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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 2 2 4 2 2 2 π a f f mk e h n n n e = F H G I K J 2 2 1 1 2 2 4 3 2 2 π a f As n approaches infinity, we have f approaching 2 2 2 2 4 3 3 π mk e h n e The classical frequency is f v r k e m r e = = 2 1 2 1 2 3 2 π π where r n h mk e e = 2 2 2 4 π Using this equation to eliminate r from the expression for f , f mk e h n e = 2 2 2 2 4 3 3 π P42.58 (a) The energy of the ground state is: E hc 1 1 240 8 16 = − = − = λ series limit eV nm 152.0 nm eV . . From the wavelength of the Lyman α line: E E hc 2 1 1 240 6 12 = = = λ nm eV 202.6 nm eV . E E 2 1 6 12 2 04 = + = . . eV eV . The wavelength of the Lyman β line gives: E E 3 1 1 240 7 26 = = nm eV 170.9 nm eV . so E 3 0 902 = . eV . Next, using the Lyman γ line gives: E E 4 1 1 240 7 65 = = nm eV 162.1 nm eV . and
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