# 1D Quantum Well - I'm getting something very basic wrong!

Ok, so this example question is in my lecture notes:

## Homework Statement

"For a 1-D quantum well, 5nm width, 2eV barrier height, calculate the number of energy levels in the well."

## Homework Equations

So the equation given is a familiar one:

$$E_n=\frac{n^2 \pi^2 \hbar^2}{2 m L^2}$$

and he also said $$E_n=2 \times 1.6 \times 10^{-19} J$$

## The Attempt at a Solution

Now, in the lecture $$n^2$$ came out to be 134, but I keep getting ~1.9, so I'm using the wrong units somewhere or something, because otherwise it's a pretty simple 'plug and chug' exercise..!

also, I'm using
$$\hbar = 6.582 \times 10^{-16} eVs$$
$$m = 0.511MeV$$

Can anyone help? :-s

-Jam

Last edited:

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Remember that the units of mass are

$$[m] = eV/c^2$$

And therefore with the values that you are using it should be

$$E_n=\frac{n^2 \pi^2 \hbar^2 c^2}{2 m L^2}$$

Unfortunately, if I put in the factor of $$c^2$$, then $$n^2$$ becomes ~$$2.13^{-17}$$, even further off the 134 that it should be :-(

Make sure that you use the same units for everything. That is if you use electron volts as you said in your first post then E = 2 eV. If you use Joules for E then you have to use Js for h, kg for m (without c^2 this time) and so on.

diazona
Homework Helper
I get 133, which seems pretty close... it's probably just a silly math mistake. If you write out the details of your calculations (and please be consistent with the units ;-), someone can probably pinpoint where it went wrong.

Ah, thankyou! Yes, it was a silly mistake, I forgot to change $$E=2 \times 1.6 \times 10^{-19}$$ to E = 2, yes it works out fine now.

Thank you both, I have an exam tomorrow where this will likely be part of a question or two, so I'm very grateful for the help!