Preview
Build a Nucleus
Assemble protons and neutrons to explore nuclear stability
Rutherford's 1911 gold-foil shock: the atom's mass lives in a kernel a femtometer across, where positive protons sit shoulder to shoulder and should blast each other apart. What holds it? By 1935 Carl Friedrich von Weizsäcker had distilled the answer into a formula that treats the nucleus like a charged liquid drop: a volume term paying every nucleon for its strong-force bonds, a surface tax on the nucleons left exposed at the edge, a Coulomb penalty that grows faster than the proton count, an asymmetry penalty for building lopsided, and a pairing bonus for even-even partnership. Five terms, one ledger — and the ledger explains the most important curve in nuclear physics. Binding energy per nucleon climbs through the light elements, crests in the iron–nickel neighborhood near 8.8 MeV — nickel-62 holds the actual summit, with famous iron-56 a razor-thin step below — then sags toward uranium. That single hump is why stars burn hydrogen toward iron and stop there, and why splitting uranium pays out the difference. The unsung hero is the neutron: it buys strong-force attraction without spending any Coulomb repulsion, which is why heavy nuclei need more neutrons than protons — iron-56 runs 30 to 26, lead-207 runs 125 to 82. And certain counts — 2, 8, 20, 28, 50, 82, 126 — close nuclear shells the way noble gases close electron shells, leaving doubly magic nuclei like calcium-40 unusually hard to shake. In this lab you are the architect: two sliders, protons and neutrons, a panel that prices every nucleus you build, and a curve that shows where the valley of stability lies. Start with a single proton — and notice it binds nothing at all.
What you'll be able to do
- Keep the A = Z + N ledger and the nuclide bookkeeping straight on the panel's own rows: the factory state is hydrogen-1 (Z=1, N=0, A=1 — a lone proton, the one nuclide with no neutrons, Chem 2e §21.1), the iron preset reads 26 + 30 = 56 with an n:p ratio of 1.15, and the uranium preset reads 92 + 143 = 235 — the isotope name, the mass number, and the neutron count are three views of the same two sliders. The ledger extends to the stability badge's Z>83 rule: 'all isotopes of elements with atomic numbers greater than 83 are unstable' (Chem 2e §21.1, Figure 21.2 caption, verbatim), which is why the uranium state reads Radioactive before any half-life is ever discussed
- Price any nucleus with the semi-empirical (Bethe-Weizsäcker) five-term ledger the sidebar prints: at the iron preset the engine's terms are volume 886.8 − surface 268.3 − Coulomb 121.3 − asymmetry 6.63 + pairing 1.50 = 492.0 MeV total, 8.79 MeV/nucleon — riding the crest of 'the most important graph in physics', which 'rises at low A, peaks very near iron (Fe, A=56), and then tapers off at high A' (OpenStax UP3 §10.2, verbatim). Ride the valley on the marker: He-4 5.46 → C-12 7.32 → Ca-40 8.54 → Fe-56 8.79 (engine world), then step to U-235 at 7.65 — the downhill side where 'repulsive electrostatic forces… begin to dominate' and fission pays out the difference (the HTML quiz Q3's keyed logic)
- Read shell structure off the Magic rows and the stability band: the magic numbers '(2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus… similar in concept to the stable electron shells observed for the noble gases' (Chem 2e §21.1, verbatim — and ⁴⁰Ca with Z=N=20 is 'double magic… particularly stable', the panel lighting both ✨ rows at once), while the valley's recipe shifts with size — light nuclei sit near n:p = 1:1 but 'heavier stable nuclei… have increasingly more neutrons than protons' (1.15 at Fe-56, 1.52 at Pb-207, verbatim) because they 'require larger numbers of neutrons to provide compensating strong forces'. Even the pairing term has its ledger: 157 of the ~260 stable nuclides are even-even against just 5 odd-odd (Chem 2e Table 21.1)
Formulas
Make a prediction
Apply the Iron-56 preset: Z = 26, N = 30 — the panel reads Total BE = 492.0 MeV, BE/A = 8.79 MeV, badge green 'Stable'. Now hold Z at 26 and drag the Neutrons slider from 30 down to 20 (building Fe-46). What does the BE/A row do?
No grading here — commit to a guess, then scroll down and test it yourself.
