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 \[{{E}_{2}}=-{{3}^{2}}\times \frac{13.6}{4}\] \[=-30.6eV\]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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