Skip to main content
Physics LibreTexts

5.4: Barrier Penetration

  • Page ID
    15022
  • [ "article:topic", "tunneling", "alpha decay", "barrier penetration", "authorname:nwalet" ]

    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 \(\PageIndex{1}\)).

    alpha_decay.png

    Figure \(\PageIndex{1}\):The potential energy for alpha decay

    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 \(v\)!