1. Limited time only! Sign up for a free 30min personal 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!

Electron beam, potential barrier

  1. Apr 24, 2010 #1
    Electron beam with kinetic energy [tex] E_{k} = 10 eV[/tex] strikes a positive potential barrier [tex] V_{0}[/tex] and the kinetic energy after the beam has passed through the barrier is [tex] E_{k} = (10 eV -V_{0})[/tex].
    How big potential [tex] V_{0}[/tex] is needed so that 40% of the electron beam is going to be reflected?
    What would happen if we now make the potential negative so the electron beam will gain the kinetic energy?

    I would say that the energy of the particles is higher than the energy of the potential barrier, thats why we observe transmission and reflection
    solving The Schrödinger equation

    [tex] \frac{\partial^2}{\partial x^2}\psi(x)+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0[/tex]
    then the solutions will be

    [tex] \psi_{1}=Ae^{ik_{1}x}+Be^{-ik_{1}x}[/tex] in the region x<0 [tex] k_{1}= \sqrt{\frac{2mE}{\hbar^2}}[/tex]
    [tex]\psi_{1}=Ce^{ik{2}x}[/tex] in the region x>0 [tex]k_{2}= \sqrt{\frac{2m[E-V_{0}]}{\hbar^2}}[/tex]

    the reflection coefficient is [tex] R= (\frac{k_{1}-k_{2}}{k_{1}+k_{2}})^2[/tex]

    Can someone help me with the solution?
    Is the reflection coefficient going to be 0.4? How to find that value of [tex] V_{0}[/tex][/QUOTE]

    I have calculted [tex] k_{1}= \sqrt{\frac{9.1\cdot10^{-31}\cdot10}{(6.582\cdot10^{-16})^{2}}[/tex]=[tex] 6.48[/tex]

    [tex]k_{1}= \sqrt{\frac{2mE}{\hbar^2}}= \sqrt{\frac{kg\cdot J}{J^2\cdot s^2}[/tex][tex]=\sqrt{\frac{kg}{N^2\cdot s^2}[/tex]=[tex] \frac{1}{m}[/tex]

    then i am trying to solve the equation with R

    [tex] R= (\frac{k_{1}-k_{2}}{k_{1}+k_{2}})^2[/tex] to calculate [tex] k_{2} [/tex] and finally calculate [tex] V_{0}[/tex] from the equation for [tex] k_{2} [/tex]

    but i am stuck here....

    [tex] R= \frac{k_{1}^2-2k_{1}k_{2}+k_{2}^2}{k_{1}^2+2k_{1}k_{2}+k_{2}^2}[/tex]
    and then i substitute for R= 04 and for [tex] k_{1}= 6.48[/tex] does anyone know i am correct so far?
    if yes, how would you solve this ?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Apr 24, 2010
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Electron beam, potential barrier
  1. Barrier Potential (Replies: 16)

  2. Potential Barrier (Replies: 0)

  3. Potential Barrier (Replies: 1)