Are there more photons than nuclei within the Sun? Is there a good way to estimate what the ratio would be?

  • $\begingroup$ There are more photons in the universe than nuclei within the sun. Or did you mean more photons emitted from the sun than nuclei? $\endgroup$
    – Zach466920
    Aug 16, 2015 at 21:48
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    $\begingroup$ Sorry I didn't specify. I mean to say that inside the Sun there are photons and nuclei. Which are there more of and in what ratio. $\endgroup$
    – Alex
    Aug 16, 2015 at 21:54
  • $\begingroup$ A photon is its own antiparticle, they carry no conserved charge so the "photon number" is not a well-defined thing. On the contrary, Baryon number is a conserved quantity, so you can count the net amount of baryons. en.wikipedia.org/wiki/…. In other words: you can count particle species that are associated to a conserved (or quasi-conserved) number of the Standard Model (e.g. "lepton number" for leptons, etc...) $\endgroup$
    – Quillo
    Sep 8, 2022 at 15:13

3 Answers 3


It's a bit of a puzzling question. I'll try to work it out, but one of the tricky parts is that atoms absorb and emit photons all the time, higher temperatures emit higher wavelengths.

Photons created in the sun (per second) can be estimated, but those are fusion gamma rays.

The sun burns about $564$ million tons of hydrogen per second. (Source), and 1 mole is $6.022 \times 10^{23}$ atoms, and you have $1$ mole of hydrogen having a mass of about $1.0794$ grams. You need about $840,000$ moles of hydrogen to equal $1$ ton, therefore $6.022 \times 10^{23} \times 8.4 \times 10^5 \times 5.64 \cdot 10^8$ hydrogen atoms get converted per second, and $1$ hydrogen as it undergoes fusion (correct me if I get this wrong) $\mathrm {H + H = D +e^+} $ (and Positron hits an electron and goes poof) and $\mathrm {D + D = H}$, so you get minimum $2.5$ photons per hydrogen (possibly a bit more as you sometimes get $\mathrm {D + H = T}$ and $\mathrm {T}$ decays into $\mathrm {^3He}$.

but doing the math, $2.5 \times 6.022 \times 8.4 \times 5.64 \times 10^{23 + 5 + 8}$ you get about $7 \times 10^{38}$ photons (gamma rays) per second created inside the sun, and if each of these spends an average of $100,000$ years inside the sun (estimates vary), multiply that by $3.1 \times 10^{12}$ seconds and you get about $2.2 \times 10^{51}$ photons in the sun that were created in the sun. The actual numbers of photons in the sun is likely a fair bit greater than that.

If we us the creation to emit ratio as an estimate, and I have no idea if that's consistent, there's a good chance it's not, but the sun creates 7 x 10 ^ 38th photons per second and it emits about 1 x 10 ^ 45th photons per second, (Source), so that's a ratio of about 1.4 x 10 ^ 6 creation to emit ratio. it makes sense that as a gamma ray works it's way through a nearly endless sea of atoms, it would lose energy into other photons as well as temperature, so 1 gamma ray photon in the middle of the sun creates many photons on it's way to the surface of the sun - perhaps as many as 1.4 million, but if I was to guess, I would say it's less.

Using the 100,000 year estimate for how long a photon stays in the sun, 1.4 x 10 ^ 6 x 2.2 x 10 ^ 51 = about 3.1 x 10 ^ 57

How many atoms are in the sun? (Source): About 1.2 x 10^57

So, perhaps, there are more photons in the sun than atoms, but that's a very rough estimate, and the numbers are too close to call.

Perhaps a better way to calculate would be to estimate how many photons an atom of a certain temperature emits per second, but I couldn't find any quick reference to that. Plasma likely behaves quite differently than the atoms closer to the surface but (perhaps) cool enough to hold onto some semblance of electron orbitals.

Another way to count might include virtual photons and photons used in communicating the electromagnetic force in which case, photons probably significantly outnumber the number of atoms. That's probably not a very good answer, but I thought I'd throw it out there.


In a dilute gas, the photon density should be the same as inside an evacuated black box of the same temperature (independent of the gas density).

Edit: In other words, the massive particles in the Sun have a temperature based on their kinetic energies. The photons must be distributed as black-body radiation at the same temperature. The density of photons in equilibrium with a black body of given temperature can be calculated.

Edit: This equilibrium is the basis of theory of the Cosmic Microwave Background. When the temperature of the Universe was above 3000K (t ∼ 380000 yr) photons were in thermal equilibrium with ionised gas. When the temperature dropped lower, the Universe became transparent and the photons became the CMB.

"Therefore, the drop in the CMB temperature by a factor of 1100 (= 3000 K/2.73 K) indicates an expansion of the universe by a factor of 1100 from the moment of decoupling until now."


The number of atoms in the sun is on the order of $10^{57}$, see here.

The number of photons emitted per second is on the order of $10^{44}$, see here.

The difference is on the order of $10^{13}$. So if photons emitted for $10^{13}$ seconds, 315,00 years, the number of photons would begin to overtake the number of atoms in the sun. The sun is much older than that, so there are more photons emitted from the sun than actual atoms in the sun.

According to askamathmatician, however, how long photons stay in the sun before being released is a tad bit unclear, so the number of photons in the sun is a bit unclear.

Basically, if you're considering all the photons inside the sun, there are more photons than atoms.

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    $\begingroup$ His question was photons inside the sun, not emitted by the sun. That makes it a lot closer. $\endgroup$
    – userLTK
    Aug 16, 2015 at 22:22
  • $\begingroup$ @userLTK I added that into my answer. $\endgroup$
    – Zach466920
    Aug 16, 2015 at 22:31
  • $\begingroup$ What the heck is a "full photon"? Or especially a "mostly full" one? You can estimate the number inside the star by using the mean free path and approximate diffusion time to an estimate the number inside the star for each that escapes. It is a large factor. $\endgroup$ Aug 16, 2015 at 22:42
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    $\begingroup$ In trying to stitch a bunch of random posts from elsewhere on the internet together you have to be carefully about filling in the gaps. You haven't done very well here and you've inserted your own non-standard nomenclature which isn't helping. The blog post you're referencing is trying to work the same estimate that I suggested, but no one says "partially released". Strictly you should use a different model for the convective layer, but surprisingly it introduces a fairly small correction. $\endgroup$ Aug 16, 2015 at 22:49
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    $\begingroup$ @dmckee for a moderator I'm surprised your being so hostile, what exactly is wrong with the first part of my answer? I've tried eliminating the "nomenclature" I introduced in the second half, but I honestly think that without it the nature of the problem is confused. In addition, the people on the blog specifically mention how different methods result in varying results, so I don't get the point of your last sentence. $\endgroup$
    – Zach466920
    Aug 16, 2015 at 22:56

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