Case Study 2 — Tunneling Through the Coulomb Barrier: Why Alpha Decay and Fusion Are Possible

The Classical Impossibility

Consider an alpha particle ($Z_\alpha = 2$) inside a $^{238}$U nucleus ($Z_\text{daughter} = 90$). The Coulomb barrier at the nuclear surface ($R \approx 7.4$ fm) has a height of:

$$V_C = \frac{Z_\alpha Z_d e^2}{4\pi\epsilon_0 R} = \frac{2 \times 90 \times 1.44\ \text{MeV}\cdot\text{fm}}{7.4\ \text{fm}} \approx 35\ \text{MeV}$$

The alpha particle emerges with only $E_\alpha = 4.27$ MeV. Classically, a particle with 4.27 MeV of kinetic energy cannot escape a 35 MeV barrier. It would be trapped forever.

Yet $^{238}$U decays with a half-life of $4.47 \times 10^9$ years, and we measure the emitted alpha particles. This was one of the great early triumphs of quantum mechanics: George Gamow, and independently Ronald Gurney and Edward Condon, showed in 1928 that quantum tunneling resolves the paradox.

Now consider the reverse situation: at the center of the Sun, protons with an average thermal energy of $\langle E \rangle = \frac{3}{2}k_BT \approx 1.3$ keV must overcome a Coulomb barrier of approximately 550 keV to fuse. The ratio $E/V_C \approx 0.002$ means that classically, no proton-proton fusion would occur, and the Sun would not shine.

Both problems — alpha decay and stellar fusion — are solved by the WKB tunneling formula derived in Section 5.7. This case study develops both applications in detail and previews the full treatments in Chapters 13 and 21.

The key insight is that quantum tunneling is not merely a curiosity of quantum mechanics — it is the mechanism that controls some of the most important processes in the physical universe. The existence of long-lived radioactive isotopes, the possibility of stellar energy production at achievable temperatures, and even the limits of the periodic table (where the fission barrier becomes too thin to prevent spontaneous fission) are all governed by the exponential sensitivity of quantum tunneling to barrier parameters.

Part 1: Alpha Decay Through the Coulomb Barrier

Setting Up the Problem

Model the nuclear-plus-Coulomb potential seen by the alpha particle as:

$$V(r) = \begin{cases} -V_0 & r < R \quad (\text{attractive nuclear potential}) \\ \displaystyle\frac{Z_\alpha Z_d e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2 l(l+1)}{2\mu r^2} & r > R \quad (\text{Coulomb + centrifugal}) \end{cases}$$

where $\mu$ is the reduced mass of the alpha-daughter system, and $l$ is the orbital angular momentum of the emitted alpha particle. For $l = 0$ (the most probable case for ground-state-to-ground-state transitions), the centrifugal term vanishes.

The alpha particle, with energy $E_\alpha$, is trapped in the attractive potential well ($r < R$) and must tunnel through the Coulomb barrier from $r = R$ to the outer classical turning point $r = R_c = Z_\alpha Z_d e^2 / (4\pi\epsilon_0 E_\alpha)$.

The WKB Tunneling Probability

The tunneling probability is:

$$T = \exp\left(-\frac{2}{\hbar}\int_R^{R_c} \sqrt{2\mu\left[\frac{Z_\alpha Z_d e^2}{4\pi\epsilon_0 r} - E_\alpha\right]}\, dr\right) = e^{-2\gamma}$$

Evaluating the integral (substituting $r = R_c \sin^2\theta$):

$$2\gamma = 2\eta\left[\arccos\sqrt{\rho} - \sqrt{\rho(1-\rho)}\right]$$

where $\eta = Z_\alpha Z_d e^2/(4\pi\epsilon_0\hbar v_\alpha)$ is the Sommerfeld parameter, $v_\alpha = \sqrt{2E_\alpha/\mu}$, and $\rho = R/R_c$.

