To get an idea of the distance dependence of the tunnelling current in STM, suppose that the wavefunction of the electron in the gap between sample and needle is given by $\psi=B \mathrm{e}^{-\kappa x}$, where $\kappa=\left\{2 m_{\mathrm{e}}(V-E) / \hbar^2\right\}^{1 / 2}$; take $V-E=2.0 \mathrm{eV}$. By what factor would the current drop if the needle is moved from $L_1=0.50 \mathrm{~nm}$ to $L_2=0.60 \mathrm{~nm}$ from the surface?
 When $L=L_1=0.50 \mathrm{~nm}$ and $V-E=2.0 \mathrm{eV}=3.20 \times 10^{-19} \mathrm{~J}$ the value of $\kappa L$ is
$$
\begin{aligned}
\kappa L_1 & =\left\{\frac{2 m_{\mathrm{e}}(V-E)}{\hbar^2}\right\}^{1 / 2} L_1 \\
& =\left\{\frac{2 \times\left(9.109 \times 10^{-31} \mathrm{~kg}\right) \times\left(3.20 \times 10^{-19} \mathrm{~J}\right)}{\left(1.054 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^2}\right\}^{1 / 2} \times\left(5.0 \times 10^{-10} \mathrm{~m}\right) \\
& =\left(7.25 \times 10^9 \mathrm{~m}^{-1}\right) \times\left(5.0 \times 10^{-10} \mathrm{~m}\right)=3.6
\end{aligned}
$$
Because $\kappa L_1>1$, we use eqn $8.19 \mathrm{~b}$ to calculate the transmission probabilities at the two distances. It follows that
$$
\text { current at } \begin{aligned}
\frac{L_2}{\text { current at } L_1} & =\frac{T\left(L_2\right)}{T\left(L_1\right)}=\frac{16 \varepsilon(1-\varepsilon) \mathrm{e}^{-2 \kappa L_2}}{16 \varepsilon(1-\varepsilon) \mathrm{e}^{-2 \kappa L_1}}=\mathrm{e}^{-2 \kappa\left(L_2-L_1\right)} \\
& =\mathrm{e}^{-2 \times\left(7.25 \times 10^{-9} \mathrm{~m}^{-1}\right) \times\left(1.0 \times 10^{-10} \mathrm{~m}\right)}=0.23
\end{aligned}
$$
