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Innuclear physics, thesemi-empirical mass formula (SEMF; sometimes also called theWeizsäcker formula,Bethe–Weizsäcker formula, orBethe–Weizsäcker mass formula to distinguish it from theBethe–Weizsäcker process) is used to approximate themass of anatomic nucleus from its number ofprotons andneutrons. As the name suggests, it is based partly ontheory and partly onempirical measurements. The formula represents theliquid-drop model proposed byGeorge Gamow,^[1] 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 physicistCarl Friedrich von Weizsäcker,^[2] and although refinements have been made to the coefficients over the years, the structure of the formula remains the same today.

The formula gives a good approximation for atomic masses and thereby other effects. However, it fails to explain the existence of lines of greaterbinding energy at certain numbers of protons and neutrons. These numbers, known asmagic numbers, are the foundation of thenuclear shell model.

The liquid-drop model was first proposed byGeorge Gamow and further developed byNiels Bohr,John Archibald Wheeler andLise Meitner.^[3] It treats thenucleus as a drop ofincompressible fluid of very high density, held together by thenuclear force (a residual effect of thestrong 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 defined purely in terms of the numbers of protons and neutrons it contains. The original Weizsäcker formula defines five terms:

• Volume energy, when an assembly of nucleons of the same size is packed together into the smallest volume, each interior nucleon has a certain number of other nucleons in contact with it. So, this nuclear energy is proportional to the volume.
• Surface energy corrects for the previous assumption made that every nucleon interacts with the same number of other nucleons. This term is negative and proportional to the surface area, and is therefore roughly equivalent to liquidsurface tension.
• Coulomb energy, the potential energy from each pair of protons. As this is a repelling force, the binding energy is reduced.
• Asymmetry energy (also calledPauli energy), which accounts for thePauli exclusion principle. Unequal numbers of neutrons and protons imply filling higher energy levels for one type of particle, while leaving lower energy levels vacant for the other type.
• Pairing energy, which accounts for the tendency ofproton pairs and neutron pairs to occur. An even number of particles is more stable than an odd number due tospin coupling.

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

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

where${\displaystyle m_{\text{n}}}$ and${\displaystyle m_{\text{p}}}$ are the rest mass of a neutron and a proton respectively, and${\displaystyle E_{\text{B}}}$ is thebinding energy of the nucleus. The semi-empirical mass formula states the binding energy is^[4]

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

The${\displaystyle \delta (N,Z)}$ term is either zero or${\displaystyle \pm {\delta }_0}$, depending on theparity of${\displaystyle N}$ and${\displaystyle Z}$, where${\displaystyle {\delta }_0=a_{\text{P}}{A}^{k_{\text{P}}}}$ for some exponent${\displaystyle k_{\text{P}}}$. Note that as${\displaystyle A=N+Z}$, the numerator of the${\displaystyle a_{\text{A}}}$ term can be rewritten as${\displaystyle (A-2Z{)}^2}$.

Each of the terms in this formula has a theoretical basis. The coefficients${\displaystyle a_{\text{V}}}$,${\displaystyle a_{\text{S}}}$,${\displaystyle a_{\text{C}}}$,${\displaystyle a_{\text{A}}}$, and${\displaystyle a_{\text{P}}}$ are determined empirically; while they may be derived from experiment, they are typically derived fromleast-squares fit to contemporary data. While typically expressed by its basic five terms, further terms exist to explain additional phenomena. Akin to how changing a polynomial fit will change its coefficients, the interplay between these coefficients as new phenomena are introduced is complex; some terms influence each other, whereas the${\displaystyle a_{\text{P}}}$ term is largely independent.^[5]

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

The basis for this term is thestrong nuclear force. The strong force affects both protons and neutrons, and as expected, this term is independent ofZ. Because the number of pairs that can be taken fromA 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 toA, 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 ofnucleons in the nucleus, the larger theirkinetic energy is, due to thePauli exclusion principle. If one treats the nucleus as aFermi ball of${\displaystyle A}$nucleons, with equal numbers of protons and neutrons, then the total kinetic energy is${\displaystyle \frac{3}{5}A{\epsilon }_{\text{F}}}$, with${\displaystyle {\epsilon }_{\text{F}}}$ theFermi energy, which isestimated as 38 MeV. Thus the expected value of${\displaystyle a_{\text{V}}}$ in this model is${\displaystyle E_{\text{b}}-\frac{3}{5}{\epsilon }_{\text{F}}\sim 17\mathrm{M}\mathrm{e}\mathrm{V},}$ not far from the measured value.

