A) 1.51 eV
B) 3.4 eV
C) 13.6 eV
D) 12.1 eV
Correct Answer: A
Solution :
Second excited state corresponds to\[n=3.\] \[\therefore \] \[E=-\frac{13.6}{{{n}^{2}}}\] \[\Rightarrow \] \[E=-\frac{13.6}{{{(3)}^{2}}}eV\] \[\Rightarrow \] \[E=-\frac{13.6}{9}eV=-1.51eV\]You need to login to perform this action.
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