Penetration of an energy step when KE > U

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Homework Help Overview

The problem involves an electron with kinetic energy (KE) of 8 eV encountering a potential step of height 7 eV, focusing on the probability of reflection from the step function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive wave functions and their derivatives to analyze the reflection probability but encounters confusion regarding the relationships between coefficients and wave numbers.
  • Some participants suggest checking for errors in differentiation and clarify the relationships between the coefficients in the wave functions.
  • Questions arise regarding the interpretation of the reflection amplitude and its relation to probability.

Discussion Status

Participants are actively discussing the mathematical relationships involved in the problem, with some guidance provided on correcting differentiation errors and clarifying the reflection amplitude. Multiple interpretations of the coefficients are being explored, but no consensus has been reached on the final probability value.

Contextual Notes

The original poster expresses frustration with the problem's complexity and the teaching style of the professor, indicating a potential lack of clarity in the problem setup or expectations.

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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:

[tex]\psi_0(x) = A sin(k_0 x) + B cos( k_0 x)[/tex]
[tex]\psi_1(x) = C sin(k_1 x) + D cos( k_1 x)[/tex]
[tex]\psi_0'(x) = A k_0 cos(k_0 x) - B k_0 sin(k_0 x)[/tex]
[tex]\psi_1'(x) = C k_1 cos(k_1 x) - D k_1 sin(k_1 x)[/tex]

Plus values for k:

[tex]k_0 = \sqrt{\frac{2m}{\hbar^2} (7eV)}[/tex]
[tex]k_1 = \sqrt{\frac{2m}{\hbar^2} (1eV)}[/tex]

The Attempt at a Solution


First I solved the two wave functions evaluated at 0, which yields:
[tex]B = D[/tex]

Next, I solved the two derivatives evaluated at 0, which yields:
[tex]A k_0 = C k_1 \rightarrow C = \sqrt{7} A[/tex]

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.

[tex]\psi_0(x) = A' e^{i k_0 x} + B' e^{-i k_0 x}[/tex]
[tex]\psi_1(x) = C' e^{i k_1 x} + D' e^{-i k_1 x}[/tex]

[tex]\psi_0'(x) = i k_0 A' e^{i k_0 x} + i k_0 B' e^{-i k_0 x}[/tex]
[tex]\psi_1'(x) = i k_1 C' e^{i k_1 x} + i k_1 D' e^{-i k_1 x}[/tex]

D' = 0, we are firing particles from right to left, not left to right.

[tex]\psi_0'(0) = \psi_1'(0) \rightarrow i k_0 A' + i k_0 B' = i k_1 C'[/tex]
[tex]\psi_0(0) = \psi_1(0) \rightarrow A' + B' = C'[/tex]

[tex]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')[/tex]
[tex]i k_0 = i k_1[/tex]

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
 
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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
 
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?
 
i know what you mean -

& yep, you're right B/A should do it
 
Thanks again. I'm off to trudge through the rain and turn this sucker in.
 

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