Chapter 33, Problem 25P

(a)

To determine

The observed wavelength of a galaxy for the given distance.

Answer to Problem 25P

The observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^6\text{ ly}$ is $656\text{ nm}$ .

Explanation of Solution

Given:

Wavelength line in the Balmer series of the hydrogen atom is $656\text{ nm}$ .

Distance from Earth is $1.0\times {10}^6\text{ ly}$ .

Formula used:

Using Doppler effect,

  $\lambda ={\lambda }_0\sqrt{\frac{1+v/c}{1-v/c}}$

Calculation:

Using Hubble’s law, velocity of the galaxy is

  $\begin{array}{c}v=Hd\\ \frac{v}{c}=\frac{Hd}{c}\\ \frac{v}{c}=\frac{(22000\text{ m/s/Mly})(1.0\text{ Mly})}{(3\times {10}^8\text{ m/s})}\\ =7.33\times {10}^{-5}\end{array}$

Now, calculating observed wavelength of the galaxy using Doppler effect

  $\begin{array}{c}\lambda =(656\text{ nm})\sqrt{\frac{1+7.33\times {10}^{-5}}{1-7.33\times {10}^{-5}}}\\ =656.05\text{ nm}\\ \approx \text{656 nm}\end{array}$

Conclusion:

Thus, the observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^6\text{ ly}$ is $656\text{ nm}$ .

(b)

To determine

The observed wavelength of a galaxy for the given distance.

Answer to Problem 25P

The observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^8\text{ ly}$ is $661\text{ nm}$ .

Explanation of Solution

Given:

Wavelength line in the Balmer series of the hydrogen atom is $656\text{ nm}$ .

Distance from Earth is $1.0\times {10}^8\text{ ly}$ .

Calculation:

Using Hubble’s law, velocity of the galaxy is

  $\begin{array}{c}v=Hd\\ \frac{v}{c}=\frac{Hd}{c}\\ \frac{v}{c}=\frac{(22000\text{ m/s/Mly})(1.0\times {10}^2\text{ Mly})}{(3\times {10}^8\text{ m/s})}\\ =7.33\times {10}^{-3}\end{array}$

Now, calculating observed wavelength of the galaxy using Doppler effect

  $\begin{array}{c}\lambda =(656\text{ nm})\sqrt{\frac{1+7.33\times {10}^{-3}}{1-7.33\times {10}^{-3}}}\\ =660.83\text{ nm}\\ \approx \text{661 nm}\end{array}$

Conclusion:

Thus, the observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^8\text{ ly}$ is $661\text{ nm}$ .

(c)

To determine

The observed wavelength of a galaxy for the given distance.

Answer to Problem 25P

The observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^{10}\text{ ly}$ is $1670\text{ nm}$ .

Explanation of Solution

Given:

Wavelength line in the Balmer series of the hydrogen atom is $\text{656 nm}$ .

Distance from Earth is $1.0\times {10}^{10}\text{ ly}$ .

Calculation:

Using Hubble’s law, velocity of the galaxy is

  $\begin{array}{c}v=Hd\\ \frac{v}{c}=\frac{Hd}{c}\\ \frac{v}{c}=\frac{(22000\text{ m/s/Mly})(1.0\times {10}^4\text{ Mly})}{(3\times {10}^8\text{ m/s})}\\ =0.733\end{array}$

Now, calculating observed wavelength of the galaxy using Doppler effect

  $\begin{array}{c}\lambda =(656\text{ nm})\sqrt{\frac{1+0.733}{1-0.733}}\\ =1671.3\text{ nm}\\ \approx \text{1670 nm}\end{array}$

Conclusion:

Thus, the observed wavelength of a galaxy whose distance from Earth is $1.0\times {10}^{10}\text{ ly}$ is $1670\text{ nm}$ .