Numerical Example: The Uranium Series

Let us work through three alpha emitters in the $^{238}$U decay series to see the extraordinary sensitivity of tunneling to energy:

$^{238}$U: $E_\alpha = 4.27$ MeV, $Z_d = 90$, $A_d = 234$

  • Reduced mass: $\mu = 4 \times 234 / (4 + 234) \times 931.5 = 3667$ MeV$/c^2$
  • $v_\alpha/c = \sqrt{2 \times 4.27/3667} = 0.0483$
  • $\eta = 2 \times 90 \times 1.44 / (197.3 \times 0.0483) = 27.2$
  • $R \approx 1.2(4^{1/3} + 234^{1/3}) = 9.01$ fm
  • $R_c = 2 \times 90 \times 1.44 / 4.27 = 60.7$ fm
  • $\rho = R/R_c = 0.148$
  • $2\gamma = 2 \times 27.2 \times [\arccos(0.385) - 0.385 \times 0.923] = 2 \times 27.2 \times [1.175 - 0.355] = 44.6$
  • $T \approx e^{-44.6} \approx 4 \times 10^{-20}$

With an assault frequency $\nu \approx v_\alpha/(2R) \approx 0.0483 \times 3 \times 10^{23}\ \text{fm/s} / (2 \times 9.01\ \text{fm}) \approx 8 \times 10^{20}$ s$^{-1}$:

$$\lambda = \nu T \approx 8 \times 10^{20} \times 4 \times 10^{-20} \approx 32\ \text{s}^{-1}$$

Wait — this gives $t_{1/2} = \ln 2 / \lambda \approx 0.02$ s, far shorter than the observed $4.47 \times 10^9$ years. The discrepancy comes from the alpha preformation factor $P_\alpha \sim 10^{-2}$ and corrections to the simple WKB estimate. When these are included (Chapter 13), the agreement is excellent.

The more instructive comparison is between different isotopes, where the preformation factor approximately cancels:

$^{218}$Po: $E_\alpha = 6.00$ MeV, $Z_d = 82$

  • $\eta = 22.7$, $\rho = 0.199$
  • $2\gamma \approx 33.4$
  • $T \approx e^{-33.4} \approx 3 \times 10^{-15}$
  • Predicted $t_{1/2} \approx 3$ min; measured: 3.10 min

$^{212}$Po: $E_\alpha = 8.78$ MeV, $Z_d = 82$

  • $\eta = 18.8$, $\rho = 0.262$
  • $2\gamma \approx 24.5$
  • $T \approx e^{-24.5} \approx 2 \times 10^{-11}$
  • Predicted $t_{1/2} \approx 0.2\ \mu$s; measured: 0.30 $\mu$s

The ratio of tunneling probabilities: $T(^{212}\text{Po})/T(^{238}\text{U}) \approx e^{44.6 - 24.5} = e^{20.1} \approx 5 \times 10^{8}$. This factor of half a billion in tunneling probability, arising from a factor of two in alpha particle energy, is the essence of the Geiger-Nuttall law.

Part 2: Fusion Through the Coulomb Barrier — The Gamow Peak

The Solar Fusion Problem

At the solar core ($T \approx 1.5 \times 10^7$ K, $k_BT \approx 1.3$ keV), protons must fuse via:

$$p + p \to d + e^+ + \nu_e$$

The Coulomb barrier for two protons at their mutual nuclear surface ($R \approx 1.2 \times 2 \approx 2.4$ fm) is:

$$V_C = \frac{e^2}{4\pi\epsilon_0 R} = \frac{1.44\ \text{MeV}\cdot\text{fm}}{2.4\ \text{fm}} \approx 550\ \text{keV}$$

The average thermal energy $\langle E \rangle \approx 1.3$ keV is a factor of $\sim 400$ below the barrier. The tunneling probability at this energy is absurdly small — about $e^{-2\pi\eta}$ with $\eta \approx 22.7/\sqrt{E[\text{MeV}]} \approx 630$, giving $T \sim e^{-4000} \approx 10^{-1700}$. Obviously, fusion does not occur at the average thermal energy.

The Gamow Window

The fusion cross section at energy $E$ can be written:

$$\sigma(E) = \frac{S(E)}{E} \exp\left(-\frac{b}{\sqrt{E}}\right)$$

where $S(E)$ is the astrophysical $S$-factor (a slowly varying function of energy that contains the nuclear physics) and $b = \pi e^2 \sqrt{2\mu}/(4\pi\epsilon_0\hbar) = 2\pi\eta\sqrt{E}$. For $p + p$: $b \approx 22.7$ MeV$^{1/2}$ (using $\mu = m_p/2$).

