According to the black-body radiation equation, the spectrum extends to infinitely high frequency (although its intensity gets small quickly towards high frequency).

(1) How do you roughly estimate the ionizing radiation power in common high power thermal sources like a 2KW heater without fan, then make sure it is safe considering it is used for years instead of 0.1 second/year in the case of DR X-ray scan? (First the portion of power getting radiated instead of convected needs to be estimated, second this seems not to be black-body and will it have a similar radiation curve that extends to infinite frequency?)

(2) In the photoelectric effect, there are things like cut-off frequency; so, why doesn't thermal radiation - which is quantum mechanical in microscopic level - have a cut-off frequency, i.e. it doesn't radiate X-ray and gamma-ray at all?

Planck's law says:

$$B_\nu(\nu, T) = \frac{2h\nu^3}{c^2} \frac1{\exp\left(\frac{h\nu}{kT}\right) - 1}.$$

This is the power emitted by a black body of temperature $$T$$ at frequency $$\nu$$. If we consider a black object (thus highest possible purely thermal emission) of temperature $$700°\mathrm C$$ (an estimate of temperature of a heater), it'll emit total power of

$$P_\text{total}(700°\mathrm C)=\pi \int\limits_0^\infty \mathrm{d}\nu B_\nu(\nu,700°\mathrm C)=13.6\,\frac{\mathrm{kW}}{\mathrm{m}^2}.$$

(This can also be calculated via the Stefan-Boltzmann law.)

If we now want to find the power of only ionizing radiation, we should instead integrate in the range of $$\nu\in[\nu_0,\infty)$$, where $$\nu_0$$ could be taken somewhere in the UV range, e.g. lower border of UVB, i.e. $$950\,\mathrm{THz}$$. Then we have:

$$P_\text{hard}(700°\mathrm C)=\pi \int\limits_{950\,\mathrm{THz}}^\infty \mathrm{d}\nu B_\nu(\nu,700°\mathrm C)=3.1\times 10^{-20}\,\frac{\mathrm{W}}{\mathrm{m}^2}.$$

So a surface of $$1\,\mathrm{m}^2$$ will emit $$~10^{-20}\,\mathrm{W}$$. This is totally negligible, even in the course of a hundred years.

• Do you have recommendation for some tools to do this calculate? I want to calculate the radiation bellow 700C too. Since the heater burn the skin as Sun, I guess radiation that fall in the range of skin sensing is above 100W. But visible light is not that far away from ionizing radiation, I wonder why the power difference is so much.
– jw_
Feb 26 '20 at 12:24
• @jw_ I did the integration in Wolfram Mathematica, using its builtin NIntegrate function. But if you need a free-software solution, you might want to look at GNU Octave's quadcc function or Scilab's integrate function. Feb 26 '20 at 12:47
• @undefined well, you can try finding the closed-form indefinite integral (although it's not elementary, see Wolfram|Alpha result), and then use Newton-Leibniz formula on it. You'll need a way to compute polylogarithm though. Or you can make a substitution like $\nu=\tan(t)$ to convert the improper integral to a proper one, and then use e.g. trapezoidal quadrature rule to find numerical approximation of the definite integral. Feb 26 '20 at 15:30
• @jw_ you can also get burns without any ionizing radiation and in the case of your space heater it is not the ionizing radiation you have to worry about. Feb 26 '20 at 16:33
• Given enough flux, you can incinerate a human with long-wavelength microwaves. This is why you don't want to stand in front of a high-power radar emitter. Feb 26 '20 at 20:38

for 1) In this link there is the black body formula of Planck's law that gives the

the power per unit solid angle and per unit of area normal to the propagation

So for a given temperature of the body one can calculate the power, in your case for high enough frequencies, take a $$Δ(ν)$$ .

Once you have the power you will have the energy per second. Divide the power with the average $$hν$$ of your $$Δ(ν)$$ and you can estimate how many photons of that approximate frequency radiate per second.

