Alpha Decay — Question 7

Question 7

The Coulomb action leads to strong inverse-square-root energy dependence

For zero orbital angular momentum, a WKB Coulomb action is $G=\hbar^{-1}\int_R^b\sqrt{2\mu(K/r-Q)}\,dr$, where $b=K/Q$. Use the substitution $r=b\sin^2\phi$.

1. Derive a closed expression for G in terms of $x=R/b$.

2. For $R/b\ll1$, obtain the leading energy dependence of $\ln T_{1/2}$ if the decay prefactor is held fixed and $P=e^{-2G}$.

Question 7 - Solution

The Coulomb action leads to strong inverse-square-root energy dependence

(a) Method and calculation

$$G=\frac{2b\sqrt{2\mu Q}}\hbar\int_{\sin^{-1}\sqrt x}^{\pi/2}\cos^2\phi\,d\phi.$$$$\boxed{G=\frac{K\sqrt{2\mu}}{\hbar\sqrt Q}\left[\cos^{-1}\sqrt x-\sqrt{x(1-x)}\right]}.$$

(b) Interpretation and further calculation

$$x\to0:\quad G\simeq\frac{\pi K\sqrt{2\mu}}{2\hbar\sqrt Q}.$$$$\boxed{\ln T_{1/2}=\text{constant}+2G\simeq\text{constant}+\frac{\pi K\sqrt{2\mu}}{\hbar\sqrt Q}}.$$

Physical check

The logarithm grows roughly with charge and inverse square root of Q. Finite radius, angular momentum and alpha preformation modify this leading approximation.