University Physics II · Introduction to Modern Physics · 15.8
Nuclei, Radioactivity, and Particles
The nucleus runs the same energy accounting as the atom at a million times the scale. Mass deficits say what can decay, a constant per-nucleus probability says how fast, and conservation laws vet every reaction.
Build the model
Connect the measurement to the mechanism.
A nucleus is a bound system, so the bookkeeping you used on atoms carries over; only the scale moves by about a million. Weigh the bound nucleus, weigh the separated nucleons, and the deficit is the binding energy: 0.030377 u for ⁴He, which is 28.30 MeV, or 7.07 MeV per nucleon against the 13.6 eV that ionises hydrogen. Divide by nucleon number and the resulting curve decides the rest.
It peaks near A ≈ 56–62, so fusing light nuclei and splitting heavy ones both move toward the peak and both release the difference in mass. A decay proceeds when the rearrangement lowers the total rest mass and the conserved quantities — charge, nucleon number, energy–momentum, lepton number — still balance. When any one nucleus goes is undetermined: each carries a fixed probability per unit time, which makes a sample fall exponentially.
Underneath sits a classification of quarks and leptons that names the pieces without explaining what holds them.
- Simple definition
- Nuclear binding energy is the mass deficit of a bound nucleus times c², and radioactive decay is any spontaneous rearrangement into a lower-mass state, each nucleus carrying a fixed decay probability λ per unit time.
- Example
- ⁴He weighs 0.030377 u less than two protons and two neutrons, so it is bound by 28.30 MeV. ²³⁸U weighs 0.004584 u more than ²³⁴Th plus an α, so it can shed one — and does, with a half-life of 4.5 × 10⁹ years.
⁴He: 4.032980 − 4.002603 = 0.030377 u, so EB = 28.30 MeV and B/A = 7.07 MeV.
1 u ↔ 931.494 MeV; atomic masses throughout, so the electrons cancel
⁴He 7.07 · ⁵⁶Fe 8.79 · ²³⁸U 7.57 MeV per nucleon, so fusion and fission both release energy.
Downhill runs toward the peak from both ends of the chart
²³⁸U → ²³⁴Th + α: Δm = 0.004584 u, so Q = 4.27 MeV.
Q in MeV; it fixes the energy released and says nothing about the rate
4.27 × 234/238 = 4.20 MeV to the α, leaving 0.07 MeV of ²³⁴Th recoil.
Parent at rest, so the two fragments carry equal and opposite momenta
¹⁴C: t½ = 5700 y gives λ = 3.85 × 10⁻¹² s⁻¹ and 0.25 Bq per gram of modern carbon.
λ in s⁻¹ · activity in becquerel, 1 Bq = 1 decay s⁻¹
n → p + e⁻ + ν̄ₑ: charge 0 = 1 − 1, A 1 = 1, lepton number 0 = 1 − 1.
Antiparticles count −1; for nuclei, A is the baryon number
Same bookkeeping as the atom, a million times the scale
A nucleus packs Z protons and N neutrons inside a radius of roughly 1.2 A¹⁄³ fm — 7.4 fm for ²³⁸U, some 10⁵ times smaller than the atom around it. Bring two protons within 2 fm and their electrostatic energy is ke²/r = 1.44 MeV·fm ÷ 2 fm = 0.72 MeV, so whatever binds a nucleus has to beat that. Weigh the result and it comes out light. Two protons and two neutrons total 4.032980 u; the ⁴He atom measures 4.002603 u; the deficit of 0.030377 u at 931.494 MeV per u is 28.30 MeV of binding energy, or 7.07 MeV per nucleon. Set that beside the 13.6 eV needed to ionise hydrogen and the ratio is 5.2 × 10⁵. Same accounting, new scale — and that is why a nucleus dropping between its own levels emits a MeV gamma ray where an atom gave you visible light.
