Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the tunneling probability is given by
$$
P_{\text{tunnel}} = e^{-2w/\eta}\tag 1
$$
where $ w $ is the width of the barrier, and $ \eta $ is the penetration distance.
So we need to find $\eta$ which is given by
$$
\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
We are given that $E_0=4.0\;\rm eV$ where $E_0$ is the work function.,which is given by $U_0-E$, so
$$
\eta = \frac{\hbar}{\sqrt{2mE_0}}
$$
Plug the known;
$$
\eta = \frac{(1.05 \times 10^{-34} )}{\sqrt{2(9.11 \times 10^{-31} )(4\times 1.6\times 10^{-19})}}= 9.72 \times 10^{-11} \;\rm m
$$
$$\eta=\bf 0.0972 \;\rm nm$$
Plug into (1);
$$
P_{\text{tunnel}} = e^{-2(0.5)/(0.0972) } =\color{red}{\bf 3.40 \times 10^{-5}}
$$
$$\color{blue}{\bf [b]}$$
When the probe passes over an atom of height $ 0.05 \; \text{nm} $, the barrier width decreases to $ w = 0.45 \; \text{nm} $.
Using (1) again;
$$
(P_{\text{tunnel}})' = e^{-2w/\eta} = e^{-2(0.45 ) / (0.0972 )} =\bf 9.52 \times 10^{-5}
$$
Comparison to Part (a);
$$\dfrac{P_{\text{tunnel}} }{(P_{\text{tunnel}})' }=\dfrac{3.40 \times 10^{-5}}{9.52 \times 10^{-5}}=\bf 0.357$$
Thus,
$$(P_{\text{tunnel}})'=\color{red}{\bf 2.8}\;P_{\text{tunnel}} $$
The tunneling probability has increased by a factor of 2.8 compared to part (a).
$$\color{blue}{\bf [c]}$$
To detect a 10% change in the tunneling probability, we calculate the new tunneling probability:
$$
P_{\text{tunnel}} = 1.10 \times 3.4 \times 10^{-5} = 3.74 \times 10^{-5}
$$
We use the formula for tunneling probability to determine the new barrier width $ w $:
$$
P_{\text{tunnel}} = e^{-2w/\eta}
$$
Substitute $ P_{\text{tunnel}} = 3.74 \times 10^{-5} $ and solve for $ w $:
$$
3.74 \times 10^{-5} = e^{-2w / 0.0972}
$$
Take the natural logarithm of both sides:
$$
\ln(3.74 \times 10^{-5}) = -2w / 0.0972
$$
$$
-10.19 = -2w / 0.0972
$$
Solve for $ w $:
$$
w = \frac{10.19 \times 0.0972}{2} =\bf 0.495 \; \text{nm}
$$
Thus, the change in width is given by
$$
\Delta w = 0.500 - 0.495 =\color{red}{\bf 0.005}\;\rm nm=\bf \dfrac{1}{20} \; \rm nm
$$
Therefore, it is capable of detecting height changes as small as $ \frac{1}{20} $ of an atomic diameter.
