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PHYS2DQuiz3Solutions

# PHYS2DQuiz3Solutions - and f df in terms of the energy...

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PHYS 2D Quiz 3 Solutions 1a.) Classical Physics (Rayleigh-Jeans Law) gives the power density of radiation increasing without limit as the wavelength goes to zero, while experimentally it goes to zero there. In the case of the Photoelectron effect, it is observed that the kinetic energy of the emitted photoelectrons scales with the frequency of the incident light, but not with the intensity of the incident radiation. In the case of the Photoelectron effect, it is observed that the time taken for the electrons to come out is much faster than that calculated from classical theory. In the case of the Photoelectron effect, it is observed that light with frequency less than a critical frequency f 0 does not emit photoelectrons no matter how great the intensity. 1b.) Derive the equivalent of Planck s function for the energy density of radiation between frequencies f
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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» =1.8x10-7 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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