Q. The $K_{\alpha}$ line of singly ionised calcium has a wavelength of 393.3nm as measured on earth. In the spectrum of one of the observed galaxies, the spectral line is located at 401.8 nm. The speed with which this galaxy is moving away from us, will be
Explanation
(c) From Hubble ‘s law
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Z\ \propto r$
Where $Z \rightarrow$ red shift, $r \rightarrow$distance of the galaxy
Also, $\ \ \ \ \ \ \ \ Z = \frac{d\lambda}{\lambda}$ $=$ $\frac{v}{c}\ =$ $\frac{speed\ of\ galaxy}{speed\ of\ light}$
Given $d\lambda$ $= \ 401.8\ - 393.3 = 8.5\ nm,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = 393.3\ nm$,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Z\ = \frac{8.5}{393.3}$ $= 0.0216$
Also$\ \ \ \ \ \ \ \ v = cZ$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3\ \times \ 10^{8} \times 0.0216$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 64.8\ \times \ 10^{5}{ms}^{- 1}\ $
Since$\ \ \ 1km\ = \ 10^{3}m\ ,\ therefore$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = 6480\ {kms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.