Respuesta :
The general expression for the ionization energy of any one-electron species is [tex]E_{I E}=1.312796 \cdot 10^{6} \mathrm{~J} / \mathrm{mol} \cdot Z^{2}\end{gathered}$[/tex]
What is ionization energy?
It is the energy needed to remove one electron from a neutral atom, which results in the formation of an ion.
The measurement is based on an isolated atom in its gaseous phase and is often expressed in kJ/mol.
a) Atom is ionized when the electron is completely removed from its electron cloud (so it being moved from first, [tex]$n_{\text {initial }}=1$[/tex] to the infinity's shell and [tex]$n_{\text {final }}=\infty$[/tex] ), and now the equation can be written as
[tex]$\Delta E=-2.18 \cdot 10^{-18} J\left(\frac{1}{n_{\text {final }}^{2}}-\frac{1}{n_{\text {initial }}^{2}}\right)\\=-2.18 \cdot 10^{-18} J\left(\frac{1}{\infty^{2}}-\frac{1}{1}\right)$[/tex]
So the ionization energy is affected by the charge of the nucleus and the general formula can be represented as:
[tex]$\Delta E_{I E}=-2.18 \cdot 10^{-18} J\left(\frac{1}{\infty^{2}}-\frac{1}{1}\right) \cdot Z^{2} \cdot\left(6.022 \cdot 10^{23}\right molecules/mol)$[/tex] and as [tex]$\frac{1}{\infty}=0$[/tex]
[tex]$\begin{gathered}\Delta E_{I E}=-2.18 \cdot 10^{-18} \mathrm{~J} \cdot Z^{2} \cdot\left(6.022 \cdot 10^{23} \text { molecules } / \mathrm{mol}\right) \\\Delta E_{I E}=1.312796 \cdot 10^{6} \mathrm{~J} / \mathrm{mol} \cdot Z^{2}\end{gathered}$[/tex]
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