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3502_15_octo13 - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 15, October 13, 2010 Solved Homework We are asked to give the minimum energy required for the ionization reaction Be 3+ Be 4+ + e - Since Be is atomic number Z = 4, this corresponds to ionization of the electron from a one-electron atom. One-electron atoms are the systems that we know how to deal with exactly based on our work of last lecture. Note that the reason the poorly worded problem presumably says the “minimum energy” is that it has in mind the energy required to just separate the electron from the nucleus and leave both particles at rest (this defines the system’s zero of energy). You could, however, provide more energy, so that not only would the particles be infinitely separated, but they’d have some kinetic energy too, instead of being at rest. However, the problem is poorly worded because the correct answer is that the minimum energy is ε , where ε is a positive number arbitrarily close to zero. How can this be? Well, what if the electronic wave function corresponds to principle quantum number n = 10 9 ? The binding energy in that case is incredibly small ( ε ), and it’s easy to remove the electron. To clean up the wording, one need only specify that Be 3+ is in its ground state . In that case, we know that the binding energy of the electron to the nucleus is E n = 1 Be 3 + ( 29 = - Z Be 2 e 4 μ Be 32 π 2 ε 0 2 1 ( 29 2 h 2 Now, we could go look up all the necessary constants and plug them in, and that would give a very nice answer. A quicker approach, however, is to use the datum provided that E n =1 (H) = - 13.6 eV. Then we can consider
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15-2 E n = 1 Be 3 + ( 29 E n = 1 H ( 29 = - Z Be 2 e 4 μ Be 32 π 2 ε 0 2 1 ( 29 2 h 2 - Z H 2 e 4 μ H 32 π 2 ε 0 2 1 ( 29 2 h 2 = 16 μ Be μ H = 16 m Be m e m Be + m e m H m e m H + m e 16 m Be m e m Be m H m e m H = 16 Thus, the final ionization potential for Be should be 16 times the ionization potential for H, or - 217.6 eV. Hydrogenic Orbitals Last lecture, we went through a fairly complicated process to derive the quantum restrictions on the energy levels of one-electron atoms, but we did not take the final step of putting our final quantum numbers back into our various differential equations to pull out the eigenfunctions themselves. If one attempts to do this in a completely general way (i.e., for arbitrary quantum numbers n and l ) one can write a general wave function in terms of what are called the associated Laguerre polynomials. This general form, however, is not terribly informative, and it is more useful to simply consider some of the specific functions associated with smaller quantum numbers. For example, what if we pick the principle quantum number n = 1? From our previous work, and in particular eq. 14-30, this requires that l and m both be zero, and β = n = 1. The requirement that
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3502_15_octo13 - Chem 3502/5502 Physical Chemistry...

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