The Semi-Empirical Mass Formula

Introduction

Could the binding energy of any nucleus be predicted from just its proton and neutron numbers? In 1935 Carl Friedrich von Weizsäcker showed that a remarkably good estimate follows from treating the nucleus like a drop of liquid with a few quantum corrections. The resulting semi-empirical mass formula (SEMF), refined by Hans Bethe and others, has five terms, each tied to a physical idea. It reproduces measured binding energies of medium and heavy nuclei to within about 1% and explains why the stable band curves as it does.

Core explanation

The formula. The binding energy of a nucleus with mass number A, proton number Z and neutron number N = A − Z is:

B = a V A − a S A^(2/3) − a C Z(Z − 1) ÷ A^(1/3) − a A (N − Z)² ÷ A + δ

Typical fitted coefficients are a V ≈ 15.8 MeV, a S ≈ 18.3 MeV, a C ≈ 0.714 MeV and a A ≈ 23.2 MeV, with δ ≈ ±12 ÷ √A MeV. Different fits give slightly different values; any consistent set works.

Volume term, a V A. Nuclear matter has a nearly constant density, so the radius grows as R ≈ 1.2 A^(1/3) fm and the volume as A. Because the strong force saturates, each nucleon contributes roughly the same binding, giving a term proportional to A.

Surface term, −a S A^(2/3). Nucleons at the surface have fewer neighbours, so they are less bound. The surface area grows as R², that is as A^(2/3). This correction is proportionally largest for light nuclei and explains the low B/A at small A. It is the nuclear counterpart of surface tension.

Coulomb term, −a C Z(Z − 1) ÷ A^(1/3). A uniformly charged sphere has electrostatic energy proportional to Z² ÷ R. Using Z(Z − 1) counts only interactions between different protons. This term grows rapidly with Z and causes the fall of B/A in heavy nuclei.

Asymmetry term, −a A (N − Z)² ÷ A. This is a quantum effect. Protons and neutrons each fill their own ladder of energy levels, and the Pauli principle allows at most two of each kind per level. For fixed A, the lowest total energy is reached with equal numbers; converting protons into neutrons forces nucleons into higher levels. The penalty grows as the square of the imbalance.

Pairing term, δ. Like nucleons pair up with opposite spins and gain extra binding. δ is positive for even-Z, even-N nuclei, zero for odd-A nuclei and negative for odd-Z, odd-N nuclei. This is why 146 of the roughly 250 stable nuclides are even–even and only four light ones (²H, ⁶Li, ¹⁰B, ¹⁴N) are odd–odd.

Most stable Z for a given A. For fixed A, the formula is a parabola in Z. Setting dB/dZ = 0 (and approximating Z − 1 ≈ Z) gives:

Z₀ ≈ A ÷ (2 + 0.0154 A^(2/3))

using a C ÷ a A with the coefficients above. For A = 56 this gives Z₀ ≈ 25.3, close to iron (Z = 26); for A = 208 it gives Z₀ ≈ 81, close to lead (Z = 82). The Coulomb term pushes Z₀ below A ÷ 2, producing the neutron excess of heavy nuclei.

Formulae

B = a V A − a S A^(2/3) − a C Z(Z − 1)/A^(1/3) − a A (N − Z)²/A + δ; δ = +a P/√A (even–even), 0 (odd A), −a P/√A (odd–odd), a P ≈ 12 MeV.

Step-by-step reasoning

To apply the formula to a nuclide:

1. Write down A, Z and N, and calculate A^(1/3) and A^(2/3). 2. Evaluate each of the five terms separately, keeping signs clear. 3. Choose δ from the parity of Z and N. 4. Sum the terms to obtain B, and divide by A for B/A. 5. Compare with the experimental value from atomic masses to judge the model's accuracy.

Visual explanation

Draw a stacked bar chart of B/A against A. Start with a flat bar at 15.8 MeV (volume). Subtract a surface slice that is large at small A and shrinks with A. Subtract a Coulomb slice that grows with A. Subtract a thin asymmetry slice for heavy nuclei. What remains is the familiar curve peaking near A = 60.

