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Question
The critical mass density needed to just halt the expansion of the universe is approximately $10^{-26}\textrm{ kg/m}^3$. (a) Convert this to $\textrm{eV/c}^2\cdot\textrm{m}^3$. (b) Find the number of neutrinos per cubic meter needed to close the universe if their average mass is $7\textrm{ eV/c}^2$ and they have negligible kinetic energies.
1. $6\times 10^{9}\textrm{ eV/c}^2\cdot\textrm{m}^3$
2. $8\times 10^{8}\textrm{ neutrinos}$
Solution Video