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In hydrogen spectrum the wavelength of $\mathrm{H}_\alpha$ line is $656 \mathrm{~nm};$ whereas in the spectrum of a distant galaxy $\mathrm{H}_\alpha$ line wavelength is $706 \mathrm{~nm}$ . Estimated speed of galaxy with respect to earth is :

Option: 1
$2 \times 10^8 \mathrm{~m} / \mathrm{s}$

Option: 2
$2 \times 10^7 \mathrm{~m} / \mathrm{s}$

Option: 3
$2 \times 10^6 \mathrm{~m} / \mathrm{s}$

Option: 4
$2 \times 10^5 \mathrm{~m} / \mathrm{s}$

Answers (1)

best_answer

Since, the wavelength $(\lambda)$ is increasing, we can say that the galaxy is receding. Doppler effect can be given by-

$\mathrm{\lambda^{\prime} =\lambda \sqrt{\frac{c+v}{c-v}} }$ ...........(1)

$\mathrm{\text { or } \quad 706 =656 \sqrt{\frac{c+v}{c-v}}}$

$\mathrm{ \text { or } \frac{c+v}{c-v} =\left(\frac{706}{656}\right)^2=1.16 }$

$\mathrm{\therefore \quad c+v =1.16 c-1.16 \mathrm{v} }$

$\mathrm{\therefore \quad v =\frac{0.16 c}{2.16} }$

$\mathrm{ =\frac{0.16 \times 3.0 \times 10^8}{2.16} \mathrm{~m} / \mathrm{s} }$

$\mathrm{ =0.22 \times 10^8 \mathrm{~m} / \mathrm{s} }$

$\mathrm{ v \approx 2.2 \times 10^7 \mathrm{~m} / \mathrm{s}}$

If we take the approximation then Eq. (1) can be written as

$\mathrm{ \Delta \lambda=\lambda\left(\frac{v}{c}\right) }$ $\mathrm{ \Delta \lambda=\lambda\left(\frac{v}{c}\right) }$

From here $\mathrm{ v=\left(\frac{706-656}{656}\right)\left(3 \times 10^8\right) \mathrm{m} / \mathrm{s} }$

$\mathrm{ v=0.23 \times 10^8 \mathrm{~m} / \mathrm{s} }$

Which is almost equal to the previous answer

Posted by

SANGALDEEP SINGH

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