# Can hot food ever emit x-rays or gamma rays?

I was just wondering, if heating food up is the result of increasing the energy of bends and stretches in the bonds of the molecules, is it ever possible for tiny amounts of x-rays and gamma rays be emitted?

When we give a molecule enough energy, its electrons can jump to higher orbitals and then return to ground state, releasing EMR with a frequency proportional to the energy gap. So if we heat food, is there a chance that some electron received enough energy to jump up to a really high energy level?

• Sure, but you'd be hard-pressed to call it "food" at that point, as it'd be dissociated into atomic substituents at best. – Todd Minehardt Jun 11 '16 at 15:15
• Agreed! However, this is worth the "Mythbuster" treatment. If we cannot generate x-rays at reasonable food temperatures, what would it take and what would we have left? See my answer below. – Ben Norris Jun 11 '16 at 15:23
• Your description in terms of bound states of an isolated molecule inside the food doesn't really work. Food is solid or liquid (condensed matter), so the molecules don't exist in isolation. If you did have, say, an isolated water molecule (in steam), it would only have bound states with energies up to some limit, which would be on the order of 1 eV. Any states with higher energies would be states in which the molecule had been ionized or dissociated. As explained in Ben Norris's answer, condensed matter emits a continuous spectrum of radiation, so it's a completely different deal. – Ben Crowell Jun 11 '16 at 21:14
• Can "hot" food ever emit x-rays or gamma rays? // Seems like a Freudian slip. I'm sure vegetables from Chernobyl and Fukushima would be "hot." – MaxW Jun 11 '16 at 22:35
• While not quite related to your question, if you really want to create X-rays in your kitchen, you'd have more success using scotch tape. Though your kitchen would have to be particularly well equipped to include the necessary vacuum chamber. – Johnny Jun 12 '16 at 3:40

In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen).

Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The energy of one photon of 30 petahertz radiation is:

$$E=h\nu = \left(6.626\times 1-^{-34}\mathrm{\ j\cdot s}\right)\left(30\times 10^{16}\ \mathrm{s^{-1}}\right) = 1.988 \times 10^{-16}\ \mathrm{J}$$

This is not a lot of energy! However, a single photon is boring. Let's consider a mole of photons. This will also ease comparison with other phenomena, whose energies are listed per mole of events.

$$1.988 \times 10^{-16}\ \mathrm{J} \times 6.022\times 10^{23}\ \mathrm{mol^{-1}}=1.197\times 10^8\ \mathrm{J\cdot mol^{-1}}$$

In theory, if you could pump that much energy into something, you should get some high energy photons out. In practice, it does not work that way. Other stuff happens first. To simplify our example, let's just consider 1 mole of water (18.0 grams) and heat it up. The fate of basically any other matter will be the same, but the energy required will vary a bit.

First, adding energy heats the water. If we start at room temperature $\left(20\ ^\circ\mathrm{C}\right)$, if takes $80\ ^\circ\mathrm{C}\times 18\ \mathrm{g}\times 4.184\ \mathrm{J\cdot g^{-1}\cdot ^\circ C^{-1}}=6025\ \mathrm{J}$ t heat that water to boiling. It takes 40.66 kJ to convert the water into gas. Neither of these puts a big dent in our energy. It takes further energy to heat the water vapor again, but let's see how far we need to take it.

Once we get enough energy into our sample of water, the molecules start to fall apart.

$$\ce{ H2O(g) -> 2H(g) + O(g)} \ \Delta H^\circ =+920\ \mathrm{kJ\cdot mol^{-1}}\ \Delta S^\circ =0.202\ \mathrm{kJ\cdot mol^{-1}\cdot K^{-1}}$$

By fixing $\Delta G=0$ at equilibrium, we can solve for a temperature at which this reaction becomes spontaneous:

$$T=\dfrac{\Delta H}{\Delta S}=\dfrac{+920\ \mathrm{kJ\cdot mol^{-1}}}{0.202\ \mathrm{kJ\cdot mol^{-1}}\cdot K^{-1}}=4596\mathrm{K}$$

We need to heat our water vapor up an additional 4218 K, which takes $19\ \mathrm{g}\times 1.996\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\times 418\ \mathrm{K}=151.5\times 10^3 \mathrm{J}$.

