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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?

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Solution

Given:- Accelerating voltage=V=50kV
=50×103V
Solution:-Using de-broglie's equation.
λ=hp(1)
λ=Wavelength of moving particle.
P=Momentum of moving particle.
h=Planck's constant=6.63×103Js
now from mechanics we know,
P=2mk=2mqV
Where,P=momentum of particle
M=mass of particle.
K=Kinetic energy of particle.
V= Electric potential of particle( Accelerating voltage).
q=charge of particle.
In the case of electron,q=e=1.6×1019C
Substitution of (1)
λ=h2meV
λ=6.63×10342×9.1×1031×1.6×1019×50×103
λ=6.63×1034120.66×1024
λ=5.467×1012m
Whereas wavelength of yellow light, λy=590nm
=5.9×107m
The wavelength of the accelerated electron is about 105 times less than the yellow light.
Since resolving power of a microscope is inversely proportional to the wavelength of light hence the resolving power of the electron microscope is about 105 times more than that of an optical microscope.

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