Q. The $k$ line of singly ionized calcium has a wavelength of 393.3 nm as measured on earth. In the spectrum of one of the observed galaxies, this spectral line is located at 401.8 nm. The speed with which the galaxy is moving away from us, will be
A) $6480\ {kms}^{- 1}$
B) $3240\ {kms}^{- 1}$
C) $4240\ {kms}^{- 1}$
D) None of the above
Explanation
(a) Here, $\lambda_{0} = 393.3\ nm$
$\lambda = 401.8\ nm$
Red shift, $z = \frac{\lambda - \lambda_{0}}{\lambda_{0}} = \frac{401.8 - 393.3}{393.3} = \frac{8.5}{393.3} = 0.0216$
Now $z = \frac{v}{c}$ then
$v = zc = 0.0216 \times 3 \times 10^{8} = 6.48 \times 10^{6}\ {ms}^{- 1} = 6480\ km\ s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.