Transmission and reflection coefficients for potential barrier

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Homework Help Overview

The discussion revolves around the transmission and reflection coefficients of an electron encountering a potential barrier, specifically focusing on the conditions under which the transmission probability reaches one hundred percent and the reflection probability reaches zero. The problem involves quantum mechanics concepts related to potential barriers and wave functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expressions for the reflection and transmission coefficients, questioning their correctness and implications for real-number outputs. There is an exploration of setting the transmission coefficient to one and the reflection coefficient to zero, with attempts to simplify the equations by manipulating trigonometric functions.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions about the expressions used for R and T. Some participants suggest specific conditions under which the coefficients can be simplified, while others caution against overlooking potential solutions. There is no explicit consensus on the validity of the initial expressions or the approach to solving for the width of the potential barrier.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information they can share or the methods they can use. The original poster's calculations and assumptions are under scrutiny, and there is an emphasis on ensuring that the coefficients yield valid physical interpretations.

mnoir
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Homework Statement



An electron with kinetic energy 5 eV (8.01E-19 J) passes through a 3 eV (4.806E-19 J) potential barrier. There are certain widths for this potential barrier in which the transmission probability will equal one hundred percent and the reflection probability will equal zero. Find the smallest non-trivial width in which this occurs.

Homework Equations



R = [i(q2-k2)sin(2qa)exp(-2ika)] / [2kqcos(2qa)-i(k2+q2)sin(2qa)]

T = [2kqexp(-2ika)] / [2kqcos(2qa)-i(k2+q2)sin(2qa)]

k = sqrt(2mE) / hbar

q = sqrt[2m(E-V0)] / hbar

width = 2a
i is imaginary
q = sqrt[2m(E-V0)] / hbar
k = sqrt(2mE) / hbar
m=9.1094E-31 kg
hbar=1.0546E-34 J*s

The Attempt at a Solution



k=1.15E10

q=7.24E9

From here I think you would set T=1 and R=0 and solve for a. I'm not sure if this method is correct, but if so, after plugging in k and q, how do I solve for a if it's in cos, sin, and exp?
 
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Are you sure your expressions for R and T are correct? They should be real numbers, but your expressions don't look like they'll yield real number answers.
 
Remember R and T are the reflection and transmission "coefficients" which are between 0 and 1 while r and t are the reflection and transmission "amplitudes" which are (k1-k2)/(k1+k2) and (2k1)/(k1+k2), respectively.
 
As far as I know these are the correct expressions, according to my professor.

For now, assuming these expressions are correct, this is what I have come up with (I could be on the wrong track though):

Set T=1. The numerator and denominator both have 2kq, so find cases in which cos=1 and sin=0 in order to get rid of i(k2+q2)sin(2qa).

T=[2kqexp(-2ika)]/[2kqcos(2qa)] =1
2kq's cancel out
when 2qa=0, 2pi, etc., cos=1
so if you set 2qa=2pi then a=pi/q
solve for q using q=sqrt[2m(E-V0)]/hbar
plug q into solve for a.
Then multiply 2*a in order to find width.
Checking the value I got for a (4.34E-10) by plugging it into cos(2qa) and sin(2qa) I get 1 and zero, respectively.
 
You don't get T=1 under those conditions. You get T=e-2ika. You should understand what dgreenheck is saying, because you're omitting some solutions right now.
 

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