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 what of an optical microscope which uses yellow light?

Answer 1

Kinetic energy gained by the electron equals the change in electric potential energy.

$K=eV..............\left(i\right)$

De-broglie wavelength of electron is given by:

$\lambda =\frac{h}{p}..............\left(ii\right)$

But $K=\frac{p^2}{2m}.........................\left(iii\right)$

From (i), (ii) and (iii)

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

$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 8\times {10}^{-15}}}$

$\lambda =5.467\times {10}^{-12}m$

This wavelength is about ${10}^5$ times less than the yellow light. Since resolving power of a microscope is inversely proportional to the wavelength of light, resolving power of electron microscope is about ${10}^5$ times more than that of a optical microscope.