HAtom3

HAtom3 - Physical Chemistry Course Number: C362 12.3...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.

Page1 / 4

HAtom3 - Physical Chemistry Course Number: C362 12.3...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online