0
$\begingroup$

I understand why treating light as particle-like photon packets leads to the observed results. I seek to understand why one would expect different results if light were treated as a wave (and thus why these results imply particle-like behaviour).

Experiment 1: Increasing the intensity. I imagine a light wave incident on the surface as Ψ = Asin(kx-ωt). I imagine increasing the intensity of light as adding multiple parallel waves incident on the surface. If each wave deposited enough energy to kick out 1 electron per second, wouldn’t adding multiple waves kick out more electrons per second without changing the energy per electron?

Is treating increased intensity as more parallel rays too much like particles? If I instead treat increased intensity as increased amplitude squared, then I think of the wave as pouring energy into the metal surface at a faster rate, like a thicker stream of water from a spout. In this case, wouldn’t the light have deposited enough energy to free an electron more quickly, and thus still eject more electrons per second?

As a corollary here, is it right to equate A² ∝ number of photons?

Experiment 2: Increased frequency. I do understand why a continuous stream of energy from a wave is inconsistent with a threshold frequency to elicit a current.

Why would increasing the frequency (and therefore energy) of a wave not increase the kinetic energy of the most energetic ejected electrons?

Wouldn’t a single, particle-like wave-packet, also have energy proportional to its amplitude² as well as its frequency, and thus kick off more energetic electrons?

$\endgroup$
  • $\begingroup$ Would this not be better placed in Physics.SE? $\endgroup$ – Tan Yong Boon Dec 26 '18 at 14:17
  • $\begingroup$ I learned about the photoelectric effect in chemistry, and my chemistry lecturers have always given the photoelectric effect as the origin of quantised electron energy levels. They give the observed results as evidence that light travels as packets of discrete energy. $\endgroup$ – Dion Silverman Dec 28 '18 at 10:16
  • $\begingroup$ I mean I don't really care where it's put, I thought that here was appropriate, but I just want an answer. $\endgroup$ – Dion Silverman Jan 6 '19 at 7:34
  • $\begingroup$ Yes, to your 1st point based on my understanding. Higher intensity means to say electrons are receiving more photons per second, so rate of ejection is higher. See here: quora.com/… $\endgroup$ – Tan Yong Boon Jan 6 '19 at 8:09
  • $\begingroup$ Regarding your second point, recall that E_photon = Work function (W) + KE of ejected electron. Since E_photon = hf, and W is constant, by increasing frequency, KE of ejected electron necessarily increases. I wonder why you said that it doesn't... $\endgroup$ – Tan Yong Boon Jan 6 '19 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.