Chapter 33, Problem 26P

To determine

The speed of a galaxy. The distance of a galaxy from Earth.

Answer to Problem 26P

The speed of a galaxy is0. 33c . The galaxy is $4.5\times {10}^9\text{ ly}$ away.

Explanation of Solution

Given:

The wavelength of a Hydrogen line at $434\text{ nm}$ is measured on Earth as $610\text{ nm}$ .

Speed of the light is

  $c\text{ = 3}\times {\text{10}}^8\text{ m/s}$

Formula used:

Red shift is calculated as

  $z=\frac{\Delta \lambda }{{\lambda }_0}=\sqrt{\frac{1+\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}{1-\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}-1$

Hubble’s law states,

  $v=Hd$

Calculation:

Difference in the apparent wavelength due to redshift is

  $\Delta \lambda =610\text{ nm }-\text{434 nm = 176 nm}$

Speed of the galaxy is measured as

  $\begin{array}{c}z=\frac{\Delta \lambda }{{\lambda }_0}\\ =\sqrt{\frac{1+\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}{1-\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}}-1\\ \frac{v}{c}=\frac{{\left(\frac{\Delta \lambda }{{\lambda }_0}+1\right)}^2-1}{{\left(\frac{\Delta \lambda }{{\lambda }_0}+1\right)}^2+1}\\ =\frac{{\left(\frac{\text{176 nm}}{\text{434 nm}}\text{+1}\right)}^2-1}{{\left(\frac{\text{176 nm}}{\text{434 nm}}\text{+1}\right)}^2+1}\\ \frac{v}{c}=\frac{0.9755}{2.9755}\\ =0.327c\\ \approx 0.33c\end{array}$

Distance of the galaxy showing red shift is

  $\begin{array}{c}v=Hd\\ d=\frac{v}{H}\\ =\frac{\frac{v}{c}}{\frac{H}{c}}\\ d\text{ =}\frac{0.3278}{\frac{22000\text{ m/s/Mly}}{\text{3}\times {\text{10}}^8\text{ m/s}}}\\ =4.47\times {10}^3\text{ Mly}\\ \approx 4.5\times {\text{10}}^9\text{ ly}\end{array}$

Conclusion:

The speed of a galaxy is $0.33c$ The galaxy is $4.5\times {10}^9\text{ ly}$ away from the Earth.