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Radius of cesium atom$0.26\text{nm}$

Expression for atomic radii $r_n$is given by,$r_n=\frac{n^2{a}_0}{\text{Z}}$

Where, Zrepresents number of protons,represents principal quantum number and$a_0$represents Bohr radius.

(a)

The equation is solved for Z, then use 1 for n, 0.26 nm for$r_n,0.0529nm$ for $a_0$, and 6 for n :

$$\begin{gathered} r_n=\frac{n^2{a}_0}{Z} \\ Z=\frac{n^2{a}_0}{r_n} \\ Z=\frac{{\left(6\right)}^2(0.0529\text{nm})}{(0.26\text{nm})} \\ Z=7.32 \end{gathered}$$

(b)

If all the inner electrons screened the nucleus, the valence electron would just see an effective of 1. But since the valence electron for cesium is in an s-shell, that means that its orbit is very elliptical, thus bypassing some of the inner electrons, and thus increasing the effective that the valence electron sees. So, an effective of 7.32 for the valence electron is plausible.

(c)

Even though the valence electron in cesium sees an effective of 7.32, whereas sodium's valence electron sees as of 2.7, the ionization energy of cesium is still lower than that of, sodium. This can be partially explained by the additional repulsive energy between the valence electron and all the other electrons in cesium as compared to sodium. With the repulsive energy between the valence electron and other electrons, it lessens the amount of energy that is needed to remove the outer electron from the atom, since the energy between the electrons adds to the valence electrons total energy.