Answer to Question #109315 in Quantum Mechanics for DD

Question

The frequency of the light emitted by a galaxy receding from the earth is measured
to be (1.5×10^3) MHz. Assuming that the wavelength of the light source is 22.5 cm,
calculate how fast the galaxy is receding from the earth?

Answer 1

The frequency of a stationary galaxy is

$f_0=\frac{c}{\lambda_0}.$

Write the equation for Doppler effect:

$f=\frac{f_0}{1-\frac{v_r}{c}}=\\ \space\\ v_r=\bigg(1-\frac{f_0}{f}\bigg)c=\bigg(1-\frac{c}{f\lambda_0}\bigg)c=\\ \space\\ =\bigg(1-\frac{3\cdot10^8}{1.5\cdot10^9\cdot0.225}\bigg)3\cdot10^8=33.3\cdot10^6\text{ m/s}=\\ =0.111c.$