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Unformatted text preview: Physical Chemistry Course Number: C362 12.3 Solution to Eq. ( 12.32 ) 1. Lets first analyze the equation. The term involving the derivatives can be interpreted as a second derivative operator and hence a radial kinetic energy (notice the angular kinetic energy was discussed above) of r e N . The two other terms are like an effective potential: h 2 l ( l +1) 2 ( r e- N ) 2 Ze 2 r e- N . This is an effec- tive potential since it involves the true potential: bracketleftbigg Ze 2 r e- N bracketrightbigg , plus an additional potential that includes the influence of the angular dependence. Hence this is a spherically averaged potential. Hence the net effective potential has the following functional form: 2. We would now like to go to reduced units, to make our differential equation simpler looking. We may first rewrite Eq. ( 12.32 ) by multiplying throughout by 2 h 2 and collecting terms on one side to obtain 1 r e N 2 r e N bracketleftBigg r e N 2 r e N bracketrightBigg l ( l + 1) ( r e N ) 2 + 2 h 2 E + Ze 2 r e N R ( r e N ) = 0 (12.33) 3. Then to make the equation look simpler we make the following substitutions: 2 = 2 E h 2 = Ze 2 h 2...
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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