A) 7:5
B) 27:20
C) 27:5
D) 20:7
Correct Answer: D
Solution :
[d] Third excited state n = 4, second excited state n= 3 \[n=4\text{ }\,\,to\,\text{ }n=3\] \[\frac{1}{{{\lambda }_{1}}}=R\left[ \frac{1}{{{(3)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]\] \[\frac{1}{{{\lambda }_{1}}}=\frac{7R}{144}\] N=3 to n=2 \[\frac{1}{{{\lambda }_{2}}}=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\] \[\frac{1}{{{\lambda }_{2}}}=\frac{5R}{36}\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{20}{7}\]
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