An electron is confined to a finite square well whose “walls” are $8.0$ eV high. If the ground-state energy is $0.5$ eV, estimate the width of the well.
My solution:
$E_1=0.5 \ eV=8\times10^{-20} \ J$
Electron mass: $m=9.11\times10^{-31} \ Kg$
$\hbar=1.055\times 10^{-34} \ Js$
$E_1=\frac{\pi^2 \hbar}{2mL^2} \ \Rightarrow \ L=\frac{\pi\hbar}{\sqrt{2mE_1}}=\frac{\pi(1.055\times 10^{-34} \ Js)}{\sqrt{2(9.11\times10^{-31} \ Kg)(8\times10^{-20} \ J)}}=8.6813\times 10^{-10} \ m\approx 0.87 \ nm$
This is a problem from Modern Physics, Paul A. Tipler / Ralph A. Llewellyn, and when I check solution in Answers section at the end of the book, I found that it is the same answer that I found.
But, my T.A. told me I can't do this because I'm apllying concepts for Infinite Square Well (equation for $E_1$). I thought I can do this because $E_1<V_0$ and does not matter if $V_0\rightarrow\infty$, $E_1$ is still under $V_0$.
What am I understanding wrong?
How can I find answer without using quation for $E_1$ that I already use?
Best Answer
As soon as $E_1 < \infty$, you cannot assume $\Psi = 0$ outside the well anymore. In other words, your formula doesn't work because the wave function "leaks into" the walls of the well (tunneling).
You need to solve the Schrödinger equation for energy eigenstates by hand here... Write it down explicitly for inside and outside the well, apply appropriate boundary conditions and you're done.
If the numerical result looks similar, this is because the ground state energy is much smaller than the well.