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Your prediction
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Answer: It falls to 7.54 MeV — ten fewer neutrons means ten fewer strong-force bonds (the volume term drops 886.8 → 728.4 MeV) while the Coulomb penalty actually grows (121.3 → 129.5 MeV); every remaining nucleon is held ~1.2 MeV more loosely, and the badge flips to amber 'Unstable'
The neutron is a paying customer, not ballast: it adds strong-force attraction (the volume term av·A) while adding ZERO Coulomb repulsion, so pulling neutrons out of a proton-rich nucleus costs real binding. The five-term ledger prices both states exactly. Fe-56 (A=56, even-even): volume 886.76 − surface 268.30 − Coulomb 121.31 − asymmetry 6.63 + pairing 1.50 = 492.0 MeV, and 492.0/56 = 8.79 per nucleon. Fe-46 (A=46): volume 728.41 − surface 235.32 − Coulomb 129.53 − asymmetry 18.16 + pairing 1.65 = 347.1 MeV, and 347.1/46 = 7.54 per nucleon. Note what moved: the volume term fell 158.4 MeV (ten nucleons' worth of bonds), the Coulomb term ROSE 8.2 MeV (the same 26 protons packed into a smaller A^{1/3}), and the asymmetry penalty tripled (6.63 → 18.16) because the build went lopsided — everything the neutrons were quietly preventing. Option B is the misconception this gate targets, and the sim prints it on its own sidebar: 'Misconception: neutrons aren't just "fillers". They dilute Coulomb repulsion among protons while adding strong force attraction.' The textbook's version: heavy nuclei 'require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together' (OpenStax Chem 2e §21.1, verbatim). Option C runs the bookkeeping backwards — the TOTAL also falls (492.0 → 347.1 MeV); there is no fixed binding budget to redistribute (had the total held at 492.0, the row would read 10.70, off the curve board's 9.0 ceiling). The badge knows too: green Stable flips to amber Unstable as the n:p ratio slides from 1.15 to 0.77, out of the valley.
Quiz (0/3)
How many neutrons does carbon-12 have? Is it stable?
What happens when you add a neutron to make carbon-13 vs carbon-14?
Why does the binding-energy-per-nucleon curve crest near iron-56?
You can now
- Keep the A = Z + N ledger and the nuclide bookkeeping straight on the panel's own rows: the factory state is hydrogen-1 (Z=1, N=0, A=1 — a lone proton, the one nuclide with no neutrons, Chem 2e §21.1), the iron preset reads 26 + 30 = 56 with an n:p ratio of 1.15, and the uranium preset reads 92 + 143 = 235 — the isotope name, the mass number, and the neutron count are three views of the same two sliders. The ledger extends to the stability badge's Z>83 rule: 'all isotopes of elements with atomic numbers greater than 83 are unstable' (Chem 2e §21.1, Figure 21.2 caption, verbatim), which is why the uranium state reads Radioactive before any half-life is ever discussed
- Price any nucleus with the semi-empirical (Bethe-Weizsäcker) five-term ledger the sidebar prints: at the iron preset the engine's terms are volume 886.8 − surface 268.3 − Coulomb 121.3 − asymmetry 6.63 + pairing 1.50 = 492.0 MeV total, 8.79 MeV/nucleon — riding the crest of 'the most important graph in physics', which 'rises at low A, peaks very near iron (Fe, A=56), and then tapers off at high A' (OpenStax UP3 §10.2, verbatim). Ride the valley on the marker: He-4 5.46 → C-12 7.32 → Ca-40 8.54 → Fe-56 8.79 (engine world), then step to U-235 at 7.65 — the downhill side where 'repulsive electrostatic forces… begin to dominate' and fission pays out the difference (the HTML quiz Q3's keyed logic)
- Read shell structure off the Magic rows and the stability band: the magic numbers '(2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus… similar in concept to the stable electron shells observed for the noble gases' (Chem 2e §21.1, verbatim — and ⁴⁰Ca with Z=N=20 is 'double magic… particularly stable', the panel lighting both ✨ rows at once), while the valley's recipe shifts with size — light nuclei sit near n:p = 1:1 but 'heavier stable nuclei… have increasingly more neutrons than protons' (1.15 at Fe-56, 1.52 at Pb-207, verbatim) because they 'require larger numbers of neutrons to provide compensating strong forces'. Even the pairing term has its ledger: 157 of the ~260 stable nuclides are even-even against just 5 odd-odd (Chem 2e Table 21.1)