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I was watching this experiment (https://www.youtube.com/watch?v=v-1zjdUTu0o) which demonstrates the photoelectric effect, but it does not make any sense to me how it proves light as a particle instead a wave. Can you please explain me? And I also want to know where exactly the electrons that are released by beam of UV light go.

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  • $\begingroup$ What do you mean, "particle?" The photoelectric effect proves that light delivers discrete quanta of energy, and it proves that the amount of energy in each quantum is a function of the wavelength. In other words, you can count photons. That sounds very particle-like, but it is not proof that photons bear any other resemblance to your idea of what "particle" means. $\endgroup$ – Solomon Slow Dec 17 '19 at 15:45
  • $\begingroup$ @Solomon Slow So energy of each quantum is dependant on wavelength/frequency of light, am i right? $\endgroup$ – user248881 Dec 17 '19 at 15:53
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    $\begingroup$ Yes. I should have said "frequency," because the wavelength can change when a photon passes through different media, but the frequency remains constant. The energy of a photon is $h\nu$ where $h$ is the Planck constant, and $\nu$ is the photon's frequency. $\endgroup$ – Solomon Slow Dec 17 '19 at 16:03
  • $\begingroup$ related: physics.stackexchange.com/questions/68147/… Note that the two answers contradict each other. $\endgroup$ – user4552 Dec 17 '19 at 19:21
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    $\begingroup$ @user248881, yes. The ideal laser would emit a beam of coherent light. That is, light in which every photon has exactly the same energy/frequency/wavelength, and in which every photon appears to have come from the same, distant, infinitessimal point. Practical lasers come very close to that ideal. $\endgroup$ – Solomon Slow Dec 18 '19 at 14:27
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According to the classical theory, light is a wave and this effect would be explained to be due to transfer of light energy on to the Zinc plate. Increasing the intensity should therefore increase the kinetic energy of the ejected electrons.

However, this did not meet up with the experimental observations which showed that light below a particular frequency did not cause any photoelectric effect. Also, increasing the intensity increased the photoelectric current instead.

Einstein then proposed that light behaved as a stream of discrete wave packets called photons which had an energy value associated with them which depended on its frequency. Both observations could be exhaustively explained with this theory.

As shown earlier in the video, the plate is negatively charged which means it has excess of electrons. When hit with UV light, the electrons leave the metal and become dissipated in the surrounding air.

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  • $\begingroup$ So according to you, “If we increase the intensity of light then it should increase the kinetic energy of the ejected electrons”. And if that happens then that proves light as a wave. My question to you is, How this proves light as a wave? It doesn’t make any sense to me! And if photons have energy (E = h ν), then EM waves also have energy (E = h/λ). For example, we shine UV light on negatively charged zinc plate, it releases electrons from it. But how exactly this proves that light coming from a UV lamp is not a wave, but a particle (photon/discrete quanta of energy)? $\endgroup$ – user248881 Dec 18 '19 at 7:26
  • $\begingroup$ That sentence was for the classical theory which states that light is a wave. But the classical theory fails to explain why normal lights will not cause photoelectric emission. Light is treated as a particle in the quantum theory which is more accurate. There is no proof as such, just better and better assumptions (theories) to explain them. $\endgroup$ – Sam Dec 18 '19 at 15:51
  • $\begingroup$ If light is treated as a wave, all wavelengths should have caused release of electrons. When a discrete particle of light has sufficient energy, it happens. $\endgroup$ – Sam Dec 18 '19 at 15:53

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