p39_034 - 34. (a) Eq. 39-11 yields = h (1 cos ) = (2.43...

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34. (a) Eq. 39-11 yields λ = h m e c (1 cos φ )=(2 . 43 pm)(1 cos 180 )=+4 . 86 pm . (b) Using the result of problem 3, the change in photon energy is E = hc λ 0 hc λ = (1240 eV · nm) µ 1 0 . 01 nm + 4 . 86 pm 1 0 . 01 nm = 41 keV . (c) From conservation of energy, ∆
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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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