A particular emission line,detected in the li | vedclass

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(A) According to the Doppler effect,the recession speed $v$ of the galaxy is given by $v = c \frac{\Delta \lambda}{\lambda}$.Given $\lambda' = 1.1\lam

Vedclass · Accepted Answer

$1.6 	imes 10^9 	ext{ ly}$

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(A) According to the Doppler effect,the recession speed $v$ of the galaxy is given by $v = c \frac{\Delta \lambda}{\lambda}$.Given $\lambda' = 1.1\lambda$,the change in wavelength is $\Delta \lambda = \lambda' - \lambda = 0.1\lambda$.Thus,$v = c 	imes \frac{0.1\lambda}{\lambda} = 0.1c$.Using Hubble's Law,$v = Hr$,where $H = 19.3 	ext{ mm/s/ly} = 19.3 	imes 10^{-3} 	ext{ km/s/ly} = 19.3 	imes 10^{-6} 	ext{ km/s/ly}$ (assuming standard units for $H \approx 19.3 	ext{ km/s per million ly}$ or similar constant).Given $c = 3 	imes 10^8 	ext{ m/s} = 3 	imes 10^5 	ext{ km/s}$.$r = \frac{v}{H} = \frac{0.1 	imes 3 	imes 10^5 	ext{ km/s}}{19.3 	imes 10^{-6} 	ext{ km/s/ly}} \approx 1.55 	imes 10^9 	ext{ ly} \approx 1.6 	imes 10^9 	ext{ ly}$.

$A$ particular emission line,detected in the light from a galaxy,has a wavelength $\lambda' = 1.1\lambda$,where $\lambda$ is the proper wavelength of the line. The distance of the galaxy from us is:

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