1196_Physics ProblemsTechnical Physics

1196_Physics ProblemsTechnical Physics - Chapter 42 P42.57...

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Chapter 42 537 P42.57 hf E mk e h n n e == F H G I K J 4 2 1 1 1 22 4 2 π af f mk e h n nn e = F H G I K J 2 21 1 4 32 2 a f As n approaches infinity, we have f approaching 2 2 4 33 mk e hn e The classical frequency is f v r ke m r e 2 1 2 1 2 ππ where r nh mk e e = 2 4 Using this equation to eliminate r from the expression for f , f mk e e = 2 2 4 P42.58 (a) The energy of the ground state is: E hc 1 1240 816 =− λ series limit eV nm 152.0 nm eV .. From the wavelength of the Lyman α line: EE hc 612 −== = nm eV 202.6 nm eV . 204 =+ = eV eV . The wavelength of the Lyman β line gives: 31 726 −= = nm eV 170.9 nm eV . so E 3 0902 . e V . Next, using the Lyman γ line gives: 41 765 = nm eV 162.1 nm eV . and E 4 0508 e V . From the Lyman δ line, 51 783 = nm eV 158.3 nm
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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