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Absorption spectrum theory says that electrons in the atoms absorb only photons of certain frequencies, which causes the dark lines in an absorption spectra. However, I understand that electrons that absorb photons of any energy greater than the work function will be free from the nuclei potential field.

Work function is typically mentioned in photoelectric effect, but any electron is able to absorb any amount of photon energy(so long that the photon energy is greater than the work function). By this logic, there shouldn't be any dark lines since most photons of frequencies larger than the work function will be absorbed.

What is wrong with this thought?

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  • $\begingroup$ What is wrong is in assuming that the bound electrons are absorbing energy or emitting it. It is the atom that does that $\endgroup$
    – anna v
    Oct 30, 2014 at 6:17

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The issue is that although electrons are capable of absorbing any energy greater than the minimum required to excite them, there are resonances - that is, the probability that a photon is absorbed is much greater when the energy of the photon corresponds to an exact transition. The reasons behind that are quite subtle - but that is the reason you see the lines. Those absorptions are much more probable, so photons with exactly those energies get absorbed more.

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  • $\begingroup$ The question wasn't very clear. To understand what the OP was really asking, you have to look in the long comment thread under my answer. $\endgroup$
    – user4552
    Oct 30, 2014 at 4:13
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What is wrong is in assuming that the bound electrons are absorbing energy or emitting it. It is the atom that does that. The system electron-nucleus is in a stable quantum mechanical state.

The system (electron/nucleus) has many available energy levels. Given a correct increment of energy by a photon, the photon will be absorbed by the system (electron/nucleus) and the electron will go to a higher energy state. Increment because we are in the realm of quantum mechanics. If the incoming photon energy does not match within the Heisenberg Uncertainty principle the energy of the energy level, the atom will not respond, and the incoming photon will be scattered off.

The point to keep in mind is that we are in the quantum domain when working with atoms, which means that for a reaction to take place the probability should be high. The probability of the atom absorbing a photon by emitting the electron is very high if the energy of the photon is within the width of the energy level the electron occupies. Otherwise the probability is very small and that is why spectra appear.

A similar view of this is in the answer by Floris.

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By this logic, there shouldn't be any dark lines since most photons of frequencies larger than the work function will be absorbed.

The dark lines are for photons whose frequences $\nu$ gives energies $h\nu$ less than the work function.

BTW, I don't think it's normal to use "work function" in this context. "Work function" usually refers to condensed matter, not individual atoms.

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  • $\begingroup$ I understand your point but my question suggests that electrons should be able to absorb photons of most frequencies rather than those that just correspond to the energy level differences. $\endgroup$
    – Standstill
    Oct 30, 2014 at 1:47
  • $\begingroup$ @Standstill: They are able to absorb photons of "most" frequencies, if "most" means the infinitely many frequencies greater than $W/h$, where $W$ is the workfunction. $\endgroup$
    – user4552
    Oct 30, 2014 at 2:10
  • $\begingroup$ Yes that's exactly what I meant. So if that is the case, wouldn't there be no dark lines anymore, since only the lowest energy electrons will absorb to produce one dark line, while it will absorb all frequencies above this, thus negating any effect of the higher electrons' absorption of photons. $\endgroup$
    – Standstill
    Oct 30, 2014 at 2:28
  • $\begingroup$ @Standstill: I don't understand your most recent comment. You might want to consider an atom with only one electron, in order to clear up the confusion about multiple electrons. Do you understand that photon energies correspond to differences of electron energies? $\endgroup$
    – user4552
    Oct 30, 2014 at 2:31
  • $\begingroup$ Let's take hydrogen. The ground state is -13.6eV. So the electron can absorb any photons equal or larger than 13.6eV. The next excited state is -3.4eV. So this electron can absorb any photon larger than 3.4eV, which includes all the energies up to 13.6eV photons and beyond. By progressing up the excited states, we will come to conclude that photons of most energies higher than the amount needed by the most excited electrons) will be absorbed, thus eliminating any absorption patterns. $\endgroup$
    – Standstill
    Oct 30, 2014 at 2:40

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