Question

A proton’s energy is $1.0MeV$ below the top of a $10fm$-wide energy barrier. What is the probability that the proton will tunnel through the barrier?

Short answer

The probability that the proton will tunnel through the barrier is$1.22\text{\%}$.

Step-by-step solution

Given Information 

We need to find the probability that the proton will tunnel through the barrier.

Simplify

Finding the penetration distance of the proton into the forbidden region:

$\eta =\frac{ħ}{\sqrt{2m\left(U_0-E\right)}}$

but we given that the proton's energy is $1.0MeV$below the top of the well, which represents the value of $\left(U_0-E\right)$. Then, $\left(U_0-E=1\text{MeV}\right)$.

Since:

$\begin{array}{c}\eta =\frac{1.05\times {10}^{-34}\text{J}\cdot \text{s}}{\sqrt{2\left(1.67\times {10}^{-27}\text{kg}\right)\left(1\times {10}^6\text{eV}\times 1.6\times {10}^{-19}\text{J}/\text{eV}\right)}}\\ \eta =4.54\times {10}^{-15}\text{m}=4.54\text{fm}\end{array}$

the probability that the proton will tunnel through the barrier is given as:

$P_{tunnel}=e^{-2w/\eta }$

There, $\left(w\right)$the width of the barrier.

Therefore:

$P_{tunnel}=\exp \left(-\frac{20\text{fm}}{4.54\text{fm}}\right)=0.0122=1.22\text{\%}$