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Relativistic Bohr Atom and MacLaurin Series

by atomicpedals
Tags: atom, bohr, maclaurin, relativistic, series
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atomicpedals
#1
May6-12, 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{1-x}=1-\frac{x}{2}-\frac{x^{2}}{8}+...[/tex]
[tex]\frac{1}{1+x}=1+...[/tex]
Do I need to make some aggressive substitutions?
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atomicpedals
#2
May7-12, 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.
atomicpedals
#3
May7-12, 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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