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,
$${V}_{\text{Coulomb}}\left(r\right)=\frac{\left(Z-2\right)2{e}^{2}}{4\pi {\mathit{\epsilon}}_{0}r}.$$ | (5.20) |
. The height of the barrier is exactly the coulomb potential at the boundary, which is the nuclear radius, ${R}_{C}=1.2{A}^{1.3}\text{fm}$, and thus ${B}_{C}=2.4\left(Z-2\right){A}^{-1\u22153}$. The decay probability across a barrier can be given by the simple integral expression $P={e}^{-2\gamma}$, with
$$\begin{array}{rcll}\gamma & =& \frac{{\left(2{\mu}_{\alpha}\right)}^{1\u22152}}{\hslash}{\int}_{{R}_{C}}^{b}{\left[V\left(r\right)-{E}_{\alpha}\right]}^{1\u22152}dr& \text{}\\ & =& \frac{{\left(2{\mu}_{\alpha}\right)}^{1\u22152}}{\hslash}{\int}_{{R}_{C}}^{b}{\left[\frac{2\left(Z-2\right){e}^{2}}{4\pi {\mathit{\epsilon}}_{0}r}-{E}_{\alpha}\right]}^{1\u22152}dr& \text{}\\ & =& \frac{2\left(Z-2\right){e}^{2}}{2\pi {\mathit{\epsilon}}_{0}\hslash v}\left[arccos\left({E}_{\alpha}\u2215{B}_{C}\right)-\left({E}_{\alpha}\u2215{B}_{C}\right)\left(1-{E}_{\alpha}\u2215{B}_{C}\right)\right],& \text{(5.21)}\text{}\text{}\end{array}$$(here $v$ is the velocity associated with ${E}_{\alpha}$). In the limit that ${B}_{C}\gg {E}_{\alpha}$ we ﬁnd
$$P=exp\left[-\frac{2\left(Z-2\right){e}^{2}}{2{\mathit{\epsilon}}_{0}\hslash v}\right].$$ | (5.22) |
This shows how sensitive the probability is to $Z$ and $v$!