Q. The $6563\ \mathring{\mathrm{A}}$ line emitted by hydrogen atom in a star is found to be red shifted by $5\ \mathring{\mathrm{A}}$. The speed with which the star is receding from the earth is
A) $17.29 \times 10^{9}m/s$
B) $4.29 \times 10^{7}m/s$
C) $3.39 \times 10^{5}m/s$
D) $2.29 \times 10^{5}m/s$
Explanation
(d) $\frac{\Delta\lambda}{\lambda} = \frac{v}{c} \Rightarrow v = \frac{\mathrm{\Delta}\lambda}{\lambda}.c = \frac{5}{6563} \times \left( 3 \times 10^{8} \right) = 2.29 \times 10^{5}m/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.