What may be the first ionisation energy of Na | vedclass

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(D) According to the Bohr model,the energy of an electron in an orbit is given by $E_n = -13.6 	imes \frac{Z_{eff}^2}{n^2} \, eV$. For $Na$ $(Z=11)$,

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$5.1$

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(D) According to the Bohr model,the energy of an electron in an orbit is given by $E_n = -13.6 	imes \frac{Z_{eff}^2}{n^2} \, eV$. For $Na$ $(Z=11)$,the valence electron is in the $3s$ orbital,so $n=3$. The ionisation energy $(I.E.)$ is the energy required to remove the electron from $n=3$ to $n=\infty$,which is equal to the magnitude of the energy of the $n=3$ state. $I.E. = 13.6 	imes \frac{Z_{eff}^2}{n^2} \, eV$. Given $Z_{eff} = 1.84$ and $n = 3$: $I.E. = 13.6 	imes \frac{(1.84)^2}{3^2} = 13.6 	imes \frac{3.3856}{9} \approx 5.11 \, eV$.

What may be the first ionisation energy of $Na$ if it has an effective nuclear charge of $1.84$,assuming the Bohr model of the $H$-atom is effective on it? (in $eV$)

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