QM - Transmission coefficient for square well

[DeltaFunction]

Homework Statement

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?

Homework 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

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.
Thanks in advance.

 

[mfb]

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.

 

[DeltaFunction]

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

 

[DeltaFunction]

I've realized 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 embarrassing