Calculating TISE for Photoelectric Effect

Consider a scenario in which the Photoelectric Effect is occurring; that is, where photons striking a metal plate are transferring their energy to the atoms in the plate and exciting and ejecting the electrons at the surface out of their orbits. The ease of such is determined by the material and the presence of any inhibiting factors on the surface of the metal, generally described as the Work Function.

However, there is also a chance that these photons tunnel through the material instead, due to the nature of quantum mechanics. This may be able to be exploited to increase photoemission rate and/or allow for lower-energy emission. This document will serve to determine if such an exploitation is possible.

The work function of zinc is typically 3.74 eV. The energy of photons in the visible spectrum are uniformly less than this, ranging from 3.26 eV (violet) to 1.66 eV (red). So, if we take the work function to be the potential barrier in the TISE, we can solve it and see what the solutions would allow for.

Our TISE, in one dimension (the dimension the photons are traveling in a line through), looks like:

$$\frac{{\hslash }^2}{2m}\frac{{\partial }^2\Psi }{\partial x^2}+(W-E)\Psi =0$$

while in the barrier. By solving the differential equation, we get solutions that look like $A{e}^{ikx}+B{e}^{-ikx}$ in one dimension (where $k=\sqrt{\frac{2m}{{\hslash }^2}(W-E)}$).

Outside the barrier, the solutions are the same, just with a $k_0$ value of $\sqrt{\frac{2m}{{\hslash }^2}(E)}$. This means the values of $A$ and $B$ are the most important for viability. This means defining the Transmission and Reflection coefficient will tell us how likely a given photon is to tunnel.

We have three states for the photon; it’s behavior before, in, and after the potential barrier, each with distinctvalues for their $A$ and $B$. We’ll identify the in-barrier constants by $C$ and $D$, and the after-barrier constants as $F$ and $G$. By solving a significant systems of equations with the boundary conditions[1], we can put $A$ in terms of $F$, giving a relation of

$$T=|\frac{F}{A}{|}^2=\frac{1}{1+\frac{{\sinh }^{2}ka}{4\frac{E}{W}(1-\frac{E}{W})}}.$$

The Reflection coefficient is $1-T$, also equal to $\frac{1}{4}(\frac{k^2+k_0^2}{k\ast k_0}{)}^2{\sinh }^{2}ka\ast T$

This doesn’t by itself tell us much. However, we can make a few simplifications. This derivation assumed the potential barrier the metal would pose would be $W$, the work function. In reality, the potential barrier the metal sheet poses is much higher than that, very much dependent on the width, making the energy of the photons negligible in comparison. This makes $k$ much larger than $k_0$, and ends up with the Transmission coefficient being a miniscule value. So, tunneling impacts are virtually nonexistant in a Photoelectric Effect-dependent scenario.