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(A) The total energy of an electron in a hydrogen-like atom is given by:$E_n = - \left( \frac{me^4}{8 \epsilon_0^2 h^2} \right) \frac{Z^2}{n^2}$Multip

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$9 \,hcR$

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(A) The total energy of an electron in a hydrogen-like atom is given by:$E_n = - \left( \frac{me^4}{8 \epsilon_0^2 h^2} \right) \frac{Z^2}{n^2}$Multiplying and dividing by $hc$ in the expression,we get:$E_n = - \left( \frac{me^4}{8 \epsilon_0^2 h^3 c} \right) hc \frac{Z^2}{n^2} = - Rch \frac{Z^2}{n^2}$where $R$ is the Rydberg constant.The ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$:$E_{ion} = E_{\infty} - E_1 = 0 - (- Rch \frac{Z^2}{1^2}) = Rch Z^2$For $Li^{++}$ ion,the atomic number $Z = 3$.Therefore,$E_{ion} = Rch (3)^2 = 9 \,hcR$.

The ionization energy of a $Li^{++}$ ion is .......

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