It is often quoted that the number of atoms in the universe is 10$^{70}$ or 10$^{80}$.

How do scientists determine this number?

And how accurate is it (how strong is the supporting evidences for it)?

Is it more likely (logically >50% chance) that the numbers are right, or is it more likely that the numbers are wrong?

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    $\begingroup$ They don't. They estimate. $\endgroup$ Feb 23, 2013 at 22:06
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    $\begingroup$ The number of atoms in the universe may be infinite, if the universe is spatially infinite. What can be estimated is the number of atoms in the observable universe. $\endgroup$
    – user4552
    May 28, 2013 at 23:17
  • $\begingroup$ Stars are fusing four hydrogens into one helium. Older stars fuse up the Periodic Table to iron. A large star will eventually suffer core collapse to a neutron star or a black hole. The number of atoms in the universe has been continuously decreasing since the Big Bang cooled. The "number" of atoms is an estimate assuming mass occurs as atoms, then observed composition abundances. $\endgroup$
    – Uncle Al
    Feb 22, 2014 at 15:41

2 Answers 2


The cosmological estimation of the number of atoms in the observable universe works as follows: one of the Friedmann equations can be written as $$ \dot{a}^2 -\frac{8\pi G}{3}\rho a^2= -kc^2, $$ where the scale factor $a(t)$ describes the expansion of the universe, $\rho$ is the total mass density (radiation, baryonic matter, dark matter, and dark energy) and the integer $k$ is the intrinsic curvature of the universe ($k$ can be 1, 0 or -1). Observations of the Cosmic Microwave Background (CMB) indicate that the spacial curvature $k/a^2$ of the universe is practically zero, so we can set $k=0$. In this case the total density is equal to the so-called critical density $$ \rho_\text{c}(t) = \frac{3H^2(t)}{8\pi G}, $$ where $$ H(t) = \frac{\dot{a}}{a} $$ is the Hubble parameter. The present-day density is then $$ \rho_\text{c,0} = \rho_\text{c}(t_0) = \frac{3H_0^2}{8\pi G}, $$ with $H_0=H(t_0)$ the Hubble constant. We can write $H_0$ in the following form $$ H_0 = 100\,h\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}, $$ with $h$ a dimensionless parameter and $1\;\text{Mpc}=3.0857\times 10^{19}\;\text{km}$ (called a megaparsec). So $$ \rho_\text{c,0} = 1.8785\,h^2\times 10^{-26}\;\text{kg}\,\text{m}^{-3}. $$ A detailed analysis of the Cosmic Microwave Background reveals what the density of ordinary matter (baryons) is: according to the latest CMB data, the present-day baryon fraction is $$ \Omega_\text{b,0}h^2 = \frac{\rho_\text{b,0}}{\rho_\text{c,0}}h^2 = 0.02205 \pm 0.00028. $$ Notice how accurately this quantity is known. The same data also yield a value of the Hubble constant: $$ H_0 = 67.3 \pm 1.2\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}, $$ in other words, $h = 0.673\pm0.012$ so that $$ \Omega_\text{b,0} = 0.0487, $$ which means that ordinary matter makes up 4.87% of the content of the universe. We don't actually need the value of $h$ to calculate the baryon density $\rho_\text{b,0}$, because the factor $h^2$ cancels out: we get $$ \rho_\text{b,0} = \Omega_\text{b,0}\rho_\text{c,0} = 0.4142\times 10^{-27}\;\text{kg}\,\text{m}^{-3}. $$ About 75% of the baryon density is in the form of hydrogen, and nearly 25% is helium; all other elements make up about 1%, so I'll ignore those. The masses of hydrogen and helium atoms are $$ \begin{align} m_\text{H} &= 1.674\times 10^{-27}\;\text{kg},\\ m_\text{He} &= 6.646\times 10^{-27}\;\text{kg}, \end{align} $$ so the number density of hydrogen and helium atoms is $$ \begin{align} n_\text{H} &= 0.75\rho_\text{b,0}/m_\text{H} = 0.1856\;\text{m}^{-3},\\ n_\text{He} &= 0.25\rho_\text{b,0}/m_\text{He} = 0.0156\;\text{m}^{-3}, \end{align} $$ and the total number density of atoms is $$ n_\text{A} = n_\text{H}+n_\text{He} = 0.2012\;\text{m}^{-3}. $$ Now, the radius of the observable universe is calculated to be $D_\text{ph} = 46.2$ billion lightyears, which is $4.37\times 10^{26}\,\text{m}$ (the subscript 'ph' stands for particle horizon; see this post for a detailed explanation). This is a derived value, which depends on all cosmological parameters; nonetheless, it is accurate to about 1%. The volume of the observable universe is thus $$ V = \frac{4\pi}{3}\!D_\text{ph}^3 = 3.50\times 10^{80}\;\text{m}^3. $$ So finally, there are about $$ N_\text{A} = n_\text{A}V = 7.1\times 10^{79} $$ atoms in the observable universe.

