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

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.

cant without the picture! ?

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray