Modern Physics 1 Question 26
23. 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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Solution:
- Since, the wavelength $(\lambda)$ is increasing, we can say that the galaxy is receding. Doppler effect can be given by
$$ \lambda^{\prime}=\lambda \sqrt{\frac{c+v}{c-v}} $$
$$ \begin{array}{rlrl} \text { or } & 706 & =656 \sqrt{\frac{c+v}{c-v}} \\ & \text { or } & \frac{c+v}{c-v} & =\frac{706}{656}^{2}=1.16 \\ & \therefore & c+v=1.16 c-1.16 v \\ & & v & =\frac{0.16 c}{2.16}=\frac{0.16 \times 3.0 \times 10^{8}}{2.16} \\ & & = & 0.22 \times 10^{8} m / s \\ & v & \approx 2.2 \times 10^{7} m / s \end{array} $$
If we take the approximation then Eq. (i) can be written as
$$ \Delta \lambda=\lambda \frac{v}{c} $$
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.
