The critical mass density needed to just halt the expansion of the universe is approximately\({\rm{1}}{{\rm{0}}^{{\rm{ - 26}}}}{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}\). (a) Convert this to\({\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\). (b) Find the number of neutrinos per cubic meter needed to close the universe if their average mass is\({\rm{7 eV/}}{{\rm{c}}^{\rm{2}}}\)and they have negligible kinetic energies.

The number of neutrinos per cubic meter needed to close the universe if their average mass is\(7\,{\rm{ev/}}{{\rm{c}}^{\rm{2}}}\).

\(N = \frac{{{\rho _c}}}{{7\,{\rm{ev/}}{{\rm{c}}^{\rm{2}}}}}\)……. (i)

Here\({\rho _c}\)is the critical mass density in\({\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\).

Conversion of \({\rm{kg/}}{{\rm{m}}^{\rm{3}}}\) to \({\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\).

\(\begin{align}{\rho _c}{\rm{ }} &= {\rm{1}}{{\rm{0}}^{{\rm{ - 26}}}}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\\ &= {\rm{1}}{{\rm{0}}^{{\rm{ - 26}}}}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\left( {\frac{{{\rm{938}}{\rm{.27}}\,{\rm{ \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{eV/}}{{\rm{c}}^{\rm{2}}}}}{{{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}\,{\rm{kg}}}}} \right)\\ &= {\rm{5}}{\rm{.62 \times 1}}{{\rm{0}}^{\rm{9}}}\,{\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\end{align}\)

Therefore the critical mass density is \({\rm{5}}{\rm{.62 \times 1}}{{\rm{0}}^{\rm{9}}}\,{\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\).

Substitute \({\rm{5}}{\rm{.62 \times 1}}{{\rm{0}}^{\rm{9}}}\,{\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}\) for \({\rho _c}\) into the equation (i)

\(\begin{align}N &= \frac{{{\rm{5}}{\rm{.62 \times 1}}{{\rm{0}}^{\rm{9}}}\,{\rm{eV/}}{{\rm{c}}^{\rm{2}}} \cdot {{\rm{m}}^{\rm{3}}}}}{{7\,{\rm{ev/}}{{\rm{c}}^{\rm{2}}}}}\\ &= 8 \times {10^8}\end{align}\)

Therefore the number of neutrons per cubic meter is \(8 \times {10^8}\).