# Electron in a potential well (numerical problem)

1. Jan 27, 2013

### abstracted6

1. The problem statement, all variables and given/known data

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 have messed up my units somewhere in the final calculation, but I have included everything before that anyways.

For eV where n=1 the book says the answer is 9.39eV, however I'm getting 2.34675(10^20).

2. Relevant equations

$$\nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0$$

3. The attempt at a solution

Started with
$$\nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0$$ Only one direction, and no potential:
$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}E\psi=0$$
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.

2. Jan 28, 2013

### CFede

You have a problem with the units of measure. Thinking only of units what you have is

$\left[\frac{n^2\pi^2\hbar^2}{2mL^2}\right]=\frac{(eV)^2s^2}{kg\cdot m^2}=\frac{eV^2}{J}$

which is obvioulsy not in eV. You should work with h bar in joules and, at the end of your calculation, go to eV.

3. Jan 28, 2013

### vela

Staff Emeritus
You can make your life a lot easier if you memorize some common combinations of the constants:
\begin{align*}
\hbar c &= 197\text{ eV nm} \\
mc^2 &= 511000\text{ eV}
\end{align*} If you multiply both the numerator and denominator of the expression for the energy by $c^2$, you get
$$E = \frac{n^2\pi^2 (\hbar c)^2}{2 (mc^2) L^2}.$$ If you specify $L$ in nm, you can see that the units of $E$ will be eV.