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8.1.5 Nuclear Models

8.1.5.2 Semi-Empirical Mass Formula

Idea of the formula

The semi empirical mass formula is a simple model that estimates the binding energy of a nucleus from its numbers of protons and neutrons. It is called semi empirical because it is based partly on physical ideas and partly on constants fitted to experimental data.

This formula belongs to the liquid drop picture of the nucleus. In that picture, the nucleus behaves in some ways like a tiny drop of nuclear fluid. The formula does not describe every detail of nuclear structure, but it captures broad trends very well, especially for medium and heavy nuclei.

The main goal is to estimate the binding energy $B(A,Z)$ of a nucleus with mass number $A$ and atomic number $Z$. Since the number of neutrons is $N = A - Z$, the nucleus is determined by $A$ and $Z$.

General form

A common form of the semi empirical mass formula is

$$
B(A,Z) = a_v A - a_s A^{2/3} - a_c \frac{Z(Z-1)}{A^{1/3}} - a_a \frac{(A-2Z)^2}{A} + \delta(A,Z)
$$

where

\[
A = Z + N
\]

and the coefficients $a_v, a_s, a_c, a_a$ are positive constants determined from experimental nuclear masses.

A typical set of approximate values is shown below.

CoefficientMeaningTypical value
$a_v$volume term coefficientabout $15.5\ \text{MeV}$
$a_s$surface term coefficientabout $16.8\ \text{MeV}$
$a_c$Coulomb term coefficientabout $0.72\ \text{MeV}$
$a_a$asymmetry term coefficientabout $23\ \text{MeV}$
$a_p$pairing scaleabout $34\ \text{MeV}$

The last term, $\delta(A,Z)$, is the pairing term. A common approximation is

$$
\delta(A,Z) =
\begin{cases}
+ a_p A^{-3/4}, & \text{even } Z \text{ and even } N \\
0, & \text{odd } A \\

Different textbooks use slightly different numerical constants and slightly different powers of $A$. The physical meaning remains the same.

The semi empirical mass formula gives the total binding energy as a sum of several competing contributions. Positive terms increase binding, negative terms reduce binding.

Physical meaning of each term

Volume term

The volume term is

$$
a_v A
$$

This says that each nucleon contributes roughly a constant amount to the binding energy. The reason is that the strong nuclear force is short ranged, so each nucleon mainly interacts with nearby nucleons. As the nucleus gets bigger, the total binding energy grows roughly in proportion to $A$.

Surface term

The surface term is

$$

Nucleons at the surface have fewer neighboring nucleons than those in the interior, so they are less strongly bound. Since the surface area of a roughly spherical nucleus scales as $A^{2/3}$, this correction subtracts an amount proportional to $A^{2/3}$.

Coulomb term

The Coulomb term is

$$

Protons repel each other electrically. This repulsion lowers the binding energy. The factor $Z(Z-1)$ counts proton pairs approximately, and the factor $A^{-1/3}$ reflects the fact that larger nuclei have larger radii, which slightly weakens the average repulsion between proton pairs.

Asymmetry term

The asymmetry term is

$$

Since $A - 2Z = N - Z$, this term penalizes a large difference between neutron number and proton number. A nucleus tends to be more stable when protons and neutrons are balanced appropriately. This effect comes from quantum statistics and the filling of nuclear energy states.

For light nuclei, stability often occurs near $N \approx Z$. For heavier nuclei, stable nuclei usually have $N > Z$, partly because extra neutrons help reduce proton proton Coulomb repulsion.

Pairing term

The pairing term is

$$
\delta(A,Z)
$$

This reflects the tendency of nucleons to form pairs. Even even nuclei, with even numbers of both protons and neutrons, are extra stable, so the term is positive. Odd odd nuclei are less stable, so the term is negative. If $A$ is odd, one kind of nucleon is unpaired and the pairing correction is often taken as zero.

Nucleus typePairing term
even $Z$, even $N$positive
odd $A$zero
odd $Z$, odd $N$negative

For the asymmetry term, note that
$$
A - 2Z = N - Z
$$
so the penalty depends on how unequal the neutron and proton numbers are.

