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I’ve been trying to understand the photoelectric effect, and, as I understand it, when electromagnetic radiation (ER) of a suitable frequency, and therefore energy, hits an atom, its electrons are ejected.

But if the incident ER is not of suitable frequency, the electrons won’t absorb it due to quantization. Does this ER just get reflected without being absorbed?

And when electrons are indeed ejected, do the leave behind positively charged atoms? How does this work?

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    $\begingroup$ The photoelectric effect happens in metals and semiconductors, not in individual atoms. The energy of the absorbed photons has to be higher than the work function of the metal but it does not have to be in resonance with an electronic state transition of individual atoms. As you noticed, the electron that leaves the metal lattice leaves behind a positive net charge. In the usual photoelectric experiment this is neutralized by the closed electric circuit. $\endgroup$
    – FlatterMann
    Commented Jun 21 at 10:43
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    $\begingroup$ @FlatterMann Ionization of an isolated atom by absorption of a photon is commonly considered to be "photoelectric". $\endgroup$
    – John Doty
    Commented Jun 21 at 13:05
  • $\begingroup$ That's not how I usually use "photoelectric effect", but you are correct, we should ask the OP for a clarification what he meant. $\endgroup$
    – FlatterMann
    Commented Jun 21 at 17:53

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The photoelectric effect liberates the electron from whatever is holding it: metal surface, valence band, atomic shell, ...

What happens next depends on what's around. In the classic textbook experiment, an electrode collects it. In a silicon diode it winds up in the N side of the junction. For x-ray photon detection, the high-energy photoelectron plows into surrounding matter, ionizing it. In all cases, some other process, like an external circuit, restores charge neutrality of the stuff involved.

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