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

Here, `V=50kV`. Therefore, energy of electrons,
`E=50kV=50xx10^(3)xx1.6xx10^(-19)=8.0xx10^(-15)J`
Now `lamda=(h)/(sqrt(2me))`
Taking `m=9.1xx10^(-31)kg`, we have
`lamda=(6.62xx10^(-31))/(sqrt(2xx9.1xx10^(-31)xx8.0xx10^(-15)))`
`=(6.62xx10^(-34))/(1.207xx10^(-22))=5.485xx10^(-12)m`
The resolving power of a microscope is inversely proportional to the wavelength of the radiation useed. Since wavelength of the radiation used. Since wavelength of the yellow light is `5990A` i.e.,`5.99xx10^(-7)m`, power of electron microscope is `10^(2)` times as large as that of the optical microscope.