A) \[\frac{5}{27}\lambda \]
B) \[\frac{36}{5}\lambda \]
C) \[\frac{27}{5}\lambda \]
D) \[\frac{5}{36}\lambda \]
Correct Answer: C
Solution :
\[\frac{1}{\lambda }={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] For first line in Lyman series, \[{{n}_{1}}=2\] and \[{{n}_{2}}=2\] \[\frac{1}{{{\lambda }_{L}}}={{R}_{H}}\left( \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right)\] \[=\frac{3{{R}_{H}}}{4}\] or \[{{R}_{H}}=\frac{4}{3\lambda }\] For first line in Balmer series, \[{{n}_{1}}=2\] and \[{{n}_{2}}=3\] \[\frac{1}{{{\lambda }_{B}}}={{R}_{H}}\left( \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right)=\frac{4}{3\lambda }\left( \frac{5}{36} \right)\] \[\frac{1}{{{\lambda }_{B}}}=\frac{5}{27\lambda }\] \[{{\lambda }_{B}}=\frac{27}{5}\lambda \]
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