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