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An atom meets one of its electrons in an excited state. The average time that this remains in that state is $1 × 10^{−8}$ s to emit a photon of wavelength $620$ nm or $2$ eV energy.

My questions are: How does this uncertainty compare to the energy of the transition? To what corresponds the variation in wavelength that it emits in average those atoms?

I already have the uncertainty in the energy of that excited state, $$E≥h/t = 4.96\times 10^{-9} eV$$

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The distribution of the emitted photons in energy centers around the natural frequency/energy at 2eV, but extends around it by roughly ΔE~ℏ/2τ, so, then, ~33neV. (Natural broadening.)

You may easily convert this energy/frequency uncertainty to wavelength uncertainty.

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