The ionization energy of He^+ is 19.6 times 1 | vedclass

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Question

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}

Vedclass · Accepted Answer

$-4.41 	imes 10^{-17} \ J \ atom^{-1}$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 	imes 10^{-18} 	imes \frac{Z^2}{n^2} \ J \ atom^{-1}$.For $He^+$ $(Z = 2)$,the ionization energy is the energy required to remove the electron from $n = 1$ to $n = \infty$,which is equal to $-E_1$. Given $IE = 19.6 	imes 10^{-18} \ J \ atom^{-1}$,so $E_1(He^+) = -19.6 	imes 10^{-18} \ J \ atom^{-1}$.Using the formula $E_n = -R_H 	imes \frac{Z^2}{n^2}$,we can find the constant $R_H$ for this specific unit system: $19.6 	imes 10^{-18} = R_H 	imes \frac{2^2}{1^2} \implies R_H = 4.9 	imes 10^{-18} \ J \ atom^{-1}$.Now,for $Li^{2+}$ $(Z = 3)$,the energy of the first stationary state $(n = 1)$ is: $E_1(Li^{2+}) = -R_H 	imes \frac{Z^2}{n^2} = -4.9 	imes 10^{-18} 	imes \frac{3^2}{1^2} = -4.9 	imes 10^{-18} 	imes 9 = -44.1 	imes 10^{-18} \ J \ atom^{-1} = -4.41 	imes 10^{-17} \ J \ atom^{-1}$.

The ionization energy of $He^+$ is $19.6 \times 10^{-18} \ J \ atom^{-1}$. What is the energy of the first stationary state $(n = 1)$ of $Li^{2+}$?

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