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Quantum Barrier Potential (should be easy)

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A 10-eV electron is incident on a potential barrier of height 25eV and width 1 nm.

    Use equation 6-76 to calculate the order of magnitude of the probablity that the electron will tunnel through the barriers.

    Repeat with width .1nm

    2. Relevant equations


    Equation 6-76 is:

    E = Electron Enegy
    V = Potential
    a = width

    T = 16 (E/V) (1- (E/V) e -2aw

    To get w

    w = sqrt ( 2m (V-E)) / hbar

    where m = mass of electron
    hbar= 6.582119 x 10-16 eV*s



    3. The attempt at a solution

    I know these are the right answers ( they are given)
    4.95 x 10-13 and 0.197

    Seems to be a simple plug and chug problem, but, However, when I plug the known values into my calculator, the expontential goes to 0. I dont know if my dimensional analysis matched up or not (im using hbar in ev*s though), or maybe my calc is limited in the size of the numbers it can display.

    Also, IM not sure what 'order of magnitude' means in the context of the question.
     
  2. jcsd
  3. Oct 17, 2008 #2
    I suspect you have an error in units when you are finding w:

    If you use m in kg, and the energies of the electron and barriers in eV, and hbar as you've given, then your units are sqrt(kg)*sqrt(eV)/(eV*s) = sqrt(kg/eV)/s.

    This does not reduce to inverse meters, which you'd need to cancel out the width of the barrier in the exponent (assuming you are using the width in meters).

    I suggest converting your energies of the electron and barrier to SI units (Joules) and using hbar in SI units (Joules*s).

    It doesn't matter when you find the factor before the exponent (because there one uses ratios of energies... units cancel since the conversion is multiplicative)... but it does matter in the exponent.

    Also: sometimes when the probabilities are low, calculators can just end up giving you zero... you should write out all your steps if this is happening and do the math with the numbers and the factors of ten separately if possible.
     
  4. Oct 17, 2008 #3
    But, if you use the mass of the electron in MeVs, everything's gravy! Why convert all the factors into Joules when you could convert the one remaining factor into eVs?
     
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