Q. The time period of rotation of the sun is 25 days and its radius is $7 \times 10^{8}m$. The Doppler shift for the light of wavelength $6000\ \mathring{\mathrm{A}}$ emitted from the surface of the sun will be
A) $0.04\ \mathring{\mathrm{A}}$
B) $0.40\ \mathring{\mathrm{A}}$
C) $4.00\ \mathring{\mathrm{A}}$
D) $40.0\ \mathring{\mathrm{A}}$
Explanation
(a) $\mathrm{\Delta}\lambda = \lambda\frac{v}{c}$ and $v = r\omega$
$v = 7 \times 10^{8} \times \frac{2\pi}{25 \times 24 \times 3600},c = 3 \times 10^{8}m/s$
$\therefore\mathrm{\Delta}\lambda = 0.04\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.