Built on Facts

Counting Photons

ScienceBlogs’ own Chad Orzel has on a number of occasions discussed the photon concept in relation to physics pedagogy. He thinks (as I do) that it’s a good concept to teach early even though formally speaking photons are considerably more complicated than the “particle of light” idea that’s often presented as an intuitive concept. But even that billiard-ball model is in a lot of ways not so far from the truth.

Since we’re speaking of intuition, it’s worth noting that the photon model is not necessarily something that at first seems to have a lot of connection to everyday reality. When was the last time you noticed individual particles of light bouncing around?

Then again, when was the last time you saw a single atom? We don’t notice them because they’re so small and there’s so many of them. A glass of water might contain a trillion trillion atoms.

Now the calculation of how many photons per second an average light bulb puts out is an easy one, and it’s been done to death. Millions of trillions of photons per second as a rough estimate, I believe. Let’s do one that’s slightly more fun. How many photons are in the room with you right now? Clearly the lights in the room are putting out a whole bunch of photons, but the speed of light is very fast and so the photons ought to nonetheless be pretty sparsely distributed. We can figure out just how this works.

Imagine a 1-square-meter solar panel. Stand outside and let sunlight shine on it. We know a certain amount of power is impinging on that solar panel – each second, a particular energy is absorbed. But light travels at about 300,000,000 meters per second. Therefore each second sees that solar panel absorb a column of photons 300,000,000 meters long. From that, it’s pretty clear that the energy density of a cubic meter of space has got to be equal to the power passing through that square meter divided by the speed of light.

Now a light bulb doesn’t fill a room uniformly but it won’t hurt to take the average. I’s estimate the room I’m in now is maybe 250 square feet, which would be about 65 cubic meters given the height of the ceiling. Given another rough estimate that the CFL primarily illuminating it is putting out 25 watts of visible light photons, that means we have a power density U = (P/s)/c, where s is going to be the surface bounding those 56 cubic meters.

The averaging we’re doing is in a lot of ways preposterously unjustified, I know. But we’re interested in an order-of-magnitude estimate, not some delicate QED experiment. Given that, how about we pretend this room is a cube. With its volume, that would give a surface area of 88 square meters. Plugging in, that’s about 9.47e-8 J/m^3. Which is 5.3e-8 J, given the volume of the room.

Now we’re almost done. We need the energy per photon. That’s just Planck’s constant times the frequency of a “typical” visible light photon – say, 3.3e-19 J. Dividing, you get about 1.6e11 photons – about a hundred billion.

As a very, very rough estimate anyway. It’s a large number, but much smaller than the number of atoms in the room.

Comments

  1. #1 _Arthur
    March 5, 2009

    Matt, how many neutrinos in your room right now ?

  2. #2 Paul
    March 5, 2009

    But light travels at about 300,000,000 meters per second.

    I think you missed by a factor of 10.

  3. #3 Paul
    March 5, 2009

    Ooops. I take that back. I think I missed by a factor of 10.

  4. #4 Uncle Al
    March 5, 2009

    1 cm^2 projected perpendicular to the sun’s rays gets 6×10^10 solar neutrinos/second (Nature 375(6526) 29 (1995)) – 323 billion becquerels/in^2 or 87 curies/in^2. Bright! (though with a small coupling constant)

    Dr. Schund’s Top Secret/Lotus Eater Neutrino Squeegee can save you. Hey DARPA, can you spare a gigadime?

  5. #5 Matthias
    March 5, 2009

    25 Watts is a lot of light. My 20 m² room is quite brightly lit by a 20 W CFL (i.e. an 100 W equivalent), which corresponds to 5 W of radiant power, assuming an efficiency of 20%.

  6. #6 Alex Besogonov
    March 5, 2009

    Even more fun.

    Imagine that you have absolutely white (100% reflecting) sphere 1mm in diameter, filled with photons with 0.555um wavelength. The insides of the sphere are lit to 0.2 lux with these photons.

    How many photons do you need to take away to reduce the luminance in the sphere by the factor of two?

  7. #7 rob
    March 5, 2009

    here’s a question: what is the size of a photon?

    given that they are excitations of an electromagnetic field, and the fields have spatial extent, you might wonder what the size of the excitation is.

  8. #8 Ned Wright
    March 5, 2009

    At room temperature there are 500 million thermal IR photons per cubic cm, so there are 3E16 photons in your 65 cubic meter room.

  9. #9 Chris
    March 5, 2009

    Millions of trillions of photons per second as a rough estimate, I believe.

    or you could say, “billions and billions”!

  10. #10 Peter Morgan
    March 6, 2009

    Yet, isn’t every atom immersed in its very own (infinite) sea of virtual photons? Do virtual photons count? What about interactions between atoms that are mediated by the electromagnetic field? How many photons are transferred during each interaction between atoms, and how many of them are real/virtual? I note also, sadly, that there is no known well-defined way in which any given photon can be said to be localized outside your room, so every photon in the universe has a toe (sorry, a tail) in your room. I’m not sure what kind of weighted average you want to use for those.

    First make sure your interpretation of interacting quantum field theory makes sense …

  11. #11 dreikin
    March 7, 2009

    So, less than a mole. Much less.. (1 mol = 6.022×10^23 ‘entities’)

  12. #12 Arjen Dijksman
    March 7, 2009

    Besides the IR and visible light photons, there are also a lot of radiofrequency and cosmic background photons swirling round.

    “But even that billiard-ball model is in a lot of ways not so far from the truth.” –> Photon needle-like model gives supplementary intuition in terms of polarization, wave-particle duality, quantum-mechanical state-vector, etc.

  13. #13 Joe Doakes
    March 7, 2009

    It seems to have been forgotten that light was proved to be a wave, and not a corpuscle, when moving from one place to another. Who regards a phonon as a particle instead of a sound wave pulse when interacting with matter, in which the only particles involved are atoms or molecules? Maybe it is more realistic to think of a photon as a light pulse when interacting with matter. The advantage of thinking of light and matter as particles instead of waves is avoiding the forbidden question: waves in what?

  14. #14 CCPhysicist
    March 7, 2009

    That is a really nice example. I’ll have to remember to do that application of the energy density when I get to EM waves this year. I introduce photons at that point in the course, even though the book defers them to modern physics, because the basic idea of a photon is so important to various modern technologies – not to mention photo-chemistry.

    I also had an interesting thought: express the number of photons per second coming out of a light bulb in terms of moles, and then compare to the volume of water per second from a hose. Need to sit down and work that out.

    But I am slightly disappointed in the depth of your analysis. When I saw the title, I figured you would be including black body radiation. Similarly, one commenter made the major error of considering only visible radiation from an incandescent light bulb. IR “heat” also consists of photons.

  15. #15 Bee
    March 9, 2009

    How many photons are on avarage absorbed and possibly reeimmetd by a single atom in, say, a metal that I put in the sun or under a lamp, or an air molecule, pick whatever you like best.