Nuclear Binding Energy: The Liquid-Drop Model

1 · Predict

The semi-empirical mass formula (liquid-drop model) estimates a nucleus's total binding energy from its proton count Z and mass number A. Holding Z fixed at 26 (iron) and changing only A, does the model predict binding energy increasing steadily with every added nucleon?

Yes — more nucleons always means more total binding energy.
No — the binding energy depends on a balance of competing terms (volume, surface, Coulomb, asymmetry) that shift differently as A grows, so it doesn't simply increase with every added nucleon.
Binding energy always decreases as A increases.

2 · Set Up

Open the iron-binding preset and press Reset. This preset models iron with Z = 26 protons.
Enable the binding-energy readout.
Set the mass number A for each trial (holding Z = 26 fixed) and record the total binding energy.

3 · Collect Data

Mass number A	Total binding energy E_B (MeV)
56
80
110

Plot binding energy E_B (y-axis) against mass number A (x-axis) for your three trials.

4 · Analyze

For one trial, compute E_B = a_vA − a_sA^(2/3) − a_cZ(Z−1)/A^(1/3) − a_a(A−2Z)²/A using a_v = 15.75, a_s = 17.8, a_c = 0.711, a_a = 23.7 (MeV), Z = 26. Compare to the table.
Your three trials rise and then fall — E_B climbs from A = 56 to A = 80, then drops again by A = 110. Explain how the volume term +a_vA, which grows with every nucleon added, is eventually overtaken by the surface and asymmetry terms, and why −a_a(A−2Z)²/A penalises a nucleus whose neutron count runs far ahead of its proton count.

5 · Extend

This same liquid-drop model explains why very heavy nuclei (large A) release energy through fission (splitting) while very light nuclei release energy through fusion (combining) — both processes move nuclei toward the peak binding-energy-per-nucleon region near iron. Explain why iron sits near this peak.
This is a simplified version of the semi-empirical mass formula (it omits the pairing term for even/odd nucleon counts), so its predictions won't exactly match a real nuclear mass table. Why might physicists still find a simplified model like this useful for teaching, even knowing it's not perfectly accurate?

The Physics Behind This Experiment

Mass–Energy Equivalence (Binding Energy)

A nucleus's mass is less than the sum of its separate protons and neutrons — that missing mass Δm converts to binding energy via B = Δmc², the energy holding the nucleus together. The value of Δm itself (and so B) comes from the semi-empirical mass formula's competing volume, surface, Coulomb, and asymmetry terms, modeling the nucleus as a liquid drop.

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