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 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

Here V=50 kV=`5xx10^(4)V,h=6.63xx10^(-34)Js`
`m_(e)=9.1xx10^(-31)`kg
`implies` Kinetic energy of electron,
K=eV
`=1.6xx10^(-19)xx5xx10^(4)`
`=8xx10^(-15)`J
`implies`de-Broglie wavelength of electron,
`lambda_(e)=(h)/(sqrt(2m_(e)K))`
`therefore lambda_(e)=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx8xx10^(-15)))` `=0.5494xx10^(-11)`
`~~5.5xx10^(-12)m`
`implies` Wavelength of yellow colour ,
`lambda_(y)=550xx10^(-9)` m
`implies` Resolving power of microscope `prop(1)/(lambda)`
`("Resolvin power of electron microscope")/("Resolving power of optical microscope")=(lambda_(y))/(lambda_(e))`
`=(550xx10^(-9))/(5.5xx10^(-12))`
`=10^(5)`
`therefore` Hence resolving power (R.P) of electron microscope is `10^(5)` time (R.P) of optical microscope.
`implies` Practically this ratio may change with geometrical changes.