# What is the most stable nuclide of an isobar?

From the semi-empirical mass formula, the mass of an atomic nucleus is $$M\left(A,Z\right)=Zm_p+(A-Z)m_n-\frac{E_b(A,Z)}{c^2}$$ I've been told (at first) that for a given mass number $A$, the most stable nuclide of the isobar is the one with the smallest mass. Then it should be the one with proton number $Z$ such that $$\frac{\partial{M}\left(A,Z\right)}{\partial{Z}}=0$$ i.e. $$m_p-m_n-\frac{1}{c^2}\frac{\partial{E_b}(A,Z)}{\partial{Z}}=0$$ but then I had a test where the professor asked for the most stable isobar and he argumented that this was the one for which the binding energy was the highest, meaning only that $Z$ satisfies $$\frac{\partial{E_b}(A,Z)}{\partial{Z}}=0$$ so I got it wrong on the test. Either way the difference is just a (small) constant, so the values $Z$ they both predict are roughly the same for low values of $A$. Let me call ${Z_m}$ the one obtained through minimizing $M$ and ${Z_E}$ the one obtained through maximizing ${E_b}$, then a plot looks like this

The values of $Z$ begin to separate for large $A$, but the good values here *supposedly** are those of $Z_m$, for example, for $A=209$: \begin{align}Z_m(209)&=83.36\approx83\\ Z_E(209)&=82.22\approx82\end{align} because the most stable isobar for $A=209$ *as given in Table A.4 in Shultis & Faw's book** corresponds to $Z=83$ (element $\text{Bi}$).

But both that the binding energy should be a maximum and that the mass should be a minimum makes sense; in the Wikipedia article, for example, the correct one is given as $Z_E$, and I'd even go with the first one, but as I've seen, the best one is the one for minimum mass, so...

How come this happens? Is it a just a fault of the theory (namely the liquid drop model) or what?

• Note that this is precisely the ambiguity that results in both Fe-56 (mass) and Ni-62 (binding energy) as being quoted as the most stable isotopes.
– user10851
Mar 7, 2014 at 3:11
• Please have a look at Elements of Nuclear Physics by Meyerhof. Mar 7, 2014 at 17:57