We wish 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. If an electron microscope is used, the minimum electron energy required is about

From de-Broglie equation, we haveλ=hp=hmvwhere λ is wavelength, ft is Planck's constant, z is mass, p is momentum and v is velocity.Given,  λ=10pm=10−11m,  m=9.1×10−31kg,  h=6.6×10−34Js∴v=hmλ=6.6×10−349.1×10−31×10−11=7.25×107m/sEnergy  of  electron=12mv2=12×9.1×10−31×7.25×10721.6×10−19=15keV