Question

The wavelength of a spectral line coming from a star is changed by the motion of the star from 6000 A to 6001 A. What is the velocity of the star with respect to earth? (The velocity of light is 3 x 10${}^8$ m/sec)

• A. 5 x 10${}^4$ m/sec receding
• B. 5 x 10${}^4$ m/sec approaching
• C. 3 x 10${}^4$ m/sec receding
• D. 3 x 10${}^4$ m/sec approaching

Answer 1

The correct option is A 5 x 10${}^4$ m/sec receding

Observed frequency for spectral lines: $f=f_0(1+\frac{\Delta v}{c})$

where $\Delta v=v_r-v_s$, $v_r$ is the velocity of the receiver and

$v_s$ is the velocity of the source wrt the medium and c is the velocity of light

$\therefore f-f_0=f_0\frac{\Delta v}{c}$

$\frac{\Delta f}{f}=\frac{\Delta v}{c}$

But, $\frac{\Delta f}{f}=\frac{\Delta \lambda }{\lambda }$

$\therefore \frac{\Delta \lambda }{\lambda }=\frac{\Delta v}{c}$

$\Rightarrow \Delta v=c\ast \frac{\Delta \lambda }{\lambda }=3\ast {10}^8\ast \frac{(6001-6000)\ast {10}^{-10}}{6000\ast {10}^{-10}}=5\ast {10}^{4}m/s$

Since, $v_r$ is zero it implies $v_s$ $=-5\ast {10}^{4}m/s$