4.4: Barrier Penetration
- Page ID
- 15022
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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),
\[V_{\text{Coulomb}}(r)= \frac{(Z-2) 2 e^2}{4\pi \epsilon_0 r}. \nonumber \]
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
\[\begin{aligned} \gamma&=\frac{(2 \mu_\alpha)^{1/2}}{\hbar} \int_{R_C}^b[V(r)-E_\alpha]^{1/2} dr\nonumber\\ &=\frac{(2 \mu_\alpha)^{1/2}}{\hbar} \int_{R_C}^b\left[\frac{2(Z-2) e^2}{4\pi \epsilon_0 r}-E_\alpha\right]^{1/2} dr\nonumber\\ &= \frac{2 (Z-2) e^2}{2\pi \epsilon_0 \hbar v} \left[ \arccos(E_\alpha/B_C)-(E_\alpha/B_C)(1-E_\alpha/B_C) \right],\end{aligned} \nonumber \]
where \(v\) is the velocity associated with \(E_\alpha\). In the limit that \(B_C \gg E_\alpha\) we find
\[P= \exp\left[ -\frac{2(Z-2) e^2}{2\epsilon_0 \hbar v}\right]. \nonumber \]
This shows how sensitive the probability is to \(Z\) and \(v\)!