A) 3.4 eV
B) 1.5 eV
C) 10.2 eV
D) 13.6 eV
Correct Answer: A
Solution :
The energy (E) needed to ionise in its \[{{n}^{th}}\]excited state is \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}\] Given, n = 2 \[\therefore \] \[{{E}_{n}}=-\frac{13.6}{{{2}^{2}}}=-3.4\,eV\]You need to login to perform this action.
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