A) 27.2 eV
B) 30.6 eV
C) 30.6 eV
D) 27.2 eV
Correct Answer: C
Solution :
For \[L{{i}^{2+}}\]ion \[E=-13.6\times \frac{{{Z}^{2}}}{{{n}^{2}}}\] \[=-13.6\times \frac{{{(3)}^{2}}}{{{(2)}^{2}}}\] \[=\frac{-13.6\times 9}{4}=-30.6eV\]You need to login to perform this action.
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