# Scanning Tunneling Microscope

#### The STM

Please find the tunneling probability for a STM (Scanning Tunneling Microscope), if the probe work function is 4.0 eV, the gap is 0.50 nm and the atom is 0.05 nm tall. Please find the increase in the STM current and what is the smallest atom that the STM can detect if the smallest current change detectable is 10%.

The electron is described by its wave function. Since it needs to go through a barrier of potential the wave function will decrease exponentially in the barrier with the barrier thickness $x$.

#### Working principle of scanning tunneling microscope

A scanning tunneling microscope is made of a sharp metallic needle (the probe) that is brought close to the (metalic) sample to investigate. A voltage potential is applied between the probe and the sample. At a certain distance (some tens or hundreds nanometers) between the two of them, an electric current appears in this circuit, not because the electric discharge, but because the tunneling of the electrons between the probe and the sample, due to the potential between them. The probe starts to scan (to move over) the surface of the sample. Because the atoms of the sample have a certain dimension, the current changes. Thus by following the changes in current, one can follow the geometry of the surface of the sample. Continue reading about this….

$\Psi(x) = \Psi(0)*exp(-kx)$ where $k$ is the wave number so that $(kx)$ is non-dimensional,

$k =1/\lambda = P/\hbar= \sqrt{2*m*E}/ \hbar$ where $E$ is the work function, $m$ is the electron mass

Therefore

$k = \sqrt{2*9.1*10^{-31}*4*1.6*10^{-19}}/10^{-34} =10^{10} m^{-1}$

The probability of tunneling is just

$P = |\psi(x)/\psi(0) |^2 = \exp(-2kx) = \exp(-2*10^{10}*5*10^{-10}) =\exp(-10) =4.54*10^{-5}$

b)

The probability of tunneling in this case is again

$P2 = \exp(-2k(x+dx)) = \exp(-2*(5-0.5)) =\exp(-9)$

Because of this:

$P2/P =\exp(-9)/\exp(-10) = \exp(1) = e^1 =2.718$

As a result the tunneling current increases with the same factor as the probability does (2.718).

c)

$I2/I = 1.1$

$\exp(-2kx_2)/\exp(-2kx) = 1.1$

$\exp(-2k(x-x_2)) =1.1$

$\Delta(x) = x_2-x =ln(1.1)/(2k) =ln(1.1)/(2*10^{10}) =4.765*10^{-12} m =0.00477 nm$