p39_078 - 78. (a) Using the result of problem 3, E= hc 1240...

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78. (a) Using the result of problem 3, E = hc λ = 1240 nm · eV 10 . 0 × 10 3 nm = 124 keV . (b) The kinetic energy gained by the electron is equal to the energy decrease of the photon: E =∆ µ hc λ = hc µ 1 λ 1 λ +∆ λ = µ hc λ ¶µ λ λ λ = E 1+ λ λ = E λ λ C (1 cos φ ) = 124 keV 10 . 0pm (2 . 43 pm)(1 cos 180 ) =4
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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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