p41_051 - f f = µ Z − 1 Z − 1 2 = µ 92 − 1 13 −...

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51. (a) According to Eq. 41-26, f ( Z 1) 2 , so the ratio of energies is (using Eq. 39-2) f/f 0 =[( Z 1) / ( Z 0 1)] 2 . (b) We refer to Appendix F. Applying the formula from part (a) to Z =92and Z 0 = 13, we obtain E E 0 =
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Unformatted text preview: f f = µ Z − 1 Z − 1 ¶ 2 = µ 92 − 1 13 − 1 ¶ 2 = 57 . 5 . (c) Applying this to Z = 92 and Z = 3, we obtain E E = µ 92 − 1 3 − 1 ¶ 2 = 2070 ....
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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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