A photon is emitted (or absorbed) by a transitioning electron. How fast is this process?
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The emission process always takes some time. How much depends on the kind of transition.
If the transition is spontaneous dipole transition, like when excited electronic state of an atom decays, the rate of spontaneous transition as found using quantum electrodynamics is
$$ \Gamma = \frac{\omega_{12}^3|\mu_{12}|^2}{3\pi\varepsilon_{0}\hbar c^3} $$ where $\omega_{12}$ is emission frequency ($(E_1-E_2)/\hbar$, $\mu_{12}$ is transition dipole matrix element magnitude $|\langle 1 |\boldsymbol{\mu}|2\rangle|$. So mean time needed for one transition is
$$ 1/\Gamma = \frac{3\pi\varepsilon_{0}\hbar c^3}{\omega_{12}^3|\mu_{12}|^2}. $$
For hydrogen transition $2P \to 1S$, this turns out tobe around 1.6 nanoseconds [1].
[1] http://farside.ph.utexas.edu/teaching/qmech/Quantum/node122.html
answered Feb 7, 2021 at 14:29
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$\begingroup$ I might be wrong but it looks like the OP asks how long does it take for the photon to be completely absorbed (or emitted) once the process begins. It is like how long does it take for an object to fall onto the earth once it starts falling. Do you give an answer to that in your post? $\endgroup$– Xfce4Commented Oct 3, 2021 at 21:15
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$\begingroup$ No, there is no clear cut end of the emission process in the standard emission model. In the model the process is infinite, and we calculate characteristic time based on decrease of probability, not based on time of some actual accomplished event of emission. $\endgroup$ Commented Oct 6, 2021 at 13:41
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$\begingroup$ @Thank you. Is this formula determined by making experiments on hydrogen gas, i.e observing how frequently the gas emits photons? Or is it derived from other equations? $\endgroup$– Xfce4Commented Oct 7, 2021 at 0:08
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2$\begingroup$ It's derived from quantum theory, and it is consistent with measurements of photomultiplier counts of radiating hydrogen beam, see W.S. Bickel, A.S. Goodman, Phys. Rev. 148, 1 (1966) $\endgroup$ Commented Oct 7, 2021 at 10:12