A) 1:3
B) 27 : 5
C) 5 : 27
D) 4:9
Correct Answer: C
Solution :
For Lyman series \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ?..(i) For Bahner series \[\frac{1}{{{\lambda }_{B}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ??.(ii) From Eqs. (i) and (ii) \[\frac{{{\lambda }_{L}}}{{{\lambda }_{B}}}=\frac{5}{27}\]
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