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

The de-Broglie wavelength of the electrons is given by:
`lambda=h^(2)m_(c)V`
where, `m="Mass of the electron"=9.1xx10^(-31)kg`
`c="Charge on the electron"=1.6xx10^(-19)c`
`V="Accelarting potential"=50 KV`
`h="Plank's constant"=6.626xx10^(-34)Js`
`lambda=(h)/(p)=(h)/sqrt(2meV)orlambda=(12.27)/(sqrtV)A^(@)`
` therefore " " lambda=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx50xx10^(3)))`
`lambda=5.53xx10^(-12)`m.
The resolving power of an electron microscope is much better than that of optical microscope.