# QM - Transmission coefficient for square well

1. Aug 23, 2013

### DeltaFunction

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

A steady stream of 5 eV electrons impinges on a square well of depth 10 eV. The width of the well is 7.65 * 10^-11 m. What fraction of electrons are transmitted?

2. Relevant equations
The following equation for the transmission coefficient, T, is given:
$$T = [1 + \frac{V_0 ^2 sinh^2 κa}{4E(V_0 - E)}]^-1$$ (**that is meant to be ^-1 for the whole bracket - apologies, this is my first time using LaTex**)
Where $$κ^2 = \frac{8mπ^2}{h^2}(V_0 - E)$$

We are also provided with a not-so-subtle hint that $$sinh~iθ = i~sinh~θ$$

3. The attempt at a solution

So I have assigned the following values based on the information:

a = 7.65 * 10^-11 m
E = 5 eV
V = - 10 eV
m = 9.11 * 10^-31 kg

It then seems like it should be very straightforward. I calculate ka and found this to be 1.52i. Then using the definition of sinh I calculate $sinh^2 κa = -4.73$.
Plugging the other values in I arrive at $$T = 0.388$$ which seemed reasonable to me, but..... The postgrad who marked my work fed back to me that the numerical answer he had was T = 0.75.
I'd be really grateful if someone can check the calculation for me, because it's really bugging me that I can't see my error.

Last edited: Aug 23, 2013
2. Aug 23, 2013

### Staff: Mentor

The h in the denominator for κ2 should be h2, I guess.

I can confirm 1.52i, but if I put this in the final formula I get 0.75.
Well, WolframAlpha does.

3. Aug 23, 2013

### DeltaFunction

Thank you, yes it should be h^2, I've now corrected that. I guess I'll go through the figures again carefully :/

4. Aug 23, 2013

### DeltaFunction

I've realised my stupid mistake. I copied down the identity incorrectly. A moments thought and I would have seen that $$sinh~iθ = i~sinh~θ$$ is nonsense :uhh: embarrassing