The binding energy of the electron in a hydro | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \, eV$.For $L

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$30.6$

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \, eV$.For $Li^{++}$,the atomic number $Z = 3$.The first excited state corresponds to $n = 2$.The binding energy (energy required to remove the electron) is the magnitude of the energy in that state: $E = 13.6 	imes \frac{Z^2}{n^2} \, eV$.Substituting the values: $E = 13.6 	imes \frac{3^2}{2^2} = 13.6 	imes \frac{9}{4} = 13.6 	imes 2.25 = 30.6 \, eV$.

The binding energy of the electron in a hydrogen atom is $13.6\, eV$. The energy required to remove the electron from the first excited state of $Li^{++}$ is ......... $eV$.

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