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I'll take number per average star, multiple by appropriate numbers , star/Galaxy, galaxies/universe, 14 billion years, and try to figure their location.

I was looking for a rough number and a distribution function.

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  • $\begingroup$ By universe you mean observable universe, right? $\endgroup$ – called2voyage Jan 4 '17 at 19:27
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    $\begingroup$ See this related question. Note that a large fraction of the photons in our universe don't come from stars, but rather are a part of the Cosmic Microwave Background (CMB) radiation. That will make your calculation woefully short of the real number of photons in the universe. $\endgroup$ – zephyr Jan 4 '17 at 19:52
  • $\begingroup$ Not only that, but single objects (e.g. active galactic nuclei) sometimes produce photon outputs larger than entire galaxies of stars! $\endgroup$ – probably_someone Jan 5 '17 at 11:13
  • $\begingroup$ There really isn't any such number. You might as well try to calculate the total mass or energy equivalent of the entire universe and pretend it's one giant black body, and calculate the spectral radiation. $\endgroup$ – Carl Witthoft Jan 5 '17 at 13:46
  • $\begingroup$ Also, how will you take into account that the number of stars in a galaxy changes with time as stars are born from infalling gas? And, what about dust grains and planets that reprocess optical photons into a different number of infrared photons? And, what about various radio sources like neutron stars and black holes? Radio waves can also be photons. And supernovae and novae? And energetic particles radiating in magnetic fields everywhere? The universe is perhaps more interesting than you think! $\endgroup$ – eshaya Jan 6 '17 at 22:29
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Extending Zephyrs comment as community wiki

See https://physics.stackexchange.com/questions/196366/number-density-of-cmb-photons

In this response, the number of photons per cubic meter is estimated as $n_\lambda = 10^8$. Since the observable universe has a volume of the order $10^{80}\ m^3$, there are about $10^{88}$ photons in the observable universe.

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