University Physics V · Nuclear Physics · 14.3
The Nuclide Chart & Mass Tables
Nuclear energetics is bookkeeping before it is physics. Learn to find a nuclide on the chart, read its mass excess off the table, and assemble any Q value from integer coefficients — so that later, when a decay or a reaction needs an energy, you already have it.
Build the model
Connect the measurement to the mechanism.
A nuclide is fixed by two integers, so all of nuclear physics is indexed by a lattice: hold Z and you have a row of isotopes, hold N a column of isotones, hold A = N + Z an anti-diagonal of isobars. Against every point sits one measured number, the neutral-atom mass — and the table does not store it. It stores the mass excess Δ = (Mₐₜ − A u)c², because Mₐₜ c² for a heavy nuclide is around 2 × 10⁵ MeV while the energies that matter are a few MeV, and Δ carries the same information at the scale of the answer.
What the shift buys is more than tidiness. Mass number is conserved in every nuclear reaction and decay, so the A u c² parts cancel identically and any Q value collapses to Q = Σ nₖ Δₖ, an integer row vector contracted with one stored array. Binding energies, separation energies, α and β Q values and reaction thresholds are that same operation with different rows: one NumPy array, one matrix product, thousands of energies.
The cost is a bookkeeping convention you must never lose track of. The tabulated masses are for neutral atoms, so each carries Z electrons and their binding. In β⁻ decay that is exactly right and the electron mass disappears from the arithmetic; in β⁺ it leaves you owing 2 mₑc² = 1022.0 keV; in electron capture it leaves a shell-binding term worth 55 eV in lithium and 88 keV in lead.
- Simple definition
- The mass excess of a nuclide is the energy equivalent of its neutral-atom mass minus its mass number times the atomic mass unit, Δ = (Mₐₜ − A u)c², and it is the quantity nuclear mass tables actually list.
- Example
- ⁴He has Mₐₜ = 4.0026033 u, so Δ = 0.0026033 u × 931.494 MeV/u = 2.4249 MeV, while ¹²C is 12 u by definition and therefore has Δ = 0 exactly.
Shifts a 2 × 10⁵ MeV number down to a 10 MeV one, so the stored value and its 0.1 keV uncertainty share a scale.
Mₐₜ is the neutral-atom mass in u, A = Z + N the integer mass number; Δ is tabulated in keV.
The single constant that turns a mass column into an energy column; the carbon zero is a definition, not a measurement.
1 u is one twelfth of a neutral ¹²C atom at rest in its ground state, so Δ(¹²C) = 0 exactly.
Mass-number conservation kills the A u c² terms, so a batch of Q values is one integer matrix times one stored array.
nₖ integer: +1 per reactant, −1 per product. b = 0, or −1022.0 keV for β⁺, or −Bₓ for capture.
³H → ³He is 14 949.81 − 14 931.22 = 18.59 keV, the tritium endpoint, with no electron mass subtracted.
In keV. No mₑ term: the daughter neutral atom carries exactly the one extra electron the decay emits.
Opens a window 0 < ΔP − ΔD < 1022 keV in which capture is the only channel; ⁷Be sits in it at 861.9 keV.
Bₓ is the captured shell's binding in the daughter atom: 55 eV in lithium, 88 keV in lead.
A chain is a vector sum, so its α and β counts come from one 2 × 2 integer solve: ²³⁸U → ²⁰⁶Pb needs 8 α and 6 β⁻.
N across, Z up. β⁻ slides along the isobar A; α drops A by 4 and leaves N − Z unchanged.
Two integers index the whole chart
A nuclide is named by two integers and everything else follows. Z, the proton number, fixes the element; N, the neutron number, distinguishes its isotopes. Plot N across and Z up and the vocabulary becomes directions on a lattice: isotopes share Z and lie in a horizontal row (¹²C, ¹³C, ¹⁴C), isotones share N and lie in a vertical column (¹⁴C and ¹⁵N both have N = 8), isobars share A = N + Z and lie on an anti-diagonal (⁴⁰Ar, ⁴⁰K, ⁴⁰Ca). A fourth label, the isomer, shares both Z and N and differs only in nuclear excitation: ⁹⁹ᵐTc sits 142.7 keV above ⁹⁹Tc and lives 6.0 hours, so an entry's identity is (Z, N, excitation), not (Z, A). About 3300 nuclides have been observed and roughly 250 are stable. The stable ones track N = Z up to ⁴⁰Ca and then bend to N > Z, until ²⁰⁸Pb runs 126 neutrons against 82 protons — the Coulomb repulsion those extra neutrons are there to dilute.
