If 'X' is the ionization energy of hydrogen,t | vedclass

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(A) The ionization energy of hydrogen is given by $(IE)_{H} = 13.6 \ eV = X$.The energy of an electron in a hydrogen-like species is given by $E_n = -

Vedclass · Accepted Answer

$\frac{3X}{4}$

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(A) The ionization energy of hydrogen is given by $(IE)_{H} = 13.6 \ eV = X$.The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV = -X 	imes \frac{Z^2}{n^2}$.For $Li^{2+}$,the atomic number $Z = 3$.The $2^{nd}$ excited state corresponds to $n_1 = 3$ and the $5^{th}$ excited state corresponds to $n_2 = 6$.The energy required for excitation is $\Delta E = E_{n_2} - E_{n_1} = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Substituting the values: $\Delta E = X 	imes 3^2 	imes (\frac{1}{3^2} - \frac{1}{6^2}) = 9X 	imes (\frac{1}{9} - \frac{1}{36})$.$\Delta E = 9X 	imes (\frac{4-1}{36}) = 9X 	imes \frac{3}{36} = 9X 	imes \frac{1}{12} = \frac{3X}{4}$.

If '$X$' is the ionization energy of hydrogen,then the energy required for the excitation of a $Li^{2+}$ electron from the $2^{nd}$ excited state to the $5^{th}$ excited state is:

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