The 6563 Å $\text{H}\alpha$ line emitted by hydrogen in a star is found to be redshifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
Exercise
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Given,
Wavelength of $\mathrm{H} \alpha$ line, $\lambda=15 \ AA=15 \times 10^{-10} \mathrm{~m}$
Redshift, $\lambda^{\prime}-\lambda=15 \ AA=15 \times 10^{-10} \mathrm{~m}$
Now, according to the formula,
$\lambda^{\prime}-\lambda=\frac{\mathrm{v}_{\mathrm{s}} \lambda}{\mathrm{c}}$
we have, $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{C}}{\lambda}\left(\lambda^{\prime}-\lambda\right)$
$ =\frac{3.0 \times 10^8}{6563 \times 10^{-10}} \times 15 \times 10^{-10} $
$ =6.8566 \times 10^5 \mathrm{~ms}^{-1}$
i.e., $V_s=6.86 \times 10^5 \mathrm{~ms}^{-1}$.
Therefore, the start is receding from the earth at a speed of $6.86 \times 10^5 \mathrm{~m} / \mathrm{s}$.
Wavelength of $\mathrm{H} \alpha$ line, $\lambda=15 \ AA=15 \times 10^{-10} \mathrm{~m}$
Redshift, $\lambda^{\prime}-\lambda=15 \ AA=15 \times 10^{-10} \mathrm{~m}$
Now, according to the formula,
$\lambda^{\prime}-\lambda=\frac{\mathrm{v}_{\mathrm{s}} \lambda}{\mathrm{c}}$
we have, $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{C}}{\lambda}\left(\lambda^{\prime}-\lambda\right)$
$ =\frac{3.0 \times 10^8}{6563 \times 10^{-10}} \times 15 \times 10^{-10} $
$ =6.8566 \times 10^5 \mathrm{~ms}^{-1}$
i.e., $V_s=6.86 \times 10^5 \mathrm{~ms}^{-1}$.
Therefore, the start is receding from the earth at a speed of $6.86 \times 10^5 \mathrm{~m} / \mathrm{s}$.
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