KCET · Physics · Dual Nature of Matter
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then, the speed of the galaxy with respect to the Earth is
- A \(500 \mathrm{~km} \mathrm{~s}^{-1}\)
- B \(50 \mathrm{~km} \mathrm{~s}^{-1}\)
- C \(200 \mathrm{~km} \mathrm{~s}^{-1}\)
- D \(20 \mathrm{~km} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(500 \mathrm{~km} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(\lambda=600 \mathrm{~nm}\)
\(=6 \times 10^{-7} \mathrm{~m}\)
\(\Delta \lambda=(601-600) \times 10^{-9} \mathrm{~m}=1 \times 10^{-9} \mathrm{~m}\)
We know that, \(\frac{\Delta \lambda}{\lambda}=\frac{v}{c} \Rightarrow v=c \cdot \frac{\Delta \lambda}{\lambda}\)
\(=3 \times 10^8 \times \frac{1 \times 10^{-9}}{6 \times 10^{-7}}\)
\(=\frac{1}{2} \times 10^6 \mathrm{~m} / \mathrm{s}=500 \times 10^3 \mathrm{~m} / \mathrm{s}\)
\(=500 \mathrm{~km} / \mathrm{s}\)
\(=6 \times 10^{-7} \mathrm{~m}\)
\(\Delta \lambda=(601-600) \times 10^{-9} \mathrm{~m}=1 \times 10^{-9} \mathrm{~m}\)
We know that, \(\frac{\Delta \lambda}{\lambda}=\frac{v}{c} \Rightarrow v=c \cdot \frac{\Delta \lambda}{\lambda}\)
\(=3 \times 10^8 \times \frac{1 \times 10^{-9}}{6 \times 10^{-7}}\)
\(=\frac{1}{2} \times 10^6 \mathrm{~m} / \mathrm{s}=500 \times 10^3 \mathrm{~m} / \mathrm{s}\)
\(=500 \mathrm{~km} / \mathrm{s}\)
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