Question

The ${6563A^\circ}$ line emitted by a hydrogen atom in a star is found to be red shifted by ${5A^\circ}$. The speed with which the star is receding from the earth is
(A) $17.3 \times {10^3}m/s$
(B) $4.29 \times {10^7}m/s$
(C) $3.39 \times {10^5}m/s$
(D) $2.29 \times {10^5}m/s$

Answer 1

Hint: Doppler shift is the change in frequency of a wave in relation to an observer who is moving relation to the wave source. Hence, wavelength is changed or shifted

Complete step by step answer:
When a body that is emitting radiation has a non-zero radial velocity relative to an observer the wavelength of the emission will be shortened or lengthened, depending upon whether the body is moving towards or away from an observer.
This change in wavelength or frequency is known as the Doppler shift.
The formula of Doppler shift for wavelength is given as
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{V}{C}$
Where
$\Delta \lambda = $ Wavelength shift
$\lambda = $ Initial value length
$C = $ Speed of light in vacuum
In question, it is given that
$\Delta \lambda ={5A^\circ}= 5 \times {10^{ - 10}}m$
$\lambda = {6563A^\circ} = 6563 \times {10^{ - 10}}m$
$C = 3 \times {10^8}m/s$
$V = ?$
So, $V = C \times \dfrac{{\Delta \lambda }}{\lambda }$
$V = 3 \times {10^8} \times \dfrac{{5 \times {{10}^{ - 10}}}}{{6563 \times {{10}^{ - 10}}}}$
$V = \dfrac{{15 \times {{10}^8}}}{{6563}}$
$V = 0.0022855 \times {10^8}$
$V \simeq 2.29 \times {10^5}m/s$

So, the correct answer is “Option D”.

Note:
In many problems of Doppler effect, they asked for frequency change which is given as
1.

Frequency $f' = \left( {1 + \dfrac{V}{C}} \right)f$
Wavelength $\lambda ' = \left( {1 - \dfrac{V}{C}} \right)\lambda $
(Blue shift)

2.

Frequency $f' = \left( {1 - \dfrac{V}{C}} \right)f$
Wavelength $\lambda ' = \left( {1 + \dfrac{V}{C}} \right)\lambda $
(Red shift)