A) \[3R\]
B) \[\frac{1}{3R}\]
C) \[\frac{4}{4R}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\frac{1}{\lambda }=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]=R\times {{2}^{2}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\] \[\Rightarrow 3R;\lambda =\frac{1}{3R}\]
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