A) \[\frac{16}{3R}\]
B) \[\frac{16}{5R}\]
C) \[\frac{5R}{16}\]
D) \[\frac{3R}{16}\]
Correct Answer: A
Solution :
Wavelength \[\frac{1}{\lambda }=R\left( \frac{1}{{{n}_{1}}}-\frac{1}{{{n}_{2}}} \right)\] \[\Rightarrow \] \[\frac{1}{\lambda }=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]=R\left( \frac{1}{4}-\frac{1}{16} \right)\] \[=R\left( \frac{4-1}{16} \right)=\frac{3R}{16}=\frac{16}{3R}\]
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