1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reflection at a Step Barrier

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle of energy E approaches a step barrier of height U0. What should be the ratio E/U0 so that the reflection coefficient is 0.43?

    2. Relevant equations

    R=(k1-k2)^2/(k1+k2)^2 <--------reflection coefficient

    3. The attempt at a solution

    I am completely stumped at how to approach this problem. I cannot find how to get a ratio of E to U0 using any equations. Any help on how to approach this problem is much appreciated.
     
  2. jcsd
  3. Oct 7, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Well, how did you define [itex]k_1[/itex] and [itex]k_2[/itex] when deriving the above reflection coefficient equation?
     
  4. Oct 7, 2009 #3
    I had k1=sqrt(2mE/(h-bar)^2) and k2=sqrt(2mK2/(h-bar)^2), where K2 is the final kinetic energy.
     
  5. Oct 7, 2009 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Okay, so doesn't that mean K2=E-U0 and k2=sqrt(2m(E-U0)/(hbar)^2)?

    Substitute k1 and k2 into your expression for R and simplify...
     
  6. Oct 7, 2009 #5
    I don't know how the K2=E-U0 comes into play. However, if I substitute in k1 and k2 into the R expression, it looks like:

    R=(k1^2-2k1k2+k2^2)/(k1^2+2k1k2+k2^2). This ends up being a complete mess and I have some E and U0 terms that stand alone and some E and U0 terms that are stuck inside square roots, so I can't get it to a ratio. I can't tell what I'm doing wrong.
     
  7. Oct 7, 2009 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    A particle with energy E passes through a potential barrier of height U0....doesn't that mean its final energy is E-U0?:wink:

    Don't expand the squares...just divide everything by k1:

    [tex]R=\left(\frac{k_1-k_2}{k_1+k_2}\right)^2=\left(\frac{1-\frac{k_2}{k_1}}{1+\frac{k_2}{k_1}}\right)^2[/tex]

    And [itex]\frac{k_2}{k_1}=[/itex]____?
     
  8. Oct 7, 2009 #7
    Okay thanks. That way really helps!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook