Question

A galaxy is moving away from the earth at a speed of $286km/s$. The shift in the wavelength of a redline at $630nm$ is $X\times {10}^{-10}m$. The value of $X$, to the nearest integer, is [Take the value of speed of light, $c=3\times {10}^{8}m{s}^{-1}$]

Answer 1

Step 1. Given data:

Wavelength of redline, $\lambda =630nm$

Speed of galaxy, $v=286km/s$

$\therefore v=286\times {10}^{3}m/s$

Speed of light, $c=3\times {10}^{8}m{s}^{-1}$

Step 2. Applying formula for the wavelength shift:

As, ${\lambda }^{\text{'}}=\lambda \left\{1-\frac{v}{c}\right\}$ $\left[v<<c\right]$

$\begin{array}{rcl}& \Rightarrow & {\lambda }^{\text{'}}=\lambda -\frac{\lambda v}{c}\\ & \Rightarrow & {\lambda }^{\text{'}}-\lambda =-\frac{\lambda v}{c}\\ & \Rightarrow & ∆\lambda =-\frac{\lambda v}{c}\\ & \Rightarrow & \left|∆\lambda \right|=\left|\frac{\lambda v}{c}\right|\\ & \Rightarrow & \left|∆\lambda \right|=\frac{630\times {10}^{-9}\times 286\times {10}^3}{3\times {10}^8}\\ \therefore \left|∆\lambda \right|& =& 6\times {10}^{-10}m\\ & & \end{array}$

The shift in the wavelength is, $\mathbf{∆}\mathit{\lambda }\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$.

So the value of $\mathit{X}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}$