Question

Because of the cosmological expansion, a particular emission from a distant galaxy has a wavelength that is $\mathbf{2}$ times the wavelength that emission would have in a laboratory. Assuming that Hubble’s law holds and that we can apply Doppler-shift calculations, what was the distance ($\mathbf{ly}$)to that galaxy when the light was emitted?

Short answer

The distance to the galaxy when the light was emitted was.$2.75\times {10}^{10}\text{\hspace{0.17em}ly}$

Step-by-step solution

Given data

Detected wavelength is twice that of emitted wavelength.

Step 2:Hubble's law and astrophysical doppler shift

According to Hubble's law, galaxies at a distance$r$ from Earth is moving away from Earth at a velocity as:

$\mathbf{v}\mathbf{=}\mathbf{Hr}$ ..... (I)

Here,$H$is the Hubble's constant with value

$\mathbf{H}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{8}\mathbf{mm}\mathbf{/}\mathbf{s}\cdot \mathbf{ly}$

According to astrophysical Doppler shift, the change in wavelength of an emission of wavelength $\lambda$ if the source is moving away with a velocity $v$ is

$\mathbf{\Delta \lambda }\mathbf{=}\frac{\mathrm{v\lambda }}{c}$ ..... (II)

Here,$c$is the speed of light in vacuum with value

$\mathbf{c}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\frac{m}{s}$

Step 3:Determinethe distance of the source galaxy

From equations (I) and (II)

$\begin{array}{c}Hr=\frac{c\Delta \lambda }{\lambda }\\ r=\frac{c\Delta \lambda }{H\lambda }\end{array}$

The detected wavelength is twice the emitted wavelength, that is .$\Delta \lambda =2\lambda$ Substitute the values in the above equation to get

$\begin{array}{c}r=\frac{3\times {10}^8\frac{\text{m}}{\text{s}}\times 2\lambda }{21.8\frac{\text{mm}}{\text{s}}\cdot \text{ly}\times \lambda }\\ =2.75\times {10}^7\cdot \left(1\text{\hspace{0.17em}ly}\right)\cdot (1\text{\hspace{0.17em}m}\times \frac{1000\text{\hspace{0.17em}mm}}{1\text{\hspace{0.17em}m}})\cdot \left(\frac{1}{1\text{\hspace{0.17em}mm}}\right)\\ =2.75\times {10}^{10}\text{\hspace{0.17em}ly}\end{array}$

Thus, the distance is.$2.75\times {10}^{10}\text{\hspace{0.17em}ly}$