Real-world analogy

A water droplet is held together by cohesion throughout its volume, but molecules at its surface are less bound, which is surface tension. If you also gave the droplet an electric charge, repulsion would try to break it apart. The nucleus behaves the same way, with two extra quantum rules for balancing and pairing its two kinds of particle.

Real-world example

The liquid-drop picture led Lise Meitner and Otto Frisch in 1939 to explain fission: a heavy nucleus, deformed by absorbing a neutron, can stretch until Coulomb repulsion overcomes the surface term, and it splits. Bohr and Wheeler then used the SEMF terms to calculate fission barriers, explaining why ²³⁵U fissions with slow neutrons while ²³⁸U does not.

Why?

Why is the asymmetry term divided by A? The spacing between nucleon energy levels shrinks as the nucleus gets larger, roughly as 1 ÷ A. Moving each surplus nucleon up costs energy proportional to that spacing, so a given imbalance N − Z costs less in a larger nucleus.

Common misconception

"The SEMF is derived from first principles." It is semi-empirical: the form of each term comes from physical models, but the coefficients are fitted to measured masses. It also ignores shell structure, so it underestimates the binding of magic nuclei by several MeV.

Worked example

Question: Use the SEMF to estimate the binding energy of ⁵⁶Fe (Z = 26, N = 30). The measured value is 492.3 MeV.

Reasoning: A^(1/3) = 3.826 and A^(2/3) = 14.64. Volume: 15.8 × 56 = 884.8 MeV. Surface: 18.3 × 14.64 = 267.9 MeV. Coulomb: 0.714 × 26 × 25 ÷ 3.826 = 121.3 MeV. Asymmetry: 23.2 × 16 ÷ 56 = 6.6 MeV. Pairing (even–even): +12 ÷ 7.48 = +1.6 MeV. B = 884.8 − 267.9 − 121.3 − 6.6 + 1.6 = 490.6 MeV.

Answer: About 491 MeV, within 0.4% of the measured 492.3 MeV.

Quick check

1. Which term of the semi-empirical mass formula is mainly responsible for the fall in binding energy per nucleon for heavy nuclei? Answer: The Coulomb term, because proton–proton repulsion grows roughly as Z² ÷ A^(1/3), faster than the volume binding.

Exam focus

Know the physical origin and A-dependence of every term and the sign convention for δ. In calculations, lay out each term on its own line; most lost marks come from sign slips or from using Z² instead of Z(Z − 1) inconsistently.

Advanced insight

Because the SEMF is quadratic in Z at fixed A, the masses of isobars lie on a parabola. For odd A there is one parabola and usually only one stable isobar. For even A the pairing term splits it into two parabolas, one for even–even and one for odd–odd nuclei, which allows two or even three stable even–even isobars, such as ⁹⁶Zr, ⁹⁶Mo and ⁹⁶Ru, and explains double beta decay.

Summary

The SEMF writes B as a volume term minus surface, Coulomb and asymmetry corrections, plus a pairing term. It captures the saturation of nuclear forces, surface effects, proton repulsion, the Pauli principle and nucleon pairing. With about five fitted coefficients it predicts binding energies to about 1%, reproduces the B/A curve and predicts the most stable Z for each A.

Practice questions

1. State the A-dependence of the volume, surface and Coulomb terms. Answer: Volume ∝ A, surface ∝ A^(2/3) and Coulomb ∝ Z(Z − 1) ÷ A^(1/3). 2. Explain why odd–odd nuclides are rarely stable. Answer: The pairing term is negative for odd–odd nuclei, making them less bound than their even–even isobars, to which they can usually decay by beta processes. 3. Calculate the asymmetry term for ²³⁸U (Z = 92, N = 146) using a A = 23.2 MeV. Answer: 23.2 × (146 − 92)² ÷ 238 = 23.2 × 2916 ÷ 238 = 284 MeV. 4. Use Z₀ ≈ A ÷ (2 + 0.0154 A^(2/3)) to predict the most stable Z for A = 127, given A^(2/3) = 25.3. Answer: Z₀ = 127 ÷ (2 + 0.390) = 53.1, so Z = 53, iodine; ¹²⁷I is indeed the only stable iodine isotope.