Barrier width for 100% transmission of 5 eV electron through 3 eV potential

[mnoir]

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?

 

[vela]

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.

 

[dgreenheck]

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.

 

[mnoir]

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.

 

[vela]

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.