Question

The wavelength of a distant star is $5700$ and the spectral light has a shift of $1.9$ towards the red end then the velocity of a star relative to the earth Will be:

• A. $5\times {10}^{5}m/s$
• B. $2\times {10}^{5}m/s$
• C. $1.8\times {10}^{5}m/s$
• D. $1\times {10}^{5}m/s$

Answer 1

The correct option is D $1\times {10}^{5}m/s$

We know that

Doppler effect in light

$\frac{V_0\sqrt{1-\frac{V_s^2}{C^2}}}{[1-\frac{V_s}{c}]}$

$V$=Observed frequency

$V_0$=source frequency$

$V_s$= source velocity

$C=$ Velocity of light

Here, $V$ *vel of star)$<<C$. So,

$\approx \frac{V_0[1-\frac{V_s^2}{{2C}^2}]}{[1-\frac{V_s}{c}]}$

So, $\frac{V-V_0}{V_0}=\frac{\Delta V}{V_0}\approx \frac{V_s}{C}....\left(i\right)$

$\frac{\Delta V}{V_0}$ frational change in frequency

We know

$v.\lambda =C\Rightarrow V=C{\lambda }^{-1}\Rightarrow \frac{\Delta V}{V^{}}=\frac{\Delta C}{C^{}}-\frac{\Delta \lambda }{{\lambda }^{}}$

As $C\Rightarrow$ speed of light is constant

$\Delta C=0;\frac{\Delta V}{V^{}}=-\frac{\Delta \lambda }{{\lambda }^{}}$

$\Delta \lambda =-1.9$ (Red shift)

$\lambda =5700$

So, $\frac{\Delta V}{V^{}}=\frac{-(-1.9)}{5700}=\frac{1}{3000}...\left(ii\right)$

From (i) an d(ii) we get

$\frac{1}{3000}=\frac{V_s}{C}\Rightarrow V_s=\frac{3\times {10}^8}{3\times {10}^3}\Rightarrow V_s={10}^{5}m/s$

option D is correct