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We need to find the probability that an electron will tunnel through a$0.50nm$ air gap from a metal to a STM probe.

First, we need to find the penetration distance of the electrons into the forbidden region.

$畏=\frac{h}{\sqrt{2m(U_0-E)}}$
but we are given that the work function is $4\text{eV}$which is equivalent to the term $\left(U_0鈭�E\right)$so,

$$\begin{gathered} 畏=\frac{1.05脳{10}^{34}\text{J}鈰�\text{s}}{\sqrt{2\left(9.11脳{10}^{鈭�31}\text{kg}\right)\left(4\text{eV}脳1.6脳{10}^{鈭�19}\text{J}/\text{eV}\right)}} \\ =9.72脳{10}^{鈭�11}\text{m} \\ =0.0972\text{nm} \end{gathered}$$

Now, the probability that the electron will penetrate the air barrier is given as:

$P_{tunnel}=e^{2w/畏}$

Here $\left(w\right)$is the width of the barrier. Since:

$P_{tunnel}=\exp \left(鈭�\frac{1\text{nm}}{0.0972\text{nm}}\right)=3.4脳{10}^5$

We need to find the factor does the tunneling current increase or not.

The tunneling current is proportional to the tunneling probability as:

$I鈭�P_{tunnel}$

so the ratio of the current for a barrier with width $\left(0.5\text{nm}\right)$ to the current for a barrier with width $\left[\left(0.5鈭�0.05\right)=0.45\text{nm}\right]$as:

role="math" localid="1650460494965" $\begin{array}{rr}\frac{I_2}{I_1}=\frac{P_{tunnel,2}}{P_{tunnel,1}}& =\frac{\exp \left(鈭�\frac{0.9\text{nm}}{0.0972\text{nm}}\right)}{\exp \left(鈭�\frac{1\text{nm}}{0.0972\text{nm}}\right)}\end{array}$

$\begin{array}{rr}\frac{I_2}{I_1}& =2.8\end{array}$

the tunneling current increases with a factor of$2.8$.

We need to find the smallest height change the STM can detect.

The height change equals the difference between the initial width and the new width that caused the increase in the current:

$h=w_i鈭�w_f$

and we are looking for the height that would make the current increase by a factor of $\left(10\%\right)or(0.1)$. Therefore:

$$\begin{gathered} \frac{I_2}{I_1}=1.1=\frac{P_{tunnel,2}}{P_{tunnel,1}} \\ =\frac{\exp \left(鈭�\frac{2w_f}{畏}\right)}{\exp \left(鈭�\frac{2w_i}{畏}\right)} \\ =\exp \left(2\frac{w_i鈭�w_f}{畏}\right) \\ =\exp \left(\frac{2h}{畏}\right) \end{gathered}$$

take to finding the minimum height $h$.

$\begin{array}{c}\ln \left(1.1\right)=\frac{2h}{畏}\\ h=\frac{\ln \left(1.1\right)畏}{2}=\frac{\ln \left(1.1\right)脳9.72脳{10}^{鈭�11}\text{m}}{2}\\ h=4.63脳{10}^{鈭�12}\text{m}=0.00463\text{nm}\end{array}$