91影视

The probability of finding an electron in some finite volume element around a point at a distance of r from the nucleus is given by the radial wave function R(r), which is simply the value of the wave function at some radius r.

$R_{30},R_{31},and{R}_{32},$We need to work out the radial wave functions and we will use:

$R_{nl}\left(r\right)=\frac{1}{r}{u}_{nl}\left(r\right)$

Where,

role="math" localid="1658143292950" $u_{nl}\left(p\right)={蚁}^{l+1}{e}^{-p}{v}_{nl}\left(p\right)$

Thus,

$$\begin{gathered} R_{nl}\left(r\right)=\frac{1}{r}{u}_{nl}\left(r\right) \\ {R}_{nl}\left(r\right)=\frac{1}{r}{p}^{l+1}{e}^{-p}{v}_{nl}\left(p\right) \end{gathered}$$ (1)

Where,

$$\begin{gathered} v_{nl}\left(p\right)=\underset{j=0}{\overset{鈭�}{鈭�}}{c}_j{p}^j \\ {c}_{j+1}=\frac{2(j+l+1)-2n}{(j+1)(j+2)(l+1))}{c}_j \end{gathered}$$ (2)

For$R_{30}$the values are$n=3$and$l=0$ , we have:

$v_{30}\left(p\right)=c_0+c_{1}p+c_2{p}^2$

where the constant can be determined using the second equation in (2) as:

$$\begin{gathered} c_1=\frac{2(1-3)}{\left(1\right)\left(2\right)}{c}_0 \\ =-2c_0 \\ {c}_2=\frac{2(2-3)}{\left(2\right)\left(3\right)}{c}_1 \\ =-\frac{1}{3}{c}_1 \\ =\frac{2}{3}{c}_0 \\ {c}_3=\frac{2(3-3)}{\left(3\right)\left(4\right)}{c}_2 \\ =0 \end{gathered}$$

Substitute into (1) to get the expression as:

$$\begin{gathered} R_{30}=\frac{1}{r}\frac{r}{3a}{e}^{-r/3a}\left[c_0-2c_0\frac{r}{3a}+\frac{2}{3}{c}_0{\left(\frac{r}{3a}\right)}^2\right] \\ {R}_{30}=\left(\frac{c_0}{3a}\right)\left[1-\frac{2}{3}\left(\frac{r}{a}\right)+\frac{2}{27}{\left(\frac{r}{a}\right)}^2\right]e^{-r/3a} \end{gathered}$$

For $R_{31}$the values are $n=3$and localid="1658144737944" $l=1$, we have:

$V_{31}\left(p\right)=c_0+c_{1}p+c_2{p}^2$

where the constant can be determined using the second equation in (2) as:

$$\begin{gathered} c_1=\frac{2(2-3)}{\left(1\right)\left(4\right)}{c}_0 \\ =-\frac{1}{2}{c}_0 \\ {c}_2=\frac{2(2-3)}{\left(2\right)\left(5\right)}{c}_1 \\ =0 \end{gathered}$$

Substitute into (1) to get:

$$\begin{gathered} R_{31}=\frac{1}{r}{\left(\frac{r}{3a}\right)}^2{e}^{-r/3a}\left(c_0-\frac{1}{2}{c}_0\frac{r}{3a}\right) \\ =\left(\frac{c_0}{9a^2}\right)r\left[1-\frac{1}{6}\left(\frac{r}{a}\right)\right]e^{-r/3a} \\ {R}_{31}=\left(\frac{c_0}{9a^2}\right)r\left[1-\frac{1}{6}\left(\frac{r}{a}\right)\right]e^{-r/3a} \end{gathered}$$

For $R_{32}$the values are $n=3$and$l=2$ , we have:

$v_{32}\left(p\right)=c_0+c_{1}p+c_2{p}^2$

where the constant can be determined using the second equation in (2) as:

$$\begin{gathered} c_1=\frac{2(3-3)}{\left(1\right)\left(6\right)}{c}_0 \\ =0 \\ {c}_2=\frac{2(3-3)}{\left(2\right)\left(5\right)}{c}_1 \\ =0 \end{gathered}$$

Substitute into (1) to get:

$$\begin{gathered} R_{32}=\frac{1}{r}{\left(\frac{r}{3a}\right)}^3{e}^{-r/3a}\left(c_0\right) \\ =\left(\frac{c_0}{27a^3}\right)r^2{e}^{-r/3a} \\ {R}_{32}=\left(\frac{c_0}{27a^3}\right)r^2{e}^{-r/3a} \end{gathered}$$

Thus the Radial wave functions are:

role="math" localid="1658145305610" $$\begin{gathered} R_{30}=\left(\frac{c_0}{3a}\right)\left[1-\frac{2}{3}\left(\frac{r}{a}\right)+\frac{2}{27}{\left(\frac{r}{a}\right)}^2\right]e^{-r/3a} \\ {R}_{31}=\left(\frac{c_0}{9a^2}\right)r\left[1-\frac{1}{6}\left(\frac{r}{a}\right)\right]e^{-r/3a} \\ {R}_{32}=\left(\frac{c_0}{27a^3}\right)r^2{e}^{-r/3a} \end{gathered}$$