A) n=1 to n=2
B) n=2 to n=6
C) n=6 to n=2
D) n=2 to n=l
Correct Answer: D
Solution :
[d] Frequency is given by \[hv=-13.6\left( \frac{1}{n_{1}^{2}}+\frac{1}{n_{2}^{2}} \right)\] For transition from n=6 to n=2, \[{{v}_{1}}=\frac{-13.6}{h}\left( \frac{1}{{{6}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{2}{9}\times \left( \frac{13.6}{h} \right)\] For transition from n=2 to n=1, \[{{v}_{2}}=\frac{-13.6}{h}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{1}^{2}}} \right)=\frac{3}{4}\times \left( \frac{13.6}{h} \right).\] \[\therefore {{v}_{1}}>{{v}_{2}}\]
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