Alpha decay as barrier penetration

Start with energetics, because it is cheap and decisive in one direction only. Q = [M(A, Z) − M(A−4, Z−2) − M(⁴He)]c², and atomic masses may be subtracted directly, because the parent's Z electrons match the daughter's Z−2 plus the helium's 2. For ²²⁶Ra: 226.025410 − 222.017578 − 4.002603 = 0.005229 u, and 0.005229 × 931.494 = 4.871 MeV. Momentum conservation then splits Q between two bodies leaving back to back, so kinetic energy divides inversely with mass: T_α = Q(A−4)/A = 4.8708 × 222/226 = 4.785 MeV, and the ²²²Rn recoil keeps 86 keV. That is why the measured principal line sits at 4.784 MeV and not at Q. The warning is the other half. Q is positive for essentially every nucleus above A ≈ 150, yet ¹⁴⁴Nd, with Q = 1.9 MeV, has a half-life of 2 × 10¹⁵ years. Energetics licenses the decay. It does not date it.

Beyond the range of the strong force the alpha and daughter see only each other's charge: V(r) = 2Z_d e²/4πε₀r = 2Z_d(1.44 MeV fm)/r. Inside, model the well as flat. The join is the contact radius R ≈ 1.2(A_d^⅓ + 4^⅓) fm — for ²²²Rn plus an α, 1.2(6.055 + 1.587) = 9.17 fm — and there the barrier stands at 2 × 86 × 1.44/9.17 = 27.0 MeV. The alpha's 4.87 MeV puts it at 18% of the top. The far side of the wall is the outer turning point b, where V(b) = Q: b = 2Z_d(1.44)/Q = 247.7/4.871 = 50.85 fm. So the classically forbidden region is 41.7 fm wide, four and a half nuclear radii — the alpha does not creep over a thin lip, it crosses a gap several times the size of the nucleus. The same barrier is visible from outside: fire 8 MeV alphas at gold and, once the gold's recoil is allowed for, their closest approach is 29 fm, more than three times the contact radius, which is exactly why Rutherford's scattering stayed pure Coulomb.

Inside a classically forbidden region the wavefunction decays instead of oscillating, and for an opaque barrier, 2γ ≫ 1, WKB gives the transmission to leading order as e^(−2γ) with γ = (1/ħ)∫ from R to b of √(2μ(V(r) − Q)) dr. Use the reduced mass μ = m_α m_d/(m_α + m_d) = 3.93 u = 3662 MeV/c², not m_α, because both bodies move. For a pure 1/r barrier the integral closes: γ = (b√(2μQ)/ħ)[arccos√x − √(x(1 − x))] with x = R/b. For ²²⁶Ra, x = 9.17/50.85 = 0.1804, so √x = 0.4247, arccos 0.4247 = 1.132 rad, √(x(1−x)) = 0.3845, and the bracket is 0.7477. The wavenumber scale is √(2μc²Q)/ħc = √(2 × 3662 × 4.871)/197.33 = 0.957 fm⁻¹. Hence γ = 0.957 × 50.85 × 0.7477 = 36.4, so 2γ = 72.8 and the transmission per approach is e^(−72.8) = 2.4 × 10⁻³². Watch what the bracket does: at x = 0 it is π/2, at x = 1 it is zero. A barrier whose turning points have met costs nothing.

e^(−2γ) is dimensionless. To reach s⁻¹ you need to know how often the alpha tries. Take its speed from its energy, v = c√(2Q/m_αc²) = c√(2 × 4.871/3727.4) = 0.0511c = 1.53 × 10²² fm/s, and let it cross the well once each way: f = v/2R = 1.53 × 10²²/18.3 = 8.4 × 10²⁰ s⁻¹. Then λ = f e^(−2γ) = 8.4 × 10²⁰ × 2.4 × 10⁻³² = 2.0 × 10⁻¹¹ s⁻¹, and t½ = ln2/λ = 3.4 × 10¹⁰ s = 1.1 × 10³ years, against a measured 1600 years. Landing within a factor of 1.5 flatters the model badly. Differentiate the exponent: d(2γ)/dR = −2√(2μc²(V(R) − Q))/ħc = −4.1 fm⁻¹, so moving R by half a femtometre swings the half-life by a factor of eight. The last honest step is to write λ = P f e^(−2γ) and let the preformation factor P — the chance that four nucleons are an α when they reach the wall — absorb the residual. For even-even ground-state transitions P lands between about 0.01 and 1, and here it is fitted, never derived.

For x ≪ 1, arccos√x − √(x(1−x)) ≈ π/2 − 2√x, which splits 2γ into two terms: π√(2μ)⋅(2Z_d e²/4πε₀)/(ħ√Q), which carries all of the Q dependence and none of R, minus 4√(2μ⋅(2Z_d e²/4πε₀)⋅R)/ħ, which carries R and no Q. Divide by ln10 and you have log₁₀(t½/s) ≈ 1.70 Z_d/√Q + c: a straight line against Q^(−1/2) whose slope is predicted rather than fitted, and which sits within 3% of the empirical Viola-Seaborg coefficient of 1.66. Test the span on real nuclides. ²³²Th has Q = 4.083 MeV and t½ = 1.4 × 10¹⁰ yr; ²¹²Po has Q = 8.954 MeV and t½ = 0.299 µs. A factor of 2.19 in Q buys 24 orders of magnitude in half-life. Do not over-trust the expansion itself, though: for ²²⁶Ra the two terms give 152.9 − 82.7 = 70.2 against the exact 72.8, and 2.6 in the exponent is a factor of 13 in rate. The line reads trends; the integral computes lifetimes.

An alpha carrying orbital angular momentum l adds ħ²l(l+1)/2μr² to the barrier. The alpha itself is 0⁺, so l is fixed by the level scheme: it must supply the whole spin change, and parity requires πₚₐᵣₑₙₜ = π_daughter(−1)^l, which forbids even l between states of opposite parity. At the contact radius of ²²⁶Ra that extra term is (ħc)²l(l+1)/(2μc²R²) = 0.063 l(l+1) MeV — just 0.38 MeV for l = 2, sitting on top of 27 MeV of Coulomb. Perturbing the WKB integral gives δ(2γ) = 0.53 for l = 2 and 2.6 for l = 5, so hindrance factors of only 1.7 and 14. Measured hindrance for high-l or odd-A transitions often runs to 10² or 10³, far more than the centrifugal term can explain. That excess is a statement about preformation and structure — an odd nucleon has to be rearranged before an alpha exists at all — and not about the barrier. Blaming it on the centrifugal wall is a common and wrong reflex.