# Darwin and relativistic kinetic energy correction for hydrogen

1. Mar 22, 2012

### kraigandrews

1. The problem statement, all variables and given/known data
Combine the Darwin correction with the relativistic kinetic energy correction for l=0 to show that the fine structure formula:

$\Delta$E$_{fs}$= - $\frac{(E^{(0)2}_{n})}{2mc^{2}}$[$\frac{4n}{j + 1/2}$-3]

remains valid for l=0

2. Relevant equations

From a previous problem the Darwin hamiltionian is shown to affected only under s-states
where for any s-state the formula $\frac{2}{na^{3/2}}$$\frac{1}{\sqrt{4\pi}}$

3. The attempt at a solution

so I know for s-states j=1/2 and l=0 (obviously). Overall, I am unsure as to what to show here, meaing:

do I combine the correction for the darwin energy to the correctionn for kinetic energy and then just show that for l=0, j=1/2 the equation still remains valid?

Thanks.

2. Mar 24, 2012