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This question is the continuation of a previous SE post of mine. Since I have $4$ questions to object to KDP answer, I decided to write this new post.

To sum things up briefly: we were concerned about the difference between 1. the case where we shine two photons side by side into a metal at a rate of $1$ photon /s, and 2. the case where we shine two photons in a line (that means one after the other) at a rate of $2$ photons/s. In both cases we have the same power of photons involved.

KDP answered that, while the power is the same in both cases, the surface power density will be different from one case to the other (if I understood well).

  1. My first question is precisely about this surface power density. KDP talks about current (not about surface power density) but I don't quite get the relation between current and surface power density. Current is in A while surface power density is in $\boldsymbol{kg\cdot s^{-3}}$ ?
  2. My second question is about the area involved in this surface power density. Why are we saying that the surface power density in the first case is half the one of the second case ? Since a photon has no diameter how can we say that two photons coming in side by side occupy an area twice bigger than the area of a single photon ?
  3. Third question is: how can we experimentally ajust the rate (in photons/s) of photons coming in in a line ? I mean how can we decrease or increase it?
  4. Last question: can we make converge two photons, exactly at the same time, to a same electron ? For example: if the work function is $\phi$, can we make converge two photons of energy $0.51 \phi $ to one electron and succeed in ejecting it ?
niobium
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1 Answers1

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KDP talks about current (not about surface power density) but I don't quite get the relation between current and surface power density.

When I was talking about current, I was talking about the current produced by the photoelectric device. This is proportional to the number of photons impinging on the photoelectric device's surface area per second.

Since a photon has no diameter how can we say that two photons coming in side by side occupy an area twice bigger than the area of a single photon?

As far as I know, photons do not have a diameter. I was just assuming that was what you were suggesting when you said the photons were coming in side by side instead of one after the other. That they were more spread out. I was simply trying to quantify things to enable analysis. I was simply making the point that if the photons are more spread out, there will be fewer photons per unit of time, hitting the surface because of the finite size of the surface.

Third question is: how can we experimentally adjust the rate (in photons/s) of photons coming in, in a line? I mean how can we decrease or increase it?

If the photons are randomly polarised (unlikely for a laser), we can reduce the rate by passing the beam through a polarising filter. We can also use parabolic mirrors with different curvature for a point source. This will alter the intensity of the light beam, but not necessarily the number of photons in a given line of them. Altering the frequency of the source will not do the trick, because that simply changes the amount of energy carried by each photon. If we have a laser source, where the photons are genuinely in a line one after the other (again unlikely) then doubling the power output of the laser and keeping the frequency constant should double the number of photons per unit length of the beam.

Last question: can we make converge two photons, exactly at the same time, to the same electron? For example: if the work function is ϕ, can we make converge two photons of energy 0.51ϕ to one electron and succeed in ejecting it?

This one is probably too deep into quantum mechanics for my liking, so I will let someone more qualified answer that.

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