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12 January, 15:13

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons would one need in an electron microscope to resolve separations of

(a) 150 Angstrom

(b) 5 Angstrom

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Answers (1)
  1. 12 January, 17:01
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    Answer: a) for 150 Angstroms 6.63 * 10^-3 eV; b) for 5 Angstroms 6.02 eV

    Explanation: To solve this problem we have to use the relationship given by De Broglie as:

    λ = p/h where p is the momentum and h the Planck constant

    if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

    Finally we have:

    eΔV=p^2/2m = h^2 / (2*m*λ^2)

    replacing we obtained the above values.
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