A) 30.6 eV
B) 13.6 eV
C) 3.4 eV
D) 122.4 eV
Correct Answer: A
Solution :
\[E=-{{z}^{2}}\frac{13.6}{{{n}^{2}}}eV\] For first excited state, \[{{E}_{2}}=-{{3}^{2}}\times \frac{13.6}{4}\] Ionization energy for first excited state of \[L{{i}^{2+}}\] is\[30.6\,\,eV\].You need to login to perform this action.
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