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# An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same

Q:11.33 An electron microscope uses electrons accelerated by a voltage of $\dpi{100} 50\hspace{1mm}kV$. 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?

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The potential difference through which electrons are accelerated(V)=50kV.

Kinetic energy(K) of the electrons would be eV where e is the electronic charge

The De Broglie wavelength associated with the electrons is

$\\\lambda =\frac{h}{\sqrt{2m_{e}K}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}\times 50000}}\\ \lambda =5.467\times 10^{-12} m$

The wavelength of yellow light = 5.9$\times$10-7 m

The calculated De Broglie wavelength of the electron microscope is about 10more than that of yellow light and since resolving power is inversely proportional to the wavelength the resolving power of electron microscope is roughly 10times than that of an optical microscope.

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