p41_050 - θ = 74 . 1 ◦ , we solve for d : d = mhc 2∆ E...

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50. We use Eq. 37-31, Eq. 40-6, and the result of problem 3 in Chapter 39, adapted to these units ( hc = 1240 eV · nm = 1240 keV · pm). Letting 2 d sin θ = = mhc/ E ,where
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Unformatted text preview: θ = 74 . 1 ◦ , we solve for d : d = mhc 2∆ E sin θ = (1)(1240 keV · nm) 2(8 . 979 keV − . 951 keV)(sin 74 . 1 ◦ ) = 80 . 3 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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