Question

Calculate the binding energy of an electron in the first excited state of $L{i}^{2+}$ ion.

• A. 35 eV
• B. 29.6 eV
• C. 30.6 eV
• D. 28.9 eV

Answer 1

The correct option is C 30.6 eV

Energy required to move an electron from any state to $n=\infty$ is called the binding energy of that state.

So, binding energy = $13.6\frac{Z^2}{n^2}$.

For, $L{i}^{2+}$, Z=3, n=2 (i.e 1st excited state)
Therefore, binding energy = $13.6\frac{3^2}{2^2}$ eV
binding energy = $13.6\frac{9}{4}$ eV
Hence binding energy = 30.6 eV