(a) Normalize${\mathit{R}}_{\mathbf{20}}$ (Equation 4.82), and construct the function${\mathit{\psi }}_{\mathbf{200}}$.

(b) Normalize${\mathit{R}}_{\mathbf{21}}$(Equation 4.83), and construct the function.

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.

First, we need to work out the radial wave functions $R_{20},and{R}_{21},$we will use:

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

Where,

$u_{nl}\left(p\right)=p^{l+1}{e}^{-p}{v}_{nl}\left(p\right)$

Thus,

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

Where,

$$\begin{gathered} V_{nl}\left(p\right)=\sum _{j=0}^{\infty }{c}_j{p}^j \\ {c}_{j+1}=\frac{2(j+l+1)-2n}{(j+1)(j+2)(l+1)}{c}_j \end{gathered}$$

For$R_{20}$ the values are$n=2$ and$l=0$ , we have:

$v_{20}\left(p\right)=c_0+c_{1}p$

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

$$\begin{gathered} c_1=\frac{2(1-2)}{\left(1\right)\left(2\right)}{c}_0 \\ =-c_0 \end{gathered}$$

Substitute into (1) to get :

$R_{20}=\frac{1}{2a}{e}^{-r/2a}{c}_0\left(1-\frac{r}{2a}\right)$

And for$R_{21}$ we have,

$$\begin{gathered} R_{21}\left(r\right)=\frac{1}{r}{u}_{21}\left(r\right) \\ =\frac{r}{{\left(2a\right)}^2}{e}^{\raisebox{1ex}{$-r$}\!\left/ \!\raisebox{-1ex}{$2a$}\right.}{c}_0 \end{gathered}$$

To$c_0$findnormalize the radial function in the equation as,

${\int }_0^{\infty }{r}^2{\left|R_{20}\left(r\right)\right|}^{2}dr={\int }_0^{\infty }{c}_0^2{r}^2{\left[\frac{1}{2a}\left(1-\frac{r}{2a}\right)\right]}^2{e}^{-r/a}dr$

Let$z=r/a$, so:

$$\begin{gathered} {\left(\frac{c_0}{2a}\right)}^2{a}^3{\int }_0^{\infty }{\left(1-\frac{z}{2}\right)}^2{e}^{-z}{z}^{2}dz=1 \\ \frac{c_0^{2}a}{4}{\int }_0^{\infty }\left(z^2-z^3+\frac{1}{4}{z}^4\right)e^{-z}dz=1 \end{gathered}$$

using the integral:

$$\begin{gathered} {\int }_0^{\infty }{x}^n{e}^{-x}={\rm T}(n+1) \\ =n! \end{gathered}$$

we get:

$$\begin{gathered} \frac{c_0^{2}a}{4}\left(2!-3!+\frac{4!}{4}\right)=1 \\ \frac{c_0^{2}a}{4}\left(2-6+\frac{24}{4}\right)=1 \\ \frac{a}{2}{c}_0^2=1 \\ {c}_0=\sqrt{\frac{2}{a}} \end{gathered}$$

the complete wave function is:

${\psi }_{200}=R_{20}\left(r\right)y_0^0(\varphi ,\theta )$

where (from table 4.3):

$y_0^0=-\frac{1}{\sqrt{4\mathrm{\pi }}}$

Then,

${\psi }_{200}=\frac{1}{\sqrt{2\mathrm{\pi a}}}\frac{1}{2a}\left(1-\frac{r}{2a}\right)e^{\raisebox{1ex}{$-r$}\!\left/ \!\raisebox{-1ex}{$2a$}\right.}$

To find$c_0$, we normalize the radial function in equation (4) following the same method in part (a) where$z=r/a$, so we get:

$$\begin{gathered} 1={\left(\frac{c_0}{4a^2}\right)}^2{a}^5{\int }_0^{\infty }{z}^4{e}^{-z}dz \\ 1=\frac{c_0^{2}a}{16}24 \\ 1=\frac{3}{2}a{c}_0^2 \\ {c}_0=\sqrt{\frac{2}{3a}} \end{gathered}$$

In this case there are 3 wave functions corresponding to$n=2,l=1$for which we need the spherical harmonics:

$$\begin{gathered} y_1^1=-{\left(\frac{3}{8\mathrm{\pi }}\right)}^{1/2}\sin \left(\theta \right)e^{i\varphi } \\ {y}_1^{-1}=-{\left(\frac{3}{8\mathrm{\pi }}\right)}^{1/2}\sin \left(\theta \right)e^{i\varphi } \\ {y}_1^0=-{\left(\frac{3}{8\mathrm{\pi }}\right)}^{1/2}\cos \left(\theta \right) \end{gathered}$$

The corresponding wave functions are:

$\left\{\begin{array}{l}{\psi }_{211}=-\frac{1}{8\sqrt{\mathrm{\pi }}}\frac{r}{a^{5/2}}{e}^{-r/2a}\sin \left(\theta \right)e^{i\varphi }\\ {\psi }_{21-1}=-\frac{1}{8\sqrt{\mathrm{\pi }}}\frac{r}{a^{5/2}}{e}^{-r/2a}\sin \left(\theta \right)e^{i\varphi }\\ {\psi }_{210}=-\frac{1}{8\sqrt{\mathrm{\pi }}}\frac{r}{a^{5/2}}{e}^{-r/2a}\cos \left(\theta \right)\end{array}\right.$