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When intuition fails: photons to the rescue!
When experiments were performed to look at the effect of light amplitude and frequency, the following results were observed:

  • The kinetic energy of photoelectrons increases with light frequency.
  • Electric current remains constant as light frequency increases.
  • Electric current increases with light amplitude.
  • The kinetic energy of photoelectrons remains constant as light amplitude increases.

In this article talking about the photoelecric effect, it notes that while kinetic energy of photoelectrons increases with light frequency and that electric current remains constant as light frequency increases.

I'm fairly sure that an increasing light frequency causes an increase in the energy of the photon (and thus the increase in the energy of the photoelectron). However, I'm not sure what electric current means in this experiment, and why it stays constant.

Wolgwang
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APerson
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1 Answers1

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This question and its answer may help explain the relationship between intensity and number of photons. As Thomas Fritsch pointed out, electric current is the number of electrons emitted per unit time. Assuming the photons' frequency and hence energy is high enough to knock an electron loose, more incident photons per unit time means more photoelectrons per unit time, which is a higher current.

EDIT: The second answer to the linked question is helpful for your comment. From the first answer we know that intensity is proportional to the product of number of photons and photon energy: $$I\propto n\nu$$ where $I$ is intensity, $n$ is number, and $\nu$ is frequency. From the second we know that intensity is proportional to the amplitude squared: $$I\propto E^2$$ where $I$ is intensity and $E$ is amplitude (I use E here because the first answer in the other question already used A for area.) Therefore $$E^2\propto n\nu$$ That shows how to relate amplitude to the other parameters.

Increasing frequency does increase intensity, but it doesn't increase $n$, the number of photons, so it doesn't increase the number of photoelectrons and hence doesn't increase the current.

EDIT 2: Above I gave the relationship between intensity and the amplitude of a classical wave. Intensity $I$ is defined as power per unit area, and power $P$ is rate of energy flow. Each photon has energy $\hbar \nu$, so if the rate at which photons land on your detector is $R$ and the area of your detector is $A$, then $$I=\frac{P}{A}=\frac{R\hbar\nu}{A}$$

BGreen
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  • Thanks! I read that answer, but I was still confused. I see v in the intensity equation, so I don't get why altering v doesn't change intensity. Also, there wasn't an amplitude in that equation. – APerson Jan 08 '19 at 23:52
  • I just added more. Does that answer your questions? – BGreen Jan 09 '19 at 16:03
  • I think it will, but I don't know what intensity is. This is the first time I've seen it used in relation to photons, and I assumed it meant the same thing as current. What is intensity? – APerson Jan 10 '19 at 23:02
  • It's not actually current. Current is defined the rate at which charge flows, whereas intensity is power per unit area, and power is the rate at which energy flows. Current is a rate of charge flow, but intensity is a rate of energy per unit area. There's a reason you've never seen it used in relation to photons - most formulae for intensity use the classical wave model of light rather than the quantum mechanical model of light. I'll edit my answer again to add how to calculate intensity for photons. – BGreen Jan 10 '19 at 23:27
  • I just added an explanation of how intensity is calculated for photons. Please let me know if you have any more questions, and don't forget to mark my answer as accepted if it's what you were looking for - thank you! – BGreen Jan 10 '19 at 23:35
  • Ah, that helps - thank you! So, increasing the amplitude, increases the intensity, which increases the number of photons, which then increases the current, correct? That makes sense numerically, but why does increasing the amplitude increase the number of photons? Sorry about all the questions. – APerson Jan 12 '19 at 18:26
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    No problem! Asking questions is a great way to learn, and you're hitting on an important point. Amplitude comes from the wave model of light, whereas photons are the particle model of light. It's not so much that "increasing amplitude increases the number of photons", but rather that "when the number of photons increases, it looks the same as if the amplitude of an electromagnetic wave had increased". Rather than a cause-and-effect relationship, they're the same thing being observed through different lenses, so to speak. – BGreen Jan 13 '19 at 19:30
  • You're welcome; I'm glad I could help! – BGreen Jan 16 '19 at 23:55
  • I would have expected that increasing the frequency of the incoming light (without changing the number of incoming photons) would increase the energy absorbed by electrons within the metal surface and increase the probability that they could break through the surface (thus increasing the measured current). – R.W. Bird Jul 18 '21 at 12:07
  • Increasing frequency does increase the energy absorbed by the electrons, but it's not probabilistic in the way you're thinking. The work function $\Phi_0$ is the minimum energy required to free a photon. If $\hbar\nu > \Phi_0$ then every photon knocks loose an electron, while if $\hbar\nu < \Phi_0$ then no photon knocks loose an electron. Therefore, as long as $\hbar\nu > \Phi_0$, increasing the frequency further won't increase the probability that an electron is knocked loose. – BGreen Jul 18 '21 at 17:37