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(A) Given,the original wavelength $\lambda = 600 \ nm = 600 	imes 10^{-9} \ m$. The observed wavelength $\lambda' = 601 \ nm = 601 	imes 10^{-9} \ m

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$500$

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(A) Given,the original wavelength $\lambda = 600 \ nm = 600 	imes 10^{-9} \ m$. The observed wavelength $\lambda' = 601 \ nm = 601 	imes 10^{-9} \ m$. The change in wavelength is $\Delta \lambda = \lambda' - \lambda = (601 - 600) 	imes 10^{-9} \ m = 1 	imes 10^{-9} \ m$. According to the Doppler effect for light,the relation is $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the speed of the galaxy and $c$ is the speed of light $(c = 3 	imes 10^8 \ m/s)$. Therefore,$v = c \cdot \frac{\Delta \lambda}{\lambda} = (3 	imes 10^8 \ m/s) \cdot \frac{1 	imes 10^{-9} \ m}{600 	imes 10^{-9} \ m}$. $v = 3 	imes 10^8 \cdot \frac{1}{600} = \frac{3 	imes 10^8}{600} = 0.5 	imes 10^6 \ m/s$. $v = 500,000 \ m/s = 500 \ km/s$.

$A$ galaxy is moving away from the Earth such that a spectral line at $600 \ nm$ is observed at $601 \ nm$. Then,the speed of the galaxy with respect to the Earth is (in $km \ s^{-1}$)

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