If the binding energy of the electron in a hy | 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 $E_n = -\frac{13.6 \ Z^2}{n^2} \ eV$.For $Li^{++}$

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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 $E_n = -\frac{13.6 \ Z^2}{n^2} \ eV$.For $Li^{++}$ (Lithium ion),the atomic number $Z = 3$.The first excited state corresponds to $n = 2$.Substituting these values into the formula,the energy of the electron in the first excited state is $E_2 = -\frac{13.6 	imes 3^2}{2^2} = -\frac{13.6 	imes 9}{4} = -30.6 \ eV$.The binding energy (energy required to remove the electron) is the magnitude of this energy,which is $30.6 \ eV$.

If 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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