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Homework Help: Minumum electron energy and microscopes

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data

    The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to "see" inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of, say, 10 pm.

    (a) If an electron microscope is used, what minimum electron energy is required?

    2. Relevant equations

    E = hc/λ
    λ = h/p (DeBroglie wavelength)
    p=sqrt(2mK) Momentum
    Δλ = (h(1-cos θ))/mc Compton Shift

    3. The attempt at a solution
    I tried the equation E = hc/λ = (6.63e-34)(3e8)/(1e-12) = 1.24e6 eV, but that isn't the answer.
  2. jcsd
  3. Dec 5, 2009 #2


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    Homework Helper

    E = hc/λ is for photons, not for electrons. What's the equation that gives an object's kinetic energy in terms of its momentum?
  4. Dec 5, 2009 #3

    So, do I use that equation in conjunction with DeBroglie's? λ = h/p This one is used for electrons, right?
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