Phy5646/An Example of spontaneous emission calculation: Difference between revisions
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\Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\ | \Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\hbarc^{3}} | ||
</math> | </math> |
Revision as of 16:45, 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
Failed to parse (unknown function "\hbarc"): {\displaystyle \Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\hbarc^{3}} }