p39_036 - 36. (a) Using the result of problem 3, we nd = hc...

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36. (a) Using the result of problem 3, we Fnd λ = hc E = 1240 nm · eV 0 . 511 MeV =2 . 43 × 10 3 nm = 2 . 43 pm . (b) Now, Eq. 39-11 leads to λ 0 = λ +∆ λ = λ + h m e c (1 cos φ ) . 43 pm + (2 . 43 pm)(1 cos 90 . 0 )=4 . 86 pm
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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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