A) 1/ 4
B) 4/9
C) 9/ 4
D) 4
Correct Answer: C
Solution :
First excited state i.e. second orbit (n = 2) Second excited state i.e. third orbit (n = 3) \[\because \ E=-\frac{13.6}{{{n}^{2}}}\] Þ \[\frac{{{E}_{2}}}{{{E}_{3}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\]You need to login to perform this action.
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