### 5.4 Barrier penetration

In order to understand quantum mechanical tunnelling in ﬁssion it makes sense to look at the simplest ﬁssion process: the emission of a He nucleus, so called $\alpha$ radiation. The picture is as in Fig.  5.12 .

Suppose there exists an $\alpha$ particle inside a nucleus at an (unbound) energy $>0$. Since it isn’t bound, why doesn’t it decay immediately? This must be tunnelling. In the sketch above we have once again shown the nuclear binding potential as a square well, but we have included the Coulomb tail,

 ${V}_{\text{Coulomb}}\left(r\right)=\frac{\left(Z-2\right)2{e}^{2}}{4\pi {𝜖}_{0}r}.$ (5.20)

. The height of the barrier is exactly the coulomb potential at the boundary, which is the nuclear radius, , and thus ${B}_{C}=2.4\left(Z-2\right){A}^{-1∕3}$. The decay probability across a barrier can be given by the simple integral expression $P={e}^{-2\gamma }$, with

$\begin{array}{rcll}\gamma & =& \frac{{\left(2{\mu }_{\alpha }\right)}^{1∕2}}{\hslash }{\int }_{{R}_{C}}^{b}{\left[V\left(r\right)-{E}_{\alpha }\right]}^{1∕2}dr& \text{}\\ & =& \frac{{\left(2{\mu }_{\alpha }\right)}^{1∕2}}{\hslash }{\int }_{{R}_{C}}^{b}{\left[\frac{2\left(Z-2\right){e}^{2}}{4\pi {𝜖}_{0}r}-{E}_{\alpha }\right]}^{1∕2}dr& \text{}\\ & =& \frac{2\left(Z-2\right){e}^{2}}{2\pi {𝜖}_{0}\hslash v}\left[arccos\left({E}_{\alpha }∕{B}_{C}\right)-\left({E}_{\alpha }∕{B}_{C}\right)\left(1-{E}_{\alpha }∕{B}_{C}\right)\right],& \text{(5.21)}\text{}\text{}\end{array}$

(here $v$ is the velocity associated with ${E}_{\alpha }$). In the limit that ${B}_{C}\gg {E}_{\alpha }$ we ﬁnd

 $P=exp\left[-\frac{2\left(Z-2\right){e}^{2}}{2{𝜖}_{0}\hslash v}\right].$ (5.22)

This shows how sensitive the probability is to $Z$ and $v$!