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Wave Particle Duality

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Microscopes are inherently limited by the wavelength of the light used. How much smaller (in order of magnitude) can we "see" using an electron microscope whose electrons have been accelerated through a potential difference of 50,000V than using red light(500nm)?

    2. Relevant equations

    k=1/2mv2


    3. The attempt at a solution

    Don't know how to start
     
  2. jcsd
  3. Nov 17, 2009 #2
    You would need to calculate the Debroglie wavelength of this highly accelerated electron. I'm going to assume that they taught you the equation for debroglie wavelength.

    Apparently wave-partiicle duality describes every massive object as exhibiting wave like behaviour at high speeds.

    You know that the kinetic of an electron = eV
    1/2mv^2 = eV

    But you need to find the momentum (p = mv) of this said electron in order to work out it's wavelength.

    Have a go at deriving this momentum of electron formula in terms of it's kinetic energy, if you can't I will help you.

    Anywho, if you manage, plug in the numbers for momentum, then plug momentum into your de broglie equation. The wavelengths will obviously be different than that of red light, you may then decide why it is better to use electrons.
     
  4. Nov 17, 2009 #3
    Thank you for your help,
    I am not sure if I'm on the right track but I used the kinetic energy formula to find velocity then I used that velocity in the lambda=h/mv formula to find the wavelength, is that right?
     
  5. Nov 17, 2009 #4
    No.

    If you simply equated eV = 1/2mv^2, you would end up getting a relativistic value for the speed which is ridiculous.

    -----------------------------------------

    Here is how to derive the momentum of an electron:

    p = mv (1)

    eV = 1/2mv^2 (2)

    These are 2 separate expressions

    What we can do to number (2) is rearrange it for v and substitute into equation number (1)#

    So number (2) becomes:

    sqrt(2eV/m) = v (where aqrt stands for square root). Substitute this into v for (1):

    p = m sqrt(2eV/m)

    Square both sides and get rid of one of the m's, and square root again. Final expression for momentum should be:

    p = sqrt(2meV)

    h (plancks constant) is 6.63E-34

    You should end up getting a wavelength of 5.49E-12m, I'm sure you can imagine why this would give a better resolution (when veiwing a microscope) than an optical microscope with wavelengths of 500nm.

    ----------------------------------

    HOWEVER

    I'm unsure whether to bring relativity into this. Because I did use KE = eV straight off the bat and the electron was moving at an extremely relativistic velocity. I decided I would calculate it's momentum by using it's rest mass, kinetic energy and total energy, but I ended up with an unuseable value. If you were supposed to apply laws of relativity to this equation then ignore what I said.

    Can a more experienced individual please advise, I don't want to give some wrong answers.
     
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