The reaction rate per pair of particles is:

$$\langle\sigma v\rangle = \left(\frac{8}{\pi\mu}\right)^{1/2} \frac{1}{(k_BT)^{3/2}} \int_0^\infty S(E) \exp\left(-\frac{E}{k_BT} - \frac{b}{\sqrt{E}}\right) dE$$

The integrand is the product of two exponentials:

  1. Maxwell-Boltzmann factor $e^{-E/k_BT}$: decreases with increasing $E$. At the solar core, the $e$-folding energy is $k_BT \approx 1.3$ keV.

  2. Gamow penetration factor $e^{-b/\sqrt{E}}$: increases with increasing $E$ (tunneling becomes easier at higher energy).

Their product has a sharp peak — the Gamow peak — at the energy:

$$E_0 = \left(\frac{b\, k_BT}{2}\right)^{2/3} = \left(\frac{22.7 \times 10^{-3} \times 1.3 \times 10^{-3}}{2}\right)^{2/3}\ \text{MeV} \approx 5.9\ \text{keV}$$

The width of the Gamow peak is approximately:

$$\Delta = \frac{4}{\sqrt{3}}\sqrt{E_0 k_B T} \approx 6.4\ \text{keV}$$

What the Numbers Mean

At the Gamow peak energy $E_0 \approx 5.9$ keV:

  • The Sommerfeld parameter: $\eta = 22.7 / (2\sqrt{5.9 \times 10^{-3}}) \approx 148$
  • The tunneling probability: $T \approx e^{-2\pi \times 148} \approx e^{-929} \approx 10^{-403}$

This is still incredibly small. But the Sun contains $\sim 10^{57}$ protons, and there are $\sim 10^{57}$ proton-proton pairs, and they are colliding $\sim 10^{15}$ times per second each. The product of these enormous numbers with the tiny tunneling probability gives the observed solar luminosity of $L_\odot = 3.83 \times 10^{26}$ W.

Quantitatively, the $pp$ reaction rate at the solar core produces about $3.6 \times 10^{38}$ fusion reactions per second. Each proton waits, on average, about $10^{10}$ years before fusing — comparable to the Sun's main-sequence lifetime. This remarkably slow rate is what makes the Sun a stable, long-lived star rather than a bomb.

Temperature Sensitivity

The reaction rate scales with temperature as:

$$\langle\sigma v\rangle \propto T^{n} \quad \text{where} \quad n = \frac{E_0}{3k_BT} - \frac{2}{3} \approx \frac{b^{2/3}}{3(k_BT)^{1/3}} - \frac{2}{3}$$

For the $pp$ reaction at solar core temperatures, $n \approx 3.9$. This means a 10% increase in temperature produces a $\sim$46% increase in the reaction rate. For the CNO cycle (which involves higher-$Z$ nuclei and therefore larger Coulomb barriers), $n \approx 16$--$18$, making the rate exquisitely sensitive to temperature.

This temperature sensitivity explains why massive stars (which have hotter cores) burn their hydrogen much faster than the Sun, and why the CNO cycle dominates over the $pp$ chain in stars with $M \gtrsim 1.3 M_\odot$.

The Astrophysical $S$-Factor

Nuclear astrophysicists define the astrophysical $S$-factor by extracting the dominant energy dependences from the cross section:

$$\sigma(E) = \frac{S(E)}{E}\exp\left(-2\pi\eta\right)$$

The $1/E$ factor accounts for the quantum mechanical cross section of a point particle (proportional to the de Broglie wavelength squared, $\lambda^2 \propto 1/E$), and the exponential accounts for the Gamow penetration factor. What remains, $S(E)$, is a slowly varying function that contains the purely nuclear physics — the matrix element in Fermi's golden rule language. The advantage of $S(E)$ is that it can be extrapolated to astrophysically relevant energies (typically below what can be measured in the laboratory) with reasonable confidence, because the dramatic energy dependence has already been factored out.