Binding energy per nucleon sets the direction of every process
Total binding energy just tracks size, so divide it by A. B/A rises steeply through the light nuclei, peaks near A ≈ 56–62 at about 8.8 MeV per nucleon — ⁵⁶Fe is bound by 492.3 MeV, so 8.79 per nucleon — then sags to 7.57 MeV per nucleon at ²³⁸U, whose total is 1801.7 MeV. Read the curve as a landscape in which downhill means toward the peak. Fuse light nuclei and the product sits higher, so mass is released: assembling one ⁴He from four hydrogen atoms sheds 0.028697 u, or 26.7 MeV, which is what the Sun runs on. Split a heavy nucleus and the fragments land near 8.5 MeV per nucleon, releasing about 0.85 MeV × 236 ≈ 200 MeV per fission. Both directions release energy because the peak sits between them. Nothing in this survey explains the shape of that curve; measuring it is enough to say which way energy flows.
A decay is allowed when the products weigh less
Take the parent's mass, subtract the products', multiply by c²: that is Q, and a decay is energetically allowed exactly when Q > 0. For ²³⁸U → ²³⁴Th + ⁴He, 238.050788 − (234.043601 + 4.002603) = 0.004584 u, so Q = 4.27 MeV. Momentum conservation then fixes the split: the parent is at rest, the fragments leave with equal and opposite momenta, so the alpha takes Q × 234/238 = 4.20 MeV and the thorium keeps 0.07 MeV — and 4.20 MeV is the line detectors actually record. Beta decay refused that pattern. For n → p + e⁻, Q = 939.565 − 938.272 − 0.511 = 0.782 MeV, yet the electrons emerge with every energy from zero up to that ceiling, which no two-body decay can produce. A third product carries the balance: the antineutrino. Gamma emission changes neither A nor Z, being a nucleus falling between its own MeV-spaced levels.
Q says whether; the decay constant says how fast
²³⁸U has Q = 4.27 MeV and a half-life of 4.5 × 10⁹ years, so λ = 4.9 × 10⁻¹⁸ s⁻¹. A free neutron has the smaller Q of 0.782 MeV and a half-life near 610 s. Released energy does not set the rate. What experiment supplies instead is a decay constant: every surviving nucleus has probability λ dt of decaying in any interval dt, independent of its age and of its neighbours. That one assumption gives dN/dt = −λN, hence N = N₀e(−λt), t½ = ln2/λ, and activity A = λN. For ¹⁴C, t½ = 5700 y means λ = 0.693/(1.80 × 10¹¹ s) = 3.85 × 10⁻¹² s⁻¹; a gram of modern carbon holds about 6.5 × 10¹⁰ ¹⁴C nuclei and so ticks at 0.25 Bq. A sample down to 0.40 of that ratio has been decaying for (5700/ln2) × ln(1/0.40) = 7.5 × 10³ years.
Every reaction balances several ledgers at once
A nuclear reaction or decay must conserve electric charge, nucleon number A, energy–momentum, angular momentum, and lepton number. Check n → p + e⁻ + ν̄ₑ: charge 0 = +1 − 1, A gives 1 = 1, and lepton number 0 = +1 − 1 once the antineutrino counts −1. That ledger is why the neutrino had to exist decades before anyone detected one. Momentum does more work than a Q value alone suggests. Rutherford's ¹⁴N + ⁴He → ¹⁷O + ¹H has Q = −1.19 MeV, so it is endothermic — but firing an alpha at a stationary ¹⁴N nucleus with exactly 1.19 MeV still fails, because the products must carry the incoming momentum and therefore keep kinetic energy. The lab threshold is |Q|(1 + mα/mN) = 1.19 × 1.286 = 1.53 MeV. Balance the ledgers first: they rule out most of what you might write down and cost nothing to check.
The particle inventory, and where this survey stops
Nucleons are not fundamental. The current inventory is six quarks and six leptons in three generations, plus the carriers — gluon, photon, W±, Z⁰ — and the Higgs. Quarks carry charge +2/3 or −1/3 and appear only inside composites: baryons of three quarks, mesons of a quark and an antiquark. A proton is uud, a neutron udd, and β-minus decay is one d becoming a u by emitting a W⁻ that turns into e⁻ + ν̄ₑ, which is why A holds while Z rises by one. The quark rest masses sum to about 9 MeV/c², near 1% of the proton's 938 MeV/c²; the rest is field energy. Now the limits. Nothing here predicts λ, which needs barrier tunnelling and transition-rate theory; nothing here says which nuclides are bound, which needs the nuclear force and shell structure; cross sections and branching ratios sit outside it too. This is conservation-law accounting on measured inputs, and it should be quoted as such.
Change one variable at a time
Make the relationship visible.
