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1181_Physics ProblemsTechnical Physics

1181_Physics ProblemsTechnical Physics - 522 Atomic Physics...

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522 Atomic Physics P42.10 Starting with 1 2 2 2 2 m v k e r e e = we have v k e m r e e 2 2 = and using r n m k e n e e = 2 2 2 = gives v k e m n m k e n e e e e 2 2 2 2 2 = = e j or v k e n n e = 2 = . P42.11 Each atom gives up its kinetic energy in emitting a photon, so 1 2 6 626 10 3 00 10 1 216 10 1 63 10 2 34 8 7 18 mv hc = = × × × = × λ . . . . J s m s m J e je j e j v = × 4 42 10 4 . m s . P42.12 The batch of excited atoms must make these six transitions to get back to state one: 2 1 , and also 3 2 and 3 1 , and also 4 3 and 4 2 and 4 1 . Thus, the incoming light must have just enough energy to produce the 1 4 transition. It must be the third line of the Lyman series in the absorption spectrum of hydrogen. The absorbing atom changes from energy E i = − = − 13 6 13 6 . . eV 1 eV 2 to E f = − = − 13 6 0 850 . . eV 4 eV 2 , so the incoming photons have wavelength λ = = × × − −
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