524 Atomic PhysicsSection 42.4The Quantum Model of the Hydrogen AtomP42.16The reduced mass of positronium is lessthan hydrogen, so the photon energy will be lessforpositronium than for hydrogen. This means that the wavelength of the emitted photon will belongerthan 656.3 nm. On the other hand, helium has about the same reduced mass but more chargethan hydrogen, so its transition energy will be larger, corresponding to a wavelength shorterthan656.3 nm.All the factors in the given equation are constant for this problem except for the reduced massand the nuclear charge. Therefore, the wavelength corresponding to the energy difference for thetransition can be found simply from the ratio of mass and charge variables.For hydrogen,µ=+≈mmmpee.The photon energy is∆EE E=−32.Its wavelength isλ=656 3. nm,where==cfhcE∆.(a)For positronium,=+=meee2so 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 .