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To clarify my question, consider the hydrogen atom. An electron on the ground state needs 10.2 eV to be excited to the second state (first excited state). If a photon with less than 10.2 eV strikes the electron, it will not be absorbed-- okay. My question is: what happens if the photon has an energy larger than 10.2 eV? Is the photon once again not absorbed, or is it partially absorbed?

I know that the same question was posted here (Can an electron jump to a higher energy level if the energy is insufficient or exceeds the $\Delta E$?) but I didn't understand the most upvoted answer, and the only other answer seemed to be conflicting. It was also posted by some other users but again, I didn't understand the answers because I'm really a beginner and they went into (what for me) is complicating QM equations. I've looked in other places too, but the answers are conflicting (some people saying photon does get absorbed, some saying it doesn't, etc).

So yeah, please use very simple language, and thank you for your time (I really appreciate it!) :)

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It depends on how much higher. Electronic transitions are not infinitesimally thin, and they all have a certain linewidth $\Delta f$, i.e. a span of frequencies below and above the central frequency $f_0$ at which the photon can be absorbed. If the photon energy is so high that it falls above the transition linewidth, then it will not be absorbed.

Generally speaking, transition linewidths are rather narrow, at least by human standards, though their precise width depends on the conditions; there are several different mechanisms which underlie them, and those produce widths on different scales.

  • For all transitions, there is an intrinsic linewidth $\Delta f = 1/\tau$ produced by the lifetime $\tau$ of the excited state, which is basically an instance of the Heisenberg uncertainty principle: you can't specify the frequency of a wave to something sharper than $\Delta f$ if its substrate is changing on timescales of order $\tau$.

    For transitions like the Lyman alpha line, $\tau$ is in the order of a few nanoseconds, giving a $\Delta f$ of the order of about a gigahertz. Here the central frequency is of the order of $f_0 \sim 1 \:\rm PHz$, which means that the line can look quite narrow: $\Delta f/f_0 \sim 10^{-9}$. (I say "look" because, by atomic precision spectroscopy standards, this is extremely wide. Everyday fare in precision spectroscopy uses lines which are narrower by many orders of magnitude, reaching down to $\Delta f/f_0 \sim 10^{-18}$.)

  • There are also external causes for nonzero linewidths, often called "inhomogeneous broadening", from a range of sources such as e.g. collisions between different atoms, or Doppler shifts coming from the different velocities in the thermal distribution of a gas. Unless you do something specific to counteract it, then inhomogeneous broadening will almost certainly dominate, giving you linewidths in the tens or hundreds of GHz or even higher. But again, since the central frequency is of the order of a petahertz, $\Delta f/f_0$ will still be quite small.

The bottom line is that photons typically do have some leeway in getting absorbed even if they have a photon energy that's a bit higher than the transition, but for most electronic transitions in gaseous atomic samples, that leeway will generally be rather limited. And, if the photon exceeds that leeway, then it won't be absorbed.

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  • $\begingroup$ Thank you for this answer, it was really helpful! Could you please just explain what a 'substrate' is? $\endgroup$ – Lolo123 Oct 15 '18 at 11:38
  • $\begingroup$ By 'substrate' I just meant the atom. The basic picture is the following: if you consider two waves at frequencies $f_0$ and $f_0\pm1/2\tau$, where $\frac1\tau\ll f_0$, then they will be pretty much identical, and they will only slip out of phase with each other after a time $\sim\tau$. (cont.) $\endgroup$ – Emilio Pisanty Oct 15 '18 at 12:45
  • $\begingroup$ Now, if you're using both of these waves to excite an atom, then by the time a period $\tau$ has passed (where we now fix $\tau$ to the lifetime of the excited state), any excited-state population caused by radiation absorbed at the start of the pulse will now have decayed, i.e. the excited state will have "forgotten" about the phase of the radiation at the start. And, as a consequence, it will absorb both waves equally well, because it cannot tell them apart. $\endgroup$ – Emilio Pisanty Oct 15 '18 at 12:45

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