Phy5645/Hydrogen Atom WKB: Difference between revisions

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<math>\text{P(r)=}\sqrt {2m(E-V(r))} =\sqrt {2m(E-\left ({\frac{\hbar ^{2}l(l+1)}{2mr^{2}}-\frac{e^{2}}{r}} \right )} )</math>
<math>\text{P(r)=}\sqrt {2m(E-V(r))} =\sqrt {2m(E-\left ({\frac{\hbar ^{2}l(l+1)}{2mr^{2}}-\frac{e^{2}}{r}} \right )} )</math>


<math>\int\limits_{r1}^{r2} {\sqrt {2m(E-\frac{\hbar ^{2}l(l+1)}{2mr^{2}}+\frac{e^{2}}{r})} }dr=(n+\frac{1}{2})\pi \hbar </math>
<math>\int\limits_{r1}^{r2} {\sqrt {2m(E-\frac{\hbar ^{2}l(l+1)}{2mr^{2}}+\frac{e^{2}}{r})} }dr=(n+\frac{1}{2})\pi \hbar </math> where r1 and r2 are turning points in this case.


<math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1-}\frac{\hbar ^{2}l(l+1)}{2mr^{2}E}+\frac{e^{2}}{Er})^{1/2}dr=(n+\frac{1}{2})\pi \hbar </math>
<math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1-}\frac{\hbar ^{2}l(l+1)}{2mr^{2}E}+\frac{e^{2}}{Er})^{1/2}dr=(n+\frac{1}{2})\pi \hbar </math>

Revision as of 20:26, 9 December 2009

Use WKB approximation to estimate energy spectrum for Hydrogen atom.



The approximation is:

where r1 and r2 are turning points in this case.