Question

Find the second excitation energy of ${\mathrm{Li}}^{2+}$

• A. $108.8\mathrm{eV}$
• B. $81.6\mathrm{eV}$
• C. $13.6\mathrm{eV}$
• D. $95.2\mathrm{eV}$

Answer 1

The correct option is A

$108.8\mathrm{eV}$

Answer (A)

Explanation for correct option:

Second excitation energy means the electron is jumping from the first shell to the third shell.

Therefore,

Second excitation energy, $\mathrm{E}={\mathrm{E}}_3-{\mathrm{E}}_1$

${\mathrm{E}}_3$ is energy in the third shell

${\mathrm{E}}_1$ is energy in the first shell

Second excitation energy,

$$\begin{gathered} \frac{{}^{-}13.6{\mathrm{Z}}^2}{{\mathrm{n}}_3^2}-\frac{{}^{-}13.6{\mathrm{Z}}^2}{{\mathrm{n}}_1^2} \\ E={}^{-}13.6\times \times Z^2\left[\frac{1}{n_3^2}-\frac{1}{n_1^2}\right] \\ E={}^{-}13.6\times 3^2\left[\frac{1}{3^2}-\frac{1}{1^1}\right] \\ E={}^{-}13.6\times 3^2\left[\frac{1}{9}-1\right] \\ E={}^{-}13.6\times 9\left[\frac{{}^{-}8}{9}\right] \\ E=13.6\times 8 \\ \mathit{E}\mathbf{=}\mathbf{108}\mathbf{.}\mathbf{8}\mathit{e}\mathit{V} \end{gathered}$$

Hence, option(A) is the correct option.