A) 27 : 5
B) 5 : 27
C) 4 : 1
D) 1 : 4
Correct Answer: A
Solution :
For Lyman series \[{{\nu }_{\text{Lymen}}}=\frac{c}{{{\lambda }_{\max }}}=Rc\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]=\frac{3RC}{4}\] For Balmer series \[{{\nu }_{\text{Balmer}}}=\frac{c}{{{\lambda }_{\max }}}=Rc\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]=\frac{5RC}{36}\] \ \[\frac{{{\nu }_{\text{Lymen}}}}{{{\nu }_{\text{Balmer}}}}=\frac{27}{5}\]
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