Use the concepts in this chapter to obtain an estimate for the number of atoms in the universe. Make the following assumptions: (a) Assume that all
of the atoms in the universe are hydrogen atoms in stars. (This is not a ridiculous assumption because over three-fourths of the atoms in the universe
are in fact hydrogen. Gas and dust between the stars represent only about 15% of the visible matter of our galaxy, and planets compose a far tinier
fraction.) (b) Assume that the sun is a typical star composed of pure hydrogen with a density of 1.4 g/cm and a radius of 7 x 108m. (c) Assume
that each of the roughly 100 billion stars in the Milky Way galaxy contains the same number of atoms as our sun. (d) Assume that each of the 10
billion galaxies in the visible universe contains the same number of atoms as our Milky Way galaxy.
Express your answer using one signififcant figure.

[tex]& \text{Mass of sun = 1}\text{.4}\times \frac{4}{3}\pi {{\left( 7\times {{10}^{10}} \right)}^{3}} \\ & \text{Mass of sun}=2.011\times {{10}^{33}}g \\ \end{align}[/tex]

Average atomic mass of hydrogen atom is [tex]\[1.008amu\][/tex]

Avogadro’s number is number of atoms in 1 mole of hydrogen.

Step 3

Calculate number of Hydrogen [tex]\[\left( {{N}_{H}} \right)\][/tex]atoms in sun

[tex]& {{N}_{H}}=\frac{\text{Mass of sun}}{\text{Molar mass of Hydrogen atom}}\times \text{Avogadro }\!\!'\!\!\text{ s }\text{Number} \\ & {{N}_{H}}=\frac{2.011\times {{10}^{33}}}{1.008}\times 6.02\times {{10}^{23}} \\ & {{N}_{H}}=1.201\times {{10}^{57}} \\[/tex]

Step 4

Calculate number of atoms[tex]\[\left( {{N}_{A}} \right)\][/tex] in our Milky Way galaxy

[tex]& {{N}_{A}}=\text{Number of stars}\times \text{Number of atoms in sun} \\ & {{N}_{A}}=1.00\times {{10}^{11}}\times 1.202\times {{10}^{57}} \\ & {{N}_{A}}=1.201\times {{10}^{68}} \\[/tex]

Step 5

Calculate number of atom in observable universe [tex]\[\left( {{N}_{Ao}} \right)\][/tex]

[tex]& {{N}_{Ao}}=\text{Number of galaxies }\times \text{ Number of atoms in our galaxy} \\ & {{N}_{Ao}}=1.00\times {{10}^{11}}\times 1.201\times {{10}^{68}} \\ & {{N}_{Ao}}=1.201\times {{10}^{79}} \\[/tex]

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