  • $\begingroup$ Most interesting indeed - a very thorough and simple explanation. "...detailed analysis of the Cosmic Microwave Background ..." is there a simple summary of how the proportion of baryons is estimated from this? That's quite a colossal paper and a colossal piece of work. Were you involved? $\endgroup$ Nov 4, 2014 at 0:03
  • $\begingroup$ @WetSavannaAnimalakaRodVance No, I wasn't involved, I'm not even a cosmologist :-) The details are complicated, but it has to do with how baryons influence the propagation of sound waves in the early universe, which eventually are imprinted on the CMB. A great online tutorial is Wayne Hu's site. The key plot can be found here. $\endgroup$
    – Pulsar
    Nov 4, 2014 at 21:57
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    $\begingroup$ A more pedestrian version of @Pulsar's answer, addressing a connected question (the number of atoms is an intermediate result) has just been released on the Numberphile youtube channel: [How many particles in the Universe? ](youtube.com/watch?v=lpj0E0a0mlU). $\endgroup$
    – user154997
    Jul 10, 2017 at 20:18

The observable universe contains about 100 billion galaxies, each containing on average close to a trillion stars. That is a total of about $10^{23}$ stars. A typical star is like our sun. Sun has a mass of about $2×10^{30}$ kg, which equates to $10^{57}$ atoms of hydrogen per star. A total of $10^{23}$ stars containing $10^{57}$ atoms each gives us a total number of atoms of $10^{80}$.

More detail, including an alternative estimation method based on cosmic microwave background observations, can be found here.

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    $\begingroup$ @Pacerier - That would be a gross violation of the cosmological principle which says 'Viewed on a sufficiently large scale, the properties of the Universe are the same for all observers.' In other words, our universe is homogeneous. Deep sky galaxy surveys, such as the Sloan Digital Sky Survey, but also cosmic microwave background observations, all confirm the cosmological principle. $\endgroup$
    – Johannes
    Dec 30, 2012 at 18:39
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    $\begingroup$ @Pacerier - what makes you say these numbers are more likely than not correct? It occurs to me that the confidence interval for the numbers of atoms in the universe to be within say $10^{78}$ and $10^{82}$ certainly exceeds 50%. $\endgroup$
    – Johannes
    Dec 30, 2012 at 18:44
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    $\begingroup$ I'd hardly consider "between $10^{70}$ and $10^{80}$" to be a very precise estimate. We could be darn certain that the number of atoms in the universe is between $10^{50}$ (roughly the number of atoms in the earth) and $10^{150}$ (a ridiculously large number). This is a range of $\pm 50$ orders of magnitude, but it is fair to say that we know with virtual certainty that the number of atoms in the universe falls within that range. Somewhere between $10^{70}$ to $10^{80}$ is sufficiently vague as to be accurate, more precise than $10^{50-150}$, and a lot more useful than "unknown". $\endgroup$
    – KDN
    Dec 30, 2012 at 21:22
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    $\begingroup$ @Johannes I think the claim that a galaxy contains on average 1 trillion stars is very bold. Our galaxy is a big one and it contains around 300 billion stars. Otherwise a good answer $\endgroup$ Feb 23, 2013 at 22:06
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    $\begingroup$ @ArturoDonJuan - I didn't forget, I ignored it. The above is an order of magnitude estimate, relatively tiny contributions to the number of atoms in the universe are better ignored as these constitute no more than a distraction. $\endgroup$
    – Johannes
    Nov 19, 2014 at 19:07

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