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Unformatted text preview: Physical Chemistry Course Number: C362 12.5 The hydrogen atom energy only depends on the “principal qauntum number” 1. But now λ depends upon E the energy as given in the first two of Eq. ( 12.34 ). Using Eqs. ( 12.34 ) n 2 = λ 2 = μ 2 Z 2 e 4 ¯ h 4 α 2 = − μZ 2 e 4 2¯ h 2 E μ (12.49) and hence E μ = − μZ 2 e 4 2¯ h 2 n 2 (12.50) And now this energy is negative!! (Recall the earlier center of mass energy levels were all greater than zero and formed a continuous spectrum.) Further more, n is an integer and hence the energy in Eq. ( 12.50 ) is quantized. Again we got quantization by enforcing the boundary conditions. Here we used the fact that the wavefunction has to be finite and that gave us a maximum i max which lead to a discrete set of states. 2. We will go ahead and note here that what we have derived is true for a one electron one nucleus system. Hence not only is the energy in Eq. ( 12.50 ) a valid expression for the hydrogen atom, but it is also from for He + , Li 2+ and so on. 3. Hence the energy levels look as follows: 4. From Eq. ( 12.48 ) we also see that the minimum value of n is 1. (That is n cannot be zero.) Hence the minimum energy for the hydrogen atom is E n =1 μ = − μZ 2 e 4 2¯ h 2 . (Note that the lowest value of the potential is r e − N → when the potential energy tends to −∞ . Hence like in the particle in a box case, the lowest energy state does not correspond to the lowest potential energy value. Zero point energy is how we referred to this phenomenon in theenergy value....
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This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.
 Winter '11
 AmarFlood

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