Why the table stores Δ and not the mass
Take the tabulated neutral-atom mass, subtract the integer mass number times u, and convert: Δ = (Mₐₜ − A u)c², with u c² = 931.494 MeV. The unit is anchored on carbon — 1 u is one twelfth the mass of a neutral ¹²C atom at rest in its ground state — so Δ(¹²C) = 0 by definition, not by measurement. For ⁴He, Mₐₜ = 4.0026033 u gives Δ = 2.4249 MeV; the neutron's 1.0086649 u gives 8071.3 keV, and ¹H's 1.0078250 u gives 7289.0 keV. The point of the subtraction is dynamic range: Mₐₜ c² for ²³⁸U is 221 743 MeV while its α decay releases 4.27 MeV, so a Q value good to a keV needs nine significant figures in Mₐₜ and only five in Δ. Read the sign pattern too. Δ is positive at both ends of the chart and negative through the middle, reaching about −91 MeV at ¹²⁰Sn. Do not mistake that for a binding curve: Δ/A is lowest at ⁵⁶Fe (−1082.2 keV) while B/A peaks at ⁶²Ni (8794.6 keV), because binding weights the proton and neutron excesses by Z/A and N/A.
Mass-number conservation makes Q a dot product
Write every mass as Mₐₜ c² = A u c² + Δ. A Q value is Σ M(in)c² − Σ M(out)c², so Q = (ΣAᵢₙ − ΣAₒᵤₜ) u c² + (ΣΔᵢₙ − ΣΔₒᵤₜ). Mass number is conserved in every nuclear reaction and every decay, so the first bracket vanishes exactly and the 931.494 MeV/u conversion never has to be done at all: Q is a difference of tabulated mass excesses and nothing else. Collect the table into a vector Δ indexed by nuclide and the recipe is one integer row, Q = n · Δ, with nₖ = +1 for each reactant, −1 for each product, and the constraint n · A = 0 built in. For ²³⁸U → ²³⁴Th + ⁴He the row is (+1, −1, −1) and Qα = 47 308.9 − 40 614.1 − 2 424.9 = 4 269.9 keV, the 4.270 MeV every table quotes. A different row on the same vector gives a separation energy or a binding energy: B = Z Δ(¹H) + N Δ(n) − Δ(Z, N), which for ⁴He is 2(7289.0) + 2(8071.3) − 2424.9 = 28 295.7 keV.
One array, one matrix product, and a free consistency check
You never compute one Q at a time. Load the mass excesses once into a float64 array, index nuclides by (Z, N), and build an integer matrix S whose rows are the reactions you care about. Then Q = S @ δ + b, one matrix–vector product, with b holding zero for reactions, α and β⁻, −1022.0 keV for each β⁺ row, and −Bₓ for each capture row. Two structural facts come free. Every admissible row must satisfy n · A = 0, so a row that fails that test is a typo rather than a reaction — the cheapest validation you will ever write. And rows can be linearly dependent, in which case Q inherits the dependence: if row 3 is row 1 plus twice row 2, then Q₃ = Q₁ + 2Q₂ identically. That is the nuclear analogue of Hess's law, and it is how an evaluation such as AME works — thousands of measured reaction and decay energies over-determine a much smaller set of masses, and the least-squares residuals are what the quoted uncertainties describe.
Neutral atoms, and where the electrons go
The tables list neutral atoms, so an entry for Z carries Z electrons and their binding. Convert to nuclear masses inside a Q value and watch what survives. Using Mnuc(Z) = Mₐₜ(Z) − Z mₑ + Bₑ(Z)/c², β⁻ decay gives Q = [Mₐₜ(Z) − Mₐₜ(Z+1)]c² = ΔP − ΔD: the daughter's neutral atom needs exactly one more electron than the parent's, and that electron is the β particle. No mₑ appears, and only the difference of total atomic electron binding, tens of eV, is left over. ³H → ³He is 14 949.81 − 14 931.22 = 18.59 keV, the tritium endpoint. For β⁺ the daughter has one electron too many in the bookkeeping and a positron is created besides, so you owe 2 mₑc² = 1022.0 keV. For capture no positron is made, but the electron came from a bound shell of the daughter, so Q(EC) = ΔP − ΔD − Bₓ. ⁷Be makes the first point: ΔP − ΔD = 861.9 keV, under 1022 keV, so it decays only by capture. ¹⁶³Ho makes the second: Q(EC) is about 2.8 keV while dysprosium's K shell is bound by 53.8 keV, so K capture is closed and the decay runs from M and higher shells.
What a mass table cannot tell you
A mass table settles energetics and nothing else. Q > 0 says a channel is open; it says nothing about the barrier, the matrix element, or the rate. ²³⁸U and ²¹²Po both have positive Qα, 4.27 MeV and 8.95 MeV, and their half-lives differ by twenty-four orders of magnitude — a fact no mass difference contains. Three further cautions. Excited states are not in the mass column, so an isomer needs its excitation energy added separately, and a ground-state Q value is not what you measure when the decay feeds an excited level. Atomic against nuclear masses is a convention never to be mixed mid-calculation; one stray Z mₑ is 511 keV per electron. And uncertainties travel through the same integer row, but uQ² = Σ nₖ² uₖ² holds only for independent entries — an evaluation determines many masses from shared linking measurements, so a defensible error bar comes from the published covariance, not from quadrature.
