# X-rays / Gamma rays "oven" vs microwave oven

Let's imagine a seller scammed you and sold you a Gamma rays / X-rays "oven" instead of a common microwave oven. The power consumption would be the same as a common microwave oven, i.e. about 1 kW, but the oven would use x-rays and/or gamma rays instead of micrometric waves. How would one notice that the oven behaves differently than a common microwave oven?

I do realize that the door of the microwave ovens have holes that let pass visible light but not microwaves, so, for the sake to make the question more interesting, let's assume that the door consists of a thick metallic wall. So essentially the oven looks like a metallic box from the outside.

Would the energy be used to heat the food as much as if microwaves were used? If not, is the energy then wasted to heat the walls of the x-rays oven?

• Microwaves have the right frequency to make water molecules oscillate. Gamma rays don’t. Jun 15, 2019 at 21:10
• @G.Smith So essentially you've fallen into the misconception that got downvoted at physics.stackexchange.com/questions/169362/… ? Jun 15, 2019 at 21:17
• See iapws.org/faq1/mwave.html and en.wikipedia.org/wiki/Microwave_oven. I did not say that microwave ovens operate at precisely the resonant frequency of water. Jun 15, 2019 at 21:36
• @G.Smith So, both x-rays and gamma rays would not be well absorbed by the food, if I understand well? I guess they would be absorbed by the walls, then? If that's so, please write up an answer, I'd accept it. Jun 15, 2019 at 21:46

In a past exam question, an experiment "performed by a student" described removing the rotating plate from the microwave and adding a "chocolate bar".

Due to standing waves being formed due to the EM waves reflecting off the opposite side to the transmitter, the "chocolate bar" would only "melt" in certain, periodic dots - at the antinodes.

The wavelength of the light can then be calculated by measuring the distance over $$2n+1$$ antinodes and then dividing this by $$2n$$.

We could then calculate the frequency of the EM waves being emitted, since we know $$c$$ and $$f=\frac{c}{\lambda}$$ and thus determine whether they are microwaves or x-rays.

Also, x-rays and microwaves have different properties - such as how they interact with organic matter so there is likely some simple test you could do in that sense.

• Following this line of thoughts, visible light have antinodes at much shorter length than microwaves, so the chocolate bar should melt almost continuously through it. But that's not what happens of course. It doesn't penetrate well into the chocolate bar at all. However X-rays certainly would! So what would happen should be investigated as you say, from the differences in how X-rays and Gamma rays interact with organic matter. Jun 15, 2019 at 21:21
• Well, plus x-rays can go completely through you with reasonable probability, making them pretty useless for heating. Jun 15, 2019 at 21:23
• @JonCuster that's what I suspected. So essentially the X-rays and Gamma rays would be mostly absorbed by the metallic walls, right? Please post this as answer if true... I would accept it. Jun 15, 2019 at 21:30
• microwave ovens have feedhorn assemblies intended to reduce the prevalence of antinodes and hot spots inside the cavity. Would the chocolate bar experiment have really worked like this in practice? Jun 15, 2019 at 21:35
• @JonCuster. wouldn't soft xrays be mostly absorbed by a human body? Jun 15, 2019 at 21:55

Gamma ray doses are measured in sieverts and grays: one joule per kilogram. A dose of 5 sievert is deadly, but the heat developed is negligible: a temperature rise in water of about 0.001 C.

So your kilowatt oven would be visibly ionizing the air, I think.

• So essentially most of the energy of the gamma rays is used to break atomic/molecular bounds instead of making them vibrate? Jun 16, 2019 at 20:07
• A gamma photon gives rise to high-energy electrons which in turn produce other ionizations. Finally, all the energy becomes heat.
– user137289
Jun 16, 2019 at 20:14
• Hmm, so in the end a part of the energy is lost to ionize (and heat up?) the air. And what is absorbed by the food translates as heat too. So, I am a bit lost with the difference(s) with the common microwave oven, with regards to the heating of the food. Jun 16, 2019 at 20:27
• You asked what you would see (extremely hypothetical question). I answered that you would likely observe an airglow in the oven.
– user137289
Jun 16, 2019 at 20:30
• Gamma rays do not care about walls, whether it is concrete or aluminum. Gamma rays will mostly pass through food too.
– user137289
Jun 16, 2019 at 20:49

A microwave oven is a resonant cavity; any microwave energy that is not heating the food is reflected by the walls, and re-enters that tasty ear of corn (or whatever). X-rays, however, do not reflect, the sole way a returning bit of radiation comes from a wall is so-called secondary radiation (fluorescence).

So, the nominal 1 kW X-ray source won't deliver its full output (or anything near it) during a single pass, and the secondary radiation is both lower energy and low in efficiency (far less than half after one wall impact).

You should expect one or two orders of magnitude diminution of cooking efficiency, and much warming of the oven walls, with an X-ray source, even if it were energy-efficient and aimed to strike the food.

It will also ionize molecules in the food, which could cause flavor changes unlike those caused by simple heating.

• Thanks a lot for your answer, I think I'll accept it soon. Aug 8 at 12:22