Question

Electron in a potential well question, please see following comment.

Answer 1

The width of the potential well of an electron can be assumed to be 2angstrom. Calculate the energy of an electron (in joules and eV) from this information for various values of n.

I believe I solved it just fine, so you can probably skip to the bottom. I think that my units are getting causing the error somehow. For eV where n=1 the book says the answer is 9.39eV, however I'm getting 2.34675(10^20).

Started with
\[\nabla^2\psi+\frac{2m}{\hbar^2}(E-v)\psi=0\] Becoming
\[\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}E\psi=0\]
Letting \[\alpha=\sqrt{\frac{2m}{\hbar^2}E}\]
\[\therefore\psi(x)=Ae^{i\alpha x}+Be^{-i\alpha x}\]
Boundary Conditions:
\[\psi(0)=0=A+B\rightarrow A=-B\]
\[\psi(L)=0=Ae^{i\alpha L}+Be^{i\alpha L}=A(e^{i\alpha L}-e^{-i\alpha L})\]From the identity:\[sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})\]
\[\therefore 0=A2isin(\alpha L)\rightarrow \alpha L=n\pi\rightarrow \alpha=\frac{n\pi}{L}\]
Putting alphas together
\[\frac{n\pi}{L}=\sqrt{\frac{2m}{\hbar^2}E}\]
\[\therefore E(L)=\frac{n^2\pi^2\hbar^2}{2mL^2}\]\[n=1,2,3...\]

These are the constants I am using for the final calculation:\[n=1\]\[\hbar=6.582(10^{-16})eVs\]\[m=9.11(10^{-31})Kg\]\[L=2(10^{-10})m\]

I get 2.34675(10^20)eV when it should be 9.39eV.

Answer 2

I forgot a minus on the second boundary condition in the B term, and another minus sign in the last term of the trig identity... fyi