University Physics IV · Nuclear Physics · 13.4
Alpha Decay as Barrier Penetration
Energetics says almost every nucleus heavier than samarium ought to spit out an alpha. Kinetics says most of them will not do it while you watch. This topic is where you learn to compute the gap between those two statements: one barrier integral, one exponential, and half-lives running from nanoseconds to ten billion years.
Build the model
Connect the measurement to the mechanism.
An alpha inside a heavy nucleus is held by the strong force out to a sharp edge near R = 1.2(Ad^⅓ + 4^⅓) fm, and outside that edge it feels nothing but the Coulomb repulsion of the daughter's charge, 2Zd e²/4πε₀r. That potential rises to about 27 MeV at the edge of ²²⁶Ra, while the decay releases only 4.87 MeV — so the alpha is born 22 MeV under a wall it has no classical way past, and must reappear 50.9 fm out, where the Coulomb curve finally falls to its own energy. WKB prices that crossing with a single integral between the turning points, and the answer arrives as an exponent: 2γ = 72.8, a transmission of about 2 in 10³² per approach.
Multiply by an assault frequency of order 10²¹ s⁻¹ and you have a rate. The model's power and its cost sit in the same place. Because 2γ carries Zd/√Q, a factor of 2.2 in Q moves half-lives across 24 orders of magnitude — the Geiger-Nuttall line, predicted rather than fitted.
But everything outside the exponent — the assault frequency, the sharp radius, and above all the probability that four nucleons were an alpha in the first place — is folded into one preformation factor of order 0.1 to 1 that this model fits rather than derives.
- Simple definition
- Alpha decay is a tunnelling problem: an alpha already inside the nucleus escapes through the Coulomb barrier its own charge builds with the daughter's, and its half-life is set almost entirely by the exponential of a WKB integral across the classically forbidden region.
- Example
- For ²²⁶Ra the alpha must get from the nuclear edge at 9.17 fm out to 50.85 fm, carrying 4.87 MeV under a barrier topping 27.0 MeV. The WKB exponent is 2γ = 72.8, so about 2 approaches in 10³² succeed — at 8×10²⁰ approaches a second, a half-life near a thousand years.
²²⁶Ra: 226.025410 − 222.017578 − 4.002603 = 0.005229 u, so Q = 4.871 MeV. A positive Q licenses the decay and says nothing at all about when.
Atomic masses in u, ×931.494 MeV/u; the Z electrons cancel because (Z−2) + 2 = Z.
²²⁶Ra: 4.8708 × 222/226 = 4.785 MeV, the line a detector sees; ²²²Rn takes the other 86 keV. Compare spectra with Tα, never with Q.
A is the parent mass number; the daughter recoil keeps the other 4Q/A.
²²⁶Ra: R = 9.17 fm, V(R) = 27.0 MeV, b = 50.85 fm. R fixes the barrier top; Q alone fixes how far out the alpha has to reappear.
Zd and Ad are the daughter's; r, R and b in fm; V and Q in MeV.
Transmission per approach is e(−2γ). For ²²⁶Ra, 2γ = 72.8 and e(−72.8) = 2.4 × 10⁻³². Every large number in this topic lives inside that exponent.
μ = mα md/(mα + md) ≈ 3.93 u ≈ 3660 MeV/c²; ħc = 197.33 MeV fm; γ is dimensionless.
²²⁶Ra: f = 8.4 × 10²⁰ s⁻¹ gives t½ = 1.1 × 10³ yr against 1.6 × 10³ measured, so P ≈ 0.7. P absorbs structure and your choice of R together.
P is the dimensionless preformation factor, fitted; f in s⁻¹; t½ = ln2/λ.
The point-Coulomb limit of the previous card. The slope is predicted, not fitted, and lands within 6% of Viola-Seaborg's 1.66; Q from 4 to 9 MeV spans 10²³ in t½.
Q in MeV, t½ in s. The intercept c ≈ −50 for the actinides, and it swallows R, f and P.
