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, `lambda = ( h )/( sqrt( 2mE))`
Given data `:`
`h = 6.62 xx 10^(-34) Js, m = 9.1 xx 10^(-31) kg`,
` = 50 keV = 50 xx 1.6 xx 10^(-19) J`
`= 8 xx 10^(-15) J`
`lambda = ( 6.62 xx 10^(-34))/( sqrt( 2xx 9.1 xx 10^(-31) xx 8 xx 10^(-15)))m`
`= ( 6.62)/( sqrt( 145.6)) xx 10^(-11)m`
`= ( 6.62)/( 12.07) xx 10^(-11) m5.48 xx 10^(-12) m `
Wavelength of yellow light,
`lambda' = 5990 xx 10^(-10)m = 5.99 xx 10^(-7) m `
Now` ( lambda)/( lambda') = 10^(-5)`
Since resolving power is inversely proportional to wavelength therefore the resolving power of electron microscope in `10^(5)` times larger than the resolving power of optical microscope.