For an observer on the earth, if a spectral line of wavelength 6600{Å} emitted by a star is found to be red shifted by 22{Å}, then the star is [AP EAPCET 21 May 2025 S2]

Correct Answer is : receding away from earth with a speed of 10×105m s−110 × 10^5 {m}\,{s}^{-1}10×105ms−1

receding away from earth with a speed of 9 × 10^5 {m}\,{s}^{-1}

receding away from earth with a speed of 10 × 10^5 {m}\,{s}^{-1}

moving towards earth with a speed of 9 × 10^5 {m}\,{s}^{-1}

moving towards earth with a speed of 10 × 10^5 {m}\,{s}^{-1}

Correct Answer is : receding away from earth with a speed of 10×105m s−110 × 10^5 {m}\,{s}^{-1}10×105ms−1

Test Index

Atoms and Nuclei

Section: Physics

Question : 12 of 33

Marks: +1, -0

6600\text{Å}

22\text{Å}

receding away from earth with a speed of $9 \times 10^5 \text{m}\,\text{s}^{-1}$
receding away from earth with a speed of $10 \times 10^5 \text{m}\,\text{s}^{-1}$
moving towards earth with a speed of $9 \times 10^5 \text{m}\,\text{s}^{-1}$
moving towards earth with a speed of $10 \times 10^5 \text{m}\,\text{s}^{-1}$

Solution:

Given data

Wavelength emitted by the star (rest wavelength):

\lambda_0=6600\text{Å}

Observed redshift (increase in wavelength):

\Delta \lambda=22\text{Å}

The redshift implies that the star is moving away from the Earth.

Formula (for non-relativistic Doppler shift)

For light at non-relativistic speeds:

\;\frac{\Delta \lambda}{\lambda_0}=\;\frac{v}{c}

where

v=

radial speed of the source relative to the observer,

c=3 \times 10^8 \text{m s}^{-1}.

Substituting values

\;\frac{22}{6600}=\;\frac{v}{3 \times 10^8}

\;v=3 \times 10^8 \times \;\frac{22}{6600}

\;v=3 \times 10^8 \times 3.33 \times 10^{-3}

\;v=1.0 \times 10^6 \text{m s}^{-1}

\;v=10^6 \text{m s}^{-1}=10 \times 10^5 \text{m s}^{-1}

Direction

Since the wavelength is red shifted, the star is receding from Earth.

Answer:

Option B

Receding away from Earth with a speed of

10 \times 10^5 \text{m/s}

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