A) 1: 3
B) 27 : 5
C) 5 : 27
D) 4 : 9
Correct Answer: C
Solution :
\[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] For first line of Lymen series n1 = 1 and n2 = 2 For first line of Balmer series n2 = 2 and n2 = 3 So, \[\frac{{{\lambda }_{Lymen}}}{{{\lambda }_{Balmer}}}=\frac{5}{27}\]
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