A) \[\frac{16}{{{\lambda }_{1}}}=\frac{9}{{{\lambda }_{2}}}\]
B) \[\frac{16}{{{\lambda }_{2}}}=\frac{9}{{{\lambda }_{1}}}\]
C) \[\frac{4}{{{\lambda }_{1}}}=\frac{1}{{{\lambda }_{2}}}\]
D) \[\frac{16}{{{\lambda }_{1}}}=\frac{3}{{{\lambda }_{2}}}\]
Correct Answer: B
Solution :
[b] \[\frac{1}{{{\lambda }_{1}}}=R{{(1)}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)\] |
and \[\frac{1}{{{\lambda }_{2}}}=R{{(1)}^{2}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] |
\[\therefore \] \[{{\lambda }_{1}}=\frac{1}{R}\] and \[{{\lambda }_{2}}=\frac{16}{3R}\] |
\[\therefore \] \[\frac{16}{{{\lambda }_{2}}}=\frac{3}{{{\lambda }_{1}}}\] |
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