If the ionization energy of He^{+} is 8.68 ti | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$ or $

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$-8.68 	imes 10^{-18} \ J$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$ or $E_n = -K 	imes \frac{Z^2}{n^2} \ J$. For $He^{+}$ $(Z=2)$,the ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$. $IE = E_{\infty} - E_1 = 0 - (-K 	imes \frac{2^2}{1^2}) = 4K = 8.68 	imes 10^{-18} \ J$. Therefore,$K = \frac{8.68 	imes 10^{-18}}{4} = 2.17 	imes 10^{-18} \ J$. Now,for $Be^{3+}$ $(Z=4)$ in the second orbit $(n=2)$: $E_2 = -K 	imes \frac{Z^2}{n^2} = -(2.17 	imes 10^{-18}) 	imes \frac{4^2}{2^2} = -(2.17 	imes 10^{-18}) 	imes \frac{16}{4} = -(2.17 	imes 10^{-18}) 	imes 4 = -8.68 	imes 10^{-18} \ J$.

If the ionization energy of $He^{+}$ is $8.68 \times 10^{-18} \ J$,then the energy of $Be^{3+}$ ion in the second orbit is:- ($Z$ of $Be = 4$)

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