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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. 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 used yellow light?


The de-Broglie wavelength of the electrons is given by: λ=h2meV Here: m = mass of the electron = 9.1×10-31 kg e = charge on the electron = 1.6×10-19 C V = accelerating potential = 50 kV h = Planck's constant = 6.626×10-34 Js ⇒λ=6.626×10-3429.1×10-311.6×10-1950×103⇒λ=0.0549 Ao Resolving power of a microscope, R=2μsin θλ This formula suggests that to improve resolution, we have to use shorter wavelength and media with large indices of refraction. For an electron microscope, μ is equal to 1(vacuum). For an electron microscope, the electrons are accelerated through a 60,000 V potential difference. Thus the wavelength of electrons is found to be, λ=12.27V=12.2760000=0.05 Å As, λ is very small (approximately 10-5 times smaller) for electron microscope than an optical microscope which uses yellow light of wavelength (5700 Å to 5900 Å). Hence, the resolving power of an electron microscope is much greater than that of optical microscope.
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