I have read this question:
where Emilio Pisanty says:
Transitions which change spin direction are electric-dipole-forbidden, and they can only take place from magnetic-dipole onwards, which means that they're suppressed (in likelihood, not in energy) compared to transitions which don't flip any spins. In both of the cases above, the photons will have a low photon energy, and they will also be dipole-forbidden, which means that they will be relatively unlikely.
where John Rennie says:
Caesium has a single electron in the outermost 6s orbital, and this electron can be aligned with or against the nuclear spin. These two configurations differ in energy by about 0.000038 eV, and transitions between them produce/absorb light with a frequency of 9,192,631,770 Hz. This is the frequency used to measure time.
Now based on this, the frequency of light is exactly 9,192,631,770 Hz, that is, each and every time the transition happens, the energy of the atom/electron system changes by 0.000038eV, and that energy should go somewhere, thus a photon should be emitted. Now this could mean that with every single transition, a single photon is emitted, meaning in one second, exactly 9,192,631,770 number of photons need to be produced. The second is defined by the number of transitions, but it does not say anything about photon emission.
But if this transition is dipole forbidden, and the emission of the photon is relatively unlikely but the frequency of light is exactly 9,192,631,770 Hz, then this could mean that some transitions emit photons, others do not (or that the transition itself is relatively unlikely), and either there are exactly 9,192,631,770 number of photons produced per second, or there are less photons produced, but the photons that are actually produced have a frequency of 9,192,631,770 Hz and energy of 0.000038eV.
Just to clarify, I am trying to ask, whether a single atom, whenever making a hyperfine transition should always emit a photon?
- Does a hyperfine transition always cause a photon emission?