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This experiment reveals metastable state and uses the fact that an electron in 22S1/2 can't go to the ground state as ∆l=0 is restricted. But I was thinking why doesn't electron go to 22P1/2 and then transition to 12S1/2.

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Electrons do go from $2^{2}S_{1/2}$ to $2^{2}P_{1/2}$ by emitting a microwave of wavelength 28.37cm. This is because $2^{2}S_{1/2}$ is slightly higher in energy compared to $2^{2}P_{1/2}$ due to self interaction of electron by exchange of a photon

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These $2^{2}P_{1/2}$ electrons will subsequently transition to $1^{2}S_{1/2}$. However, in order to experimentally observe this it the $2^{2}P_{1/2}$ state is first transitioned to $2^{2}P_{3/2}$ using a Microwave of frequency 2395 MHz and then the subsequent transition to $1^{2}S_{1/2}$ is experimentally observed as was done in the Lamb Retherford experiment.

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  • $\begingroup$ If they do go from 2<sup>2</sup>S<sub>1/2</sub> to 2<sup>2</sup>P<sub>1/2</sub> then how does 2<sup>2</sup>S<sub>1/2</sub> become a metastable state? And it is a stable state because galvanometer gives a reading. Is there some probability thing going on? $\endgroup$ Feb 19, 2019 at 16:16
  • $\begingroup$ That comes from Heisenberg's uncertainty principle. As $\Delta T = \frac{\hbar}{\Delta E}$, the transition from $2^{2}S_{1/2}$ to $2^{2}P_{1/2}$ has a very small energy difference so it spends a lot time in $2^{2}S_{1/2}$ before it transitions to $2^{2}P_{1/2}$. Hence, it is metastable. $\endgroup$
    – Invariance
    Feb 19, 2019 at 17:17

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