You will find that one has to wait for a loooo..ng time for an X-ray photon to come our of your radiator. Better worry about the Xrays created by the muons passing your body at the rate of 1 every $$cm^2$$ every minute, much worse when you fly in an airplane.

for 2) Quantum mechanics gives probabilities for limits. When the probability gets very very small, that is a limit.To get enough energy for an x-ray the material would have to synchronize so that random vibrations would quantum mechanically allow for a quantum level to exist to have enough energy for an x-ray photon to come out.How improbable that is is reflected in the Planck formula.

Also a limit is given by energy conservation laws , also included in the formula.

The basic line is that it is very improbable, as the other answers state.

• Now I'm wondering how much the metal frame of an airplane actually blocks such radiation.... Feb 26 '20 at 16:00
• @Michael barely, but if I remember from last time I did the calculation you would have to be at 30,000 feet for about 6 months straight to reach the 20 mSv European annual limit for radiation workers (the American limit is 50 mSv) Feb 26 '20 at 16:57
• Considering that gold is one of the best X-ray reflectors, and that it only achieves total reflection at less than 3 degree incident angle, you can rest assured that X-rays and above are not the least bit concerned with puny things like atoms. Feb 26 '20 at 20:35

(1) The element in most heaters appears to be a cooler color/temperature than the sun - and only similarly warm. So, it is safer than the sun.

(2) The photo-electric effect reveals that below a cut-off frequency there is not enough energy to release an electron, so the energy eventually ends up as heat. Above the cut-off an electron may be freed. But all frequencies are involved and can do something.

• The outer part of the sun is largely in thermal equilibrium and shields you from the inner part of the sun - a classic black body case. And the distance does not matter as long as the warmth you feel is the same - unless you think the atmosphere is shielding you from the sun. Feb 26 '20 at 13:22
• Good to learn that, and in fact the warmth argument is good, now get it.
– jw_
Feb 26 '20 at 13:56
• Most of the sun is so opaque that it takes a photon thousands of years or more to escape from the core to the surface. Gamma rays from fusion are absorbed by the intermediate layers and re-emitted at lower frequencies, which is why we observe very few of them at Earth. Feb 26 '20 at 20:42
• “unless you think the atmosphere is shielding you from the sun” This really make the Sun comparison invalid. The atmosphere could shield most high energy radiation.
– jw_
Feb 27 '20 at 1:34
• The atmosphere does shield high-energy radiation, specifically ozone blocks UV (which is right at the edge of ionizing radiation). Makes sense: if the radiation is high enough energy to ionize, why wouldn't it ionize the atoms in the atmosphere? Feb 27 '20 at 11:09

If you plug in the values of $$T=1000$$K, $$\nu=10^{15}$$Hz into the spectral radiance formula of a blackbody (Plank’s law): $$B_{\nu}(\nu, T) = \cfrac{2h \nu^3}{c^2} \frac{1}{\exp\left(\frac{h \nu}{k_BT}\right) - 1}$$ where $$B_{\nu}(\nu, T)$$ is the power delivered per unit area per solid angle at a frequency range centred at $$\nu$$ by a blackbody at temperature $$T$$ and do the following integral $$P\left(T,\nu_0\right)=\int_{\nu_0}^{\infty} \cfrac{2h \nu^3}{c^2} \frac{d\nu}{\exp\left(\frac{h \nu}{k_BT}\right) - 1}$$ Here since the argument of the exponential is $$\sim 48$$, we can approximate the integral to

$$P\left(T,\nu_0\right)=\int_{\nu_0}^{\infty} \cfrac{2h \nu^3}{c^2}\exp\left(-\frac{h \nu}{k_BT}\right)d\nu$$

This is a straightforward integral which can be done by parts to get the total ionising radiation power to be $$P\left(1000\text K,10^{15}\text{Hz}\right)=1.69\times10^{-16}\frac{\text{W}}{\text{m}^2\cdot\text{st}}$$ This as you can see, is minuscule.

