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

`lambda = (1.227)/(sqrt(50, 000)) nm`.
`5 xx487 xx 10^(-3) nm`
Resolving power `prop(1)/(lambda)` [for other factors being fixed] `:. (RP_(em))/(RP_(y,m))=(550 nm)/(5.487xx10^(-3)nm)`
`~~(5500xx10^(3))/(5.5)`
`~~ 100xx10^(3)`
So, `RP` of electron microscope is `10^(5)` times of microscope. Wave length of accelerated electron in electron microscpoe is very small as compared with wavelength of yellow light.