Change one variable at a time
Make the relationship visible.
Eight alpha steps alone reach A = 206 but land on Z = 76, six protons short. Add beta steps and watch the neutron-excess readout: it ignores the alpha slider entirely, because (−2, −2) leaves N − Z alone. Set nα = 8 and nβ = 6 to hit lead exactly.
MASS NUMBER A222
PROTON NUMBER Z86
NEUTRON EXCESS N − Z50
STEPS FROM Pb-20616
Live interpretationMASS NUMBER A: 222. PROTON NUMBER Z: 86. NEUTRON EXCESS N − Z: 50. STEPS FROM Pb-206: 16
Catch the common trap
Explain before calculating.
A table of neutral-atom masses gives Δ(²²Na) = −5181.6 keV and Δ(²²Ne) = −8024.7 keV. What is the Q value of the β⁺ branch of ²²Na decay?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA table gives mass excesses Δ(³H) = 14 949.81 keV and Δ(³He) = 14 931.22 keV. Find the Q value of the β⁻ decay ³H → ³He + e⁻ + ν̄ₑ, and say why no electron mass appears in the arithmetic.
- Write the decay as an integer row on the mass-excess vector: n = +1 on ³H, −1 on ³He. Check the constraint n · A = 3 − 3 = 0, so the A u c² terms cancel and nothing has to be converted from u.
- The masses are for neutral atoms. ³H carries one electron, ³He carries two, and that extra electron in the daughter atom is exactly the β particle the decay emits — so the electron correction b is zero for every β⁻ row.
- Q = Δ(³H) − Δ(³He) = 14 949.81 − 14 931.22 = 18.59 keV.
- The energy is shared between the electron and the antineutrino, so 18.59 keV is the endpoint of the β spectrum, reached only when the neutrino takes nothing.
AnswerQ = 18.59 keV. Subtracting mₑc² = 511 keV as well would give a negative Q and make tritium stable — the electron is already counted in the neutral-atom masses.
Medium⁷Be (Δ = 15 769.0 keV) decays to ⁷Li (Δ = 14 907.1 keV). Decide which of β⁺ emission and electron capture is energetically allowed, and give the Q value of the open channel. The K-shell binding energy in lithium is 55 eV.
- Both channels start from the same atomic mass-excess difference: D = Δ(⁷Be) − Δ(⁷Li) = 15 769.0 − 14 907.1 = 861.9 keV.
- For β⁺ the neutral-atom bookkeeping leaves one surplus electron on the daughter side and a positron is created as well, so Q(β⁺) = D − 2 mₑc² = 861.9 − 1022.0 = −160.1 keV.
- A negative Q closes the channel outright: ⁷Be cannot emit a positron, whatever the nuclear matrix element.
- For capture no positron is made; the only correction is the binding of the electron taken from the daughter's K shell, so Q(EC) = D − BK = 861.9 − 0.055 = 861.8 keV.
- Q(EC) > 0, so capture is the only open channel — which is why ⁷Be decays 100% by electron capture, with a half-life of 53.2 days.
Answerβ⁺ is forbidden (Q = −160.1 keV); electron capture is allowed with Q(EC) = 861.8 keV. Any nuclide with 0 < D < 1022.0 keV is a capture-only decayer.
HardThe ²³⁸U series ends at stable ²⁰⁶Pb. Find how many α and β⁻ decays the chain contains, then its total energy release, given Δ(²³⁸U) = 47 308.9 keV, Δ(²⁰⁶Pb) = −23 785.4 keV and Δ(⁴He) = 2 424.9 keV.
- Place both ends on the lattice: ²³⁸U is (N, Z) = (146, 92) and ²⁰⁶Pb is (124, 82), so the net displacement is (ΔN, ΔZ) = (−22, −10).
- Each α contributes (−2, −2) and each β⁻ contributes (−1, +1), so nα and nβ solve the pair −2nα − nβ = −22 and −2nα + nβ = −10.
- Adding the two equations gives −4nα = −32, so nα = 8, and substituting back gives nβ = 22 − 16 = 6. Check A: 238 − 4(8) = 206 ✓. Check Z: 92 − 16 + 6 = 82 ✓.
- Now one integer row on the mass-excess vector. The net reaction is ²³⁸U → ²⁰⁶Pb + 8 ⁴He, and the six β⁻ steps need no electron correction because the neutral-atom masses already carry those electrons.
- Q = Δ(²³⁸U) − Δ(²⁰⁶Pb) − 8 Δ(⁴He) = 47 308.9 + 23 785.4 − 8(2 424.9) = 71 094.3 − 19 399.2 = 51 695.1 keV.
Answer8 α and 6 β⁻, releasing 51.70 MeV in total. Because α changes A by 4 and β⁻ not at all, every member of the chain has A ≡ 2 (mod 4) — the 4n + 2 series.