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I don't understand something in the photoelectric effect experiment (which consists in shining light to a metallic material and then measuring the kinetic energy of the ejected electrons, as well as the amount of ejected electrons).

From what I have understood we are considering photons shined at a certain rate (say $10$ photons per second) and, if their energy is greater than a certain work function, they eject electrons, at a rate proportional (or maybe equal?) to the $10$ photons per second.

What I don't understand, is that these photons surely come together encompassing a certain area (whose plane is orthogonal to the direction of propagation of the photons). Does the cross-sectional area of the beam affect the experiment?

Experimentally, how did they measure the amount of ejected electrons? Did they simply divided the total number of ejected electrons by the area (that encompasses the outcoming electrons)?

EDIT

I guess the point of my post boils down to this question:

How do we differentiate experimentally the case where photons come $2$ by $2$ (side by side) at a rate of $1$ photon/s and the case where photons come in a crocodile line at a rate of $2$ photons/s? In both cases, the beam has the same power but different areas. It seems that $2$ electrons will be ejected per second.

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  • $\begingroup$ If the light only hits the metal, and you have a reasonable chance of detecting the emitted electrons, the area that is illuminated has nothing to do with, well, anything. I'm unclear what your question is here. $\endgroup$
    – Jon Custer
    Commented May 29 at 13:59
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    $\begingroup$ I have edited to make your question clearer. $\endgroup$ Commented May 29 at 17:22

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This answer was for version 1 of the question.

Experimentally, how did they measure the amount of ejected electrons ?

Typically you just measure the current produced by the ejected electrons, which is an indication of how many electrons were released. The area over which the ejected electrons come from is irrelevant.

Did they simply divided the total number of ejected electrons by the area

That would give a number of electrons ejected per area, so that isn't a number of electrons.

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Classically, it is expected that the EM field will deliver energy into a spherical volume at a rate proportional to its intensity. This in turn implies some delay in emission of particles absorbing that energy. This should happen regardless of frequency.

Shining a monochromatic light of configurable frequency onto a metal plate would produce a current almost as soon as the light was turned on. The current was measured by producing a potential difference(V) to cancel out the current. At some critical potential, current would cease.

So the light source, angle of incidence, and detector could be kept at constant position relative to the plate while varying the more salient features.

It was discovered that so long as frequency exceeded a minimum, a current would flow regardless of intensity, and it would start about the same time the beam hit the plate.

Experimentally, a linear relationship was established, $eV=kf-\phi$.

A line fit establishes $k=h$, Planck's constant for the slope, $\phi$ is the work function, the minimum energy needed for an electron to escape.

As to counting electrons, I think the current itself would suffice. Figure a current implies a current density, $J=I/A$ where I is current and A is cross-sectional area depending on geometry of the system. If area of detector is sufficiently small compared to cross section of the incident beam, it could serve as a good Area there.

$J=ne \sqrt{(2eV/m_e)}$, where $n$ is electrons per volume.

So determine $n$ from the current and graph it against the intensity of the light source. There's likely an optimal angle for detection, but if the position of detector is held constant, the electron density to light intensity relationship should still hold. I believe a linear relationship was established historically.

I'd think the phase variance would make your two by two side by side scenario impossible to perform in the early 20th century. Sufficiently collimated, the in-a-line scenario might have been possible, but flux into the collimator would equal flux out of the collimator and that's all dependent on the source.

So I think area of a sufficiently small detector placed in the path of outgoing electrons would suffice to give you a sense of how many electrons are emitted. Since everything is reckoned in terms of a per/area anyway I think it cancels out

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How do we differentiate experimentally the case where photons come 2 by 2 (side by side) at a rate of 1 photon/s and the case where photons come in a crocodile line at a rate of 2 photons/s? In both cases, the beam has the same power but different areas. It seems that 2 electrons will be ejected per second.

You are right than in both cases the power is the same and the case where the photons are coming in side by side will have half as many photons per unit cross sectional area of the incoming beam relative to the inline case. For a receptor with a given area, the side by side photons will generate half the current of the inline beam. The inline case is more concentrated somewhat like a laser beam. It has twice the "intensity" for a given power. This is how you could in principle, distinguish between the two cases. The current generated by the photoelectric device is proportional to the number of of electrons emitted per second which is the same as the number of photon impinging on the surface per second. Electric current can be defined as Coulombs per second which is the amount of charge in a current passing a point per unit time which is proportional to the number of electrons generated per second.

The area of the receptor is relevant. The greater the area, the more photons per second that are intercepted. This is similar to why the power generated by solar panels id proportional to the area of the panels.

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  • $\begingroup$ Your answer tackles exactly right the point of my question. I have made a separate post to ask more questions, and extend the discussion on this subject: physics.stackexchange.com/questions/816571/… $\endgroup$
    – niobium
    Commented May 31 at 7:25

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