So, we now have pumped nearly 200,000 J into our water sample, atomized it, and heated it to approximately 5000 K. We are now close to the temperature of the outer layers of the sun! Surely we have enough energy at this temperature to produce x-rays. Nope. At 5000 K, we produce minimal x-rays. Most of the radiation is in the visible, UV, and IR (think about what we get from the sun). Below is a plot of black-body radiation as a function of temperature (image by Wikipedia user Darth Kule and released into the public domain):

Okay, so we are far beyond the reality of what can happen in a conventional oven (or almost any reasonable heat source used for food). At this temperature, we can use the Planck Law to calculate the power output ($I$) of x-rays at the temperature. We can also do this at some normal temperatures and for gamma rays. This model is a little goofy, since food is not a black body, but we will at least calculate the max x-ray and gamma ray output.

Rather than grinding through all the maths, I'll just put in a table of some temperatures and watts. 1 watt is not a lot of power. Most lightbulbs produce light in the kilowatts.

$$\begin{array}{|c|c|c|c|}\hline \mathrm{T\ (K)} & \mathrm{P_{x-ray}\ (W)} & \mathrm{P_{gamma}\ (W)} & \mathrm{notes} \\ \hline 378 & \approx 0 & \approx 0 & \text{boiling point of water} \\ \hline 550 & \approx 0 & \approx 0 & \text{approximate common highest temperature on residential ovens}\\ \hline 700-800 & \approx 0 & \approx 0 & \text{temperature range for wood-fired ovens, tandoors, etc.}\\ \hline 5770 & 4.26\times 10^{-129} & \approx 0 & \text{temperature of the photosphere of the sun}\\ \hline 1.57\times 10^7 & 10.4 & 7.87\times 10^{-54} & \text{estimated temperature of the center of the sun} \\ \hline \end{array}$$

So, if you could heat your food to the temperature of the sun, it would produce minuscule x-ray radiation. It would also no long resemble food.

• "This model is a little goofy, since food is not a black body" It'll be pretty damned black at the kinda temperatures you're talking about! *baddum-tsh* – David Richerby Jun 11 '16 at 21:36
• @DavidRicherby not anymore. Hydrogen as well as oxygen are transparent, and any hydrogen-oxygen plasma will be too dilute to significantly interact with photons as well. – John Dvorak Jun 11 '16 at 21:50
• I think that, very strictly speaking, any black body above absolute zero has a chance of emitting an x-ray/gamma photon, because the Planck distribution is never zero for any finite wavelength. Of course, due to exponential suppression, the probability of emitting such a photon is ridiculously small. I believe I have read that a black body at $\mathrm{37\ ^oC}$ can be expected to emit a visible photon every 1000 years. – Nicolau Saker Neto Jun 11 '16 at 22:13
• "Most lightbulbs produce light in the kilowatts." - What kind of lights do you have in your house? o_O – marcelm Jun 11 '16 at 22:26
• Does the table at the end list power emitted per unit mass? Unit area? I first understood that whole center of the sun produces 10 W of x-rays, which doesn't seem right. :P – ntoskrnl Jun 12 '16 at 14:32

It has nothing to do with what you were going for, but there is a small, but non-trivial amount of x- and gamma-ray output for most food and so the answer is trivially "yes".

In particular any food containing potassium will have the usual admixture of K-40 (with it's 1.3 and 1.5 MeV gamma lines).

When I worked in a lab with a low-background, high-sensitivity germanium detector I used to try to bring a banana in my lunch on days we were expecting a visitor so I could demonstrate the detector without having to do any paperwork concerning possible exposure of the visitor to radiation: just put the banana in a clean ziplock bag, put that in the detector, start the DAQ, and watch a nice peak grow on the screen. Broccoli works just as well, of course, but it's broccoli.

But of course none of this has anything to do with heating the food.

