# Rectangular finite potential well problem

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## Homework Statement

An electron enters in a finite rectangular potential well of length 4 angstroms. When the entering electrons have a kinetic energy of 0.7 eV they can travel through the region without having any reflection. Use this information to calculate the depth of the potential well.

## Homework Equations

Not sure because I don't know if the kinetic energy of the electron is greater or lesser than the potential well.
I know that inside the well, $\Psi _{II}(x)= C \sin (kx)+ D \cos (kx)$. I also know the form of the wave function outside the well (thanks to https://www.physicsforums.com/showthread.php?t=540406&page=3) but I don't have the normalization, etc. I tried to search in google, wikipedia and hyperphysics and can't find the whole detailed solution.
I was thinking of calculating the flux of probability (I need the exact wavefunctions, normalized) in order to calculate the coefficient of transmission.

## The Attempt at a Solution

I'm confused. I don't know how to tackle the problem, too many unknowns to me.
Any tip is appreciated.

## Answers and Replies

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Ah I think I just realized how to solve the problem.
The electron has a greater kinetic energy that the potential well.
The coefficient of transmission is the one given by a barrier of potential when the energy of the particle exceed the potential barrier.
It is worth $T=\frac{1}{1+ \frac{V_0 ^2 \sin ^2 (k_1 a)}{4E(E-V_0)}}$ where $k_1 = \sqrt { \frac{2m(V_0-E)}{\hbar ^2}}$. But unfortunately $k_1$ is complex.
So I set T=1 and I try to solve for $V_0$... but this doesn't go well. This gives me $V_0=0$.
By the way I used the formula given in http://en.wikipedia.org/wiki/Rectangular_potential_barrier#Analysis_of_the_obtained_expressions.
Does someone understand better than me what's going on?

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Hmm, I've watched the first part of and he reaches a slightly different result than wikipedia for the transmission coefficient.
His $k_1$ is worth $\sqrt {\frac{2m (E-V_0)}{\hbar ^2} }$.
So for my exercise, I must solve for $V_0$ and $n=1$ in the expression a $\sqrt {\frac{2m (E-V_0)}{\hbar ^2}}=n \pi$. I've checked out the units, all seems to work.
When I plug and chug the given values for the problem, I reach that $V_0\approx -1.65 eV$ or $-1.6 eV$ if I round correctly.
Can someone check out my work? That would after all make wikipedia "slightly wrong" for a formula.

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