The deBroglie wavelength of a particle, lambda = h/p, describes the wave particl

Question

The deBroglie wavelength of a particle, lambda = h/p, describes the wave particle duality for small particles such as electrons. (a.)What is the deBroglie wavelength (in nanometers) of an electron with 10 eV energy? (b.)What is the deBroglie wavelength for a 10 keV electron -- typical of the electron beam energy in an electron microscope? (c.) How does the wavelength of the 10 keV electron compare with the wavelength of visible light that would be used in an optical microscope? (d.) What does this tell you about the usefulness of an electron microscope versus an optical microscope in studying nanoscale structures? Note: 10 eV and 10 keV electrons are nonrelativistic, so you can use classical physics to calculate their momentum.

Explanation / Answer

a)

The wavelength of a 10 eV photon is given by:

l = h c / Eph = 6.625 x 10-34 x 3 x 108/(10 x 1.6 x 10-19 ) = 124.2 nm.

b)

The wavelength of a 10K eV photon is given by:

l = h c / Eph = 6.625 x 10-34 x 3 x 108/(104 x 1.6 x 10-19 ) = 0.1242 nm.

c) An electron microscope is a microscope that uses a beam of accelerated electrons as a source of illumination. Because the wavelength of an electron can be up to 100,000 times shorter than that of visible light photons, the electron microscope has a higher resolving powerthan a light microscope and can reveal the structure of smaller objects.

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