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I read somewhere that most things1 emits all kinds of radiation, just very few of some kinds. So that made me wondering whether there is a formula to calculate how many X-rays an 100W incandescent light bulb would emit, for example in photons per second. For example, we already know that it emits infrared and visible light.

I find it hard to describe what I have tried. I searched on the internet for a formula, but couldn't find it. Yet I thought this was an interesting question, so I posted it here.


1 Black holes don't emit any radiation excepted for Hawking radiation if I get it right.

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    $\begingroup$ I was about to rant about wrong grammar for using "many" instead of "much" to describe intensity of radiation until I realized that photons are technically countable. $\endgroup$ – slebetman Aug 17 '15 at 13:58
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    $\begingroup$ @annav, is there really any upper energy limit to what we can call "X-rays"? I always thought that "X-ray" referred to high-energy photons that emanate from electron interactions, and "Gamma Ray" referred to high-energy photons that emanate from atomic nuclei. I have worked with medical machines generating photons with energies as high as 25 MeV---way higher than most gammas---and the manuals always said "X-ray". $\endgroup$ – Solomon Slow Aug 17 '15 at 16:10
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    $\begingroup$ About 25 bursts of gamma rays per year due to natural alpha decay of Tungsten producing secondary gamma rays. $\endgroup$ – Count Iblis Aug 17 '15 at 19:21
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    $\begingroup$ @ErikE Note that in some dialects (specifically in the UK) hundred sounds like 'undred, and the decision to use 'a' or 'an' depends on whether the initial sound is a vowel or consonant, so this is a rare case where the dialect affects orthography. I have seen a similar case with "a historical" / "an historical", depending on whether you voice the initial h. $\endgroup$ – Mario Carneiro Aug 18 '15 at 6:07
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    $\begingroup$ @MarioCarneiro I see! I was thinking only "one hundred". I would never omit the "one". $\endgroup$ – ErikE Aug 18 '15 at 14:04
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The formula you want is called Planck's Law. Copying Wikipedia:

The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency.

$$ B_\nu(\nu, T) = \frac{ 2 h \nu^3}{c^2} \frac{1}{e^\frac{h\nu}{k_\mathrm{B}T} - 1} $$

Now to work out the total power emitted per unit area per solid angle by our lightbulb in the X-ray part of the EM spectrum we can integrate this to infinity:

$$P_{\mathrm{X-ray}} = \int_{\nu_{min}}^{\infty} \mathrm{B}_{\nu}d\nu, $$

where $\nu_{min}$ is where we (somewhat arbitrarily) choose the lowest frequency photon that we would call an X-ray photon. Let's say that a photon with a 10 nm wavelength is our limit. Let's also say that 100W bulb has a surface temperature of 3,700 K, the melting temperature of tungsten. This is a very generous upper bound - it seems like a typical number might be 2,500 K.

We can simplify this to:

$$ P_{\mathrm{X-ray}} = 2\frac{k^4T^4}{h^3c^2} \sum_{n=1}^{\infty} \int_{x_{min}}^{\infty}x^3e^{-nx}dx, $$

where $x = \frac{h\nu}{kT}$. wythagoras points out we can express this in terms of the incomplete gamma function, to get

$$ 2\frac{k^4T^4}{h^3c^2}\sum_{n=1}^{\infty}\frac{1}{n^4} \Gamma(4, n\cdot x) $$

Plugging in some numbers reveals that the n = 1 term dominates the other terms, so we can drop higher n terms, resulting in

$$ P \approx 10^{-154} \ \mathrm{Wm^{-2}}. $$

This is tiny. Over the course of the lifetime of the universe you can expect on average no X-Ray photons to be emitted by the filament.

More exact treatments might get you more exact numbers (we've ignored the surface area of the filament and the solid angle factor for instance), but the order of magnitude is very telling - there are no X-ray photons emitted by a standard light bulb.

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  • $\begingroup$ It's a great answer, thank you. But the number is much, much lower than I had expected. By the way, I know a way to solve the integral and the series, if you would like to know how I can write an answer. $\endgroup$ – wythagoras Aug 17 '15 at 13:31
  • $\begingroup$ By all means! I'd be really interested to see what you think. griffin175's answer physics.stackexchange.com/a/200883/81404 seems to roughly agree that there are basically no photons. $\endgroup$ – Chris Cundy Aug 17 '15 at 13:35
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    $\begingroup$ I'm having trouble with the closed form due to a stupid mistake. Doing the substitution $u=nx$, we get $$ \sum_{n=1}^{\infty} \int_{\nu_{min} \cdot n}^{\infty} \frac{1}{n}\left(\frac{u}{n}\right)^3e^{-u}\mathrm{d}u$$ $$ \sum_{n=1}^{\infty} \frac{1}{n^4} \Gamma(4,\nu_{min}\cdot n)$$ where $\Gamma$ is the upper complete gamma function. But even if $\Gamma(4,\nu_{min})$ is extremely small, rather something in the order of $e^{-\nu_{min}}\nu_{min}^3$, and $\nu_{min} = 3 \times 10^{16}$ if I didn't misunderstood you. $\endgroup$ – wythagoras Aug 17 '15 at 14:32
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    $\begingroup$ Worth saying that Tungsten melts at 3695K. Assuming that's where you got the upper temperature bound from. $\endgroup$ – OrangeDog Aug 17 '15 at 14:47
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    $\begingroup$ Why did you say "we can simply this" and then make another equation that's like twice as big? Man, math is crazy. $\endgroup$ – corsiKa Aug 17 '15 at 23:07
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The wavelengths of light emitted can be calculated using planks law and the temperature of the object. For your average 100W incandescent light bulb, the filament is 2823 kelvin according to google.

