Phy5646/hydrogen atom lifetime lifetime: Difference between revisions

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'''Excited Hydrogen Atom Lifetime.'''
'''Excited Hydrogen Atom Lifetime.'''
(Submitted by group 3)


We start with the wavefunctions of the ground and first excited state of the hydrogen atom.
We start with the wavefunctions of the ground and first excited state of the hydrogen atom.
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<math>\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}</math>
<math>\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}</math>
The transistion rate is given by the Fermi Golden rule;
<math> R=\dfrac{\pi}{\epsilon \hbar^2} |p|^2 \rho(\omega) </math>


We must evaluate equations of the form <math> <\psi_{100}|r| \psi_{2ab}> </math>
We must evaluate equations of the form <math> <\psi_{100}|r| \psi_{2ab}> </math>
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Exploiting the symmetry of the wavefunctions we find that the only non-zero element for the z compoent is,
Exploiting the symmetry of the wavefunctions we find that the only non-zero element for the z compoent is,


<math> <\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) dr d\theta d\phi
<math> <\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) dr d\theta d\phi </math>
 
Integrating over all space we find;
 
<math> <\psi_{100} |z| \psi_{210}>= -(\dfrac{2^8}{3^5 \sqrt{2}})a_o </math>
 
For the integrations over x and y we note that all the wavefunctions are even in these variables except for <math>\psi_{21 \pm 1}</math>
 
<math><\psi_{100}|x|\psi_{21 \pm1}>=\mp \dfrac{1}{8 pi a_o^4}\int r^4 e^{-3r/2a} \sin(\theta)^3 (\cos(\phi)\pm i \sin(\phi))\cos(\phi)dr d\theta d\phi = \mp \dfrac{2^7}{3^5}a_o </math>
 
<math><\psi_{100}|y|\psi_{21 \pm1}>= \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 </math>
 
Our equation for <math>\omega</math> is as follows;
 
<math>\omega= \dfrac{E_{2}-E_{1}}{\hbar}= -\dfrac{3 E_{1}}{4 \hbar}</math>
 
This yields
 
<math> R= -\dfrac{2^(10)}{3^8}(\dfrac{E_{1}}{m c^2})^2 \dfrac{c}{a_o}= 6.27x10^8 1/s </math>
 
This gives a value fore the lifetime of the <math> \psi_{210} </math>and <math>\psi_{21 \pm1}</math> states as <math> \tau= \dfrac{1}{r}= 1.60\times10^{-9}s </math>
 
The <math> \psi_{200}</math> state had matrix elements of 0, this implies that the lifetime is;
 
<math> \tau= \dfrac{1}{0}= \infty </math>
 
This implies that the <math>\psi_{200}</math> state is stable.

Latest revision as of 16:27, 27 April 2010

Excited Hydrogen Atom Lifetime.

(Submitted by group 3)

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

The transistion rate is given by the Fermi Golden rule;

We must evaluate equations of the form

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

Integrating over all space we find;

For the integrations over x and y we note that all the wavefunctions are even in these variables except for

Our equation for is as follows;

This yields

This gives a value fore the lifetime of the and states as

The state had matrix elements of 0, this implies that the lifetime is;

This implies that the state is stable.