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

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(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -K \frac{Z^2}{n^2}$,where $K = 2.18 	imes 10^{-18} \ J$. For $He^+$ $(Z=2)$,th

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

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(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -K \frac{Z^2}{n^2}$,where $K = 2.18 	imes 10^{-18} \ J$. For $He^+$ $(Z=2)$,the ionization energy is the energy required to remove the electron from $n=1$ to $n=\infty$. $IE = E_{\infty} - E_1 = 0 - (-K \frac{2^2}{1^2}) = 4K$. Given $4K = 19.6 	imes 10^{-18} \ J$,so $K = 4.9 	imes 10^{-18} \ J$. For $Li^{2+}$ $(Z=3)$,the energy of the first stationary state $(n=1)$ is $E_1 = -K \frac{Z^2}{n^2} = -4.9 	imes 10^{-18} 	imes \frac{3^2}{1^2} = -4.9 	imes 10^{-18} 	imes 9 = -4.41 	imes 10^{-17} \ J \ atom^{-1}$.

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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