The nuclide chart and mass tables

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 has a half-life of 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.

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 = 4.00260325 u gives Δ = 2.4249 MeV; the neutron's 1.0086649 u gives 8071.3 keV, and ¹H's 1.00782503 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, bottoming out near −91.7 MeV at ¹¹⁸Sn. Do not mistake that for a binding curve: Δ/A is lowest at ⁵⁶Fe (−1082.3 keV) while B/A peaks at ⁶²Ni (8794.6 keV), because binding weights the proton and neutron excesses by Z/A and N/A.

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 Q_α = 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.

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.

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 M_nuc(Z) = Mₐₜ(Z) − Z mₑ + Bₑ(Z)/c², β⁻ decay gives Q = [Mₐₜ(Z) − Mₐₜ(Z+1)]c² = Δ_P − Δ_D: the daughter's neutral atom needs exactly one more electron than the parent's, and that electron is the β particle. No mₑ appears; the only residue of the conversion is the difference of total atomic electron binding, Bₑ(Z + 1) − Bₑ(Z), which is 65 eV for tritium but roughly 18 keV near lead, and the neutral-atom masses already contain it. 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 daughter atom is left with a vacancy in the shell the electron was taken from, so Q(EC) = Δ_P − Δ_D − Bₓ, with Bₓ that shell's binding in the daughter. ⁷Be makes the first point: Δ_P − Δ_D = 861.9 keV, under 1022 keV, so it decays only by capture. ¹⁶³Ho makes the second: Δ_P − Δ_D is about 2.8 keV while dysprosium's K shell is bound by 53.8 keV, so Q(EC) for K capture would be negative, K capture is closed, and the decay runs from M and higher shells.

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 Q_α, 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 σ_Q² = Σ nₖ² σₖ² 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.