An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors ( such as numerical aperture, etc. ) are taken roughly to be the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light (`lambda_(y) = 5.9 xx 10^(-7) m ) `.

Explanation

de Broglie wavelength asscoiated with electron
`lambda = ( 12.27)/( sqrt( V)) xx 10^(-10) m`
Here `V = 50 kV = 50 xx 10^(3) V`
`lambda= ( 12.27 )/( sqrt( 50 xx 10^(3))) = 5. 5 xx 10^(-12) m`
Wavelength of yellow light,
`lambda_(y) = 5.9 xx 10^(-7) m `
The resolving power of an electron microscope is given by
`RP =( 1)/( d_("min")) = ( 2mu sin beta)/( 1.22 lambda)`
Where `d_("min")` = minimum separation
For constant numerical aperture
Resolving power of microscope `prop (1)/( lambda)`
`:. ("Resolving power of electron microscope")/("Resolving power of optical microscope") = ( lambda_(y))/( lambda)`
`= ( 5.9 xx 10^(-7))/( 5.5 xx 10^(-12)) ~~10^(5)`
That is, resolving power of electron microscope is `10^(5)` tmes the resolving power of opical microscope