Question

In the hydrogen spectrum, the wavelength of the line is $656$nm, whereas in a spectrum of a distant galaxy , the line wavelength is $706$ nm. Estimated speed of the galaxy with respect to the earth is
$\begin{align}
  & a)2\times {{10}^{8}}m/s \\
 & b)2\times {{10}^{7}}m/s \\
 & c)2\times {{10}^{6}}m/s \\
 & d)2\times {{10}^{5}}m/s \\
\end{align}$

Answer 1

Hint: When there is relative motion between the source of light and the observer, the frequency of light observed, is different as compared to the actual frequency of the source. The extent to which the difference in the frequency of light observed depends on the relative velocity. So, using Doppler’s expression we can find the velocity of the observer and the source.

Complete answer:
Formula used:
$\lambda -\overline{\lambda }=\pm \dfrac{v}{c}\lambda $
Let us call this as equation 1
The above equation can be used to calculate the velocity of the galaxy with respect to the earth.
 Before that let us understand the above equation i.e.
$\lambda $= $656$nm, represents the actual wavelength of the source,
$\overline{\lambda }$= $706$ nm, represents the actual wavelength of the source observed from earth,
 (in this particular question it is the wavelength of the light observed from the galaxy),
$v$ is the relative velocity of the source and the observer.
$c$= $3\times {{10}^{8}}$ m/s -is the speed of light (constant).
$\pm $ gives an idea whether the source (galaxy ) is moving towards the observer(earth) or moving away.
In the above question
$v$ is equal to the velocity of the galaxy with respect to earth
$\because $ $v=$ velocity of the observer(i.e. on earth) -velocity of the source(i.e. the galaxy)
From equation 1 we have,
$\lambda -\overline{\lambda }=\pm \dfrac{v}{c}\lambda $
After substitution of the quantities in the above equation,
$656nm-706nm$ = $\dfrac{v}{3\times {{10}^{8}}m/s}656nm$
$\begin{align}
  & -50\times {{10}^{-9}}=\dfrac{v}{3\times {{10}^{8}}}656\times {{10}^{-9}} \\
 & \\
\end{align}$
After simplifying by cross multiplication we get,
$v=-2\times {{10}^{7}}m/s$
Hence the correct answer to the question is option b.

Additional information:
The minus sign indicates that the galaxy is moving away with respect to the earth.
The above phenomenon is also known as red shift i.e.
Red shift: When the wavelengths in the middle part of the visible spectrum shifts towards the red region that is $\overline{\lambda }\rangle \lambda $, the shift is termed as red shift.
Blue shift: When the wavelengths in the middle part of the visible spectrum shifts towards the blue region that is $\lambda \langle \overline{\lambda }$, the shift is termed as blue shift.

Note:
1)Always first subtract the actual and the observed wavelength and then use the sign accordingly on the opposite side of the equation 1.
2)If the frequency is given in the question use the expression $c=\lambda \gamma $ (where $\lambda $ is the wavelength and $\gamma $ is the frequency of the source ), and make the necessary changes by substituting $\gamma =\dfrac{c}{\lambda }$ in equation.