• This is a for-sure reason to use to say that our food emits high energy EMR. – user314901 Jul 3 '16 at 4:33

Not even close! Your food would likely be vaporized, and leave an awful mark on your kitchen table, before it could contain enough energy to be producing xrays.

When I was a youth, I used to do a lot of 'microwave experiments', as in domestic microwave ovens. I would be known for buying a used microwave from a thrift shop for around \$10 and use it to produce 'microwave plasma' with (which was essentially ionizing carbon, analogous to the way neon gas would ionize if you stuck a neon gas filled tube in the microwave and turned it on).

During the course of this, I witnessed a lot of different materials arcing, melting, vaporizing, or ionizing in domestic microwave ovens. At some point, probably after cooking up filamented (incandescent) light bulbs, I started thinking and also wondered about what kind of energies produce xrays. Could I be irradiating myself with with harmful rays or high-energy particles being emitted off of all these arcs? Some of the lightbulbs glowed a crazy green color right before the end of its life, and I remembered reading reading that old-timey xray bulbs produced a distinct green light when producing xrays.

So I looked into it, and it turns out... There is no way that my 1000 watt domestic microwave oven was producing xrays, even if you consider that most lightbulbs are evacuated (under a mild vacuum).

Apparently, if I wanted to start creating x-rays in my own back yard (i didn't), I was going to need at least 2 things:

• More power. A lot more, namely voltage. I dont know what kind of voltage was being induced during arcing, but xray tubes range from 30 to 150 kilovolts (kV)!
• A much stronger vacuum. Xray tubes have whats called a 'high vacuum', which is about 10^-4 Pa (or 0.00009869233 atm) . In comparison, standard incandescent lightbulbs sit at around 70 kPa (0.7 atm), or 'low vacuum'.

Now people will love to start in with 'Well technically, it could', point out that we can calculate this, and proceed to work out a value that requires hundreds of decimal places before you see a significant digit, but just because the equations don't hit a limit and break down at some point, does not mean its ability to describe something physically real doesn't.

Other quick interesting facts about microwaves:

• As stated above, black carbon will ionize in the microwave, such as that from burnt wood or black toner from copy machines.
• While it is true that when the magnetron tube is on, the radio waves are at 2.45 Ghz, the magnetron itself turns on and off at 60 cycles per second. This explains the noise that the plasma makes. When the magnetron tube i on, the plasma expands, when the magnatron tube is switched off, shrinks. The plasma expanding and contracting at 60 Hz is why it makes a loud, 60 Hz noise.
• Glass is not very good at absorbing microwaves... until its molten. Once glass becomes molten, it will absorb microwaves appreciably, causing it to not only stay molten, but grow, eventually turning all the glass molten. Also this is how some glass manufactures keep their glass molten more efficiently, once already melted.
• Microwave assisted chemical synthesis is the new 'green' way to synth, requiring, less to no solvents, sometimes allowing for two homogenized, dry solids to react, higher yields, lower activation energies or no catalysts and in only a fraction of the time to boot.
• Re, "More power... xray tubes range from 30 to 150 kilovolts" Voltage is not power, and you can generate harmful levels of X radiation with equipment that uses no more power than your microwave oven. – Solomon Slow Jun 13 '16 at 18:33
• Re, "Xray tubes have... In comparison, standard incandescent lightbulbs..." The operating principle of an x-ray tube is completely different from the operating principle of an incandescent light bulb. A light bulb works by heating a tiny bit of wire until it is so hot that it's black-body radiation is visible. An X-ray tube works by forming a beam of high-energy electrons, and colliding them with a metal target to produce Bremsstrahlung radiation. The high vacuum is needed for the electron gun in the X-ray tube. A light bulb doesn't need vacuum, it just needs an inert gass fill. – Solomon Slow Jun 13 '16 at 18:43
• @Adam White, would you mind explaining more why the equation breaks down in describing real phenomena? "Just because the equations don't hit a limit and break down at some point, does not mean its ability to describe something physically real doesn't". – user314901 Jul 3 '16 at 4:41