Q from the mass table — and the alpha does not keep it all
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: Tα = 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.
The barrier is Coulomb outside a sharp edge
Beyond the range of the strong force the alpha and daughter see only each other's charge: V(r) = 2Zd e²/4πε₀r = 2Zd(1.44 MeV fm)/r. Inside, model the well as flat. The join is the contact radius R ≈ 1.2(Ad^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 = 2Zd(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 their closest approach is 28 fm, three times the nuclear radius, which is exactly why Rutherford's scattering stayed pure Coulomb.
WKB turns the whole barrier into one number
Inside a classically forbidden region the wavefunction decays instead of oscillating, and WKB gives the transmission as e(−2γ) with γ = (1/ħ)∫ from R to b of √(2μ(V(r) − Q)) dr. Use the reduced mass μ = mα md/(mα + md) = 3.93 u = 3662 MeV/c², not mα, 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.
A probability per approach is not a rate
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/mαc²) = 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.
Expand the integral and Geiger-Nuttall drops out
For x ≪ 1, arccos√x − √(x(1−x)) ≈ π/2 − 2√x, which splits 2γ into two terms: π√(2μ)⋅(2Zd e²/4πε₀)/(ħ√Q), which carries all of the Q dependence and none of R, minus 4√(2μ⋅(2Zd e²/4πε₀)⋅R)/ħ, which carries R and no Q. Divide by ln10 and you have log₁₀(t½/s) ≈ 1.70 Zd/√Q + c: a straight line against Q(−1/2) whose slope is predicted rather than fitted, and which sits within 6% 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.
The centrifugal term hinders, but less than you expect
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 = πdaughter(−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.
Change one variable at a time
Make the relationship visible.
Drag Q from 4.0 to 9.0 MeV and watch b sweep in from 62 fm to 28 fm while the half-life readout falls from 10¹⁷ s to 10⁻⁶ s. Then nudge R by 0.5 fm: the same readout moves by a factor of seven, which is why the preformation factor has to be fitted rather than derived.
BARRIER TOP V(R)26.9 MeV
OUTER TURNING POINT b51.1 fm
GAMOW EXPONENT 2γ73.0
log10 HALF-LIFE IN s10.6
Live interpretationBARRIER TOP V(R): 26.9 MeV. OUTER TURNING POINT b: 51.1 fm. GAMOW EXPONENT 2γ: 73.0. log10 HALF-LIFE IN s: 10.6
Catch the common trap
Explain before calculating.
Two even-even alpha emitters have nearly the same daughter charge, Zd ≈ 84, and nearly the same contact radius, R ≈ 9.1 fm. One releases Q = 4.0 MeV, the other Q = 8.0 MeV. Roughly what does that factor of two in Q do to the half-life?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
Easy²²⁶Ra α-decays to ²²²Rn. Using atomic masses M(²²⁶Ra) = 226.025410 u, M(²²²Rn) = 222.017578 u and M(⁴He) = 4.002603 u, with 1 u = 931.494 MeV/c², find Q and the kinetic energy the alpha actually carries away.
- Atomic masses may be subtracted directly here: the parent carries 88 electrons and the products carry 86 + 2 = 88, so they cancel to within their own binding energies, a few tens of eV.
- Sum the products: 222.017578 + 4.002603 = 226.020181 u. Mass lost: 226.025410 − 226.020181 = 0.005229 u.
- Convert: Q = 0.005229 × 931.494 = 4.871 MeV. Positive, so the decay is energetically allowed.
- The two fragments leave back to back with equal and opposite momentum, so kinetic energy divides inversely with mass: Tα = Q(A − 4)/A = 4.8708 × 222/226 = 4.785 MeV.
- The ²²²Rn recoil keeps the remainder: 4.871 − 4.785 = 0.086 MeV, or 86 keV.
AnswerQ = 4.871 MeV; the alpha carries 4.785 MeV and the ²²²Rn recoil 86 keV. The measured principal line of ²²⁶Ra is 4.784 MeV, so comparing a spectrum against Q rather than Tα would be wrong by 86 keV.