The spectral radiance, $B$, is equal to $$\frac{1.2\cdot10^{52}}{\mathrm{wavelength}^{5}\cdot e^{\frac{1.99\cdot10^{43}}{\mathrm{wavelength}\cdot4\cdot10^{26}}}-1}$$

Math to solve for spectral radiance is hard, so this online calculator will do all the work. X-rays are between 0.01nm and 10nm. The total radiance at 10nm is $2.7\cdot10^{-187}$photons/s/m2/sr/µm. That's so unbelievably small, It would take a very long time for that bulb to emit an xray photon. The calculator wont give the spectral radiance of the smaller wavelength xrays so we'll just use the biggest X-rays.

In order to figure out how many photons per second are emitted you would need to know the surface area of the filament. It's a tiny metal sting, that would be hard to find out, but if you really want to, break open a bulb and measure its length and diameter with a caliper. Estimate surface area using the surface area of a cylinder formula A=πdh. Forget the ends, they're too small to bother with.

If you don't want to go through the trouble of breaking a bulb, make a wild guesstimate. 0.6m length and $5\cdot10^{-4}$ diameter, being generous. area of 0.001 m2. So $2.7\cdot10^{-187}$photons/s/m^2/sr/µm, then with the given surface area, $2.7*10^{-190}$ photons/s/sr/µm. That's 8.5 photons every $10^{186}$ years. Maybe if you watch 100,000,000,000,000 light bulbs you might catch an X-ray within your lifetime.

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  • $\begingroup$ From tempurature you have its emmissions curve per $m^3$, and from specs (1600 lumens for 100 W bulb) you have the amount of visible light it emits. From that, you should be able to calculate surface area, no? $\endgroup$ – Yakk Aug 17 '15 at 18:01
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    $\begingroup$ 100,000,000,000,000 is nowhere NEAR enough bulbs. You need more like 100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 bulbs. Give or take a couple of orders of magnitude. $\endgroup$ – Kyle Oman Aug 17 '15 at 18:48
  • $\begingroup$ @KyleOman You are right. Exepted that if out take all these bulbs as one big sphere, that the pressure will be much higher, so that the temperature will rise. $\endgroup$ – wythagoras Aug 17 '15 at 18:51
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    $\begingroup$ Well, that's more than the estimated number of atoms in the observable Universe, so you'll also have an appreciable effect on cosmology (even before turning on the radiation field), and the whole thing will likely collapse gravitationally and heat that way (shining brightly in the X-ray, I would bet), and probably form the mother of all supermassive black holes. But I was mostly being (and still am being) somewhat facetious. Though a hundred trillion lightbulbs is a somewhat plausible number, whereas $10^{186}$ is on a totally different scale. Important to make that distinction imho. $\endgroup$ – Kyle Oman Aug 17 '15 at 18:59
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    $\begingroup$ Fun fact: Most modern filaments are double-coiled tungsten alloy strings -- basically wound in a very tight coil, which is then wound in a looser coil... LOTS of surface area to be had! See upload.wikimedia.org/wikipedia/commons/0/08/Filament.jpg $\endgroup$ – Doktor J Aug 17 '15 at 21:05
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I will give a closed form for the integral in Chris Cundy's answer.

Doing the substitution $u=nx$, we get $$ \sum_{n=1}^{\infty} \int_{x_{min} \cdot n}^{\infty} \frac{1}{n}\left(\frac{u}{n}\right)^3e^{-u}\mathrm{d}u$$

$$ \sum_{n=1}^{\infty} \frac{1}{n^4} \Gamma(4,x_{min}\cdot n)$$

where $\Gamma$ is the upper complete gamma function. We write $a=x_{min}$ as it will be used a lot so a short name is more useful. Using the reduction formula for the gamma function when the first argument is an integer, we get: $$ \sum_{n=1}^{\infty} \left(\frac{1}{n^4}e^{-an}\left(6+6an+3a^2n^2+a^3n^3\right)\right) $$

$$ 6\sum_{n=1}^{\infty} \frac{1}{n^4}e^{-an} + 6a\sum_{n=1}^{\infty} \frac{1}{n^3}e^{-an}+ 3a^2\sum_{n=1}^{\infty} \frac{1}{n^2}e^{-an}+a^3\sum_{n=1}^{\infty} \frac{1}{n}e^{-an}$$

Now note that $$\frac{\mathrm{d}}{\mathrm{d}a} \sum_{n=1}^{\infty} \frac{1}{n}e^{-an} = \sum_{n=1}^{\infty} \frac{\mathrm{d}}{\mathrm{d}a} \left[\frac{1}{n}e^{-an}\right]=\sum_{n=1}^{\infty}-e^{-an}=-\sum_{n=1}^{\infty}(e^{-a})^n=1-\frac{1}{1-e^{-a}}$$

$$\sum_{n=1}^{\infty} \frac{1}{n}e^{-an} = \int 1-\frac{1}{1-e^{-a}} \mathrm{d}a = -\ln|1-e^{-a}|$$

We'll get the other terms in a similiar way. The final result is:

$$\sum_{n=1}^{\infty} \int_{a}^{\infty} x^3e^{-nx}dx = -6\mathrm{Li}_4(e^a)+6a\mathrm{Li}_3(e^a)+6a^2 \mathrm{Li}_2(1-e^{-a})-9a^2\mathrm{Li}_2(e^a)\\+2a^3\ln|1-e^{-a}|-9a^3\ln|1-e^{a}|+5\frac{3}4 a^4$$

I used a Computer Algebra System to find this form. $\mathrm{Li}_n$ is the polylogarithm function.

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protected by Qmechanic Aug 17 '15 at 22:45

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