Chapter 46, Problem 47P

(a)

To determine

Find the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth.

Answer to Problem 47P

The wavelength of the sodium line emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth is $590.09\text{ nm}$.

Explanation of Solution

A sodium line is emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth with the wavelength of $590\text{ nm}$ observed by the observer on earth.

Write the formula Hubble’s law

  $v=Hr$                                                                                                               (I)

Write the formula for wavelength [relativistic Doppler effect]

    $\lambda =\lambda \text{'}\frac{\sqrt{1-v/c}}{\sqrt{1+v/c}}$                                                                                                      (II)

Here, $v$ is the recessional velocity, $H$ is Hubble’s constant $H=22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}$ and $r$ is the distance, is the actual wavelength, is the wavelength observed by the observer on Earth, is the velocity and is the speed of light $\left[c=2.998\times {10}^8\text{m}/\text{s}\right]$ .

Conclusion:

Substitute $22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}$ for $H$ and $2.00\times {10}^6\text{ ly}$ for $r$ in equation (I) to find the value of $v$

$\begin{array}{l}v=22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}\times 2.00\times {10}^6\text{ ly}\\ v=4.4\times {10}^4\text{m}/\text{s}\end{array}$

The recessional velocity is $4.4\times {10}^4\text{m}/\text{s}$.

Substitute $4.4\times {10}^4\text{m}/\text{s}$ for $v$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$ and $590\text{ nm}$ for $\lambda$ in equation (II) to find the value of $\lambda \text{'}$

$\begin{array}{l}\lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+\left(4.4\times {10}^4\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}{\sqrt{1-\left(4.4\times {10}^4\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}\\ \lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+0.0001468}}{\sqrt{1-0.0001468}}\\ \lambda \text{'}=590.09\text{ nm}\end{array}$

Thus, the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth is $590.09\text{ nm}$.

(b)

To determine

Find the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^8\text{ ly}$ away from the earth.

Answer to Problem 47P

The wavelength of the sodium line emitted from the galaxy $2.00\times {10}^8\text{ ly}$ away from the earth is $599\text{ nm}$.

Explanation of Solution

A sodium line is emitted from the galaxy $2.00\times {10}^8\text{ ly}$ away from the earth with the wavelength of $590\text{ nm}$ observed by the observer on earth.

Conclusion:

Substitute $22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}$ for $H$ and $2.00\times {10}^8\text{ ly}$ for $r$ in equation (I) to find the value of $v$

$\begin{array}{l}v=22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}\times 2.00\times {10}^8\text{ ly}\\ v=4.4\times {10}^6\text{m}/\text{s}\end{array}$

The recessional velocity is $4.4\times {10}^6\text{m}/\text{s}$.

Substitute $4.4\times {10}^6\text{m}/\text{s}$ for $v$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$ and $590\text{ nm}$ for $\lambda$ in equation (II) to find the value of $\lambda \text{'}$

$\begin{array}{l}\lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+\left(4.4\times {10}^6\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}{\sqrt{1-\left(4.4\times {10}^6\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}\\ \lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+0.01468}}{\sqrt{1-0.01468}}\\ \lambda \text{'}=599\text{ nm}\end{array}$

Thus, the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth is $599\text{ nm}$.

(c)

To determine

Find the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^9\text{ ly}$ away from the earth.

Answer to Problem 47P

The wavelength of the sodium line emitted from the galaxy $2.00\times {10}^9\text{ ly}$ away from the earth is $684\text{ nm}$.

Explanation of Solution

A sodium line is emitted from the galaxy $2.00\times {10}^9\text{ ly}$ away from the earth with the wavelength of $590\text{ nm}$ observed by the observer on earth.

Conclusion:

Substitute $22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}$ for $H$ and $2.00\times {10}^9\text{ ly}$ for $r$ in equation (I) to find the value of $v$

$\begin{array}{l}v=22\times {10}^{-3}\text{m}/\text{s}\cdot \text{ly}\times 2.00\times {10}^9\text{ ly}\\ v=4.4\times {10}^7\text{m}/\text{s}\end{array}$

The recessional velocity is $4.4\times {10}^7\text{m}/\text{s}$.

Substitute $4.4\times {10}^7\text{m}/\text{s}$ for $v$, $2.998\times {10}^8\text{m}/\text{s}$ for $c$ and $590\text{ nm}$ for $\lambda$ in equation (II) to find the value of $\lambda \text{'}$

$\begin{array}{l}\lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+\left(4.4\times {10}^7\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}{\sqrt{1-\left(4.4\times {10}^7\text{m}/\text{s}\right)/\left(2.998\times {10}^8\text{m}/\text{s}\right)}}\\ \lambda \text{'}=590\text{ nm}\times \frac{\sqrt{1+0.1468}}{\sqrt{1-0.1468}}\\ \lambda \text{'}=684\text{ nm}\end{array}$

Thus, the wavelength of the sodium line emitted from the galaxy $2.00\times {10}^6\text{ ly}$ away from the earth is $684\text{ nm}$.