Question

(a) What is the expectation value of the distance from the proton of an electron in a 3p state? (b) How does this compare with the expectation value in the 3 d state, calculated in Example 7.7? Discuss any differences.

Short answer

a) The expectation value of the distance between proton and electron in 3p state is $3a_o,6a_o,12a_o.$

b) The expectation value of the distance between proton and electron in 3d state is $12a_o.$.

Step-by-step solution

 Given data

At 3p state, the radius function is given by

$R_{3,1}\left(r\right)=\frac{1}{{\left(3a_o\right)}^{3/2}}\frac{4\sqrt{2}}{9a_0}\left(1-\frac{r}{6a_o}\right)r^{-r\left(3a_o\right)}$

Concept

The probability of finding an electron in a given region of an atom is given by the square of the wave function of that electron integrated over the given region. The orbital in an atom is the region where the probability of finding an electron is 90% .

The probability density is

$$\begin{gathered} p\left(r\right)\propto r^2{R}_{3,1}^2\left(r\right) \\ =r^2{\left(r\left(r-\frac{r}{6a_o}\right)e^{-r/\left(3a_o\right)}\right)}^2 \\ =r^4{e}^{-2r/\left(3a_0\right)}{\left(1-\frac{r}{6a_o}\right)}^2 \\ =\left(r^4-\frac{r^5}{3a_o}+\frac{r^6}{36a_0^2}\right)e^{-2r/\left(3a_o\right)} \end{gathered}$$

Calculation

To calculate maximum value, derivate probability density function and equate to 0

$\frac{dp\left(r\right)}{dr}=0$

Using the value of p(r) calculated above

role="math" localid="1659183593664" $$\begin{gathered} \frac{d}{dr}\left(r^4-\frac{r^5}{3a_o}+\frac{r^6}{36a_0^2}\right)e^{-2r/\left(3a_o\right)}=0 \\ \left(4r^3-\frac{5r^4}{3a_o}+\frac{r^5}{6a_0^2}\right)e^{-2r/\left(3a_o\right)}+\left(r^4-\frac{r^5}{3a_o}+\frac{r^6}{36a_0^2}\right)\frac{-2}{3a_o}{e}^{-2r/\left(3a_o\right)}=0 \\ {e}^{-2r/\left(3a_o\right)}\left(\left(4r^3-\frac{5r^4}{3a_o}+\frac{r^5}{6a_0^2}\right)+\left(r^4-\frac{r^5}{3a_o}+\frac{r^6}{36a_0^2}\right)\frac{-2}{3a_o}\right)=0 \\ {r}^3{e}^{-2r/\left(3a_o\right)}\left(4-\frac{7r}{3a_o}+\frac{7r^2}{18a_0^2}-\frac{r^3}{54a_0^3}\right)=0 \end{gathered}$$

The above equation holds for$r=0andr=\infty$ , but they are points of minima. For maxima the term in parentheses must be zero.

$4-\frac{7r}{3a_o}+\frac{7r^2}{18a_0^2}-\frac{r^3}{54a_0^3}=0$

The expectation values are$3a_o,6a_o,12a_o$

b)

The electron is in a 3d state. Therefore, the most probable location is at $r=12a_o$. This value is more than the most probable radius for a 3d electron.

The graph for the 3p electron is more spread and has more radial energy but the 3d electron has more rotational energy.

The expectation value in 3 d state is$12a_o$ .