The term${\displaystyle a_{\text{S}}{A}^{2/3}}$ is known as thesurface 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 ofA. 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 createssurface tension in liquids.

If the volume of the nucleus is proportional toA, 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 theCoulomb orelectrostatic term.

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

${\displaystyle E=\frac{3}{5}\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{R},}$

whereQ is the total charge, andR 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 anempirical nuclear radius of${\displaystyle R\approx r_0{A}^{\frac{1}{3}}}$ andQ =Ze. However, because electrostatic repulsion will only exist for more than one proton,${\displaystyle Z^2}$ becomes${\displaystyle Z(Z-1)}$:

${\displaystyle E=\frac{3}{5}\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{R}=\frac{3}{5}\frac{1}{4\pi {\epsilon }_0}\frac{(Ze{)}^2}{r_0{A}^{1/3}}=\frac{3e^2{Z}^2}{20\pi {\epsilon }_0{r}_0{A}^{1/3}}\approx \frac{3e^{2}Z(Z-1)}{20\pi {\epsilon }_0{r}_0{A}^{1/3}}=a_{\text{C}}\frac{Z(Z-1)}{A^{1/3}},}$

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

${\displaystyle a_{\text{C}}=\frac{3e^2}{20\pi {\epsilon }_0{r}_0}.}$

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

${\displaystyle a_{\text{C}}=\frac{3}{5}\frac{\hslash c\alpha }{r_0}=\frac{3}{5}\frac{R_{\text{P}}}{r_0}\alpha m_{\text{p}}{c}^2,}$

where${\displaystyle \alpha }$ is the fine-structure constant, and${\displaystyle r_0{A}^{1/3}}$ is theradius of a nucleus, giving${\displaystyle r_0}$ to be approximately 1.25 femtometers.${\displaystyle R_{\text{P}}}$ is the protonreduced 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.

The term${\displaystyle a_{\text{A}}\frac{(N-Z{)}^2}{A}}$ is known as theasymmetry term (orPauli term).

The theoretical justification for this term is more complex. ThePauli exclusion principle states that no twoidenticalfermions can occupy exactly the samequantum 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

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

where${\displaystyle {\epsilon }_{\text{F,p}}}$ and${\displaystyle {\epsilon }_{\text{F,n}}}$ are theFermi 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 constantC.

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

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

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

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

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 anegative sign).

${\displaystyle {\epsilon }_{\text{F}}}$ is 38MeV, 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 theshell model, a proton and a neutron with overlappingwavefunctions will have a greaterstrong 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 thanisospin), 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 theabsolute difference${\displaystyle |N-Z|}$, and the form${\displaystyle (N-Z{)}^2}$ is simple anddifferentiable, which is important for certain applications of the formula. In addition, small differences betweenZ andN do not have a high energy cost. TheA in the denominator reflects the fact that a given difference${\displaystyle |N-Z|}$ is less significant for larger values ofA.

The term${\displaystyle \delta (A,Z)}$ is known as thepairing term (possibly also known as the pairwise interaction). This term captures the effect ofspin coupling. It is given by^[7]

where${\displaystyle {\delta }_0}$ is found empirically to have a value of about 1000 keV, slowly decreasing with mass number A. Odd-odd nuclei tend to undergobeta decay to an adjacent even-even nucleus by changing a neutron to a proton or vice versa. The pairs have overlapping wave functions and sit very close together with a bond stronger than any other configuration.^[7] When the pairing term is substituted into the binding energy equation, for evenZ,N, the pairing term adds binding energy, and for oddZ,N the pairing term removes binding energy.

The dependence onmass number is commonly parametrized as

${\displaystyle {\delta }_0=a_{\text{P}}{A}^{k_{\text{P}}}.}$

The value of the exponentk_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 thePauli 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 bothZ andN 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 theshell model, two protons with the same quantum numbers (other thanspin) will have completely overlappingwavefunctions and will thus have greaterstrong 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.