Drag the daughter's mass number down towards 4 and watch the two energy bars level off — the momentum arrows stay equal at every setting, so the energy only splits evenly when the fragments have equal mass, while at A = 234 the α already takes 98% of Q.
DECAY ENERGY Q4.28 MeV
ALPHA ENERGY4.21 MeV
DAUGHTER RECOIL0.072 MeV
SHARED MOMENTUM p177 MeV/c
Live interpretationDECAY ENERGY Q: 4.28 MeV. ALPHA ENERGY: 4.21 MeV. DAUGHTER RECOIL: 0.072 MeV. SHARED MOMENTUM p: 177 MeV/c
Catch the common trap
Explain before calculating.
A ²³⁸U nucleus at rest decays to ²³⁴Th and an alpha particle. The atomic-mass difference gives Q = 4.27 MeV, released as kinetic energy of the two products. Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFind the binding energy of ⁷Li and its binding energy per nucleon. Atomic masses: ⁷Li = 7.016003 u, ¹H = 1.007825 u, neutron = 1.008665 u. Take 1 u ↔ 931.494 MeV.
- Count the pieces: ⁷Li has Z = 3, so N = 7 − 3 = 4. Weigh three hydrogen atoms and four neutrons: 3(1.007825) + 4(1.008665) = 3.023475 + 4.034660 = 7.058135 u.
- Subtract the measured atomic mass: Δm = 7.058135 − 7.016003 = 0.042132 u. Atomic masses on both sides, so the three electrons cancel.
- Convert the deficit: EB = 0.042132 u × 931.494 MeV/u = 39.25 MeV.
- Divide by A: B/A = 39.25 ÷ 7 = 5.61 MeV per nucleon. That is far below the 8.8 MeV peak near A ≈ 56, so ⁷Li still sits on the rising side of the curve — it even undercuts the anomalously tight ⁴He spike at 7.07.
AnswerEB = 39.25 MeV, or 5.61 MeV per nucleon
Medium²¹⁰Po at rest decays by alpha emission to ²⁰⁶Pb. Atomic masses: ²¹⁰Po = 209.982874 u, ²⁰⁶Pb = 205.974465 u, ⁴He = 4.002603 u. Find Q and the kinetic energy of each product.
- Weigh the products: 205.974465 + 4.002603 = 209.977068 u, against the parent's 209.982874 u.
- Δm = 209.982874 − 209.977068 = 0.005806 u. It is positive, so the decay is allowed: Q = 0.005806 × 931.494 = 5.41 MeV.
- The parent is at rest, so the fragments leave with equal and opposite momenta and the energy splits inversely with mass: Kα = Q × 206/210 = 5.4083 × 0.98095 = 5.31 MeV.
- The lead keeps the remainder: 5.4083 − 5.3052 = 0.103 MeV. A detector records the 5.31 MeV line, not 5.41 MeV — the recoil is already spent.
AnswerQ = 5.41 MeV; the alpha carries 5.31 MeV and the ²⁰⁶Pb recoil 0.103 MeV
HardA sealed source holds 2.00 μg of ¹³¹I, half-life 8.02 days, molar mass 131 g mol⁻¹. Find the decay constant, the initial activity, and the activity 30.0 days later.
- λ = ln2/t½, with t½ = 8.02 × 86400 = 6.929 × 10⁵ s, so λ = 0.6931 ÷ (6.929 × 10⁵) = 1.00 × 10⁻⁶ s⁻¹.
- Count the nuclei: N₀ = (2.00 × 10⁻⁶ g ÷ 131 g mol⁻¹) × 6.022 × 10²³ mol⁻¹ = 9.194 × 10¹⁵.
- A₀ = λN₀ = 1.0003 × 10⁻⁶ s⁻¹ × 9.194 × 10¹⁵ = 9.20 × 10⁹ Bq.
- 30.0 days is 30.0/8.02 = 3.74 half-lives, so A = A₀ × 2(−3.74) = 9.20 × 10⁹ × 0.0748 = 6.88 × 10⁸ Bq — a factor of 13.4 down, with no reference to Q anywhere.
Answerλ = 1.00 × 10⁻⁶ s⁻¹, A₀ = 9.20 × 10⁹ Bq, and A(30.0 d) = 6.88 × 10⁸ Bq