Electrons in an electron microscope have a kinetic energy of

Electrons in an electron microscope have a kinetic energy of 5.11 105 eV(a) Find the de Broglie wavelength of the electrons.(b) Find the ratio of this wavelength to the wavelength of light at the middle of the visible spectrum (550 nm). (c) How many times greater magnification is theoretically possible with this microscope than with a light microscope?

Solution

a) lambda = hc / energy

lambda = 1240 eV-nm / (5.11 x 10^5 eV ) = 2.43 x 10^-3 nm   = 2.427 x 10^-12 m

b) Ratio = 2.427 x 10^-12 / (550 x 10^-9)   = 4.41 x 10^-6

c) m = 1/ratio = 226653.23