Question

A distant star which is moving away with a velocity of ${10}^6$ m/sec is emitting red line of frequency$4.5\times {10}^{1}4$ Hz. The observed frequency of this spectral line is

• A. $4.5\times {10}^{8}Hz$
• B. $4.485\times {10}^{14}Hz$
• C. $4.515\times {10}^{14}Hz$
• D. $4.5\times {10}^{14}Hz$

Answer 1

The correct option is B $4.485\times {10}^{14}Hz$

The source is moving with a velocity, $u={10}^{6}m/s$
The frequency will be Doppler shifted.
The relationship between observed frequency (${\nu }^{\prime }$) and emitted frequency (${\nu }_0$) is given by: ${\nu }^{\prime }=\frac{v+u_r}{v+u_s}{\nu }_0$
where, $v$ is the velocity of waves in the medium.
$u_r$ is the velocity of the receiver relative to the medium; positive if the receiver is moving towards the source (and negative in the other direction);
$u_s$ is the velocity of the source relative to the medium; positive if the source is moving away from the receiver (and negative in the other direction).

Here $u_r=0$ and $u_s={10}^6$ $m/s$
${\nu }^{\prime }=\frac{3\times {10}^8}{3\times {10}^8+{10}^6}4.5\times {10}^{14}=4.485\times {10}^{14}$ $Hz$