Medium²³⁸U α-decays to ²³⁴Th with Q = 4.270 MeV. Take R = 1.2(Ad^⅓ + 4^⅓) fm and e²/4πε₀ = 1.440 MeV fm. Find the contact radius, the barrier height there, the outer turning point, and the width of the classically forbidden region. What alpha beam energy would be needed to touch that ²³⁴Th surface head-on?
- Contact radius: 234^⅓ = 6.162 and 4^⅓ = 1.587, so R = 1.2 × 7.749 = 9.30 fm.
- Coulomb strength: the daughter has Zd = 90, so 2Zd e²/4πε₀ = 2 × 90 × 1.440 = 259.2 MeV fm.
- Barrier top: V(R) = 259.2/9.30 = 27.9 MeV. The alpha's 4.270 MeV is 15% of that, so it starts 23.6 MeV under the wall.
- Outer turning point, where V(b) = Q: b = 259.2/4.270 = 60.7 fm.
- Forbidden width: b − R = 60.7 − 9.30 = 51.4 fm, about 5.5 times the contact radius. The alpha has to cross that, not climb the 23.6 MeV.
- Read the same barrier from outside: a beam alpha needs 27.9 MeV to reach r = R head-on, and Rutherford's 8 MeV alphas stop at 259.2/8 = 32.4 fm, nowhere near the surface.
AnswerR = 9.30 fm, V(R) = 27.9 MeV, b = 60.7 fm, and the forbidden region is 51.4 fm wide. A beam would need 27.9 MeV to touch the surface; the escaping alpha manages it with 4.27 MeV by going through rather than over.
Hard²¹²Po α-decays to ²⁰⁸Pb with Q = 8.954 MeV. Take R = 9.01 fm, μ = 3.927 u = 3658 MeV/c², ħc = 197.33 MeV fm and mαc² = 3727.4 MeV. Estimate the WKB exponent, the transmission, the assault frequency and the half-life, then compare with the measured 0.299 µs and quote the preformation factor.
- Turning point and ratio: b = 2 × 82 × 1.440/8.954 = 236.2/8.954 = 26.38 fm, so x = R/b = 9.01/26.38 = 0.3416.
- Wavenumber scale: √(2μc²Q)/ħc = √(2 × 3658 × 8.954)/197.33 = √65507/197.33 = 255.9/197.33 = 1.297 fm⁻¹.
- Bracket: √x = 0.5846, arccos 0.5846 = 0.9459 rad, and √(x(1 − x)) = √(0.3416 × 0.6584) = 0.4743, so the bracket is 0.9459 − 0.4743 = 0.4716.
- Exponent: 2γ = 2 × 1.297 × 26.38 × 0.4716 = 32.27, so the transmission is e(−32.27) = 9.7 × 10⁻¹⁵ per approach — seventeen orders of magnitude easier than ²²⁶Ra's 2.4 × 10⁻³², for a Q less than twice as large.
- Assault frequency: v = c√(2 × 8.954/3727.4) = 0.0693c = 2.08 × 10²² fm/s, so f = v/2R = 2.08 × 10²²/18.03 = 1.15 × 10²¹ s⁻¹.
- Rate and half-life: λ = f e(−2γ) = 1.15 × 10²¹ × 9.7 × 10⁻¹⁵ = 1.12 × 10⁷ s⁻¹, so t½ = ln2/λ = 6.2 × 10⁻⁸ s = 62 ns. Measured 299 ns, so the model runs 4.8 times fast and P = 0.21.
Answer2γ = 32.3, transmission 9.7 × 10⁻¹⁵, t½ ≈ 62 ns against a measured 299 ns, so P ≈ 0.2. The same recipe on ²³²Th (Q = 4.083 MeV) gives 2γ = 89 and t½ ≈ 1.4 × 10¹⁰ yr — 24 orders away, on 2.2 times less Q.