Semi-empirical mass formula

The term ${\displaystyle a_{\text{V}}A}$ 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 ${\displaystyle A(A-1)/2}$, one might expect a term proportional to ${\displaystyle A^{2}}$. 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 ${\displaystyle a_{\text{V}}}$ is smaller than the binding energy possessed by the nucleons with respect to their neighbors (${\displaystyle E_{\text{b}}}$), 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 ${\displaystyle A}$ nucleons, with equal numbers of protons and neutrons, then the total kinetic energy is ${\displaystyle {\tfrac {3}{5}}A\varepsilon _{\text{F}}}$, with ${\displaystyle \varepsilon _{\text{F}}}$ the Fermi energy, which is estimated as 38 MeV. Thus the expected value of ${\displaystyle a_{\text{V}}}$ in this model is ${\displaystyle E_{\text{b}}-{\tfrac {3}{5}}\varepsilon _{\text{F}}\sim 17~\mathrm {MeV} ,}$ not far from the measured value.

The term ${\displaystyle a_{\text{S}}A^{2/3}}$ 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 ${\displaystyle A^{1/3}}$ and the surface area to ${\displaystyle A^{2/3}}$. This explains why the surface term is proportional to ${\displaystyle A^{2/3}}$. It can also be deduced that ${\displaystyle a_{\text{S}}}$ should have a similar order of magnitude to ${\displaystyle a_{\text{V}}}$.

The term ${\displaystyle a_{\text{C}}{\frac {Z(Z-1)}{A^{1/3}}}}$ or ${\displaystyle a_{\text{C}}{\frac {Z^{2}}{A^{1/3}}}}$ 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

${\displaystyle E={\frac {3}{5}}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q^{2}}{R}},}$

where Q is the total charge, and R is the radius of the sphere. The value of ${\displaystyle a_{\text{C}}}$ can be approximately calculated by using this equation to calculate the potential energy, using an empirical nuclear radius of ${\displaystyle R\approx r_{0}A^{\frac {1}{3}}}$ and Q = Ze. However, because electrostatic repulsion will only exist for more than one proton, ${\displaystyle Z^{2}}$ becomes ${\displaystyle Z(Z-1)}$:

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

where now the electrostatic Coulomb constant ${\displaystyle a_{\text{C}}}$ is

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

Using the fine-structure constant, we can rewrite the value of ${\displaystyle a_{\text{C}}}$ as

${\displaystyle a_{\text{C}}={\frac {3}{5}}{\frac {\hbar c\alpha }{r_{0}}}={\frac {3}{5}}{\frac {R_{\text{P}}}{r_{0}}}\alpha m_{\text{p}}c^{2},}$

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

Illustration of basis for asymmetric term

The term ${\displaystyle a_{\text{A}}{\frac {(N-Z)^{2}}{A}}}$ is known as the asymmetry term (or Pauli term).

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 modeling the nucleus as a Fermi ball of protons and neutrons. Its total kinetic energy is

${\displaystyle E_{\text{k}}={\frac {3}{5}}(Z\varepsilon _{\text{F,p}}+N\varepsilon _{\text{F,n}}),}$

where ${\displaystyle \varepsilon _{\text{F,p}}}$ and ${\displaystyle \varepsilon _{\text{F,n}}}$ are the Fermi energies of the protons and neutrons. Since these are proportional to ${\displaystyle Z^{2/3}}$ and ${\displaystyle N^{2/3}}$ respectively, one gets

${\displaystyle E_{\text{k}}=C(Z^{5/3}+N^{5/3})}$ for some constant C.

The leading terms in the expansion in the difference ${\displaystyle N-Z}$ are then

${\displaystyle E_{\text{k}}={\frac {C}{2^{2/3}}}\left(A^{5/3}+{\frac {5}{9}}{\frac {(N-Z)^{2}}{A^{1/3}}}\right)+O{\big (}(N-Z)^{4}{\big )}.}$

At the zeroth order in the expansion the kinetic energy is just the overall Fermi energy ${\displaystyle \varepsilon _{\text{F}}\equiv \varepsilon _{\text{F,p}}=\varepsilon _{\text{F,n}}}$ multiplied by ${\displaystyle {\tfrac {3}{5}}A}$. Thus we get

${\displaystyle E_{\text{k}}={\frac {3}{5}}\varepsilon _{\text{F}}A+{\frac {1}{3}}\varepsilon _{\text{F}}{\frac {(N-Z)^{2}}{A}}+O{\big (}(N-Z)^{4}{\big )}.}$

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).

${\displaystyle \varepsilon _{\text{F}}}$ is 38 MeV, so calculating ${\displaystyle a_{\text{A}}}$ 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 ${\displaystyle |N-Z|}$, and the form ${\displaystyle (N-Z)^{2}}$ 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 ${\displaystyle |N-Z|}$ is less significant for larger values of A.

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 ${\displaystyle \delta (A,Z)}$ 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[5]

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

where ${\displaystyle \delta _{0}}$ is found empirically to have a value of about 1000 keV, slowly decreasing with mass number A. The binding energy may be increased by converting one of the odd protons or neutrons into a neutron or proton, so the odd nucleon can form a pair with its odd neighbour forming and even Z, N. The pair have overlapping wave functions and sit very close together with a bond stronger than any other configuration.[5] When the pairing term is substituted into the binding energy equation, for even Z, N, the pairing term adds binding energy, and for odd Z, N the pairing term removes binding energy.

The dependence on mass number is commonly parametrized as

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

The value of the exponent k_P 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:

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

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 ${\displaystyle A^{k_{\text{P}}}}$ 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 ${\displaystyle A^{-1}}$ 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.