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(B) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $v$

Vedclass · Accepted Answer

receding away from earth with a speed of $10 	imes 10^5 \, 	ext{m} \, 	ext{s}^{-1}$

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(B) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 	imes 10^8 \, 	ext{m} \, 	ext{s}^{-1})$.Given: $\lambda = 6600 \, 	ext{Å}$, $\Delta \lambda = 22 \, 	ext{Å}$.Since the light is red-shifted, the wavelength increases, which implies the star is receding away from the earth.Using the formula: $v = c 	imes \frac{\Delta \lambda}{\lambda}$.Substituting the values: $v = (3 	imes 10^8 \, 	ext{m} \, 	ext{s}^{-1}) 	imes \frac{22 \, 	ext{Å}}{6600 \, 	ext{Å}}$.$v = (3 	imes 10^8) 	imes \frac{1}{300} = 10^6 \, 	ext{m} \, 	ext{s}^{-1} = 10 	imes 10^5 \, 	ext{m} \, 	ext{s}^{-1}$.Thus, the star is receding away from the earth with a speed of $10 	imes 10^5 \, 	ext{m} \, 	ext{s}^{-1}$.

For an observer on the earth, if a spectral line of wavelength $6600 \, \text{Å}$ emitted by a star is found to be red shifted by $22 \, \text{Å}$, then the star is

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