In photoelectric effect, we are knocking off an electron from the bulk using the energy of photons. The excess energy is converted into kinetic energy of the electron. Now consider the reverse. High energy electron is getting captured by the bulk. So now a photon is released depending on how much energy the electron lost. Since the electron before capture can have arbitrary energy, the emitted photon can have arbitrary energy. Thus there is no cutoff for radiation.

The cutoff comes in the case of photoelectric effect due to the restriction on allowed energies of the bound electron. However there is no such restriction on free electron.

• (1) some error to fix? Bv(v,t) is per Hz value, it need to be integrated with frequency, this also means "so all subsequent powers for higher frequencies will be less than the above value" does not work since the integrage converge but a multiplication will result in infinite. (2) "Now consider the reverse. High energy electron is getting captured by the bulk" An electron with high energy, how can it be captured in one step? - the whole reasoning may be right, but this detail is the key to decide whether high energy photon can be emitted right?
– jw_
Feb 26 '20 at 7:16
• Yes! You’re right. Fixed my error. As for the electron capture, it’s a probabilistic process. We’d have to calculate the overlap if the free electron wavefunction with the bound states of the atom. Feb 26 '20 at 7:28
• It seems that you don't want to do the calculation as I did:) , could you make some strict estimation of the upper bound? Why is everybody talking about the Sun, that is not a right comparison! For the second question, I suddenly thing it is just like brake radiation, when electron collide with something and brake or bounce back, during this process photon is emitted - never elastic.
– jw_
Feb 26 '20 at 7:34
• You’re right about the calculation part :P, maybe I’ll try it after my classes. The overlaps only give you the probabilities of capture. But yes, you can have radiation without capture, like brake radiation. That probably is the dominant mechanism for radiation in plasmas (like sun). I compared it with the sun because that’s one constant source of radiation! Feb 26 '20 at 7:42
• Google is not being very helpful, but from en.wikipedia.org/wiki/Infrared_heater "Ceramic elements operate in the temperature of 300 to 700 °C producing infrared wavelengths in the 2 to 10 μm range". I assume the OP is not asking about "modern" ceramic heaters, though, and they're more interested in the style of heater shown at the top of the article. IIRC, such heaters radiate at roughly 1000 °C, and they (obviously) radiate a significant amount of visible light as well as infrared. Feb 26 '20 at 11:07

How to roughly estimate the ionizing radiation power in common high power thermal source like 2KW heater without fan then make sure it is safe considering it is used for years instead of 0.1 second/year in case of DR Xray scan?

A common 2 kW resistance heater does not emit ionizing radiation. It's radiant heat is primarily infra red. The range for infrared radiation is about 430 x 10$$^{12}$$Hz to 300 x 10$$^9$$ Hz. It is considered non-ionizing radiation.

Ionizing radiation begins around 3 x 10$$^{15}$$ Hz. X-radiation is in the range of 3 x 10$$^{16}$$ to 2 x 10$$^{19}$$ Hz.

Bottom line: Infrared heaters do not pose the health risks associated with ionizing radiation like X-radiation.

Hope this helps.

• It does not, or very little? If doesn not, then what is the theory behind - the radiation curve says it does? If very little, can a roughly reasoning estimation be provided?
– jw_
Feb 26 '20 at 3:40
• @jw_ what "radiation curve"? The energy of an infrared photon is insufficient to ionize an atom. Feb 26 '20 at 3:42
• Anything have a radiation curve - the power spectrum of its radition - power vs frequency.
– jw_
Feb 26 '20 at 3:44
• @jw_ Can you provide a reference for the "radiation curve" ? The energy of an infrared photon anywhere in the frequency range I cited is insufficient to ionize atoms as are x-rays. Feb 26 '20 at 3:48
• @BobD I believe he is referring to the black body spectra. Feb 26 '20 at 4:48