# Why can't the electrons in Lamb-Retherford Experiment transition from 2S to 2P(j=1/2)

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.

## 1 Answer

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

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.

• 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? Feb 19, 2019 at 16:16
• 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. Feb 19, 2019 at 17:17