Number of Atoms in Observable Universe

Theorem

The number of atoms in the observable universe is approximately $10^{80}$.

Proof

According to our current observations and understanding of the Universe, the age of the universe is estimated to be $13.787$ billion years.

That means the light from the most distant galaxies has traveled $13.787$ billion years to reach us.

The speed of light is $299 \, 792 \, 458 \ \text {m s}^{-1}$

This implies that the radius of the observable universe (assuming no expansion) is:

\(\ds R_U\)

\(=\)

\(\ds \paren {13.787 \times 10^9 \text { years} } \times \paren {365 \cdotp 24219 \, 878 \text { days per year} } \times \paren { 86 \, 400 \text { seconds per day} } \times \paren {299 \,792 \, 458 \ \text {m s}^{-1} }\)

\(\ds \)

\(=\)

\(\ds 1.304 \times 10^{26} \ \mathrm m\)

However, contemplating expansion at an average rate of $8.8 \%$ per billion years, we obtain:

\(\ds R_U\)

\(=\)

\(\ds 1.304 \times 10^{26} \times e^{0.088 \times 13.787} \ \mathrm m\)

\(\ds \)

\(=\)

\(\ds 4.39 \times 10^{26} \ \mathrm m\)

This in turn implies that the volume of the observable universe is:

\(\ds V_U\)

\(=\)

\(\ds \dfrac {4 \pi {R_U}^3} 3\)

Volume of Sphere

\(\ds \)

\(=\)

\(\ds \dfrac {4 \pi \paren {4.39 \times 10^{26} } \ \mathrm m^3} 3\)

\(\ds \)

\(=\)

\(\ds 3.54 \times 10^{80} \ \mathrm m^3\)

Finally, assuming there is approximately $1$ atom per cubic metre brings us to our estimate of approximately $10^{80}$ atoms in the observable universe.

$\blacksquare$