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=50xx10^(3)`Volt
`:.` K.E. of electron, `E=50xx10^(3)eV=50xx10^(3)xx1.6xx10^(-19)J=50xx1.6xx10^(-16)J`
`:.` de -Broglie wavelength of electron is `lambda=h/(sqrt(2mE))=(6.63xx10^(-34))/(sqrt(2xx9.11xx10^(-31)xx50xxxx1.6xx10^(-16))) =5.5xx10^(-12)m`
For yellow light, wavelength `lambda=5.9xx10^(-7)m`
Since resolving power (R.P) is inversely proportional to wave length, therefore, R.P. of an electron microscope is about `10^(5)` times that of an opticalk microscope. In practice, differences in other (geometrical) factors can change their comparison somewhat.