Chapter 44: Q31P (page 1364)

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.

Short Answer

Expert verified

Thus, the distance of the galaxy is. $90.1 Mpc$

Step by step solution

Use relativistic Doppler shift formula.

The relativistic Doppler shift formula is:

$λ = λ_{0} \sqrt{\frac{(1 + \frac{v}{c})}{(1 - \frac{v}{c})}}$

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.

$\begin{matrix} \sqrt{\frac{(1 + \frac{v}{c})}{(1 - \frac{v}{c})}} = \frac{λ}{λ_{0}} \\ = \frac{602.0}{590.0} \\ = 1.0203 \end{matrix}$

Solve further,

$\begin{matrix} \frac{(1 + \frac{v}{c})}{(1 - \frac{v}{c})} = 1.04109 \\ 1 + \frac{v}{c} = 1.04109 - 1.04109 \frac{v}{c} \\ \frac{v}{c} = \frac{1.04109}{2.04109} \\ = 2.013 \times 10^{- 2} \end{matrix}$

The Hubble’s law is:

$\begin{matrix} v = H d \\ H = 67 \frac{km/s}{Mpc} \end{matrix}$

$\begin{matrix} 1 Mpc = 3.26 \times 10^{6} light-years \\ = 3.084 \times 10^{19} km \end{matrix}$ $\begin{matrix} 1 Mpc = 3.26 \times 10^{6} light-years \\ = 3.084 \times 10^{19} km \end{matrix}$

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

$\begin{matrix} d = \frac{v}{H} \\ = \frac{2.013 \times 10^{- 2} \times 3 \times 10^{5} {kms}^{-1}}{67 {kms}^{-1}} Mpc \\ = 90.1 Mpc \end{matrix}$

Hence, the distance of the galaxy is. $90.1 Mpc$

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Short Answer

Step by step solution

Use relativistic Doppler shift formula.

Evaluate the distance of galaxy.

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