An electron microscope uses electrons accelerated by a voltage of `50kV`. Determine the De Broglie wavelength associated with the 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

Given , V = 50 KV = 50000 V
` lambda = (12.27 ) lambda = (12.27 )/(sqrt(V)) = (12.27 )/(sqrt50000)= 0.055 rArr lambda = 5.5 xx 10^(-12) m , ` For yellow leight `(lambda) = 5.9 xx10^(-7)` m
As resolving power (RP) `prop 1/lambda rArr ("RP of electron microscope ")/(" RP of optical microscope ") = (lambda_(r))/(lambda_(e)) = (5.9 xx 10^(-7))/(5.5 XX10^(12)) = 10^(5)`
` :. ` RP of electron microscope = `105 xx "RP of optical microscope "`