Question

The binding energy of the second excited state of $L{i}^{2+}$ ion is:

• A. 13.60 eV
• B. 27.20 eV
• C. 24.17 eV
• D. 18.20 eV

Answer 1

The correct option is A 13.60 eV

Since we are calculating the binding energy of the second excited state of $L{i}^{2+}$ ion, we need to calculate, $\Delta E_{3to\infty }$.

$\Delta E_{3to\infty }=E_{\infty }-E_3$
$=0-(-13.6\frac{3^2}{3^2})eV$
$=13.6eV$