p39_044

# p39_044 - 44. (a) From Eq. 39-11 = h (1 cos ) = (2.43 pm)(1...

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44. (a) From Eq. 39-11 λ = h m e c (1 cos φ )=(2 . 43 pm)(1 cos 90 )=2 . 43 pm . (b) The fractional shift should be interpreted as ∆ λ divided by the original wavelength: λ λ = 2 . 425 pm 590 nm =4 . 11 × 10 6 . (c) The change in energy for a photon with λ = 590 nm is given by E ph =∆ µ hc λ ≈− hc λ λ 2 = (4 . 14 × 10 15 eV · s)(2 . 998 × 10 8 m / s)(2 . 43 pm) (590 nm) 2 = 8 . 67 × 10 6 eV . For an x ray photon of energy E ph =50keV,∆ λ remains the same (2 . 43 pm), since it is independent of E ph . The fractional change in wavelength is now λ λ = λ hc/E ph = (50 × 10 3 eV)(2 . 43 pm) (4 . 14 × 10 15 eV · s)(2 . 998 × 10 8 m / s) =9 . 78 × 10 2 , and the change in photon energy is now E ph = hc µ 1 λ +∆ λ 1 λ
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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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