A) coming closer with a velocity of \[\text{5463}\,\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ 7858}\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) moving away with a velocity of \[\text{1315}\,\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ 1530}\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) coming closer with a velocity of \[\text{5}.\text{67}\times \text{l}{{0}^{\text{7}}}\text{m}/\text{s}\]
D) moving away with a velocity of \[0.\text{95 }\times \text{ 1}{{0}^{\text{7}}}\text{ m}/\text{s}\]
Correct Answer: C
Solution :
Using the relation for Dopplers shift \[0.\text{95 }\times \text{ 1}{{0}^{\text{7}}}\text{ m}/\text{s}\] (Given) Since, \[\text{1}.\text{89 }\times \text{ 1}{{0}^{7}}\text{ m}/\text{s}\] \[\text{3}.\text{78 }\times \text{l}{{0}^{\text{7}}}\text{ m}/\text{s}\] \[2.5\times {{(5000)}^{2}}eV\] Or \[\text{2}.\text{5 }\times \text{ 5}000\text{ eV}\] \[\frac{2.5}{5000}eV\] \[\frac{2.5}{{{(5000)}^{2}}}eV\] \[\text{1}.\text{5 }\times \text{ 1}{{0}^{\text{4}}}\text{ m}/\text{s}\] Since, \[\text{1}.\text{5 }\times \text{ 1}{{0}^{\text{4}}}\text{ m}/\text{s}\] decreases, the star is approaching the observer.
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