A) -122.4 eV
B) 30.6 eV
C) -30.6 eV
D) 13.6 eV
Correct Answer: C
Solution :
\[{{E}_{n}}=\frac{{{E}_{1}}}{{{n}^{2}}}\times {{Z}^{2}}\] For \[\text{L}{{\text{i}}^{\text{2+}}}\text{,}\]the excited state, \[n=2\]and \[Z=3\] \[\therefore \] \[{{E}_{n}}=\frac{-13.6}{{{2}^{2}}}\times {{(3)}^{2}}\] \[=\frac{-13.6}{4}\times 9\] \[=-30.6\,eV\]
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