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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In the laboratory, one of the l;ines of sodium is emitted at a wavelength of 590.0 nm. In the light from a particular galaxy, however, this line is seen at a wavelength of 602.0 nm. Calculate the distance to the galaxy, assuming that Hubble’s law holds and that the Doppler shift of Eq 37-36 applies.

Thus, the distance of the galaxy is.$90.1\text{\hspace{0.17em}Mpc}$

See the step by step solution

## Step 1: Use relativistic Doppler shift formula.

The relativistic Doppler shift formula is:

${\mathbit{\lambda }}{\mathbf{=}}{{\mathbit{\lambda }}}_{{\mathbf{0}}}\sqrt{\frac{\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{\right)}}{\mathbf{\left(}\mathbf{1}\mathbf{-}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{\right)}}}$

## Step 2: Evaluate the distance of galaxy.

Light of wavelength 590.0 nm is observed to be 602.0 nm. Therefore, the speed of the source can be obtained from the equation.

$\begin{array}{c}\sqrt{\frac{\left(1+\frac{v}{c}\right)}{\left(1-\frac{v}{c}\right)}}=\frac{\lambda }{{\lambda }_{0}}\\ =\frac{602.0}{590.0}\\ =1.0203\end{array}$

Solve further,

$\begin{array}{c}\frac{\left(1+\frac{v}{c}\right)}{\left(1-\frac{v}{c}\right)}=1.04109\\ 1+\frac{v}{c}=1.04109-1.04109\frac{v}{c}\\ \frac{v}{c}=\frac{1.04109}{2.04109}\\ =2.013×{10}^{-2}\end{array}$

$\begin{array}{c}\frac{\left(1+\frac{v}{c}\right)}{\left(1-\frac{v}{c}\right)}=1.04109\\ 1+\frac{v}{c}=1.04109-1.04109\frac{v}{c}\\ \frac{v}{c}=\frac{1.04109}{2.04109}\\ =2.013×{10}^{-2}\end{array}$

The Hubble’s law is:

$\begin{array}{c}v=Hd\\ H=67\frac{\text{km/s}}{\text{Mpc}}\end{array}$

$\begin{array}{c}1\text{\hspace{0.17em}Mpc}=3.26×{10}^{6}\text{\hspace{0.17em}light-years}\\ =3.084×{10}^{19}\text{km}\end{array}$$\begin{array}{c}1\text{\hspace{0.17em}Mpc}=3.26×{10}^{6}\text{\hspace{0.17em}light-years}\\ =3.084×{10}^{19}\text{km}\end{array}$

Therefore, the distance of the galaxy as calculated using the Hubble’s law will be:

$\begin{array}{c}d=\frac{v}{H}\\ =\frac{2.013×{10}^{-2}×3×{10}^{5}{\text{kms}}^{\text{-1}}}{67{\text{kms}}^{\text{-1}}}\text{Mpc}\\ =90.1\text{\hspace{0.17em}Mpc}\end{array}$

Hence, the distance of the galaxy is.$90.1\text{\hspace{0.17em}Mpc}$

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