A) 122.4 eV
B) 30.6 eV
C) 13.6 eV
D) 3.4 eV
Correct Answer: B
Solution :
\[{{E}_{n}}=\frac{13.6}{{{n}^{2}}}\times {{Z}^{2}}.\]For first excited state n = 2 and for \[L{{i}^{++}},\ z=3\]Þ \[E=\frac{13.6}{4}\times 9=30.6\ eV\]You need to login to perform this action.
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