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!

Elektron beam strikes in potential barrier

  1. Apr 20, 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]
  2. jcsd
  3. Apr 20, 2010 #2
    Yes, the reflection coefficient will be 0.4. So just solve for [tex]V_0[/tex] using that equation and knowing E = 10 eV.
  4. Apr 23, 2010 #3
    I have calculted [tex] k_{1}= \sqrt{\frac{9.1\cdot10^{-31}\cdot16.02\cdot10^{-19}}{(6.626\cdot10^{-34})^{2}}[/tex]=[tex] 0.2728[/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}= 0.2728[/tex] does anyone know i am correct so far?
    if yes, how would you solve this ?
  5. Apr 23, 2010 #4
    Hmm... I wouldn't use meters as the units. Try to stick in eV and Angstroms. A couple of useful constants are:

    [tex]m_e c^2 = 0.511\,\text{MeV}[/tex]
    [tex]\hbar c \approx 2000\,\text{eV}\cdot\text{\AA}[/tex]

    But I get a very different answer than your [tex]k_1[/tex]. Remember that the wavelength of a particle is inversely related to the [tex]k[/tex] vector. So you are saying this electron has a wavelength on the order of meters. I think that is something you'd be able to see, which doesn't seem quite right.
  6. Apr 24, 2010 #5
    hello you are right! I forgot that [tex] \hbars [/tex] is not the same as h!!!so here is my correct calculation
    [tex] k_{1} = \sqrt{\frac{2mE}{\hbar^2}}=\sqrt{\frac{2\cdot9.1\cdot10^{-31}\cdot10eV}{(6.582\cdot10^{-16})^2}[/tex] =[tex] 6.48[/tex] is this value correct

    i have substituted m in [kg] and [tex] \hbar[/tex] in eVs

    why do you mention about Ångström? Do i need to use this unit in any of these calculations?
  7. Apr 24, 2010 #6
    You converted all the units to SI units except for energy. Why did you leave it as eV? Also what are the units of your [tex]k_1[/tex] in that problem.
  8. Apr 24, 2010 #7
    the unit for k should be [tex] \frac{1}{m}[/tex]can i express energy in J instead and do the same thing with [tex] \hbar[/tex] and express it in Js?
  9. Apr 24, 2010 #8
    If you want to solve it in SI units then you need to keep everything in SI units. So yes, convert eV to J, and make sure [tex]\hbar[/tex] is in SI units.
  10. Apr 24, 2010 #9
    in that case i get very big number.....
    [tex] \sqrt{\frac{2\cdot9.1\cdot10^{-31}\cdot1.602\codot10^{-18}}{(1.054\cdot10^{-34})^2}[/tex]= [tex]\sqrt{\frac{2.91564\cdot10^{-48}}{1.110916\cdot10^{-68}}[/tex] =[tex]\sqrt{2.624\cdot10^{20}}[/tex] = [tex]1.62\cdot10^{10}[/tex]

    which is extraordinary big.....dont you think?
  11. Apr 24, 2010 #10
    It seems very reasonable. Remember that the wavelength of the electron is inversely proportional to the k-vector. So that looks like a wavelength on the order of an angstrom. So that is better looking than meters.
  12. Apr 24, 2010 #11
    okey i am glad to hear that.
    do you have any idea how to solve the equation for R to obtain [tex] k_{2}[/tex] (which should be the next step to calculate [tex] V_{0}[/tex])?

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








    but then i get 2 values for [tex] k_{2}[/tex]
    Last edited: Apr 25, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook