1183_Physics ProblemsTechnical Physics

1183_Physics ProblemsTechnical Physics - 524 Atomic Physics...

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524 Atomic Physics Section 42.4 The Quantum Model of the Hydrogen Atom P42.16 The reduced mass of positronium is less than hydrogen, so the photon energy will be less for positronium than for hydrogen. This means that the wavelength of the emitted photon will be longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge than hydrogen, so its transition energy will be larger , corresponding to a wavelength shorter than 656.3 nm. All the factors in the given equation are constant for this problem except for the reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the transition can be found simply from the ratio of mass and charge variables. For hydrogen, µ = + mm m pe e . The photon energy is EE E =− 32 . Its wavelength is λ = 656 3 . nm, where == c f hc E . (a) For positronium, = + = m ee e 2 so the energy of each level is one half as large as in hydrogen, which we could call
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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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