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(D) The Doppler shift in wavelength for a source moving away from the observer is given by the formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$

Vedclass · Accepted Answer

$6$

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(D) The Doppler shift in wavelength for a source moving away from the observer is given by the formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.Given:Speed of the galaxy $v = 286 \, km/s = 286 	imes 10^{3} \, m/s$.Speed of light $c = 3 	imes 10^{8} \, m/s$.Wavelength $\lambda = 630 \, nm = 630 	imes 10^{-9} \, m$.Rearranging the formula to find the shift $\Delta \lambda$:$\Delta \lambda = \frac{v}{c} 	imes \lambda$.Substituting the values:$\Delta \lambda = \frac{286 	imes 10^{3}}{3 	imes 10^{8}} 	imes 630 	imes 10^{-9}$.$\Delta \lambda = \frac{286}{3 	imes 10^{5}} 	imes 630 	imes 10^{-9}$.$\Delta \lambda = \frac{286 	imes 630}{3} 	imes 10^{-14} = 286 	imes 210 	imes 10^{-14} = 60060 	imes 10^{-14} = 6.006 	imes 10^{-10} \, m$.Comparing this with $x 	imes 10^{-10} \, m$,we get $x \approx 6$.

$A$ galaxy is moving away from the Earth at a speed of $286 \, km/s$. The shift in the wavelength of a red line at $630 \, nm$ is $x \times 10^{-10} \, m$. The value of $x$,to the nearest integer,is........
[Take the value of speed of light $c$ as $3 \times 10^{8} \, m/s$]

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$A$ galaxy is moving away from the Earth at a speed of $286 \, km/s$. The shift in the wavelength of a red line at $630 \, nm$ is $x \times 10^{-10} \, m$. The value of $x$,to the nearest integer,is........[Take the value of speed of light $c$ as $3 \times 10^{8} \, m/s$]