OEP
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Homework Statement
An electron with KE = 8eV is incident on the potential step of height 7eV. What's the probability the electron will be reflected from the step function?
Homework Equations
The wave equations and their derivatives:
\psi_0(x) = A sin(k_0 x) + B cos( k_0 x)
\psi_1(x) = C sin(k_1 x) + D cos( k_1 x)
\psi_0'(x) = A k_0 cos(k_0 x) - B k_0 sin(k_0 x)
\psi_1'(x) = C k_1 cos(k_1 x) - D k_1 sin(k_1 x)
Plus values for k:
k_0 = \sqrt{\frac{2m}{\hbar^2} (7eV)}
k_1 = \sqrt{\frac{2m}{\hbar^2} (1eV)}
The Attempt at a Solution
First I solved the two wave functions evaluated at 0, which yields:
B = D
Next, I solved the two derivatives evaluated at 0, which yields:
A k_0 = C k_1 \rightarrow C = \sqrt{7} A
From here I'm lost.
I also did the same thing with complex exponentials... but I think I ended up with a nonsensical k_0 = sqrt(7) k_0.
\psi_0(x) = A' e^{i k_0 x} + B' e^{-i k_0 x}
\psi_1(x) = C' e^{i k_1 x} + D' e^{-i k_1 x}
\psi_0'(x) = i k_0 A' e^{i k_0 x} + i k_0 B' e^{-i k_0 x}
\psi_1'(x) = i k_1 C' e^{i k_1 x} + i k_1 D' e^{-i k_1 x}
D' = 0, we are firing particles from right to left, not left to right.
\psi_0'(0) = \psi_1'(0) \rightarrow i k_0 A' + i k_0 B' = i k_1 C'
\psi_0(0) = \psi_1(0) \rightarrow A' + B' = C'
i k_0 A' + i k_0 B' = i k_1 C' \rightarrow i k_0 A' + i k_0 B' = i k_1 (A' + B')
i k_0 = i k_1
Which is nonsensical. Any ideas? This seems like one of those good ol' problems that you first learn to solve, but... our professor is just a bit zany and I can't follow him. :P