Question

What is the observed wavelength ofthe $\mathbf{656}\mathbf{.}\mathbf{3}\mathit{n}\mathit{m}$(first Balmer) line of hydrogen emitted by a galaxy at a distance of $\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$ly? Assume that the Doppler shift of Eq. 37-36 and Hubble’s law apply.

Short answer

Thus, the value of wavelength is 668 mm.

Step-by-step solution

Frequency and wavelength expression.

Let$f$represent the frequency detected by an observer moving with velocity$\overrightarrow{v}$relative to that rest frame. Then, when the direction of $\overrightarrow{v}$is directly away from the source.

Then the expression for the observed frequency is given by,

$f=f_0\sqrt{\frac{1-\beta }{1+\beta }}$

The wavelength is inversely proportional to the frequency$(\lambda \propto f^{-1})$$(\lambda \propto f^{-1})$

. Thus,

$\lambda ={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$…… (i)

The universe is expanding, with the distant galaxies moving away from us at a rate$v$ is given by Hubble’s law as,

$v=Hr$

Here$H$ is Hubble’s constant whose value is equal to,$0.0218\text{\hspace{0.17em}m/s}\cdot \text{\hspace{0.17em}ly}$,$r$is the distance to the galaxy and $v$is the recessional speed.

As, $v=\beta c$

$\Rightarrow \beta =\frac{Hr}{c}$…… (ii)

Evaluate the wavelength. 

Substitute the values into the equation (ii)

$\begin{array}{c}\beta =\frac{(0.0218)(2.40\times {10}^8)}{2.998\times {10}^8}\\ =0.017452\end{array}$

Substitute $656.3\text{\hspace{0.17em}nm}$for ${\lambda }_o$and$0.017452$for$\beta$into the equation (i)

$\begin{array}{c}\lambda =656.3\sqrt{\frac{1+0.017452}{1-0.017452}}\\ =668\text{\hspace{0.17em}n}m\end{array}$width="192">λ=656.31+0.0174521-0.017452=668 nm

Hence, the value of wavelength is 668 mm.