How many photons does my remote control garage opener emit? Every time I drive up to my house I imagine all the photons spitting out of the remote control garage opener when I press the button. And I imagine the door opener in the garage receiving them. There must be a ton of these particles going in all directions if the door always opens, right? 
I'm just curious: 
How many photons does the remote control send out, roughly? 
And how many photons must my garage door opener receive to know it is time to open the door? Just one? 
 A: Most likely, your garage door opener operates at a frequency of 315 MHz.  Multiplying by Planck's constant, that means each photon has energy of about $2\times 10^{-25}$ joules.  
Most likely, your garage door opener operates at about $1/10$ of a watt (or less, per comments below).  So each second, it emits 1/10 of a joules of energy.  
That's $(2/10) \times 10^{25}$ photons per second or (very roughly) $5\times 10^{24}$.  In other words, $5,000,000,000,000,000,000,000,000$ photons per second.  
To see how many of those hit the receiver, let $r$ be the distance from the transmitter to the receiver.  The surface area of a sphere of radius $r$ is $4\pi r^2$.  So if the reciever has area $A$, a fraction $A/(4 \pi r^2)$ of the photons hit the receiver.  That's going to be a pretty small fraction, but still a whole lot of photons.
A: The concept of a photon comes from quantum field theory, but the radiation from your remote control is well in the classical regime, so we need to figure out how to go from the quantum field theory to the classical field theory.
Now, the way we do that is by forming so called minimum uncertainty states. There's a Heisenberg uncertainty principle for fields analogous to that for particles. The uncertainty principle can be stated as saying that an electromagnetic wave can't have a definite amplitude and phase, and making one more certain necessarily increases the uncertainty in the other.
The amplitude of the wave is related to number of photons. You can realize this because each photon carries energy, and classically the amplitude is related to the energy. Thus Heisenberg says that the state produced by your remote control doesn't really have a definite number of photons.
But we can still estimate the most contributing states if the uncertainty is minimal. If the radiation generated contains only one frequency, the probability of finding $n$ photons in a minimal uncertainty state is proportional to $|a|^{2n}/n!$ where $a$ is a complex number. It is related to the amplitude by that we should have $\hbar \omega |a|^2 = \epsilon_0|E|^2 V$, these being two expressions for the total energy emitted by the remote control.
Now since the energy is much larger than the energy of a single photon, $|a|$ must be (very) large. Then the most significant contribution will come from $n$ also large. For large $n$, we have that $n!$ is approximately proportional to $(n/e)^n$, so we have $(e|a|^2/n)^n$ to maximize. This is equivalent to maximizing the logarithm, $$n(2e\ln |a| - \ln n)$$
which is found to have the maximum $|a|^2 = n$. (Who would've thought!?)
So, to the extent that the radiation from your remote control can be said to have a definite number of photons, that number is indeed the energy divided by the energy per photon. But you should note that this calculation is for a monochromatic wave. No device can emit such radiation, all radiation is a mix of frequencies. For real radiation you should take the energy at each frequency, and calculate the number of photons at that frequency.
The above is sort of hunting sparrows with 8.8 cm FlaK since in the end I just did the obvious thing of taking energy divided by energy per photon. But I wanted to stress that because of the Heisenberg principle, and that the classical limit of a quantum field theory isn't as obvious as you might think, it's not so obvious that this is actually the correct thing to do.
Now, I said, to the extent that the radiation can be said to have a definite number of photons. How large is that extent? Well, when $|a|$ is very large, the quantity $|a|^{2n}/n!$ is extremely peaked, and if you work it out, $|a|$ is indeed extremely large. So, for "ordinary" radiating devices, not much is gained by thinking about their radiation in terms of photons, for they are well, well within the classical regime.
