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Homework Help: 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]

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

    [tex](\sqrt{R}k_{1}+\sqrt{R}k_{2})^2-(k_{1}-k_{2})^2=0[/tex]

    [tex]\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}-k_1+k_2\right)\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}+k_1-k_2\right)=0[/tex]

    [tex]\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}-k_1+k_2=0\right)\quad\vee\quad\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}+k_1-k_2=0\right)[/tex]

    [tex]\left(-k_1(1-\sqrt{R})+k_2(1+\sqrt{R})=0\right)\quad\vee\quad\left(k_1(1+\sqrt{R})-k_2(1-\sqrt{R})=0\right)[/tex]

    [tex]\left(k_2(1+\sqrt{R})=k_1(1-\sqrt{R})\right)\quad\vee\quad\left(k_1(1+\sqrt{R})=k_2(1-\sqrt{R})\right)[/tex]

    [tex]\left(k_2=k_1\frac{(1-\sqrt{R})}{(1+\sqrt{R})}\right)\quad\vee\quad\left(k_1\frac{(1+\sqrt{R})}{(1-\sqrt{R})}=k_2\right)[/tex]

    but then i get 2 values for [tex] k_{2}[/tex]
     
    Last edited: Apr 25, 2010
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