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Unformatted text preview: and f+df in terms of the energy density between wavelengths and +d We are given: ? ? , ?? = 8 ? 3 3 ?? ? ? / 1 Now: ? , ? = ? ( ? , ) ?? ? ? Substitute in these expressions: ? = and ?? ? = 2 This is the end result: ? , ? = 8 5 ? ? / 1 2 A photon, =1.8x107 m, is absorbed on a Molybdenum surface, = 4.2 eV, and emits an electron. a.) Calculate the momentum of the incident photon in kg m/s. = = 3.68 10 27 kg m s b.) Calculate the velocity of the emitted electron in m/s. First, convert the work function to joules. 1 2 2 = = 2 = 9.72 10 5 m s...
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This note was uploaded on 05/07/2009 for the course PHYS 2D taught by Professor Hirsch during the Spring '08 term at UCSD.
 Spring '08
 Hirsch
 Power, Radiation

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