Question

Assume the average density of the universe is\({\rm{0}}{\rm{.1}}\)of the critical density needed for closure. What is the average number of protons per cubic meter, assuming the universe is composed mostly of hydrogen?

Short answer

The average number of protons per cubic metre is obtained as: \({\rm{0}}{\rm{.6 }}{{\rm{m}}^{{\rm{ - 3}}}}\).

Step-by-step solution

Expression for the average number of protons per cubic meter

The average number of protons per cubic meter is calculated by,

\(\overline N = \frac{{{\rho _c}\left( {0.10} \right)}}{{{m_p}}}\)

Here\(\overline N \) is the number of protons per cubic meter,\({\rho _c}\)is the critical density,\({m_p}\)is the mass of proton.

Evaluating the number of protons

With the help of the previous question, infer:

\({\rho _{closure}}{\rm{ }} = {\rm{1}}{{\rm{0}}^{{\rm{ - 26}}}}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\)

Then, the density will be:

\(\begin{align}\rho &= {\rm{0}}{\rm{.1 }}{\rho _{closure}} \\ &= {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{ kg/}}{{\rm{m}}^3}\end{align}\)

The mass per cubic meter is\(m{\rm{ }} = {\rm{ 1}}{{\rm{0}}^{{\rm{ - 27}}}}\,{\rm{kg/}}{{\rm{m}}^3}\).

The number of protons is then evaluated as:

\(\begin{align}\overline N {\rm{ }} &= {\rm{ }}\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{ kg/}}{{\rm{m}}^3}}}{{{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{kg}}}}{\rm{ }}\\ &= {\rm{ 0}}{\rm{.60 per cubic metre}}\end{align}\)

Therefore, the average number of protons per cubic metre is \({\rm{0}}{\rm{.6 }}{{\rm{m}}^{{\rm{ - 3}}}}\).