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A galaxy moves with respect to the earth, so that sodium line of $589.0 \mathrm{~nm}$ is observed at $589.6 \mathrm{~nm}$. The speed of the galaxy is
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Verified Answer
The correct answer is:
$306 \mathrm{kms}^{-1}$
Given, wavelength of sodium line,
$$
\lambda=589.0 \mathrm{~nm}=5.89 \times 10^{-7} \mathrm{~m}
$$
Observed wavelength,
$$
\lambda^{\prime}=589.6 \mathrm{~nm}=5.896 \times 10^{-7} \mathrm{~m}
$$
$\therefore$ Change in wavelength,
$$
\begin{aligned}
\Delta \lambda &=\lambda^{\prime}-\lambda \\
&=5.896 \times 10^{-7}-5.89 \times 10^{-7} \\
&=0.006 \times 10^{-7} \mathrm{~m}=6 \times 10^{-10} \mathrm{~m}
\end{aligned}
$$
According to Doppler's shift equation,
Speed of galaxy is given as
$$
\begin{aligned}
v &=\frac{\Delta \lambda}{\lambda} \times c \\
&=\frac{6 \times 10^{-10}}{5.89 \times 10^{-7}} \times 3 \times 10^{8} \\
&=3.06 \times 10^{5} \mathrm{~m} / \mathrm{s} \\
&=306 \times 10^{3} \mathrm{~m} / \mathrm{s}=306 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$
$$
\lambda=589.0 \mathrm{~nm}=5.89 \times 10^{-7} \mathrm{~m}
$$
Observed wavelength,
$$
\lambda^{\prime}=589.6 \mathrm{~nm}=5.896 \times 10^{-7} \mathrm{~m}
$$
$\therefore$ Change in wavelength,
$$
\begin{aligned}
\Delta \lambda &=\lambda^{\prime}-\lambda \\
&=5.896 \times 10^{-7}-5.89 \times 10^{-7} \\
&=0.006 \times 10^{-7} \mathrm{~m}=6 \times 10^{-10} \mathrm{~m}
\end{aligned}
$$
According to Doppler's shift equation,
Speed of galaxy is given as
$$
\begin{aligned}
v &=\frac{\Delta \lambda}{\lambda} \times c \\
&=\frac{6 \times 10^{-10}}{5.89 \times 10^{-7}} \times 3 \times 10^{8} \\
&=3.06 \times 10^{5} \mathrm{~m} / \mathrm{s} \\
&=306 \times 10^{3} \mathrm{~m} / \mathrm{s}=306 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$
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