Go to the U of M home page
School of Physics & Astronomy
School of Physics and Astronomy Wiki

User Tools

Trace:
classes:2009:fall:phys4101.001:lec_notes_1111

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
classes:2009:fall:phys4101.001:lec_notes_1111 [2009/11/16 17:01] – ludemanclasses:2009:fall:phys4101.001:lec_notes_1111 [2009/11/19 10:21] (current) – ludeman
Line 20: Line 20:
  
 ====Final Words on the Radial Equation==== ====Final Words on the Radial Equation====
-For bound state, E<0 with l=0, equation [4.37] reduces from 3-D to 1-D with <math>=-k^2=\frac{2m(E+V_0)} \hbar^2</math> +For bound state, E<0 with l=0, equation [4.37] reduces from 3-D to 1-D with <math>-k^2=\frac{2m(E+V_0)} \hbar^2</math> 
  
 Then for r<a: <math>u"=-k^2u</math> Therefore <math>u(r)=Asin(kr)</math> Then for r<a: <math>u"=-k^2u</math> Therefore <math>u(r)=Asin(kr)</math>
Line 34: Line 34:
 //The key point here is that tan(z) doesn't exist in 3-D. In descriptive terms, if the well is to shallow there will not be a solution. There needs to be a wide enough and deep enough well for a bound state.//  //The key point here is that tan(z) doesn't exist in 3-D. In descriptive terms, if the well is to shallow there will not be a solution. There needs to be a wide enough and deep enough well for a bound state.// 
  
-//Also there are no allowed energies at n=0. Recall that for cot(z) the lowest allowed energies are <math>\frac{\pi} 2<z<\pi</math>//+//Also there are no allowed energies at n=0. Recall that for cot(z) the lowest allowed energies are// <math>\frac{\pi} 2<z<\pi</math>
  
 ====Radial Wave Function for the Hydrogen Atom==== ====Radial Wave Function for the Hydrogen Atom====
Line 54: Line 54:
 6. Replace test function gives <math>u(\rho)=C_0\rho^{l+1}e^{\rho}</math>. This requires that <math>C_{(j_{max}+1)}=0</math> and <math>n=j_{max}+l+1</math> 6. Replace test function gives <math>u(\rho)=C_0\rho^{l+1}e^{\rho}</math>. This requires that <math>C_{(j_{max}+1)}=0</math> and <math>n=j_{max}+l+1</math>
  
-Therefore, <math>\rho_0=2n</math> and the allowed energies are in concurrence with Eq. [4.70] +Therefore, <math>\rho_0=2n</math> and the allowed energies of the Hydrogen Atom are in concurrence with Eq. [4.70] 
  
 ------------------------------------------ ------------------------------------------
classes/2009/fall/phys4101.001/lec_notes_1111.1258412461.txt.gz · Last modified: 2009/11/16 17:01 by ludeman

Page Tools