A) -52,4eV
B) -27.2eV
C) -68eV
D) -3.4 eV
Correct Answer: D
Solution :
The energy of \[{{n}^{th}}\]excited state is given by \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}\] For first excited state, n = 2 \[\therefore \] \[{{E}_{n}}=-\frac{13.6}{4}=-3.4eV\]You need to login to perform this action.
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