# Homework Help: Penetration of an energy step when KE > U

1. Feb 26, 2009

### OEP

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

2. Relevant 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:

$$k_0 = \sqrt{\frac{2m}{\hbar^2} (7eV)}$$
$$k_1 = \sqrt{\frac{2m}{\hbar^2} (1eV)}$$

3. 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

2. Feb 26, 2009

### lanedance

I prefer the exponential notation as it is clear what is moving right & left

you have missed a negative in you differentiation

should get
A + B = C

and
k0(A-B) = k1C

then solve these for reflection amplitude is r = A/B, probability r^2

3. Feb 26, 2009

### OEP

Thanks a lot! I guess it's being so over my head that I tend to overlook the smaller things... I have R being at 2.64 which I don't think is right for its square being a probability. We aren't looking for B / A are we?

4. Feb 27, 2009

### lanedance

i know what you mean -

& yep, you're right B/A should do it

5. Feb 27, 2009

### OEP

Thanks again. I'm off to trudge through the rain and turn this sucker in.