A) \[\frac{20\lambda }{27}\]
B) \[\frac{3\lambda }{16}\]
C) \[\frac{5\lambda }{36}\]
D) \[\frac{3\lambda }{4}\]
Correct Answer: A
Solution :
The first line corresponds to \[n=3.\]Therefore, \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{4}-\frac{1}{{{3}^{2}}} \right)=R\left( \frac{1}{4}-\frac{1}{9} \right)=\frac{5R}{36}\]or \[{{\lambda }_{1}}=\frac{36}{5R}.\] For the second line \[n=4\] \[\therefore \] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{4}-\frac{1}{{{4}^{2}}} \right)\,\,\Rightarrow {{\lambda }_{2}}=\frac{16}{3R}\] \[\therefore \]\[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16}{3R}.\frac{5R}{36}=\frac{20}{27}\]or \[{{\lambda }_{2}}=\frac{20}{27}\lambda \] Hence, the correction option is (a).
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