Phy5646/An Example of spontaneous emission calculation: Difference between revisions

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<math>
<math>
\Gamma_{1s2p} = \frac{4}{3}\frac{e^{2}\omega^{3}_{21}}{\hbar\c^{3}}
\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}} }