Answer to Question #105734 in Quantum Mechanics for Reeshabh Kumar

Question

The frequency of the light emitted by a galaxy receding from the earth is measured
to be 1.5×103 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

We are given

$\nu=1.5 \cdot 10^{3} MHz$

$\lambda_0=0.225 m$

Define the radiation frequency of a stationary galaxy

$\nu_0=\frac{c}{\lambda_0}=\frac{3 \cdot 10^{8}m/s}{0.225 m}=1.333 \cdot 10^{3}MHz$

Using the formula for the Doppler effect

$\nu=\frac{\nu_0}{1-\frac{V_r}{c}}$

then we will find $V_r$

$V_r=\frac{\nu-\nu_0}{\nu}\cdot c$

$V_r=\frac{1.5 \cdot 10^{3}-1.333 \cdot 10^{3}}{1.5 \cdot 10^{3}}\cdot 3 \cdot 10^{8}=3.34 \cdot 10^{7}m/s=0.111 \cdot c$