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.