Chapter 33, Problem 40GP

(a)

To determine

Neutrino rest mass to close the universe for the given value.

Answer to Problem 40GP

Neutrino rest mass to close the universe when nucleons make up only $2\%$ of the mass needed is $46{\text{ eV/c}}^2$ .

Explanation of Solution

Given:

The neutrino density is, like photons, about ${10}^9$ times that of nucleons. Nucleons make up $2\%$ of the mass needed.

Formula used:

Nucleon number density is the ratio of the nucleon mass density and mass of a nucleon.

Calculation:

Nucleon mass density is $0.02\times {10}^{-26}{\text{ kg/m}}^3$

Nucleon number density is $\frac{0.02\times {10}^{-26}{\text{ kg/m}}^3}{1.67\times {10}^{-27}\text{ kg/nucleon}}=0.12{\text{ nucleon/m}}^3$

Neutrino number density is ${10}^9$ times nucleon number density that is, $1.2\times {10}^8{\text{ nucleon/m}}^3$

Neutrino mass density is

  $\begin{array}{c}\frac{0.98\times ({10}^{-26}{\text{ kg/m}}^3)}{1.2\times {10}^8{\text{ nucleon/m}}^3}=8.17\times {10}^{-35}\frac{\text{kg}}{\text{neutrino}}\times \frac{9.315\times {10}^8{\text{ eV/c}}^2}{1.66\times {10}^{-27}\text{ kg}}\\ =46{\text{ eV/c}}^2\end{array}$

Conclusion:

Thus, neutrino rest mass density is $46{\text{ eV/c}}^2$ .

(b)

To determine

Neutrino rest mass to close the universe for the given value.

Answer to Problem 40GP

Neutrino rest mass to close the universe when nucleons make up only $5\%$ of the mass needed is ${\text{18 eV/c}}^2$ .

Explanation of Solution

Given:

The neutrino density is, like photons, about ${10}^9$ times that of nucleons. Nucleons make up $5\%$ of the mass needed.

Formula used:

Nucleon number density is the ratio of the nucleon mass density and mass of a nucleon.

Calculation:

Nucleon mass density is $0.05({10}^{-26}{\text{ kg/m}}^3)$

Nucleon number density is $\frac{0.05({10}^{-26}{\text{ kg/m}}^3)}{1.67\times {10}^{-27}\text{ kg/nucleon}}=0.30{\text{ nucleon/m}}^3$

Neutrino number density is ${10}^9$ times nucleon number density that is, $3.0\times {10}^8{\text{ nucleon/m}}^3$

Neutrino mass density is

  $\frac{0.95\times ({10}^{-26}{\text{ kg/m}}^3)}{3.0\times {10}^8{\text{ nucleon/m}}^3}=3.17\times {10}^{-35}\frac{\text{kg}}{\text{neutrino}}\times \frac{9.315\times {10}^8{\text{ eV/c}}^2}{1.66\times {10}^{-27}\text{ kg}}=18{\text{ eV/c}}^2$

Conclusion:

Thus, neutrino rest mass density is ${\text{18 eV/c}}^2$ .