Question

An electron microscope uses eletrons accelerated by a voltage of $50kV$. Determine the deBroglie 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?

Answer 1

Consider the question

$v=50kV=50\times {10}^{3}volt$

K.E of electron $E=50\times {10}^{3}eV$

$\begin{array}{c}50\times {10}^3\times 1.6\times {10}^{-19}\\ 50\times 1.6\times {10}^{-16}J\\ de-Brogliewavelengthofelectronis\lambda =\frac{h}{\sqrt{2mE}}\\ =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.11\times {10}^{-31}\times 50\times 1.6\times {10}^{-16}}}\\ =5.5\times {10}^{-12}m\end{array}$

For yellow light , wave length $\lambda =5.9\times {10}^{-7}m$ since the revolving power is inversely proportional to wave length ,an electron microscope is about ${10}^5$ times that of optical microscope.