p41_042 - 50 . 0 keV = 24 . 8 pm . (b) and (c) The values...

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42. (a) We use eV = hc/λ min (see Eq. 41-23 and Eq. 39-4). The result of problem 3 in Chapter 39 is adapted to these units ( hc = 1240 eV · nm = 1240 keV · pm). λ min = hc eV = 1240 keV · pm
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Unformatted text preview: 50 . 0 keV = 24 . 8 pm . (b) and (c) The values of λ for the K α and K β lines do not depend on the external potential and are therefore unchanged....
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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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