Why the formula works

The formula works because it combines a few simple large scale effects.

The volume term favors larger binding energy as more nucleons are added. The surface term reduces binding because surface nucleons are less surrounded. The Coulomb term becomes more important as the number of protons increases, making very heavy nuclei less stable. The asymmetry term discourages a large imbalance between neutrons and protons. The pairing term accounts for extra stability patterns that appear in real nuclei.

Together, these effects reproduce the main trends in nuclear masses and binding energies across the periodic table.

Relation to nuclear mass

If we know the binding energy, we can estimate the nuclear mass. The idea is that the mass of a bound nucleus is less than the sum of the masses of its separate nucleons.

For a nucleus with $Z$ protons and $N$ neutrons,

$$
M_{\text{nucleus}} = Z m_p + N m_n - \frac{B}{c^2}
$$

If atomic masses are used instead of bare nuclear masses, electron masses must be treated consistently. In practical nuclear mass calculations, one must be careful about whether the formula is being applied to nuclear mass or atomic mass.

A larger binding energy means a smaller nuclear mass, because
$$
M = \text{sum of free nucleon masses} - \frac{B}{c^2}
$$

Binding energy per nucleon

An especially useful quantity is the binding energy per nucleon,

$$
\frac{B}{A}
$$

This gives a measure of how tightly bound the nucleus is on average. The semi empirical mass formula helps explain why $\frac{B}{A}$ rises for light nuclei, reaches a maximum for medium mass nuclei, and then slowly decreases for very heavy nuclei.

This trend is central for understanding why fusion of light nuclei and fission of heavy nuclei can both release energy, although those topics belong elsewhere.

Example of term competition

Consider what happens as $A$ becomes large.

The volume term grows like $A$, which strongly increases binding. The surface term grows more slowly, like $A^{2/3}$. The Coulomb term becomes increasingly important because the number of proton pairs rises strongly with $Z$. For very heavy nuclei, Coulomb repulsion lowers the binding enough that the nucleus becomes less stable.

This competition between attraction and repulsion is one of the key successes of the formula.

Stability and the valley of beta stability

For a fixed mass number $A$, different values of $Z$ give different binding energies. The most stable isobar is the one with the largest binding energy, or equivalently the smallest mass.

Using the semi empirical mass formula without the pairing term for simplicity, one can find the most stable $Z$ for given $A$. The result is approximately

$$
Z \approx \frac{A}{2 + \frac{a_c}{2a_a}A^{2/3}}
$$

This shows that for small $A$, the stable value is near $Z \approx A/2$, so $N \approx Z$. For larger $A$, the stable nucleus has relatively fewer protons and more neutrons.

This helps explain the observed band of stable nuclei.

Valley of stability idea

Strengths of the formula

The semi empirical mass formula is very useful because it can

predict approximate binding energies, explain broad nuclear stability trends, estimate which nuclei are more tightly bound, and provide a first understanding of why some nuclei are unstable.

It is one of the most important approximate formulas in nuclear physics.

Limitations

The formula is not exact. It does not fully describe shell effects, magic numbers, or detailed nuclear spectra. Those features require more refined models. In particular, nuclei with closed shells often show extra stability beyond what the liquid drop style formula predicts.

So the formula is best viewed as a powerful average description, not a complete microscopic theory.

The semi empirical mass formula explains average trends in nuclear masses and stability, but it does not fully capture shell structure and other fine details.

Summary interpretation

The semi empirical mass formula says that nuclear binding is controlled by five main effects. Bulk nuclear attraction increases the binding, surface nucleons reduce it, proton repulsion reduces it further, neutron proton imbalance reduces it, and nucleon pairing adds or removes a smaller correction.

In compact form,

$$
B(A,Z) = a_v A - a_s A^{2/3} - a_c \frac{Z(Z-1)}{A^{1/3}} - a_a \frac{(N-Z)^2}{A} + \delta(A,Z)
$$

with $N = A - Z$.

This formula gives a remarkably successful first estimate of nuclear masses and stability across the chart of nuclides.

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8.1.5 Nuclear Models

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