The 6563 mathring{A} line emitted by a hydro | vedclass

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(D) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelen

Vedclass · Accepted Answer

$2.29 	imes 10^{5} \ m/s$

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(D) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.Given: $\lambda = 6563 \ \mathring{A}$,$\Delta \lambda = 5 \ \mathring{A}$,and $c = 3 	imes 10^{8} \ m/s$.Rearranging the formula to solve for $v$: $v = \frac{\Delta \lambda}{\lambda} 	imes c$.Substituting the values: $v = \frac{5}{6563} 	imes 3 	imes 10^{8} \ m/s$.$v \approx 0.0007618 	imes 3 	imes 10^{8} \ m/s$.$v \approx 2.2855 	imes 10^{5} \ m/s \approx 2.29 	imes 10^{5} \ m/s$.

The $6563 \ \mathring{A}$ line emitted by a hydrogen atom in a star is found to be red-shifted by $5 \ \mathring{A}$. The speed with which the star is receding from the Earth is:

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