A) \[30.6\] eV
B) \[13.6\] eV
C) \[3.4\] eV
D) \[122.4\] eV
Correct Answer: A
Solution :
Orbital energy, \[E=-{{Z}^{2}}\frac{13.6}{{{n}^{2}}}eV\] For first excited state, \[{{E}_{2}}=-{{3}^{2}}\times \frac{13.6}{4}\] = - 30.6 eV lonisation 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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