# How many photons are in a microwave oven?

I have always had trouble connecting the picture of classical electromagnetism with the idea of photons. To make this connection better I'd like to ask the following question. How many photons, at an instant in time, are there inside a microwave while you are heating your food? Since microwaves are so large (~12cm) I figure it could either be one photon or many photons that together form the microwave.

• Do you mean the photons related to microwave radiation heating the food or all photons (including heat radiation inside, photons serving the chemical bonds in the oven and food, etc) ? The question may be very hard. Including or excluding the virtual photons ? Commented May 3, 2020 at 18:08

Household microwave ovens operate at 2.45 GHz, apparently.

So the energy of a photons is $$E=hf=6.63\times 10^{-34} \times 2.45 \times {10^9}=1.62 \times 10^{-24}$$ Joules

So if you take an 800 Watt unit, that is putting $$4.92\times 10^{26}$$ photons per second into the microwave unit. Which is indeed `many photons'.

(That could be an overestimate by a factor of about 2, depending on whether your '800 Watt' microwave is one that consumes 800 W wall plug power, or one that puts a useful 800 W into the oven. I don't know the exact specification and a quick google search was unhelpful. )

• When a microwave oven starts an internal fan kicks in. Clearly the air flow is necessary to convect away heat from the electronics. I don't know whether most of the heat is generated in the emitter (the magnetron) or in other electronic components. I wonder:it could be that the emitter will inevitably re-absorb microwave radiation, generating heat. Commented May 3, 2020 at 11:47
• @Cleonis I've always been told that this is why one should never use a microwave empty. The radiation simply has no other place to go, except back into the magnetron which is conveniently tuned to the right sequence. But then the cathode used in the magnetron is also heated and who knows what else. But the order of magnitude is right. Much less than 800 W of radiation results cold food and much more in a trip to the circuit breaker.
– mlk
Commented May 3, 2020 at 17:48
• The usual household microwave oven is 600-800W microwave power derived from 1000-1200W wall power. There is no 800W wall power microwave from factory. Commented May 3, 2020 at 18:03
• Thanks for the enlightenment! Commented May 3, 2020 at 18:59

I agree with @RogerJBarlow's calculation arriving at a production rate of $$5\cdot 10^{26}$$ photons per second. However, this doesn't yet answer the question

How many photons, at an instant in time, are there inside a microwave?

So let's continue the calculation from here.

Each photon is reflected from the metal walls several times until it finally is absorbed by the food. So let's assume it travels around $$l=1\text{ m}$$ in a zig-zag way. Using the speed of light ($$c=3\cdot 10^8 \text{m/s}$$) this will a take a time of $$t=\frac{l}{c}=\frac{1\text{ m}}{3\cdot 10^8\text{ m/s}} =3\cdot 10^{-9}\text{ s}.$$ Hence, we find the number of photons inside the oven: $$N=5\cdot 10^{26}{\text{ photons/s}}\cdot 3\cdot 10^{-9}\text{ s} =1.5 \cdot 10^{18}\text{ photons}.$$

• Fair enough - I ducked out of that point as it involves a guess as to how long the photons hang around, or how far they travel. And I thought the questioner basically wanted to know whether the number was "few" or "many" Commented May 3, 2020 at 11:25
• I do not understand the logic of this. Where does the energy go? Actually I would count the reflected photons as extra photons adding to the power input by their delay in disappearing. If there were no walls all the wattage would be the first number.Reflections add to that number imo., Commented May 3, 2020 at 13:34
• @annav I'm not sure I understand your objection. The energy ends up as heat in the food. A reflection at the wall (i.e. absorption + reemission of a photon) doesn't change the EM field energy and number of photons. Absorption by the food reduces the EM field energy and number of photons. Commented May 3, 2020 at 16:41
• The walls are designed to be reflective of microwaves so as to increase the energy per second within the cavity. The simple wattage from input count gives the energy per second from the primary source, if you want to include the reflected energy it should increas the number of photons in the chamber with respect to having no walls as the reflection is like new sources . Commented May 3, 2020 at 17:18
• The energy fills up the volume of the microwave. ;) Instead of reflective walls, imagine that the microwave is simply a very long tunnel. The power rating of the microwave tells you approximately how many photons per second enter the tunnel (cross the plane describing the "start"), while the calculation above estimates how many are in flight inside the volume of the tunnel, accounting for the "extra energy" you describe. Commented May 3, 2020 at 20:13

Classical electromagentic waves are characterized by their frequency. They are an emergent state from zillions of photons. Photons are not waves, they are zero mass elementary point particles, characterized by their spin and energy .

The number of photons making up the classical wave can be estimated for specific cases. The energy of an individual photon contributing to the classical wave of frequency $$ν$$ is $$E=hν$$, where h is the Planck constant . The energy of the photons in the microwave range is is $$1.24 µeV$$ –> $$12.4 feV$$ .

If you know the power of your microwave oven, in watts, which are per second, you can get it to electron volts with this calculator and then divide by the energy of a photon for the frequency of your oven.

Try it and you will understand my saying that the classical beam is composed out of zillions of photons.

By the way, the photons do not stay in the oven, they are absorbed and scattered and end up as heat in the food and surroundings.

• This was really useful. Also in term of the theoretical aspect. But what do you mean photons are not waves? Aren't they described by some wavefunction? Commented May 3, 2020 at 10:18
• Ah, but the wavefunction $Ψ$ are not measureable, because they are complex numhers, $Ψ^*Ψ$, the complex conjugate square gives the probability of finding a particle at (x,y,z) See this answer of mine physics.stackexchange.com/questions/90646/… Commented May 3, 2020 at 10:24
• Yes I'm familiar with quantum mechanics (relatively ofcourse). I mean that since the probability density $\rho=\Psi^*\Psi$ isn't a delta function but extends in space, isn't it accurate to call it a wave? A point particle whose probability function looks like a wave. Commented May 3, 2020 at 10:51
• if you looked at the double slit single photon experiment in the link I gave, you would understand the distinction between waves due to probability distributions and waves in a medium. The footprint of each photon is a dot. Commented May 3, 2020 at 11:52