The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo uld one need in an electron microscope to resolve separations of (a) 150 Angstrom
(b) 5 Angstrom
1 answer:
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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