Relativistic Bohr Atom and MacLaurin Seriesby atomicpedals Tags: atom, bohr, maclaurin, relativistic, series 

#1
May612, 09:53 PM

P: 196

1. The problem statement, all variables and given/known data
By expanding a MacLaurin Series show that [tex]E_{n}=\epsilon_{n}  \mu c^{2} =  \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}\frac{3}{4n})][/tex] 2. Relevant equations Through a lengthy derivation I arrived at [tex]\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n_{r}+\sqrt{l^{2}Z^{2}\alpha^{2}}}}}[/tex] I should add that k is what the text is using for the azimuthal quantum number, I used l in my derivation out of habit. 3. The attempt at a solution I've got no ideas where to go with this thing. I should take advantage of identites [tex]\sqrt{1x}=1\frac{x}{2}\frac{x^{2}}{8}+...[/tex] [tex]\frac{1}{1+x}=1+...[/tex] Do I need to make some aggressive substitutions? 



#2
May712, 01:33 PM

P: 196

I think I can make the following justifiable substitution
[tex]\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n'}}}[/tex] Where [itex]n' = n_{r} + \sqrt{l^{2}Z^{2} \alpha^{2}}[/itex]. But this still doesn't get me any closer to arriving at [itex]E_{n}=\epsilon_{n}  \mu c^{2} =  \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}\frac{3}{4n})][/itex] through a Maclaurin series. 



#3
May712, 03:53 PM

P: 196

Ok, I may be closer... using the substitution above to put [itex]E_{n} = \epsilon_{n}  \mu c^{2}[/itex] in terms of n' I can calculate a Maclaurin series as follows
[tex] \mu c^{2} + \frac{\mu c^{2} \sqrt{n'}}{\alpha Z}  \frac{n^{3/2} (\mu c^{2} \alpha Z)}{2(\alpha^{4} Z^{4})} + \frac{n^{5/2} (3 \mu c^{2} \alpha Z)}{8 \alpha^{6} Z^{6}}  ...[/tex] Am I on the right track to the final [itex]E_{n}[/itex]? 


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