With what probability does nuclear fusion occur at energies far below the Coulomb barrier?

Question

Even at the core of the sun, the temperature of $\sim 10^7$ K only results in $kT\sim1$ keV, which is about a thousand times less than the electrical potential energy of $\sim1$ MeV needed in order to bring two hydrogen nuclei to within the ~1 fm range of the strong nuclear force. Therefore nuclear fusion reactions can only occur inside the sun, or in any other normal star, through the process of quantum-mechanical tunneling. The low probability of this tunneling, along with the need for a weak interaction in order to fuse two protons into a deuterium nucleus, are the two factors that make stars have lifetimes billions of years long.

How is the tunneling probability calculated?

Answer 1

The WKB approximation states that in one dimension, the tunneling probability $P$ can be approximated as

$\ln P=-\frac{2\sqrt{2m}}{\hbar}\int_a^b \sqrt{V-E} dx$ ,

where the limits of integration $a$ and $b$ are the classical turning points, $m$ is the reduced mass, the electrical potential $V$ is a function of $x$, and $E$ is the total energy. Setting $V=kq_1q_2/x$, we have for the integral

$I=\int_a^b \sqrt{V-E} dx$

$=\frac{kq_1k_2}{\sqrt{E}}\int_{A}^1\sqrt{u^{-1}-1} du$ ,

where $A=a/b$. The indefinite integral equals $-u\sqrt{u^{-1}-1}+\tan^{-1}\sqrt{u^{-1}-1}$, and for $A\ll 1$ the definite integral is then $\pi/2$. The result is

$\ln P=-\frac{\pi kq_1q_2}{\hbar}\sqrt{\frac{2m}{E}} $ .

This result was obtained in Gamow 1938, and $G=-(1/2)\ln P$ is referred to as the Gamow factor or Gamow-Sommerfeld factor.

The fact that the integral $\int_A^1\ldots$ can be approximated as $\int_0^1\ldots$ tells us that the right-hand tail of the barrier dominates, i.e., it is hard for the nuclei to travel through the very long stretch of $\sim 1$ nm over which the motion is only mildly classically forbidden, but if they can do that, it's relatively easy for them to penetrate the highly classically forbidden region at $x\sim1$ fm. Surprisingly, the result can be written in a form that depends only on $m$ and $E$, but not on $a$, i.e., we don't even have to know the range of the strong nuclear force in order to calculate the result.

The generic WKB expression depends on $E$ through an expression of the form $V-E$, which might have led us to believe that with a 1 MeV barrier, it would make little difference whether $E$ was 1 eV or 1 keV, and fusion would be just as likely in trees and houses as in the sun. But because the tunneling probability is dominated by the tail of the barrier, not its peak, the final result ends up depending on $1/\sqrt{E}$.

Because $P$ increases extremely rapidly as a function of $E$, fusion is dominated by nuclei whose energies lie in the tails of the Maxwellian distribution. There is a narrow range of energies, known as the Gamow window, in which the product of $P$ and the Maxwell distribution is large enough to contibute significantly to the rate of fusion.

Gamow and Teller, Phys. Rev. 53 (1938) 608

Answer 2

$ E_{out} = 4 \pi n_1 n_2 (m_0-m_a) c^2{(\frac {m}{2k_B T})}^{3/2} \int_0^ \infty \sigma(v) v^2 e^{\frac {-mv^2}{2k_B T}} dv$

$m_0$ is the sum masses of the reacting nuclei, $m_a$ is the sum masses of the product nuclei, $n_1$ is the number density of one of the reacting nucleus and $n_2$ is the number density of the other reacting nucleus, $\sigma(v)$ is the probability of fusion reaction happening, if the energy is less than the coulomb barrier, it is the probability of quantum tunneling through the coulomb barrier. If we could find that probability we could calculate the total energy output of a nuclear fusion reaction.