Which of the SEMF terms account for low A binding energy?

[Flucky]

Homework Statement

Which of the semi-empirical mass formula terms accounts for a drop in stability at low A and why?

Homework Equations

B = a_vA - a_sA^{$\frac{2}{3}$} -a_cZ(Z-1)A^{$\frac{-1}{3}$} -a_a(A-2Z)²A^-1 ± a_pA^{$\frac{3}{4}$}

The Attempt at a Solution

I'm struggling getting my head around this one.

The Coulomb term will decrease the binding energy as Z increases so I've ruled that out.

The asymmetric term decreases the binding energy as N>Z (more stable when N ≈ Z) so I've ruled that out.

All the rest are only dependent on A, so I would guess that it is the first term (volume term) that is the cause of low A instability. However the volume term increases linearly with A and it's such a dramatic change when looking at the binding energy per nucleon curve.

I can't explain why it would be the volume term if it is.

Any ideas?

 

[anathema]

I would like to offer my perspective on this question. The semi-empirical mass formula is used to calculate the binding energy of a nucleus, which is a measure of its stability. The formula takes into account several terms, including the volume term, the surface term, the Coulomb term, the asymmetric term, and the pairing term.

Based on the given formula, the term that accounts for a drop in stability at low A is the surface term (as). This is because the surface term is proportional to A^(2/3), which means that as A decreases, the value of the term also decreases. This results in a decrease in the binding energy and therefore a decrease in stability.

The volume term, on the other hand, is proportional to A and therefore does not change significantly as A decreases. The Coulomb term also does not change significantly as A decreases, since it is proportional to Z^2. The asymmetric term and the pairing term also do not have a significant effect on stability at low A.

In conclusion, it is the surface term in the semi-empirical mass formula that accounts for a drop in stability at low A. This is because as the number of nucleons decreases, the surface-to-volume ratio of the nucleus increases, leading to a decrease in stability. I hope this explanation helps in understanding this concept better.