# Relativistic Bohr Atom and MacLaurin Series

by atomicpedals
Tags: atom, bohr, maclaurin, relativistic, series
 P: 196 1. The problem statement, all variables and given/known data By expanding a MacLaurin Series show that $$E_{n}=\epsilon_{n} - \mu c^{2} = - \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}-\frac{3}{4n})]$$ 2. Relevant equations Through a lengthy derivation I arrived at $$\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n_{r}+\sqrt{l^{2}-Z^{2}\alpha^{2}}}}}$$ 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 $$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^{2}}{8}+...$$ $$\frac{1}{1+x}=1+...$$ Do I need to make some aggressive substitutions?
 P: 196 I think I can make the following justifiable substitution $$\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n'}}}$$ Where $n' = n_{r} + \sqrt{l^{2}-Z^{2} \alpha^{2}}$. But this still doesn't get me any closer to arriving at $E_{n}=\epsilon_{n} - \mu c^{2} = - \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}-\frac{3}{4n})]$ through a Maclaurin series.
 P: 196 Ok, I may be closer... using the substitution above to put $E_{n} = \epsilon_{n} - \mu c^{2}$ in terms of n' I can calculate a Maclaurin series as follows $$- \mu c^{2} + \frac{\mu c^{2} \sqrt{n'}}{\alpha Z} - \frac{n^{3/2} (\mu c^{2} \alpha Z)}{2(\alpha^{4} Z^{4})} + \frac{n^{5/2} (3 \mu c^{2} \alpha Z)}{8 \alpha^{6} Z^{6}} - ...$$ Am I on the right track to the final $E_{n}$?

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