Phy5646/hydrogen atom lifetime lifetime

Excited Hydrogen Atom Lifetime.

We start with the wavefunctions of the ground and first excited state of the hydrogen atom.

$\psi _{100}={\dfrac {e^{-r/a_{o}}}{\sqrt {\pi a_{o}^{3}}}}$

$\psi _{200}={\dfrac {e^{-r/2a_{o}}}{\sqrt {32\pi a_{o}^{3}}}}\left(2-{\dfrac {r}{a_{o}}}\right)$

$\psi _{210}={\dfrac {e^{-r/2a_{o}}}{\sqrt {32\pi a_{o}^{3}}}}\left({\dfrac {r}{a_{o}}}\right)cos(\theta )$

$\psi _{21\pm 1}={\dfrac {e^{-r/2a_{o}}}{\sqrt {64\pi a_{o}^{3}}}}\left({\dfrac {r}{a_{o}}}\right)sin(\theta )e^{\pm i\phi }$

The transistion rate is given by the Fermi Golden rule;

$R={\dfrac {\pi }{\epsilon \hbar ^{2}}}|p|^{2}\rho (\omega )$

We must evaluate equations of the form $<\psi _{100}|r|\psi _{2ab}>$

Exploiting the symmetry of the wavefunctions we find that the only non-zero element for the z compoent is,

$<\psi _{100}|z|\psi _{210}>=-{\dfrac {1}{4{\sqrt {2}}\pi a_{o}^{4}}}\int r^{4}e^{-3r/2a}\cos(\theta )^{2}\sin(\theta )drd\theta d\phi$

Integrating over all space we find;

$<\psi _{100}|z|\psi _{210}>=-({\dfrac {2^{8}}{3^{5}{\sqrt {2}}}})a_{o}$

For the integrations over x and y we note that all the wavefunctions are even in these variables except for $\psi _{21\pm 1}$

$<\psi _{100}|x|\psi _{21\pm 1}>=\mp {\dfrac {1}{8pia_{o}^{4}}}\int r^{4}e^{-3r/2a}\sin(\theta )^{3}(\cos(\phi )\pm i\sin(\phi ))\cos(\phi )drd\theta d\phi =\mp {\dfrac {2^{7}}{3^{5}}}a_{o}$

$<\psi _{100}|y|\psi _{21\pm 1}>=\mp {\dfrac {1}{8\pi a_{o}^{4}}}4!({\dfrac {2a_{o}}{3}})^{5}{\dfrac {4}{3}}\int (\cos(\phi )\pm i\sin(\phi ))\sin(\phi )d\phi =-i{\dfrac {2^{7}}{3^{5}}}a_{o}$

Our ewuation for \omega is as follows;

$\omega ={\dfrac {E_{2}-E_{1}}{\hbar }}=-{\dfrac {3E_{1}}{4\hbar }}$

This yeilds'

$R=-{\dfrac {2^{(}10)}{3^{8}}}({\dfrac {E_{1}}{mc^{2}}})^{2}{\dfrac {c}{a_{o}}}=6.27x10^{8}1/s$

This gives a value fore the lifetime of the $\psi _{210}$and $\psi _{21\pm 1}$ states as $\tau ={\dfrac {1}{r}}=1.60x10^{-}9s$

The $\psi _{200}$ state had matrix elements of 0, this implies that the lifetime is;

$\tau ={\dfrac {1}{0}}=\infty$

This implies that the $\psi _{200}$ state is stable.