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 uses yellow light?

Answer

$\lambda =\frac{\text{h}}{\text{p}} ={\frac{\text{h}}{\sqrt{2\text{meV}}}} \text{ or } \lambda = \frac{12.27}{\sqrt{\text{V}}}\text{A}^{o}$
$\therefore\lambda = \frac{6.63\times10^{-34}}{\sqrt{(2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50\times10^{3}})}$
$\lambda = 5.33\times10^{-12}\text{m}$
The resolving power of an electron microscope is much better than that of optical microscope.

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