A) \[\frac{16}{R}\]
B) \[\frac{16}{3R}\]
C) \[\frac{16}{5R}\]
D) \[\frac{16}{7R}\]
Correct Answer: B
Solution :
\[\vec{v}=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]where \[{{n}_{1}}=2,{{n}_{2}}=4\] \[\vec{v}=R\left( \frac{1}{4}-\frac{1}{16} \right)\] \[\frac{1}{\lambda }=R\left( \frac{12}{4\times 6} \right)\Rightarrow \lambda =\frac{16}{3R}\]
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