With what speed should a galaxy move outward | vedclass

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

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(C) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the original wavelength,$v$ is the relative velocity,and $c$ is the speed of light.Given: $\lambda = 5890 \ \mathring{A}$,observed wavelength $\lambda' = 5896 \ \mathring{A}$.Change in wavelength $\Delta \lambda = \lambda' - \lambda = 5896 - 5890 = 6 \ \mathring{A}$.Speed of light $c = 3 	imes 10^8 \ 	ext{m/s} = 3 	imes 10^5 \ 	ext{km/s}$.Using the formula: $v = c 	imes \frac{\Delta \lambda}{\lambda}$.$v = (3 	imes 10^5 \ 	ext{km/s}) 	imes \frac{6 \ \mathring{A}}{5890 \ \mathring{A}}$.$v = \frac{18 	imes 10^5}{5890} \ 	ext{km/s} \approx 305.6 \ 	ext{km/s}$.Rounding to the nearest integer,the speed is approximately $306 \ 	ext{km/s}$.

With what speed should a galaxy move outward with respect to Earth so that the sodium-$D$ line at wavelength $5890 \ \mathring{A}$ is observed at $5896 \ \mathring{A}$? (in $\text{km/s}$)

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