A) 7/5
B) 27/20
C) 27/5
D) 20/7
Correct Answer: C
Solution :
We know, \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] Where R is the Rydberg constant For case\[(I)\,{{n}_{1}}=3\]and \[{{n}_{2}}=2\] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{2}^{2}}} \right)=-\frac{5R}{36}\] \[\Rightarrow \]\[{{\lambda }_{1}}=-\frac{36}{5R}\] ?(i) For case (II) \[{{n}_{1}}=2\]and \[{{n}_{2}}=1\] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{1}^{2}}} \right)=-\frac{3R}{4}\] \[\Rightarrow \] \[{{\lambda }_{2}}=-\frac{4}{3R}\] From (i) and (ii), \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{-\frac{36}{5R}}{-\frac{4}{3R}}=\frac{27}{5}\]
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