# How does the semi-empirical mass formula prove or detract from the existence of the “valley of stability”?

I came across this passage in my lecture notes.

The semi-empirical mass formula (SEMF) is a better estimator of atomic mass than the standard mass formula, which neglects spatial arrangement of nucleons and resulting interactions. It follows directly from the SEMF that the observed "valley of stability" at $Z=\frac{A}{2}$, is in fact only a good estimator of stability at small $A$, though we will not concern ourselves with the details.

I understand that the SEMF accounts for nuclear interactions due to spatial arrangement of the nucleons; however, I'm not sure where the second part is coming from. How does the SEMF prove or detract from the existence of the "valley of stability" at $Z=\frac{A}{2}$?

• Hi W. Ryan, I guess that the SEMF is the Weizsäcker parametrization (the one I wrote), but I'm not sure of what is meant by "standard mass formula", so my answer might be totally on the wrong way. Maybe you could include the formulas of the two parametrizations? – pppqqq Jan 31 '17 at 15:50
• Yes, the parametrization I'm using is the same one you use. By standard mass term, all I meant was $M(A,Z)=Z(m_p+m_e)+(A-Z)m_n$, so your answer totally works. Thanks. – W. Ryan Jan 31 '17 at 16:15

The semi-empirical mass formula is given by a series of corrections to the standard mass formula. The standard mass term is given by, $$M(Z,A)=Z(M_p+m_E)+(A-Z)M_n,$$

while the corrections are given by the following:

$$f_1(Z,A)=-a_vA$$ $$f_2(Z,A)=a_sA^{2/3}$$ $$f_3(Z,A)=a_c\frac{Z(Z-1)}{A^{1/3}}\approx a_c\frac{Z^2}{A^{1/3}}$$ $$f_4(Z,A)=a_a\frac{(Z-\frac{A}{2})^2}{A}$$ $$f_5(Z,A) = \begin{cases} -f(A) & Z\text{ even, } A-Z=N \text{ even} \\ 0 & Z\text{ even, } N\text{ odd, or, } Z\text{ odd, } N\text{ even}\\ f(A) & Z\text{ odd, } N\text{ odd} \end{cases}$$

From this, the semi-empirical mass formula becomes,

$$M'(Z,A)=M(Z,A)+f_1(Z,A)+f_2(Z,A)+f_3(Z,A)+f_4(Z,A)+f_5(Z,A).$$

Now, to answer your question. Stability refers to minimizing the mass. That is, the lower the mass is for some $(Z,A)$ pair, with either fixed, the greater the mass deficit, and hence greater the binding energy. We will hence attempt to minimize $M'(Z,A)$ for fixed $A$, to find the constraint on $Z$ such that $M$ is a minimum. This amounts to nothing more than taking the first derivative of $M'(Z,A)$ with respect to $Z$ while $A$ is held constant. We can ignore the corrections $f_1$, and $f_2$ and $f_5$, as they have no dependence in $Z$, and differentiate the other terms.

We have,

$$\frac{\partial M(Z,A)}{\partial Z}=M_p+m_e-M_n,$$ $$\frac{\partial f_3}{\partial Z}=\frac{2a_c}{A^{1/3}}Z,$$ $$\frac{\partial f_4}{\partial Z}=\frac{2a_a}{A}Z-a_a,$$

So that the minimum is given by,

$$Z=\frac{a_a+M_n-M_p-m_e}{\frac{2a_c}{A^{1/3}}+\frac{2a_a}{A}}.$$

Plugging in the constant terms, we get,

$$Z=\frac{93.92}{2(\frac{0.697}{A^{1/3}}+\frac{93.14}{A})},$$

where everything is in MeV.

We can now graph this function, versus the standard $Z=\frac{A}{2},$ to get the following plot.

As can be observed, the red line, $Z=\frac{A}{2}$ deviates from the true $Z$ which minimizes mass for fixed $A$, and the error increases for larger and larger $A$. For example the error at $A=20$ is about $4.5$ %, whereas the error is much larger, $\tilde{}25$ % for $A=208$.

I also made a plot of absolute error of $\frac{Z}{2}-\text{SEMF-}Z$ in $Z$ against $A$ on the same axes if you're interested.

If I didn't misunderstand your quote, this means that the semi-empirical mass formula predicts a deviation from $Z=\frac{A}{2}$ for heavy nuclei (more neutrons). In fact, from the standard parametrization (cfr. Ref. [1]): $$m(A,Z)=Zm_p + (A-Z)m_n -a_V A +a_S A^{\frac{2}{3}}+\\+a_C Z^2 A^{-\frac{1}{3}} +a_A \frac{(Z-\frac{A}{2})^2}{A}-\frac{(-1)^Z+(-1)^N}{2}a_P A^{-\frac{1}{2}}$$ follows that the most stable configuration, which is obtained (neglecting the pairing term) for $$\left(\frac{\partial m}{\partial Z}\right) _A=0,$$ corresponds to: $$Z=\frac{A}{2}\frac{m_n-m_p +a_A}{a_CA^{\frac{2}{3}}+a_A}.$$ The formula shows, as intuition suggests, that for $A\gg 1$ (and therefore $Z\gg 1$), the Coulomb term becomes more and more important, so that more neutrons are necessary to keep the nucleus together. Moreover, Ref. [1] gives: $$a_A = 92.86\, \text {MeV},\\a_C=0.71\, \text{MeV},$$ so that, for small enough (but not too small!) $A$, since $m_n-m_p \approx 1.5 \text{MeV}$, the SEMF actually predicts $Z=\frac{A}{2}$.

[1] Bertulani C.A., "Nuclear Physics in a Nutshell", page 121.

Recall the form of the SEMF: the asymmetry term has a term with $A-2Z$. The effect of this term is $0$ when $Z=\frac{A}{2}$, thus contributing to the valley of stability - looking at the form of the rest of the Semi-Empirical Mass Formula will allow you to deduce how it contributes.