# Semi-empirical mass formula

In nuclear physics, the semi-empirical mass formula (SEMF) (sometimes also called Weizsäcker's formula, or the Bethe–Weizsäcker formula, or the Bethe–Weizsäcker mass formula to distinguish it from the Bethe–Weizsäcker process) is used to approximate the mass and various other properties of an atomic nucleus from its number of protons and neutrons. As the name suggests, it is based partly on theory and partly on empirical measurements. The formula represents the liquid drop model proposed by George Gamow, which can account for most of the terms in the formula and gives rough estimates for the values of the coefficients. It was first formulated in 1935 by German physicist Carl Friedrich von Weizsäcker, and although refinements have been made to the coefficients over the years, the structure of the formula remains the same today.

The SEMF gives a good approximation for atomic masses and several other effects, but does not explain the appearance of a greater binding energy and therefore stability at so-called magic numbers of protons and neutrons.

## The liquid drop model and its analysis

A graphical representation of the semi-empirical binding energy formula. The binding energy per nucleon in MeV is plotted for various nuclides as a function of Z, the atomic number (on the y-axis), and N, the neutron number (on the x-axis). A dashed line is included to show experimentally-known nuclides are typically above approximately 7.6 MeV per nucleon– the highest binding energies, in excess of 8.5 MeV, are seen for Z = 26, iron.

The liquid drop model, first proposed by George Gamow and then developed by Niels Bohr and John Archibald Wheeler, treats the nucleus as a drop of incompressible fluid of very high density. Given that the nucleus, composed of nucleons (protons and neutrons), is held together by the nuclear force (a residual effect of the strong force), there is a similarity to the structure of a spherical liquid drop. While a crude model, the liquid drop model accounts for the spherical shape of most nuclei, and makes a rough prediction of binding energy.

The corresponding mass formula is purely in terms of the numbers of protons and neutrons it contains, and contains five terms:

## The formula

The mass of an atomic nucleus, for $N$  neutrons, $Z$  protons, and therefore $A=N+Z$  nucleons, is given by

$m=Zm_{p}+Nm_{n}-{\frac {E_{B}(N,Z)}{c^{2}}}$

where $m_{p}$  and $m_{n}$  are the rest mass of a proton and a neutron, respectively, and $E_{B}$  is the binding energy of the nucleus. The semi-empirical mass formula states the binding energy is:

$E_{B}=a_{V}A-a_{S}A^{2/3}-a_{C}{\frac {Z(Z-1)}{A^{1/3}}}-a_{A}{\frac {(A-2Z)^{2}}{A}}\pm \delta (A,Z)$ 

Each of the terms in this formula has a theoretical basis, as will be explained below. The coefficients $a_{V}$ , $a_{S}$ , $a_{C}$ , $a_{A}$  and a coefficient that appears in the formula for $\delta (A,Z)$  are determined empirically.

## Terms

### Volume term

The term $a_{V}A$  is known as the volume term. The volume of the nucleus is proportional to A, so this term is proportional to the volume, hence the name.

The basis for this term is the strong nuclear force. The strong force affects both protons and neutrons, and as expected, this term is independent of Z. Because the number of pairs that can be taken from A particles is ${\frac {A(A-1)}{2}}$ , one might expect a term proportional to $A^{2}$ . However, the strong force has a very limited range, and a given nucleon may only interact strongly with its nearest neighbors and next nearest neighbors. Therefore, the number of pairs of particles that actually interact is roughly proportional to A, giving the volume term its form.

The coefficient $a_{V}$  is smaller than the binding energy possessed by the nucleons with respect to their neighbors ($E_{b}$ ), which is of order of 40 MeV. This is because the larger the number of nucleons in the nucleus, the larger their kinetic energy is, due to the Pauli exclusion principle. If one treats the nucleus as a Fermi ball of $A$  nucleons, with equal numbers of protons and neutrons, then the total kinetic energy is ${3 \over 5}A\varepsilon _{F}$ , with $\varepsilon _{F}$  the Fermi energy which is estimated as 28 MeV. Thus the expected value of $a_{V}$  in this model is $E_{b}-{3 \over 5}\varepsilon _{F}\sim 17\;\mathrm {MeV}$ , not far from the measured value.

