# p137bp8sol - Physics 137B, Fall 2007, Moore Problem Set 8...

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Unformatted text preview: Physics 137B, Fall 2007, Moore Problem Set 8 Solutions 1. (Bransden 11.1) Number of photons per second = 1W Energy per photon = 1W hc/ = 5 . 034 10 24 s ( )m- 1 Photon flux = Number of photon per second 4 R 2 = 4 . 006 10 21 m 2 s ( )m- 1 10m wavelength: 5 . 10 25 photons second , 4 . 10 22 photons m 2 s 10cm wavelength: 5 . 10 23 photons second , 4 . 10 20 photons m 2 s 5890 A wavelength: 2 . 965 10 18 photons second , 2 . 360 10 15 photons m 2 s 1 A wavelength: 5 . 10 14 photons second , 4 . 10 11 photons m 2 s 1 (Cross section 1s 2p) The rate of absorption from 1s to 2p is given by: (Bransden 11.57) W = I ( ) 3 h 2 c 2 | D | 2 (1) Where I ( ) is the intensity per frequency at the transition frequency. (The assump- tion in the above expression is that bandwidth of the incident light is large compare to the inverse time scale of the system) The cross section is given by: (Bransden 11.58) = hW I (2)...
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## This note was uploaded on 08/01/2008 for the course PHYSICS 137B taught by Professor Moore during the Fall '07 term at University of California, Berkeley.

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p137bp8sol - Physics 137B, Fall 2007, Moore Problem Set 8...

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