1181_Physics ProblemsTechnical Physics

1181_Physics ProblemsTechnical Physics - 522 Atomic Physics...

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522 Atomic Physics P42.10 Starting with 1 22 2 2 mv ke r e e = we have v mr e e 2 2 = and using r n mke n ee = 2 = gives v mn n e e 2 2 2 = = ej or v n n e = 2 = . P42.11 Each atom gives up its kinetic energy in emitting a photon, so 1 2 6626 10 300 10 1216 10 163 10 2 34 8 7 18 mv hc == ×⋅× × λ .. . . Js ms m J e j v 442 10 4 . m s . P42.12 The batch of excited atoms must make these six transitions to get back to state one: 21 , and also 32 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 0850 . . eV 4 eV 2 , so the incoming photons have wavelength = = −− × F H G I K J = hc EE fi 136
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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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