The energy required to remove an electron fro | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like ion is given by the formula: $E_n = -\frac{13.6 	imes Z^2}{n^2} \ eV$.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 ion is given by the formula: $E_n = -\frac{13.6 	imes Z^2}{n^2} \ eV$.For $Li^{++}$ (Lithium ion),the atomic number $Z = 3$.The first excited state corresponds to $n = 2$.Substituting these values into the formula:$E_2 = -\frac{13.6 	imes 3^2}{2^2} \ eV$$E_2 = -\frac{13.6 	imes 9}{4} \ eV$$E_2 = -30.6 \ eV$.The energy required to remove the electron (ionization energy from this state) is the energy needed to bring the electron to $n = \infty$ (where $E = 0$).Therefore,the required energy is $E_{\infty} - E_2 = 0 - (-30.6 \ eV) = 30.6 \ eV$.

The energy required to remove an electron from the first excited state of $Li^{++}$ is ......... $eV$.

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