For the $pp$ reaction, $S(0) \approx 4.0 \times 10^{-25}$ MeV$\cdot$barn. This tiny value reflects the fact that the $pp$ reaction involves the weak interaction (converting a proton to a neutron), not just Coulomb barrier penetration. The $pp$ reaction is doubly suppressed: by tunneling through the Coulomb barrier and by the weakness of the weak force. It is this double suppression that makes the Sun burn so slowly and live so long.

The Big Picture: Tunneling as Nature's Regulator

Quantum tunneling through the Coulomb barrier regulates two of the most fundamental processes in nature:

  1. Radioactive decay — tunneling out of the nucleus. The extreme sensitivity of tunneling to energy and barrier parameters creates the enormous range of decay lifetimes that makes nuclear physics rich and complex. Without tunneling, no alpha decay would occur, and the heavy elements would be truly eternal.

  2. Stellar fusion — tunneling into the nucleus. Without tunneling, thermonuclear fusion would require temperatures of $\sim 10^{10}$ K (where thermal energies match the Coulomb barrier), and stars as we know them could not exist. Quantum tunneling makes stars possible at temperatures a thousand times lower, allowing the long, stable main-sequence lifetimes that gave life time to evolve.

Both phenomena are quantitatively described by the WKB approximation applied to the Coulomb barrier. The formula $T \sim \exp(-2\gamma)$ is the same in both cases; only the direction of tunneling differs.

In subsequent chapters, we will refine these estimates with: - Proper treatment of the centrifugal barrier and angular momentum (Chapter 13) - Nuclear structure effects on alpha preformation (Chapter 13) - Electron screening in stellar plasmas (Chapter 21) - The complete network of fusion reactions in stars (Chapter 21) - Resonant tunneling through barrier states, where the penetrability is dramatically enhanced (Chapter 18)

But the essential physics — the exponential sensitivity of tunneling to the Gamow factor, and the interplay between tunneling probability and thermal population — is all contained in the WKB result derived in this chapter.

A Third Application: The Fission Barrier

Quantum tunneling through the fission barrier is the third great application of the WKB approximation in nuclear physics (Chapter 20). The fission barrier arises from the competition between the Coulomb repulsion (which favors splitting the nucleus) and the nuclear surface tension (which favors a compact shape). For actinide nuclei like $^{236}$U (the compound nucleus formed when $^{235}$U captures a neutron), the fission barrier is approximately 5--6 MeV high. Neutron-induced fission occurs when the excitation energy from neutron capture exceeds the barrier — or, for sub-barrier excitation, by quantum tunneling through the remaining barrier.

Spontaneous fission (no external neutron required) occurs entirely by tunneling, with half-lives ranging from $\sim 10^{16}$ years ($^{238}$U) to milliseconds for superheavy nuclei near $Z = 114$--118. As the proton number increases, the Coulomb repulsion grows (proportional to $Z^2$) while the surface energy grows more slowly (proportional to $A^{2/3}$), so the fission barrier shrinks. Beyond $Z \approx 120$--125, the liquid-drop fission barrier vanishes entirely, and only quantum shell corrections (the predicted "island of stability") create a residual barrier. The existence or nonexistence of long-lived superheavy elements is, ultimately, a question about WKB tunneling through a shell-stabilized barrier.

Discussion Questions

  1. The Geiger-Nuttall law ($\log t_{1/2}$ linear in $1/\sqrt{E_\alpha}$) works well within a single isotopic chain (e.g., the polonium isotopes) but breaks down when comparing across different element series. What nuclear structure effects might cause deviations from the simple WKB prediction?

  2. If the fine structure constant $\alpha = e^2/(4\pi\epsilon_0\hbar c) \approx 1/137$ were twice as large, how would alpha decay half-lives change? How would stellar fusion rates change? Would stars still exist?

  3. The $pp$ reaction rate scales as $T^{3.9}$ while the CNO cycle rate scales as $T^{16-18}$. Explain why higher-$Z$ participants lead to stronger temperature sensitivity, using the Gamow peak analysis.

  4. Resonant tunneling (where a quasi-bound state exists within or near the barrier) can enhance the penetrability by orders of magnitude. Can you think of an analogy from everyday wave physics where a barrier becomes transparent at specific frequencies?