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. \begin{equation}m_p-m_n-\frac{1}{c^2}\frac{\partial{E_b}(A,Z)}{\partial{Z}}=0\end{equation} 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

Most stable isobars through minimizing $M$ and through maximizing $E_b$

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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – user10851
    Mar 7, 2014 at 3:11
  • $\begingroup$ Please have a look at Elements of Nuclear Physics by Meyerhof. $\endgroup$
    – rainman
    Mar 7, 2014 at 17:57

1 Answer 1


The liquid drop model (LDP) is an approximate description of the mass of nuclei. It is a parametric formula that is fitted to the experimental values. The formula for binding energy is expressed in a similar way and comes from the same assumption. Therefore, both being parametrizations, they are approximate. And indeed if you examine nuclei by nuclei it will fluctuate largely from the experimental values. The small differences you see raise from small differences in the parameters of the formula, due to diversity in the fitting data and ultimately, due to numerical approximations. But bare in mind, non of these formulas have a predictive character, there values are off for most of the nuclei known.

Furthermore, the model itself is known to be approximative since it does not consider many of the features observed in nuclei. This is the ultimate reason why these formulas will give very approximate results, in many cases only valid for order of magnitude.


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