Semi-empirical mass formula


The semi-empirical mass formula, also known as the Weizsäcker formula, provides a theoretical estimate of the mass and binding energy of an atom exclusively through its nuclear properties:

m(Z,A)=Nmn+Zmp+ZmeB(Z,A)c2m(Z,A)=N m_n+Z m_p+Z m_e-\frac{B(Z,A)}{c^{2}}

where mnm_n, mpm_p, and mem_e are the masses of the neutron, proton, and electron. B(Z,A)B(Z,A) is the nuclear binding energy, given by

B(Z,A)c2=avAasA2/3aCZ(Z1)A1/3aa(NZ)24A+δ\frac{B(Z,A)}{c^{2}}=a_v A-a_s A^{2/3}-a_{C}\frac{Z(Z-1)}{A^{1/3}}-a_a\frac{(N-Z)^2}{4A}+\delta

where

The numerical values of the parameters (Povh et al., Particles and Nuclei, 7th ed.) are1

av=15.67MeVc2as=17.23MeVc2aC=0.714MeVc2aa=93.15MeVc2ap=34MeVc2\begin{align} a_v &= 15.67\,\tfrac{\text{MeV}}{c^{2}}\\ a_s &= 17.23\,\tfrac{\text{MeV}}{c^{2}}\\ a_{C} &= 0.714\,\tfrac{\text{MeV}}{c^{2}}\\ a_a &= 93.15\,\tfrac{\text{MeV}}{c^{2}}\\ a_p &= 34\,\tfrac{\text{MeV}}{c^{2}} \end{align}

This formula is derived from the liquid drop model of nuclear structure, which posits that the nucleus behaves like a quantum fluid of neutrons and protons, a Fermi liquid.

Since electronic binding energy is quite negligible compared to nuclear scales of energy, this formula also provides the mass of the atomic nucleus by simply omitting the mass of the electrons.

Parameterization

The binding energy formula contains five experimentally determined constants, each with a specific interpretation. The two major forces at play are the strong interaction, which is what keeps the nucleons bound, and electromagnetism, which seeks to repel positively charged protons.

  1. avAa_v A is the volume term, proportional to AA. If every nucleon interacted with every other nucleon, the proportionality would be BA(A1)A2B\propto A(A-1)\sim A^{2}. Experimentally, the nuclear radius scales as RA1/3R\propto A^{1/3}, yet in practice one finds BAB\propto A, indicating that nucleons interact only with their close neighbors.
  2. asA2/3a_s A^{2/3} is the surface term. The neighbor-interaction picture holds for nucleons inside the nucleus, but those on the surface have fewer neighbors and are less tightly bound, thus contributing less to the total. This term corrects for that deficit by subtracting an amount proportional to the nuclear surface, which experimentally goes like R2A2/3R^{2}\propto A^{2/3}.
  3. aCZ(Z1)A1/3a_{C} \dfrac{Z(Z-1)}{A^{1/3}} is the Coulomb term. Protons are all positively charged and repel one another via electromagnetism, weakening their binding. The proportionality is derived from the behavior of the electric potential energy in a nucleus modeled as a uniformly charged sphere of electric charge ZeZe:
Ue.m.Z(Z1)αcRZ(Z1)A1/3U_{\text{e.m.}}\propto \frac{Z(Z-1)\alpha\hbar c}{R}\propto \frac{Z(Z-1)}{A^{1/3}}
  1. aa(NZ)24Aa_a\dfrac{(N-Z)^2}{4A} is the asymmetry term. It accounts for the fact that in heavier nuclei, the neutron–proton symmetry is broken to offset the increasing Coulomb repulsion among protons. This is because neutrons do not contribute to repulsion (they are electrically neutral), but do contribute to strong attraction (they are baryons). When protons and neutrons are equal in number this term vanishes, since they are the most stable. Greater asymmetry leads to greater instability (less binding energy).
  2. δ\delta is the pairing term. Nuclei are more stable when they contain even numbers of protons and neutrons, which suggests that "pair" with each other in some sort of stable configuration. Indeed, among all known stable nuclei, only four of them have odd NN and ZZ (they are \ce2H\ce{^{2}H}, \ce6Li\ce{^{6}Li}, \ce10B\ce{^{10}B} and 14N^{14}N). This term is discontinuous and works as follows:
δ=+apA3/4if A,Z,N evenδ=0if A oddδ=apA3/4if A even and Z,N odd\begin{aligned} \delta &= +a_p A^{-3/4} &&\text{if }A,Z,N\text{ even}\\ \delta &= 0 &&\text{if }A\text{ odd}\\ \delta &= -a_p A^{-3/4} &&\text{if }A\text{ even and }Z,N\text{ odd} \end{aligned}

It should be noted that unlike with other terms, the 3/4-3/4 exponent of AA is purely empirical, obtained from a fit. In fact, there is debate regarding whether 1/2-1/2 is a more accurate exponent, as modern data seems to suggest so.

Footnotes

  1. This value of apa_{p} assumes an exponent of 3/4-3/4. Povh et al. use an exponent 1/2-1/2 and give ap=11.2 MeV/c2a_{p}=11.2\text{ MeV}/c^{2} instead.