p39_013

# p39_013 - time The energy of each photon is less so it must...

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13 .( a )L e t R be the rate of photon emission (number of photons emitted per unit time) and let E be the energy of a single photon. Then, the power output of a lamp is given by P = RE if all the power goes into photon production. Now, E = hf = hc/λ ,where h is the Planck constant, f is the frequency of the light emitted, and λ is the wavelength. Thus P = Rhc/λ and
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Unformatted text preview: time. The energy of each photon is less, so it must emit photons at a greater rate. (b) Let R be the rate of photon production for the 700 nm lamp. Then, R = λP hc = (700 nm)(400 J / s) (1 . 60 × 10 − 19 J / eV)(1240 eV · nm) = 1 . 41 × 10 21 photon / s . The result hc = 1240 eV · nm developed in Exercise 3 is used....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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