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A few questions about Quantum stuff

  1. Feb 3, 2005 #1
    I have the eqns, and have started out on these problems...but I'm not sure where to go at certain points.

    1.) It takes 1.9 eV to free an electron from cesium.

    a.) Find the longest wavelength light which can free the electron.


    Well, we have the general equation from the photoelectric effect

    Ephoton = Eeject + kinetic

    Ephoton >= Eeject in order to free an electron.

    Therefore:

    hc/lambda >= Eeject

    [h/(1.609*10^-19)]*c/Eeject >= lambda

    I'm not sure if this is lambda max though.

    b.) Find the potential required to stop th eelectron if the incident light is 250 nm, and 350 nm.

    I'm basically using the same equation as before, but solving for energy, by plugging in lambda. Is this correct?

    _ _ _ _ _ _

    2.) In the double slit experiment (slits separated by distance d), with a light bulb behind the slits, what momentum photons will disturb the interference pattern? What type of light is this (e.g. visibile, X-Ray), and what about if d is atomic size?

    Now the interference is disturbed IF d > lambda. But what equation should I use for this?

    hc/lambda again?

    When doing that, I got:

    hc/lambda >= 1ev (because it's one electron). Is this the way to go?

    Once again, thanks for any and all help, I have a few more questions, but I want to work on them some more.

    I would go to office hours as well...but it's snowing HARD!
     
  2. jcsd
  3. Feb 3, 2005 #2
    22 views and no reply? Somebody tell me if I'm on the right track! :cool:
     
  4. Feb 3, 2005 #3

    Doc Al

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    Staff: Mentor

    Right (assuming you use correct units). This is the smallest energy photon that can free an electron, so it has the greatest wavelength.

    You can use that equation to find the energy of each wavelength photon, then use that photon energy to calculate the KE of the freed electron.


    I do not know what they mean by "disturb" the pattern.
     
  5. Feb 3, 2005 #4
    Disturb as in, momentum of the photon is greater than the momentum of the electron, therefore messing up the normal path of the electron.

    Thanks for your help on the others!
     
  6. Feb 4, 2005 #5

    Doc Al

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    Staff: Mentor

    Ah.... you're talking about the double slit experiment with electrons where you shine light on the electrons after they pass the slits. So you know the approximate wavelength that will disturb the pattern. What kind of light has a wavelength of atomic size? Look on a chart of the electromagnetic spectrum.
     
  7. Feb 4, 2005 #6
    Do I use the formula

    hc/lambda >= Eeject once again?

    And I assume that Eeject = 1eV since it's one electron.
     
  8. Feb 4, 2005 #7

    Doc Al

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    Staff: Mentor

    If you talking about problem 2, no. You don't need any formulas for problem 2.

    That makes no sense. (It's always one electron!) Eeject is the energy needed to free an electron from some bound state, like an electron bound in an atom. The electrons in the double slit are not bound.
     
  9. Feb 4, 2005 #8
    Oh, I see. But in that case, how do I calculate the required momentum, if I don't need a formula?

    Also, I have one other problem:

    3.) Say we trap an electron in a 1-d box. What is it's lowest velocity if the box is 1mm, what if the box were atomic size?


    I'm using the uncertainty principle, so:

    deltaP*deltax >= h/4pi

    so i get deltax = .001 and so

    deltaP >= 5.27*10^-32

    but what feasible equation is there for relating momentum and velocity, since p=mv, will not work in this case?

    and what do they mean by atomic size?
     
  10. Feb 4, 2005 #9

    Doc Al

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    momentum of a photon

    Oops... my bad: I didn't see that you were asked to calculate the momentum of the photons. The momentum of a photon is [itex]p = E/c = h/\lambda[/itex].
     
  11. Feb 4, 2005 #10

    Doc Al

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    OK.

    Why won't it? Use [itex]\Delta p = m \Delta v[/itex]. (The entire problem is something of a hand-waving argument, but I'm sure that's what they want you to do.)

    What if the size of the box were on the order of an atomic diameter. Just use [itex]10^{-10}[/itex] m. (Close enough!)
     
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