Modern Physics Ques 34
- In hydrogen spectrum, the wavelength of $H _{\alpha}$ line is $656$ $ nm$; whereas in the spectrum of a distant galaxy $H _{\alpha}$ line wavelength is $706 $ $nm$. Estimated speed of galaxy with respect to earth is
$(1999,2 M)$
(a) $2 \times 10^{8} $ $m / s$
(b) $2 \times 10^{7} $ $m / s$
(c) $2 \times 10^{6} $ $m / s$
(d) $2 \times 10^{5} $ $m / s$
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Answer:
Correct Answer: 34.(b)
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- Since, the wavelength $(\lambda)$ is increasing, we can say that the galaxy is receding. Doppler effect can be given by
$ \lambda^{\prime}=\lambda \sqrt{\frac{c+v}{c-v}} $ $\quad$ …….(i)
$\text { or } \quad 706 =656 \sqrt{\frac{c+v}{c-v}} $
$ \text { or }\quad \frac{c+v}{c-v} =(\frac{706}{656})^{2}=1.16 $
$ \therefore \quad c+v=1.16 c-1.16 $ $v $
$ v =\frac{0.16 c}{2.16}=\frac{0.16 \times 3.0 \times 10^{8}}{2.16} $
$ = 0.22 \times 10^{8} $ $m / s $
$v \approx 2.2 \times 10^{7} $ $m / s$
If we take the approximation then Eq. (i) can be written as
$ \Delta \lambda=\lambda (\frac{v}{c}) $ $\quad$ …….(ii)
From here $v=(\frac{\Delta \lambda}{\lambda}) \cdot c=(\frac{706-656}{656})\left(3 \times 10^{8}\right)$
$ v=0.23 \times 10^{8} $ $m / s $
which is almost equal to the previous answer. So, we may use Eq. (ii) also.
BETA
