In studying electron screening in multi electron atoms, you begin with the alkali metals. You look up experimental data and find the results given in the table.

The ionization energy is the minimum energy required to remove the least-bound electron from a ground-state atom. (a) The units kJ/mol given in the table are the minimum energy in kJ required to ionize 1 mol of atoms. Convert the given values for ionization energy to the energy in eV required to ionize one atom. (b) What is the value of the nuclear charge Zfor each element in the table? What is the nquantum number for the least-bound electron in the ground state? (c) Calculate Z_eff for this electron in each alkali-metal atom. (d) The ionization energies decrease as Zincreases. Does Z_eff increase or decrease as Zincreases? Why does Z_effhave this behaviour?

The conversion factor is as follows,

$$\begin{gathered} \mathrm{E}\left(\mathrm{eV}/\mathrm{atom}\right)=\mathrm{E}\left(\mathrm{kJ}/\mathrm{mol}\right)\left(\frac{1\mathrm{mol}}{6.02214\times {10}^{23}\mathrm{atoms}}\right)\left(\frac{1000\mathrm{J}}{1\mathrm{kJ}}\right)\left(\frac{1\mathrm{eV}}{1.602\times {10}^{-19}\mathrm{J}}\right) \\ =0.010364\mathrm{E}\left(\mathrm{kJ}/\mathrm{mol}\right) \\ \mathrm{Li}: \\ \mathrm{E}=\left(0.010364\right)\left(520.2\mathrm{kJ}/\mathrm{mol}\right) \\ =5.391\mathrm{eV} \\ \mathrm{Na}: \\ \mathrm{E}=\left(0.010364\right)\left(495.8\mathrm{kJ}/\mathrm{mol}\right) \\ =5.139\mathrm{eV} \\ \mathrm{K}: \\ \mathrm{E}=\left(0.010364\right)\left(418.8\mathrm{kJ}/\mathrm{mol}\right) \\ =4.341\mathrm{eV} \\ \mathrm{Rb}: \\ \mathrm{E}=\left(0.010364\right)\left(403.0\mathrm{kJ}/\mathrm{mol}\right) \\ =4.177\mathrm{eV} \\ \mathrm{Cs}: \\ \mathrm{E}=\left(0.010364\right)\left(357.7\mathrm{kJ}/\mathrm{mol}\right) \\ =3.894\mathrm{eV} \\ \mathrm{Fr}: \\ \mathrm{E}=\left(0.010364\right)\left(380\mathrm{kJ}/\mathrm{mol}\right) \\ =3.9\mathrm{eV} \end{gathered}$$

From reference to the periodic table,

$$\begin{gathered} \mathrm{Li}:\mathrm{Z}=3,\mathrm{n}=2 \\ \mathrm{Na}:\mathrm{Z}=11,\mathrm{n}=3 \\ \mathrm{K}:\mathrm{Z}=19,\mathrm{n}=4 \\ \mathrm{Rb}:\mathrm{Z}=37,\mathrm{n}=5 \\ \mathrm{Cs}:\mathrm{Z}=55,\mathrm{n}=6 \\ \mathrm{Fr}:\mathrm{Z}=87,\mathrm{n}=7 \end{gathered}$$

The expression of ionization energy is,

${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{{\mathrm{n}}^2}\left(13.6\mathrm{eV}\right)$

Substitute all the values calculated above,

$$\begin{gathered} \mathrm{Li}: \\ 5.391\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{2^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=1.26 \\ \mathrm{Na}: \\ 5.139\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{3^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=1.84 \\ \mathrm{K}: \\ 5.391\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{3^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=2.26 \\ \mathrm{Rb}: \\ 5.391\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{4^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=2.77 \\ \mathrm{Cs}: \\ 5.391\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{5^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=3.21 \\ \mathrm{Fr}: \\ 5.391\mathrm{eV}=\frac{{\mathrm{Z}}_{\mathrm{eff}}^2}{6^2}\left(13.6\mathrm{eV}\right) \\ {\mathrm{Z}}_{\mathrm{eff}}=3.8 \end{gathered}$$

The inference that can be drawn is that the Z_eff increases with an increase in Z. This is because the outer shell electrons has increasing probability density in the inner shells as Z increases thereby increasing the effective nuclear charge.