Barrier penetration

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 $α$ radiation. The picture is as in Fig. 5.12.

alpha˙decay

Figure 5.12: The potential energy for alpha decay

Suppose there exists an $α$ 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_{Coulomb} (r) = \frac{(Z - 2) 2 e^{2}}{4 π 𝜖_{0} r} .

(5.20)

. The height of the barrier is exactly the coulomb potential at the boundary, which is the nuclear radius, $R_{C} = 1.2 A^{1.3} fm$ , 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 $P = e^{- 2 γ}$ , with

\begin{array}{rcl} γ & = & \frac{{(2 μ_{α})}^{1 ∕ 2}}{ℏ} \int_{R_{C}}^{b} {[V (r) - E_{α}]}^{1 ∕ 2} d r \\ = & \frac{{(2 μ_{α})}^{1 ∕ 2}}{ℏ} \int_{R_{C}}^{b} {[\frac{2 (Z - 2) e^{2}}{4 π 𝜖_{0} r} - E_{α}]}^{1 ∕ 2} d r \\ = & \frac{2 (Z - 2) e^{2}}{2 π 𝜖_{0} ℏ v} [arccos (E_{α} ∕ B_{C}) - (E_{α} ∕ B_{C}) (1 - E_{α} ∕ B_{C})], & (5.21) \end{array}

(here $v$ is the velocity associated with $E_{α}$ ). In the limit that $B_{C} ≫ E_{α}$ we ﬁnd

P = exp [- \frac{2 (Z - 2) e^{2}}{2 𝜖_{0} ℏ v}] .

(5.22)

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

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