A) 13.1 eV
B) 3.4 eV
C) 1.51 eV
D) 1.9 eV
Correct Answer: B
Solution :
For first excited state \[n=2\] \[\therefore \] \[{{E}_{2}}=-\frac{13.6}{{{(2)}^{2}}}\] \[=-3.4eV\]
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