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3502_27_nove16_LG

3502_27_nove16_LG - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 27, November 16, 2009 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 116-118.) Solved Homework In the ground-state He atom at the first step, each electron is assumed to occupy a hydrogenic 1s orbital. Recall that, with Z = 2 for He, this orbital is defined as ! 100 r , " , # ( ) = 8 \$ % & ( 1/2 e ) 2 r The square of this wave function defines the charge cloud that creates the repulsive potential. Thus, at any point r 0 the total repulsion E felt by a point charge q at point r 0 will be E r 0 ( ) = q 1 r " r 0 # 100 r , \$ , % ( ) 0 2 & 0 & 0 ( 2 r 2 dr sin \$ d \$ d % where somewhat sloppy notation has been used to try to keep things simpler. The distance between point r 0 and any point in the integration depends on all three polar coordinates (not just r ), so the integral is not necessarily easy to solve. What is easier to notice is that, since the electron we are interested in is itself also delocalized (and not a point charge q ), the total repulsion is a double integral, where we integrate over the space of the first electron the repulsion felt by the fractional charge at any point defined by the equation above that involves integrating over the space of the second electron. The result is just our old friend the Coulomb integral E = ! 100 1 ( ) 2 "" 1 r 12 ! 100 2 ( ) 2 dr 1 dr 2 where each dr now stands for a complete spherical polar differential volume element. The solution to this integral is fairly painful, but can be achieved through some heroic coordinate transformations. We’ll forego the pleasure. Since 1s orbitals have their maximum amplitudes at the nucleus, electron 1 will be most repelled by electron 2 near the nucleus, and less so further away. So, electron 1 will tend to localize its density further from the nucleus than would be true were electron 2

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27-2 not to be there (which is what ψ 100 assumes, since it is a one-electron orbital, not a two- electron orbital). We've seen this previously in discussion of the variational process where we treated the He atomic number as a parameter and determined it to be 1.69—this drop from 2 causes the electron to localize further away. Of course, each electron does this to the other, so in the next step, the repulsion near the nucleus felt by each will be reduced (because each electron is now spread out further), and after step 2 the electron will contract back in a bit. Now, at step 3, since charge has reconcentrated near the nucleus, the electronic orbitals will expand out a little bit more, and this will continue in smaller and smaller incremental steps until we declare convergence to have been achieved based on the increment being sufficiently small. If the original second electron had been in a 2s orbital instead of a 1s, it would have repelled electron 1 considerably less, since it would have much smaller amplitude near the nucleus. The original electron 1 would not have expanded as far outward at step
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3502_27_nove16_LG - Chem 3502/5502 Physical Chemistry...

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