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Quantum Mechanic. penetration distance and probability Density

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Assume that a typical electron in a piece of metallic sodium has energy [itex]- E_{0}[/itex] compared to a free electron, where [itex]E_{0}[/itex] is the 2.7 eV work function of sodium.

    At what distance beyond the surface of the metal is the electron's probability density 20% of its value at the surface?

    2. Relevant equations

    [itex]η=\frac{\hbar}{\sqrt{2m( U_{0} - E)}}[/itex]

    [itex]ψ_{(x)}=ψ_{edge} e^{-(x-L)/η}[/itex]

    3. The attempt at a solution

    I've assumed the surface of the sodium metal is the barrier. I have also assumed [itex]U_{0}-E= 2.7eV[/itex]


    Given that the probability density must be 20% at a distance from the surface I've used the following method to get my answer and want to check if my assumptions/work are correct.

    I've worked [itex]η≈1.19*10^{-10}m[/itex]

    Because I need probability density to be 20% of what it would be at the barrier I've done the following;

    [itex]0.2=|ψ|^{2}[/itex]

    [itex]ψ_{(x)}=ψ_{edge} e^{-(x-L)/η}[/itex] ∴ [itex]|ψ_{(x)}|^{2}=(ψ_{edge} e^{-(x-L)/η})^{2}[/itex]

    [itex]\sqrt{0.2}=ψ_{edge} e^{-(x-L)/η}[/itex]

    Given x value is going to be x value of barrier + zη

    [itex]0.447=ψ_{edge} e^{-(zη)/η}[/itex]

    [itex]ln 0.447 =ψ_{edge} -(zη)/η[/itex]

    [itex]-0.805=ψ_{edge} -(zη)/η[/itex]

    [itex]-0.805=-z[/itex]

    [itex]z=0.805[/itex]

    Would the answer be 0.805η (0.096 nm)

    Any help would be greatly appreciated :)
     
    Last edited: May 25, 2012
  2. jcsd
  3. May 26, 2012 #2
    Problem is now solved, the answer and method was correct
     
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