An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with he electrons. If other factors ( such as numerical aperture etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Explanation

Wavelength of electron,
`lamda_(e) = (h)/(sqrt(2meV))`
`= (6.626 xx 10^(-34))/((2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx 50 xx 10^(3))^((1)/(2)))m`
`= 0.0549 Å`
Resolving power of a microscope `prop (1)/(lamda)`
Wavelength of yellow light, `lamda_(y) = 5900 Å`
`:. ("resolving power of electron microscope")/("resolving power of optical microscope")`
`= (lamda_(y))/(lamda_(e)) = (5900)/(0.0549)= 1.07 xx 10^(5)`