The K line of singly ionised calcium has a wa | vedclass

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

Vedclass · Accepted Answer

$6480$

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(A) The Doppler shift formula for light is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,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 $(3 	imes 10^8 \, m/s)$.Given: $\lambda = 393.3 \, nm$,$\lambda' = 401.8 \, nm$.$\Delta \lambda = \lambda' - \lambda = 401.8 - 393.3 = 8.5 \, nm$.Substituting the values into the formula:$\frac{8.5}{393.3} = \frac{v}{3 	imes 10^8}$.$v = \frac{8.5 	imes 3 	imes 10^8}{393.3} \approx 0.064836 	imes 10^8 \, m/s$.$v \approx 6.4836 	imes 10^6 \, m/s = 6483.6 \, km/s$.Rounding to the nearest provided option,the speed is approximately $6480 \, km/s$.

The $K$ line of singly ionised calcium has a wavelength of $393.3 \, nm$ as measured on Earth. In the spectrum of one of the observed galaxies,this spectral line is located at $401.8 \, nm$. The speed with which the galaxy is moving away from us will be..... $km/s$.

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