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 and which unit is used to express the mass. Several examples are as shown below.

	Eisberg & Resnick[8]	Least-squares fit (1)	Least-squares fit (2)[9]	Rohlf[10]	Wapstra[11]
unit	Da	MeV	MeV	MeV	MeV
${\displaystyle a_{\mathrm{V}}}$	0.01691	15.8	15.76	15.75	14.1
${\displaystyle a_{\mathrm{S}}}$	0.01911	18.3	17.81	17.8	13
${\displaystyle a_{\mathrm{C}}}$	0.000763[α]	0.714	0.711	0.711	0.595
${\displaystyle a_{\mathrm{A}}}$	0.10175[β]	23.2	23.702	23.7	19
${\displaystyle a_{\mathrm{P}}}$	0.012	12	34	11.18	33.5
${\displaystyle k_{\mathrm{P}}}$	−1/2	−1/2	−3/4	−1/2	−3/4
${\displaystyle {\delta }_0}$ (even-even)	$\frac{{\displaystyle -0.012}}{A^{1/2}}$	$\frac{{\displaystyle +12}}{A^{1/2}}$	$\frac{{\displaystyle +34}}{A^{3/4}}$	$\frac{{\displaystyle +11.18}}{A^{1/2}}$	$\frac{{\displaystyle +33.5}}{A^{3/4}}$
${\displaystyle {\delta }_0}$ (odd-odd)	$\frac{{\displaystyle +0.012}}{A^{1/2}}$	$\frac{{\displaystyle -12}}{A^{1/2}}$	$\frac{{\displaystyle -34}}{A^{3/4}}$	$\frac{{\displaystyle -11.18}}{A^{1/2}}$	$\frac{{\displaystyle -33.5}}{A^{3/4}}$
${\displaystyle {\delta }_0}$ (even-odd, odd-even)	0	0	0	0	0
^This model uses${\displaystyle Z^2}$ in the numerator of the Coulomb term. ^This model uses${\displaystyle (Z-A/2{)}^2}$ in the numerator of the Asymmetry term.

The formula does not consider the internalshell 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, especially⁴He. For light nuclei, it is usually better to use a model that takes this shell structure into account.

By maximizingE_b(A,Z) with respect toZ, one would find the bestneutron–proton ratioN/Z for a given atomic weightA.^[10] We get

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

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

By substituting the above value ofZ back intoE_b, one obtains the binding energy as a function of the atomic weight,E_b(A).MaximizingE_b(A)/A with respect toA gives the nucleus which is most strongly bound, i.e. most stable. The value we get isA = 63 (copper), close to themeasured values ofA = 62 (nickel) andA = 58 (iron).

The liquid-drop model also allows the computation offission barriers for nuclei, which determine the stability of a nucleus againstspontaneous fission. It was originally speculated that elements beyond atomic number104 could not exist, as they would undergo fission with very short half-lives,^[12] though this formula did not consider stabilizing effects of closednuclear shells. A modified formula considering shell effects reproduces known data and the predictedisland of stability (in which fission barriers and half-lives are expected to increase, reaching a maximum at the shell closures), though also suggests a possible limit to existence of superheavy nuclei beyondZ = 120 andN = 184.^[12]

• Freedman, R.; Young, H. (2004).Sears and Zemansky's University Physics with Modern Physics (11th ed.). Pearson Addison Wesley. pp. 1633–1634.ISBN 978-0-8053-8768-1.
• Liverhant, S. E. (1960).Elementary Introduction to Nuclear Reactor Physics.John Wiley & Sons. pp. 58–62.LCCN 60011725.
• Choppin, G.; Liljenzin, J.-O.; Rydberg, J. (2002)."Nuclear Mass and Stability"(PDF).Radiochemistry and Nuclear Chemistry (3rd ed.).Butterworth-Heinemann. pp. 41–57.ISBN 978-0-7506-7463-8.

• Nuclear liquid drop model in thehyperphysics online reference atGeorgia State University.
• Liquid drop model with parameter fit fromFirst Observations of Excited States in the Neutron Deficient Nuclei^160,161W and¹⁵⁹Ta, Alex Keenan, PhD thesis,University of Liverpool, 1999 (HTML version).

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