91Ó°ÊÓ

$n=1,2,3,4$For the electron in potassium.

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

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

(a)

For the $n=1$electrons, if they orbit the 19 protons in the potassium nucleus and half of the other $n=1$electron, the effective Z that the electron would see would be 18.5, since 19-0.5=18.5. That can then be used in the radii equation, along with being 1:

$$\begin{gathered} r_1=\frac{{\left(1\right)}^2{a}_0}{(18.5)} \\ {r}_1=0.054a_0 \end{gathered}$$

For the $n=2$electrons, if they orbit the 19 protons in the potassium nucleus, both 1 s electrons, and half of the seven others$n=2$ electrons, the effective Z that the electron would see would be 13.5 (since 19-2 -3.5=18.5). that can then be used in the radii equation, along with being 2:

$$\begin{gathered} r_2=\frac{{\left(2\right)}^2{a}_0}{(13.5)} \\ {r}_2=0.296a_0 \end{gathered}$$

For the$n=3$electrons, if they orbit the 19 protons in the potassium nucleus, both 1s electrons, all eight of the n=2 electrons, and half of the seven other electrons, the effective Z that the electron would see would be 5.5 (since 19-2-8-3.5=5.5 ). that can then be used in the radii equation, along with n being 3:

$$\begin{gathered} r_3=\frac{{\left(3\right)}^2{a}_0}{(5.5)} \\ {r}_3=1.636a_0 \end{gathered}$$

For the $â–\mu n=4$electron, if it orbits the 19 protons in the potassium nucleus, both 1s electrons, all eight of the n=2 electrons, and all eight of the n=3 electrons, the effective that the electron would see would be 1 (since 19-2-8-8=1). That can then be used in the radii equation, along with being 4:

$$\begin{gathered} r_4=\frac{{\left(4\right)}^2{a}_0}{I} \\ {r}_4=16a_0 \end{gathered}$$

(b)

Calculate the radii

$$\begin{gathered} r_1=\left(0.054a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_1=2.86×{10}^{-3}\text{nm} \\ {r}_2=\left(0.3a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_2=1.59×{10}^{-2}\text{nm} \end{gathered}$$

Similarly, calculate further:

$$\begin{gathered} r_3=\left(1.64a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_3=8.67×{10}^{-2}\text{nm} \end{gathered}$$

(c)

The atomic orbit radius equation can be used to estimate the effective Z that the valence electron sees, using $0.22\text{nm}$for the$r_n$for the n and $0.0529\text{nm}$for the$a_0$ after solving for Z :

$$\begin{gathered} Z=\frac{n^2{a}_0}{r_n} \\ Z=\frac{{\left(4\right)}^2(0.0529\text{nm})}{(0.22\text{nm})} \\ Z=3.85 \end{gathered}$$

So since in the valence electron orbits an effective Z of 1 in the original model, and it would need to orbit an effective $Z$ of $3.85$ based just on the experimental radius, there would need to be $3.8$ more protons be "unscreened"

The valence electron of potassium is in the s-shell, and consequently will have a roughly elliptical orbit. Therefore, it will spend part of its time near the nucleus, inside the orbits of the lower n electrons.