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(A) The energy of an electron in the $n$-th orbit of a hydrogen-like ion is given by $E_n = -13.6 \frac{Z^2}{n^2} \, eV$.The first excitation state co

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$54.4$

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(A) The energy of an electron in the $n$-th orbit of a hydrogen-like ion is given by $E_n = -13.6 \frac{Z^2}{n^2} \, eV$.The first excitation state corresponds to $n=2$,and the ground state corresponds to $n=1$.The excitation energy is $\Delta E = E_2 - E_1 = -13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{1^2} ight) = -13.6 Z^2 \left( \frac{1}{4} - 1 ight) = 13.6 Z^2 	imes \frac{3}{4}$.Given $\Delta E = 40.8 \, eV$,we have $40.8 = 13.6 	imes Z^2 	imes 0.75$.$40.8 = 10.2 	imes Z^2 \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.The energy required to remove the electron from the ground state is the ionization energy,$E_{ion} = |E_1| = 13.6 	imes Z^2 \, eV$.$E_{ion} = 13.6 	imes (2)^2 = 13.6 	imes 4 = 54.4 \, eV$.

Excitation energy of a hydrogen-like ion in its first excitation state is $40.8 \, eV$. Energy needed to remove the electron from the ion in the ground state is ........ $eV$.

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