Alpha Decay and Quantum Tunnelling

Introduction

Some heavy nuclei can lower their total mass-energy by emitting an alpha particle, yet one may decay quickly while another survives for geological times. The difference is not explained by the Q-value sign alone. An alpha-like group must escape the nuclear potential through a repulsive Coulomb barrier. Quantum tunnelling gives it a small but nonzero chance, and that chance responds dramatically to the alpha energy.

Core explanation

In alpha decay , a parent with mass number A and proton number Z produces a daughter with A − 4 and Z − 2 plus an alpha particle, a ⁴₂He nucleus. For neutral atomic masses, Qα ≈ [Mparent − Mdaughter − M(⁴He atom)]c². When Q is positive, energy is available to the alpha particle and daughter recoil. A positive Q does not mean the alpha can simply climb over the electrostatic barrier by classical motion. The nuclear force strongly attracts nucleons at short range; beyond that range, the positively charged alpha and daughter repel each other.

The potential-energy picture has a deep nuclear well and an outer Coulomb barrier . An alpha-like cluster inside the nucleus has an energy below the top of that barrier. Classical mechanics would forbid escape. Quantum mechanics assigns the cluster a wavefunction with nonzero amplitude through a finite-width forbidden region. If it emerges beyond the outer turning point, Coulomb repulsion accelerates it away. OpenStax's tunnelling treatment uses this picture to explain alpha emission and the strong dependence on barrier thickness.

For a simplified electrostatic model outside the nuclear radius, potential energy decreases roughly as the inverse of distance. An alpha with higher available energy meets the outer turning point closer to the nucleus, so it tunnels through a narrower region . Tunnelling probability depends exponentially on the integral of the barrier's forbidden momentum, making a modest energy change produce a very large decay-rate change. This is why it is incorrect to estimate alpha half-life by saying that twice the Q-value gives twice the speed. The relationship is much more sensitive and also depends on nuclear structure.

The Geiger–Nuttall relationship captures a broad empirical pattern within related alpha-emitting families: higher alpha-particle energy tends to correspond to a shorter half-life, often described by a roughly linear relation between log half-life and inverse square root of alpha energy for a fixed or similar daughter charge. It follows qualitatively from Coulomb-barrier tunnelling. One should not fit all alpha emitters to one universal straight line because daughter charge, orbital angular momentum, alpha preformation and shell effects shift the pattern. The relation describes a rate trend, not a new conservation law.

Two probabilities contribute to a simplified decay constant: formation of an alpha-like cluster inside the parent and penetration of the barrier on an attempted escape. The barrier is central, but preformation and angular-momentum barriers can vary among nuclei. If the daughter is left excited, some Q is spent on excitation and the alpha emerges with less kinetic energy. A later gamma transition may release that excitation. The IAEA decay-data review distinguishes alpha kinetic energy, daughter excitation and recoil in the alpha-energy balance.

The emitted alpha energy is not exactly Q. For a two-body decay of a parent initially at rest, alpha and daughter recoil have equal momentum magnitudes. Their kinetic energies are inversely related to mass in the nonrelativistic approximation, so the lighter alpha receives most of Q. If daughter mass is roughly 50 times alpha mass, the alpha takes about 50/51 of the kinetic energy and the daughter about 1/51. This matters when comparing a measured alpha line with a mass-table Q-value.

Alpha particles deposit energy densely over a short path in matter, but their environmental or biological impact requires a separate exposure question. The nuclear mechanism explains emission rate; detector response and dose depend on what material is irradiated and whether the alpha source is outside or inside the body. Do not use alpha half-life or energy alone as a complete safety assessment.

Step-by-step reasoning

For a proposed alpha emitter, first balance A and Z to identify the daughter. Calculate Q from consistent masses to establish energetic permission. Sketch the nuclear well and Coulomb barrier, marking the alpha energy below the barrier top and the two turning points. Predict qualitatively that a higher alpha energy usually narrows the barrier and raises penetrability, shortening half-life within comparable nuclides. Then qualify the trend with daughter charge, shell structure and cluster preformation before discussing a specific measured lifetime.

