2
$\begingroup$

For an electron incident a one-dimensional finite square well the transmission probability is $\approx$1 for electron energies $E_1=0.6 \textbf{ eV}$, $E_2=1.9 \textbf{ eV}$ and $E_3=3.4 \textbf{ eV}$. How can the potential well width and depth be calculated?

I know that an expression for the transmission coefficient $T$ can be found from deriving the wave-equation for the finite potential well, where the probability of transmission is the probability amplitude ratio of the incident wave to the transmitted wave, i.e. $T=\frac{|F|^2}{|A|^2}$ where $A$ is the amplitude of the incident wave and $F$ is the amplitude of the transmitted wave. By Properly normalizing these amplitudes from the boundary conditions (continuity in regard to $\psi(x)$ and $\frac{d}{dt}\psi(x)$ at the boundaries) the ratio can be described as \begin{equation} T^{-1} = 1 + \frac{V_0 ^2}{4E(E+V_0)} \sin^2(\frac{2a}{\hbar}\sqrt{2m(E+V_0)}) \end{equation} where $a$ is the width of the well, $m$ the mass of the electron and $V_0$ the absolute value of the well depth.

The transmission probability is equal to one if the sine is zero, which is for \begin{equation} \frac{2a}{\hbar}\sqrt{2m(E+V_0)}=n\pi \end{equation} Where $n$ is any integer.

The solution seems simple enough, just plug in the values for the energies and solve for the width $a$ and the depth $V_0$. However this leads me to an unsolvable system of equations since I do not know which $n$ corresponds to which resonant energy value. How can one approach this problem more successfully?

$\endgroup$
3
  • $\begingroup$ have you tried with the $n$ s corresponding to the $E_n$ s? $\endgroup$
    – danimal
    Commented Apr 6, 2015 at 16:42
  • $\begingroup$ @danimal That was my initial approach. This however led to a negative $V_0$ value. The resonance at $E_1$ does not have to be given by $n=1$. $\endgroup$ Commented Apr 6, 2015 at 16:52
  • $\begingroup$ ok i see... can you make an approximation using the infinite square well $E\propto (n/a)^2$? $\endgroup$
    – danimal
    Commented Apr 6, 2015 at 16:54

1 Answer 1

1
$\begingroup$

The problem can be solved by introducing the relations \begin{equation} n_2-n_1=1,\:\:\:n_3-n_2=1 \end{equation} Since the resonance occurs for the energies $E_1$,$E_2$, and $E_3$ in a sequence the argument for the sine $n \pi$ also occur in a sequence in this case.

This leads to one of possible solvable equations for the well potential \begin{equation} 2 \sqrt{E_2+V_0} = \sqrt{E_1+V_0}+\sqrt{E_3+V_0} \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.