Chapter 18, Problem 8P

To determine

The number of joules of energy in each liter of space and the number of $1\text{-mm-}$ wavelength microwave photons the energy would correspond to in each liter.

Answer to Problem 8P

The number of joules of energy in each liter of space is $1.85\times {10}^{-13}\text{ J}$ and the energy would correspond to $1\text{ billion}$  $1\text{-mm-}$ wavelength microwave photons in each liter.

Explanation of Solution

The critical density of the Universe is $3\times {10}^{-30}\text{g}/{\text{cm}}^{\text{3}}$ and the dark energy in the universe currently is $68.5\%$.

Therefore, the dark energy is

$0.685\times 3\times {10}^{-30}\text{g}/{\text{cm}}^{\text{3}}=2.055\times {10}^{-30}\text{g}/{\text{cm}}^{\text{3}}$

Convert $\text{g}/{\text{cm}}^{\text{3}}$ to $\text{kg}/\text{L}$

$2.055\times {10}^{-30}\text{g}/{\text{cm}}^{\text{3}}\times \frac{1000{\text{ cm}}^3}{1\text{ L}}\times \frac{1\text{ kg}}{1000\text{ g}}=2.055\times {10}^{-30}\text{kg}/\text{L}$

The mass of dark energy contains in $1\text{ L}$  of space is $2.055\times {10}^{-30}\text{ kg}$.

Write the formula to find the value of energy

    $E=m{c}^2$        (I)

Write the expression to find the energy of photon

    $E=\frac{hc}{\lambda }$        (II)

Here, $E$ is the energy, $h$ is the Planck’s constant, $\lambda$ is the wavelength, $m$ is the mass and $c$ is the speed of light.

Substitute $2.055\times {10}^{-30}\text{ kg}$ for $m$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the equation (I) to find the value of $E$

$\begin{array}{c}E=\left(2.055\times {10}^{-30}\text{ kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =1.85\times {10}^{-13}\text{ J}\end{array}$

The dark energy present in $1\text{ L}$ is $1.85\times {10}^{-13}\text{ J}$.

Substitute $6.63\times {10}^{-34}\text{ Js}$ for $h$, $1\text{ mm}\left(={10}^{-3}\text{ m}\right)$ for $\lambda$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the equation (II) to find the value of $E$

$\begin{array}{c}E=\frac{\left(6.63\times {10}^{-34}\text{ Js}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{{10}^{-3}\text{ m}}\\ =1.99\times {10}^{-22}\text{ J}\end{array}$

The energy of $1\text{-mm-}$ wavelength microwave photons is $1.99\times {10}^{-22}\text{ J}$.

The  $1\text{-mm-}$ wavelength microwave photons in $1\text{ L}$ of space is

$\begin{array}{c}=\frac{1.85\times {10}^{-13}\text{ J}}{1.99\times {10}^{-22}\text{J}/\text{photon}}\\ =0.93\times {10}^9\text{ photons}\\ \approx 1\text{ billion photons}\end{array}$

Conclusion:

The number of joules of energy in each liter of space is $1.85\times {10}^{-13}\text{ J}$ and the energy would correspond to $1\text{ billion}$  $1\text{-mm-}$ wavelength microwave photons in each liter.