Illustration of the terms of the semi-empirical mass formula in the liquid drop model of the atomic nucleus.

### Surface term

The term $a_{S}A^{2/3}$  is known as the surface term. This term, also based on the strong force, is a correction to the volume term.

The volume term suggests that each nucleon interacts with a constant number of nucleons, independent of A. While this is very nearly true for nucleons deep within the nucleus, those nucleons on the surface of the nucleus have fewer nearest neighbors, justifying this correction. This can also be thought of as a surface tension term, and indeed a similar mechanism creates surface tension in liquids.

If the volume of the nucleus is proportional to A, then the radius should be proportional to $A^{1/3}$  and the surface area to $A^{2/3}$ . This explains why the surface term is proportional to $A^{2/3}$ . It can also be deduced that $a_{S}$  should have a similar order of magnitude as $a_{V}$ .

### Coulomb term

The term $a_{C}{\frac {Z(Z-1)}{A^{1/3}}}$  or $a_{C}{\frac {Z^{2}}{A^{1/3}}}$  is known as the Coulomb or electrostatic term.

The basis for this term is the electrostatic repulsion between protons. To a very rough approximation, the nucleus can be considered a sphere of uniform charge density. The potential energy of such a charge distribution can be shown to be

$E={\frac {3}{5}}\left({\frac {1}{4\pi \varepsilon _{0}}}\right){\frac {Q^{2}}{R}}$

where Q is the total charge and R is the radius of the sphere. Identifying Q with $Ze$ , and noting as above that the radius is proportional to $A^{1/3}$ , we get close to the form of the Coulomb term. However, because electrostatic repulsion will only exist for more than one proton, $Z^{2}$  becomes $Z(Z-1)$ . The value of $a_{C}$  can be approximately calculated using the equation above:

$R\approx r_{0}A^{\frac {1}{3}}.$

Quantum charge integers:

$Q=Ze\$
$Z^{2}\approx Z(Z-1)\ .$

Solving by substitution:

$E={\frac {3}{5}}\left({\frac {1}{4\pi \varepsilon _{0}}}\right){\frac {Q^{2}}{R}}={\frac {3}{5}}\left({\frac {1}{4\pi \varepsilon _{0}}}\right){\frac {(Ze)^{2}}{(r_{0}A^{\frac {1}{3}})}}={\frac {3e^{2}Z^{2}}{20\pi \varepsilon _{0}r_{0}A^{\frac {1}{3}}}}\approx {\frac {3e^{2}Z(Z-1)}{20\pi \varepsilon _{0}r_{0}A^{\frac {1}{3}}}}=a_{C}{\frac {Z(Z-1)}{A^{1/3}}}$

Potential energy of charge distribution:

$E={\frac {3e^{2}Z(Z-1)}{20\pi \varepsilon _{0}r_{0}A^{\frac {1}{3}}}}$

Electrostatic Coulomb constant:

$a_{C}={\frac {3e^{2}}{20\pi \varepsilon _{0}r_{0}}}$

The value of $a_{C}$  using the fine structure constant:

$a_{C}={\frac {3}{5}}\left({\frac {\hbar c\alpha }{r_{0}}}\right)={\frac {3}{5}}\left({\frac {R_{P}}{r_{0}}}\right)\alpha m_{p}c^{2}$

where $\alpha$  is the fine structure constant and $r_{0}A^{1/3}$  is the radius of a nucleus, giving $r_{0}$  to be approximately 1.25 femtometers. $R_{P}$  is the proton Compton radius and $m_{p}$  the proton mass. This gives $a_{C}$  an approximate theoretical value of 0.691 MeV, not far from the measured value.

$a_{C}=0.691{\text{ MeV}}$

### Asymmetry term

The term $a_{A}{\frac {(A-2Z)^{2}}{A}}$  or $4a_{A}{\frac {((A/2)-Z)^{2}}{A}}$  is known as the asymmetry term (or Pauli term). Note that as $A=N+Z$ , the parenthesized expression can be rewritten as $(N-Z)$ . The form $(A-2Z)$  is used to keep the dependence on A explicit, as it will be important for a number of uses of the formula.