Visual explanation

Plot potential energy vertically against alpha–daughter separation horizontally. Draw a deep well inside radius R, a peak at the nuclear surface and a falling Coulomb curve outside. Add a horizontal alpha-energy line below the peak; shade the forbidden region between the inner and outer turning points. A second, slightly higher energy line meets the outer curve sooner, so its shaded region is narrower. Label the transmitted wave on the far side “small but nonzero probability,” not “particle borrowed energy.”

Real-world analogy

A ball trapped behind a hill cannot leave if its energy is below the crest in classical physics. A quantum particle differs: its wavefunction can extend through a finite barrier. The analogy helps identify why ordinary escape seems impossible, but a literal ball does not tunnel, and the alpha does not temporarily violate conservation of energy. Energy conservation holds before and after decay.

Real-world example

Imagine two related heavy nuclides with positive alpha-decay Q-values, one producing a 4 MeV alpha and another a 6 MeV alpha. The higher-energy alpha faces a thinner Coulomb barrier and is generally more likely to escape per opportunity, so a shorter half-life is plausible. Without their charges, preformation factors and measured level schemes, the energy comparison alone cannot give a numerical half-life ratio.

Why?

Why does a positive Q not guarantee rapid alpha emission? Q compares initial and final rest energies, whereas decay rate depends on the probability of reaching those final products. The alpha-like cluster remains confined by an effective barrier even when the final state is energetically lower. Tunnelling can be exceedingly rare, producing long half-lives for energetically allowed decays.

Common misconception

“The alpha particle borrows energy to cross the barrier.” Quantum tunnelling is not an exception to energy conservation; the particle's state has nonzero transmission amplitude through a finite forbidden region at its actual energy. Another misconception says a measured alpha kinetic energy equals Q exactly. Daughter recoil, and sometimes daughter excitation, take part of the released energy.

Worked example

An alpha decay to a ground-state daughter releases Q = 5.00 MeV. Approximate the daughter's mass as 200 u and the alpha mass as 4 u. Two-body momentum sharing gives Tα ≈ Q × Md/(Md + Mα) = 5.00 × 200/204 ≈ 4.90 MeV . Daughter recoil receives about 0.098 MeV . The small recoil energy is real; using 5.00 MeV as the exact measured alpha line would overstate it under this approximation.

Quick check

1. Why does a higher alpha energy usually shorten half-life within a related nuclide family? Answer: Higher energy moves the outer turning point inward, narrows the Coulomb barrier and greatly increases quantum-tunnelling probability.

Exam focus

Separate Q-value from decay rate. Write the daughter change A − 4 and Z − 2 correctly. Use a barrier diagram to explain tunnelling and the Geiger–Nuttall trend; do not present one universal slope for all nuclei. In energy calculations include daughter recoil and possible excitation. Avoid describing tunnelling as an energy-conservation violation.

Advanced insight

In a semiclassical WKB picture, barrier penetration is approximately exponential in minus twice an integral across the forbidden region. The integrand depends on the square root of the gap between potential energy and alpha energy. This gives the dramatic energy sensitivity underlying Geiger–Nuttall plots. Changes in orbital angular momentum add a centrifugal barrier, and shell closures can change alpha preformation, producing departures from a simple line even for similar Q-values.

Summary

Alpha decay can be energetically allowed while remaining slow because an alpha-like cluster must tunnel through a Coulomb barrier. Higher alpha energy usually means a narrower barrier and shorter half-life, expressed empirically by the Geiger–Nuttall trend within related nuclei. Preformation, daughter charge, angular momentum, recoil and excitation also matter. Energy conservation is maintained throughout.

Practice questions

1. What are the daughter A and Z after alpha decay of ²³⁸₉₂U? Answer: A becomes 234 and Z becomes 90, giving ²³⁴₉₀Th plus ⁴₂He.

2. Why can two positive-Q alpha decays have very different half-lives? Answer: Their barrier widths, daughter charges, alpha preformation probabilities and angular-momentum constraints can differ greatly.

3. Does the alpha particle receive all of Q in a ground-state two-body decay? Answer: No. The recoiling daughter carries some kinetic energy to conserve momentum.

4. What does the Geiger–Nuttall relationship connect qualitatively? Answer: Within comparable alpha emitters, greater alpha energy is generally associated with a faster decay and shorter half-life.