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# 4.4: 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 (Figure $$\PageIndex{1}$$).

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 Figure $$\PageIndex{1}$$) we have once again shown the nuclear binding potential as a square well (red curve), but we have included the Coulomb tail (blue curve),

$V_{\text{Coulomb}}(r)= \frac{(Z-2) 2 e^2}{4\pi \epsilon_0 r}.$

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 $$B_C=2.4 (Z-2)A^{-1/3}$$. The decay probability across a barrier can be given by the simple integral expression $$P=e^{-2\gamma}$$, with

\begin{aligned} \gamma&=\frac{(2 \mu_\alpha)^{1/2}}{\hbar} \int_{R_C}^b[V(r)-E_\alpha]^{1/2} dr\nonumber\\ &=\frac{(2 \mu_\alpha)^{1/2}}{\hbar} \int_{R_C}^b\left[\frac{2(Z-2) e^2}{4\pi \epsilon_0 r}-E_\alpha\right]^{1/2} dr\nonumber\\ &= \frac{2 (Z-2) e^2}{2\pi \epsilon_0 \hbar v} \left[ \arccos(E_\alpha/B_C)-(E_\alpha/B_C)(1-E_\alpha/B_C) \right],\end{aligned}

where $$v$$ is the velocity associated with $$E_\alpha$$. In the limit that $$B_C \gg E_\alpha$$ we find

$P= \exp\left[ -\frac{2(Z-2) e^2}{2\epsilon_0 \hbar v}\right].$

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