The theoretical justification for this term is more complex. The Pauli exclusion principle states that no two identical fermions can occupy exactly the same quantum state in an atom. At a given energy level, there are only finitely many quantum states available for particles. What this means in the nucleus is that as more particles are "added", these particles must occupy higher energy levels, increasing the total energy of the nucleus (and decreasing the binding energy). Note that this effect is not based on any of the fundamental forces (gravitational, electromagnetic, etc.), only the Pauli exclusion principle.

Protons and neutrons, being distinct types of particles, occupy different quantum states. One can think of two different "pools" of states, one for protons and one for neutrons. Now, for example, if there are significantly more neutrons than protons in a nucleus, some of the neutrons will be higher in energy than the available states in the proton pool. If we could move some particles from the neutron pool to the proton pool, in other words change some neutrons into protons, we would significantly decrease the energy. The imbalance between the number of protons and neutrons causes the energy to be higher than it needs to be, for a given number of nucleons. This is the basis for the asymmetry term.

The actual form of the asymmetry term can again be derived by modelling the nucleus as a Fermi ball of protons and neutrons. Its total kinetic energy is

$E_{k}={3 \over 5}(N_{p}{\varepsilon _{F}}_{p}+N_{n}{\varepsilon _{F}}_{n})$

where $N_{p}$ , $N_{n}$  are the numbers of protons and neutrons and ${\varepsilon _{F}}_{p}$ , ${\varepsilon _{F}}_{n}$  are their Fermi energies. Since the latter are proportional to ${N_{p}}^{2/3}$  and ${N_{n}}^{2/3}$ , respectively, one gets

$E_{k}=C(N_{p}^{5/3}+N_{n}^{5/3})$  for some constant C.

The leading expansion in the difference $N_{n}-N_{p}$  is then

$E_{k}={C \over 2^{2/3}}\left((N_{p}+N_{n})^{5/3}+{5 \over 9}{(N_{n}-N_{p})^{2} \over (N_{p}+N_{n})^{1/3}}\right)+O((N_{n}-N_{p})^{2}).$

At the zeroth order expansion the kinetic energy is just the Fermi energy $\varepsilon _{F}\equiv {\varepsilon _{F}}_{p}={\varepsilon _{F}}_{n}$  multiplied by ${3 \over 5}(N_{p}+N_{n})^{2/3}$ . Thus we get

$E_{k}={3 \over 5}\varepsilon _{F}(N_{p}+N_{n})+{1 \over 3}\varepsilon _{F}{(N_{n}-N_{p})^{2} \over (N_{p}+N_{n})}+O((N_{n}-N_{p})^{4})={3 \over 5}\varepsilon _{F}A+{1 \over 3}\varepsilon _{F}{(A-2Z)^{2} \over A}+O((A-2Z)^{4}).$

The first term contributes to the volume term in the semi-empirical mass formula, and the second term is minus the asymmetry term (remember the kinetic energy contributes to the total binding energy with a negative sign).

$\varepsilon _{F}$  is 38 MeV, so calculating $a_{A}$  from the equation above, we get only half the measured value. The discrepancy is explained by our model not being accurate: nucleons in fact interact with each other, and are not spread evenly across the nucleus. For example, in the shell model, a proton and a neutron with overlapping wavefunctions will have a greater strong interaction between them and stronger binding energy. This makes it energetically favourable (i.e. having lower energy) for protons and neutrons to have the same quantum numbers (other than isospin), and thus increase the energy cost of asymmetry between them.

One can also understand the asymmetry term intuitively, as follows. It should be dependent on the absolute difference $|N-Z|$ , and the form $(A-2Z)^{2}$  is simple and differentiable, which is important for certain applications of the formula. In addition, small differences between Z and N do not have a high energy cost. The A in the denominator reflects the fact that a given difference $|N-Z|$  is less significant for larger values of A.

### Pairing term

Magnitude of the pairing term in the total binding energy for even-even and odd-odd nuclei, as a function of mass number. Two fits are shown (blue and red line). The pairing term (positive for even-even and negative for odd-odd nuclei) was derived from the binding energy data in: G. Audi et al., 'The AME2012 atomic mass evaluation', in Chinese Physics C 36 (2012/12) pp. 1287–1602.

