Unformatted text preview: and f+df in terms of the energy density between wavelengths Î» and Î» +d Î» We are given: ? ? , Â°Â±?? = 8 Â²â„Ž? 3 Â³ 3 ?? ? â„Ž? / Â´Â° âˆ’ 1 Now: ? Î» , Â°Â±?Î» = ? ( ? , Â° ) Âµ ?? ?Î» Âµ ?Î» Substitute in these expressions: ? = Â³ Î» and Âµ ?? ?Î» Âµ = Â³ Î» 2 This is the end result: ? Î» , Â°Â±?Î» = 8 Â²â„ŽÂ³ Î» 5 ?Î» ? â„ŽÂ³ / Î» Â´Â° âˆ’ 1 2 A photon, Î» =1.8x107 m, is absorbed on a Molybdenum surface, Ï† = 4.2 eV, and emits an electron. a.) Calculate the momentum of the incident photon in kg m/s. Â¶ = â„Ž Î» = 3.68 âˆ™ 10 âˆ’ 27 kg âˆ™ m s b.) Calculate the velocity of the emitted electron in m/s. First, convert the work function to joules. 1 2 Â·Â¸ 2 = â„ŽÂ³ Î» âˆ’ Ï† â†’ Â¸ = Â¹ 2 Â· Âº â„ŽÂ³ Î» âˆ’ Ï†Â» = 9.72 âˆ™ 10 5 m s...
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 Spring '08
 Hirsch
 Energy, Power, Photon, Light, Radiation, Fundamental physics concepts, Photoelectron effect

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