding energy per nucleon

[logearav]

From the graph plotted between Mass Number and Binding Energy per Nucleon, we infer the BE/A is stable for mass numbers between 20 and 60 (approximately) and decreases for A>60. Similarly the BE/A is very less for lighter nuclei, say, Helium.
Nuclear forces are short range forces. So it is understandable for decrease in BE for heavier nuclei. But, I can't understand why BE is less for lighter nuclei, because there is only a few nucleons and being short range forces, nuclear forces should be predominant. But how BE is less for lighter nuclei which have a very few nucleon?

 

[kurros]

I have a vague memory that it can be though of in terms of `nearest neighbours'. If you have only a couple of nucleons then there are not as many adjacent nuclei to bind to, so you cannot have a larger binding energy per nucleon than if there were more nucleons around. There is a kind of combinatorical benefit to having more nucleons around at this point. As you say, at some point you have so many nucleons that the ones `on opposite sides of the nucleus' are hardly directly communicating at all, at which point adding more nucleons only affects the binding energy of a small fraction of nucleons, rather than all of them. The total binding energy of course goes up still, but less fast than the number of nuclei is going up.

Of course this is a very crude semi-classical sort of picture, but I think it is vaguely true.

 

[jtbell]

Think of spheres like ball-bearings packed together in a container. Each of the spheres in the interior of the container has about the same number of "neighbors." The strong nuclear force is very short range so it basically acts only between "neighboring" nucleons. So each nucleon in the interior of a nucleus contributes about the same amount to the total binding energy.

Google for "semi-empirical binding energy formula" and note the "volume term" which this gives rise to.