Q. The periodic time of rotation of a certain star is 22 days and its radius is $7 \times 10^{8}\ metres$. If the wavelength of light emitted by its surface be $4320\ \mathring{\mathrm{A}}$, the Doppler shift will be (1 day = $86400\ sec$)
A) $0.033\ \mathring{\mathrm{A}}$
B) $0.33\ \mathring{\mathrm{A}}$
C) $3.3\ \mathring{\mathrm{A}}$
D) $33\ \mathring{\mathrm{A}}$
Explanation
(a) $\mathrm{\Delta}\lambda = \lambda.\frac{v}{c}$ where $v = r\omega = r \times \left( \frac{2\pi}{T} \right)$
$\therefore\mathrm{\Delta}\lambda = \frac{4320 \times 7 \times 10^{8} \times 2 \times 3.14}{3 \times 10^{8} \times 22 \times 86400} = 0.033\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.