If the ionization potential of Li^{2+} is 122 | vedclass

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(C) The ionization energy $(I.E.)$ for a hydrogen-like species is given by the formula: $I.E. = 13.6 	imes Z^{2} 	imes (\frac{1}{n_1^{2}} - \frac{1}

Vedclass · Accepted Answer

$489.6$

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(C) The ionization energy $(I.E.)$ for a hydrogen-like species is given by the formula: $I.E. = 13.6 	imes Z^{2} 	imes (\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}) \ eV$. For ionization from ground state $(n=1)$ to infinity $(n=\infty)$,$I.E. = 13.6 	imes Z^{2} \ eV$. For $Li^{2+}$,$Z=3$,so $I.E. = 13.6 	imes 3^{2} = 13.6 	imes 9 = 122.4 \ eV$. For the $5^{th}$ ionization potential of Carbon $(Z=6)$,we are removing the $5^{th}$ electron from the $C^{4+}$ ion. The electronic configuration of $C^{4+}$ is $1s^{2}$. The $5^{th}$ electron is removed from the $n=1$ shell. $I.E._{5} = 13.6 	imes Z^{2} 	imes \frac{1}{n^{2}} = 13.6 	imes 6^{2} 	imes \frac{1}{1^{2}} = 13.6 	imes 36 = 489.6 \ eV$.

If the ionization potential of $Li^{2+}$ is $122.4 \ eV$,what is the $5^{th}$ ionization potential of carbon in $eV$?

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