Consider $Z=19$potassium. As a rough approximation assume that each of its$n=1$electron s orbits 19 pro. tons and half an electron-that is, on average, half its fellow$\mathit{n}\mathbf{=}\mathbf{1}$electron. Assume that each of its$\mathit{n}\mathbf{=}\mathbf{2}$electrons orbits 19 protons, two Is electrons. and half of the seven other$\mathit{n}\mathbf{=}\mathbf{2}$electrons. Continue the process, assuming that electrons at eachorbit a correspondingly reduced positive charge. (At each, an electron also orbits some of the electron clouds of higher. but we ignore this in our rough approximation.)

(a) Calculate in terms of$a_0$the orbit radii of hydrogenlike atoms of these effective Z,

(b) The radius of potassium is often quoted at around$\mathbf{0}\mathbf{.}\mathbf{22}\mathbf{}\mathbf{\text{nm}}$. In view of this, are your$\mathit{n}\mathbf{=}\mathbf{1}$through$\mathit{n}\mathbf{=}\mathbf{3}$radii reasonable?

(c) About how many more protons would have to be "unscreened" to the$\mathit{n}\mathbf{=}\mathbf{4}$electron to agree with the quoted radius of potassium? Considering the shape of its orbit, should potassium's$\mathit{n}\mathbf{=}\mathbf{4}$electron orbit entirely outside all the lower-electrons?

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

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

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

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 n 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 n 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 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 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 then=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 n being 4:

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

(b)

Calculate the radii

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

Similarly, calculate further:

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

(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