A) Moving away with a velocity of \[1.5\times {{10}^{5}}m/s\]
B) Coming closer with a velocity of \[1.5\times {{10}^{5}}m/s\]
C) Moving away with a velocity of \[1.5\times {{10}^{4}}m/s\]
D) Coming closer with a velocity of \[1.5\times {{10}^{4}}m/s\]
Correct Answer: B
Solution :
\[\frac{\Delta \lambda }{\lambda }=\frac{v}{c}\,\Rightarrow \,\frac{0.05}{100}=\frac{v}{3\times {{10}^{8}}}\] Þ v = 1.5 ´ 105 m/s (Since wavelength is decreasing, so star coming closer)
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