Question

A hydrogen line in a star's spectrum

Answer 1

Answer:

Riegel's velocity relative to us is: $v_{r} = 2.07 * 10^4 m/s$

Explanation:

Frequency of Riegel's spectrum, $f_{r} = f - f'$

$f' = f - f_{r} = 4.26 * 10^{10} Hz$

From the relativistic doppler effect, the wavelength of Riegel's spectrum is:

$\lambda = \lambda_{r} (1 - \frac{v_{r} }{c} )\\\frac{c}{f} = \frac{c}{f_r} (1 - \frac{v_{r} }{c} )\\v_{r} = (\frac{f - f_r}{f} )c\\Since, f' = f - f_r\\v_{r} = (\frac{f'}{f} )c\\v_{r} = (\frac{4.26 * 10^{10}}{6.17 * 10^{14}} ) * (3 * 10^8)\\v_{r} = 2.07 * 10^4 m/s$