Electron in a Finite Square Well

[Potatochip911]

Homework Statement

An electron in a finite square well has 6 distinct energy levels. If the finite square well is 10nm long determine:

a) Approximate the possible values for the depth of the finite square well $V_0$.
b) Using a well depth value in the middle of the results obtained from part a) find the energy when the electron is in the $n=3$ state.
c) For the $n=3$ state determine the un-normalized wave function
d)For $-20nm<x<20nm$ draw the $n=3$ wave function
e) For $-20nm<x<20nm$ draw the probability function

Homework Equations

$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$

The Attempt at a Solution

Part a) of the question is somewhat tedious and involves solving transcendental equations, I essentially just followed how they solved it on this site which gave me the the possible range for the depth of the finite square well to be $V_0=0.0237eV$ to $V_0=0.0333eV$

Now for part b), taking the average of these gives $V_0=0.0285eV$ although I'm not sure how I can use this to find the energy of the electron in the n=3 state. In my textbook they seem to use the equation $E_n=n^2\frac{\pi^2\hbar^2}{2mL^2}$ which doesn't seem to mention the potential at all.

From the plot I've listed the coordinates of the n=3 intersection in case that's useful. The lines are: $y=-\mbox{cot}(x)$, $y=\tan(x)$, and $y=\sqrt{(\frac{8.648}{x})^2-1}$ where $8.648=\mu=L\sqrt{\frac{2m}{\hbar^2}V_0}$

 

[drvrm]

Part a) of the question is somewhat tedious and involves solving transcendental equations, I essentially just followed how they solved it on this site which gave me the the possible range for the depth of the finite square well to be V0=0.0237eVV0=0.0237eVV_0=0.0237eV to V0=0.0333eVV0=0.0333eVV_0=0.0333eV

Now for part b), taking the average of these gives V0=0.0285eVV0=0.0285eVV_0=0.0285eV although I'm not sure how I can use this to find the energy of the electron in the n=3 state. In my textbook they seem to use the equation En=n2π2ℏ22mL2En=n2π2ℏ22mL2E_n=n^2\frac{\pi^2\hbar^2}{2mL^2} which doesn't seem to mention the potential at all.

Well i saw your refereed treatment and i think its exact and Transcendental equations are are not a problem as it solves graphically the two relations obtained from applying the boundary conditions at the walls of the well-
As you wish to have six possible eigen states the well depth will have a bounding value that you have to calculate and taking half does not mean thay one can get the three-
The above calculations are well known in case of nuclear potentials also-say deutreron in a finite well - however if well depth is large compared to energies involved then approximations can be made giving simpler results for energy values.
my advise is to follow your referred site-