# Ionizing radiation in thermal radiation

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.

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.

(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.