Phy5646/An Example of spontaneous emission calculation: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
Line 13: | Line 13: | ||
\Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\hbar c^{3}}|\langle 100|\mathbf{r}|21m'\rangle|^{2} | \Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\hbar c^{3}}|\langle 100|\mathbf{r}|21m'\rangle|^{2} | ||
</math> | </math> | ||
where <math>|21m'\rangle</math> |
Revision as of 17:04, 24 April 2010
(Submitted by Team 1)
This example was taken from "Theory and Problems of Quantum Mechanics", Yoaf Peleg, et al, p. 298.
Problem: Find the transition rate of spontaneous emission for a hydrogen atom in the first excited state.
Solution:
The transition rate for is given by
where