A) \[\frac{3Rc}{27}\]
B) \[\frac{Rc}{25}\]
C) \[\frac{8Rc}{9}\]
D) \[\frac{5Rc}{36}\]
Correct Answer: D
Solution :
\[\frac{1}{\lambda }=R\left[ \frac{1}{{{n}^{2}}_{2}}-\frac{1}{{{n}^{2}}_{1}} \right]\] \[\frac{1}{\lambda }=R\left[ \frac{1}{{{\left( 2 \right)}^{2}}}-\frac{1}{{{\left( 3 \right)}^{2}}} \right]\] \[\frac{1}{\lambda }=R\left[ \frac{1}{4}-\frac{1}{9} \right]=\frac{5R}{36}\] \[\operatorname{f}=\frac{c}{\lambda }\] \[\operatorname{f}=c\times \frac{5R}{36}=\frac{5Rc}{36}\]
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