A) \[-30.6eV\]
B) \[-13.6eV\]
C) \[-24.6eV\]
D) \[-28.6eV\]
Correct Answer: A
Solution :
Energy of election \[=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}eV\]for \[Li,\,\,Z=3,\,\,n=2\] \[\therefore \] energy of last \[{{e}^{-}}=-\frac{13.6\times {{3}^{2}}}{{{2}^{2}}}\]\[=-30.6eV\]
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