Even the smoothest mirror finishes are “rough” when viewed at a scale of $100nm$. When two very smooth metals are placed in contact with each other, the actual distance between the surfaces varies from$0nm$at a few points of real contact to $\approx 100nm$. The average distance between the surfaces is$\approx 50nm$. The work function of aluminum is $4.3eV$. What is the probability that an electron will tunnel between two pieces of aluminum that are $50nm$apart? Give your answer as a power of$10$rather than a power of$e$.

We need to find the probability that an electron will tunnel between two pieces of aluminum that are $50nm$apart.

Finding the distance of the Aluminum's electrons into the forbidden region.

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

but we that the work function of aluminum is $4.3eV$, which is the energy that must be given to an electron in order to go out of the aluminum and it represents the value of $\left(U_0-E\right)$

Since,

$\begin{array}{c}\eta =\frac{1.05\times {10}^{-34}\text{J}\cdot \text{s}}{\sqrt{2\left(9.11\times {10}^{-31}\text{kg}\right)\left(4.3\text{eV}\times 1.6\times {10}^{-19}\text{J}/\text{eV}\right)}}\\ \eta =9.38\times {10}^{-11}\text{m}=0.0938\text{nm}\end{array}$

Now, the probability that the electron will penetrate the air barrier between the two Aluminum surfaces is given as:

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

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

Therefore:

$P_{tunnel}=\exp \left(-\frac{100\text{nm}}{0.0938\text{nm}}\right)=e^{-1066}$

In order to write the answer as a power of $10$, we need to find in terms of $10$.

$e={10}^n$

to finding $n$ , take the logarithm of both sides:

$n=\log \left(e\right)=0.4343$

Now, replace $e$with ${10}^{0.4343}$

$$\begin{gathered} P_{tunnel}={\left({10}^{0.4343}\right)}^{-1666} \\ {P}_{tunnel}={10}^{-463} \end{gathered}$$