Question

An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc., to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which used yellow light?

Answer 1

The de-Broglic wavelength of the electrons is given by :

$\lambda =\frac{h}{\sqrt{2meV}}$

Where,

m = mass of the electron = $9.1\times {10}^{-31}$ kg

e = charge on the electron = $1.6\times {10}^{-19}C$

V = accelerating potential = $50kV$

h = Planck's constant = $6.626\times {10}^{-34}$Js

$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2(9.1\times {10}^{-31})(1.6\times {10}^{-19})(50\times {10}^3)}}$

$\Rightarrow \lambda =0.0549\stackrel{\circ }{A}$

Resolving power of a microscope, $R=\frac{2\mu sin\theta }{\lambda }$

This formula shows that to enhance resolution, we have to use shorter wavelength and media with large indices of refraction.

For an electron microscope, $\mu$ is equal to 1 (vacuum).

For an electron microscope, the electrons are accelerated through a 60,000 V potential difference.

Thus, the wavelength of electrons is given by,

$\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{60000}}=0.05\stackrel{\circ }{A}$

As, $\lambda$ is very little (roughly ${10}^{-5}$ times smaller ) for electron microscope than an optical microscope which uses yellow light of wavelength $\left(5700\stackrel{\circ }{A}to5900\stackrel{\circ }{A}\right)$. So, the resolving power of electron microscope is about ${10}^5$ greater than that of optical microscope.