The term $\delta (A,Z)$  is known as the pairing term (possibly also known as the pairwise interaction). This term captures the effect of spin-coupling. It is given by:

$\delta (A,Z)={\begin{cases}+\delta _{0}&Z,N{\text{ even }}(A{\text{ even}})\\0&A{\text{ odd}}\\-\delta _{0}&Z,N{\text{ odd }}(A{\text{ even}})\end{cases}}$

where $\delta _{0}$  is found empirically to have a value of about 1000 keV, slowly decreasing with mass number A. The dependence on mass number is commonly parametrized as

$\delta _{0}={a_{P}}{A^{k_{P}}}.$

The value of the exponent kP is determined from experimental binding energy data. In the past its value was often assumed to be −3/4, but modern experimental data indicate that a value of −1/2 is nearer the mark:

$\delta _{0}={a_{P}}{A^{-1/2}}$  or $\delta _{0}={a_{P}\prime }{A^{-3/4}}$ .

Due to the Pauli exclusion principle the nucleus would have a lower energy if the number of protons with spin up were equal to the number of protons with spin down. This is also true for neutrons. Only if both Z and N are even can both protons and neutrons have equal numbers of spin up and spin down particles. This is a similar effect to the asymmetry term.

The factor $A^{k_{P}}$  is not easily explained theoretically. The Fermi ball calculation we have used above, based on the liquid drop model but neglecting interactions, will give an $A^{-1}$  dependence, as in the asymmetry term. This means that the actual effect for large nuclei will be larger than expected by that model. This should be explained by the interactions between nucleons; For example, in the shell model, two protons with the same quantum numbers (other than spin) will have completely overlapping wavefunctions and will thus have greater strong interaction between them and stronger binding energy. This makes it energetically favourable (i.e. having lower energy) for protons to form pairs of opposite spin. The same is true for neutrons.

## Calculating the coefficients

The coefficients are calculated by fitting to experimentally measured masses of nuclei. Their values can vary depending on how they are fitted to the data. Several examples are as shown below, with units of megaelectronvolts.

Least-squares fit (1) Least-squares fit (2) Rohlf Wapstra
$a_{V}$  15.8 15.76 15.75 14.1
$a_{S}$  18.3 17.81 17.8 13
$a_{C}$  0.714 0.711 0.711 0.595
$a_{A}$  23.2 23.702 23.7 19
$a_{P}$  12 34 11.18 33.5
exponent of A in δ0 −1/2 −3/4 −1/2 −3/4
$\delta _{0}$  (even-even) $+12 \over A^{1/2}$  $+34 \over A^{3/4}$  $+11.18 \over A^{1/2}$  $+33.5 \over A^{3/4}$
$\delta _{0}$  (odd-odd) $-12 \over A^{1/2}$  $-34 \over A^{3/4}$  $-11.18 \over A^{1/2}$  $-33.5 \over A^{3/4}$
$\delta _{0}$  (even-odd, odd-even) 0 0 0 0

The formula does not consider the internal shell structure of the nucleus. The semi-empirical mass formula therefore provides a good fit to heavier nuclei, and a poor fit to very light nuclei, especially 4He. For light nuclei, it is usually better to use a model that takes this shell structure into account.

## Examples of consequences of the formula

By maximizing Eb(A,Z) with respect to Z, one would find the best neutron–proton ratio N/Z for a given atomic weight A. We get

$N/Z\approx 1+{\frac {a_{C}}{2a_{A}}}A^{2/3}.$

This is roughly 1 for light nuclei, but for heavy nuclei the ratio grows in good agreement with experiment.

By substituting the above value of Z back into Eb, one obtains the binding energy as a function of the atomic weight, Eb(A). Maximizing Eb(A)/A with respect to A gives the nucleus which is most strongly bound, i.e. most stable. The value we get is A = 63 (copper), close to the measured values of A = 62 (nickel) and A = 58 (iron).

The liquid drop model also allows the computation of fission barriers for nuclei, which determine the stability of a nucleus against spontaneous fission. It was originally speculated that elements beyond atomic number 104 could not exist, as they would undergo fission with very short half-lives, though this formula did not consider stabilizing effects of closed nuclear shells. A modified formula considering shell effects reproduces known data and the predicted island of stability, though also suggests a limit to existence of superheavy nuclei beyond Z = 120 and N = 184.