Explicitly verify that the simple function $R\left(r\right)=A{e}^{br}$can be made to satisfy radial equation (7-31), and in so doing, demonstrate what its angular momentum and energy must be.

$R\left(r\right)=A^{e^{-br}}$

Here, A and b constants.

The property of any rotating body, which is described as the product of the moment of inertia and the angular velocity of rotation of the body, is known as angular momentum. It is analogous to the linear momentum for rotatory motion.

The radial equation is,

$-\frac{h^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{h^{2}I\left(I+1\right)}{2m{r}^2}R\left(r\right)-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}R\left(r\right)=ER\left(r\right)......\left(1\right)$(1)

Where,

m is the mass

e is the magnitude of the charge

I is the angular momentum

${\epsilon }_0$is the permittivity of free space.

Substitute $A{e}^{-br}$for $R\left(r\right)$in equation (1).

$-\frac{h^2}{2m}\frac{1}{r^2}\left(r^2\frac{d}{dr}\right)A{e}^{-br}+\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}A{e}^{br}=EA{e}^{br}$

Now, solve the localid="1659322395860" $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)A{e}^{-br}$term as follows:

localid="1659322573731" $$\begin{gathered} \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)A{e}^{-br}=\frac{1}{r^2}\frac{d}{dr}\left(r^2\left(-b\right)\right)A{e}^{-br} \\ =\frac{-b}{r^2}\left(2r\right)+\frac{-b}{r^2}\left(r^2\right)\left(-bA{e}^{-br}\right) \\ =\frac{-b}{r^2}\left(A{e}^{-br}\right)\left(2r\right)+b^{2}A{e}^{-br} \\ =-bA{e}^{-br}\left(\frac{1}{r}\left(2-br\right)\right) \end{gathered}$$

Substitute $-bA{e}^{-br}\left(\frac{1}{r}\left(2-br\right)\right)$for $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)A{e}^{-br}$in the equation (1).

$$\begin{gathered} -\frac{h^2}{2m}\left(-bA{e}^{-br}\left(\frac{1}{r}\left(2-br\right)\right)\right)+\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}-\frac{e^2}{r}A{e}^{br}=EA{e}^{br} \\ \left(\begin{array}{c}-\frac{h^2\left(-bA{e}^{-br}\right)}{2m}\left(\frac{2}{r}\right)\left(-\frac{h^2\left(-bA{e}^{-br}\right)}{2m}\right)\left(-b\right)+\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}A{e}^{-br}\\ =EA{e}^{br}\end{array}\right) \\ -\frac{h^{2}b\left(A{e}^{-br}\right)}{mr}-\frac{h^2\left(A{e}^{-br}\right)b^2}{2m}+\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}A{e}^{-br}=EA{e}^{-br}.............\left(2\right) \end{gathered}$$

The last three terms in the above equation are proportional to the terms in the radial equation given by equation (1).

Comparing the coefficients of the terms in the above equation with equation (1), we get-

$\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}=\frac{h^{2}I(I+1)}{2m{r}^2}A{e}^{-br}$

On solving we get,

I =0

Therefore, the angular momentum is zero.

The potential energy term in equation (2) is equal to $-\frac{h^2\left(A{e}^{-br}\right)b}{2mr}$

The potential energy term in equation (1) is equal to $-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}\left(A{e}^{-br}\right)$

Thus equate the above two expressions and solve as follows:

$$\begin{gathered} \frac{h^2\left(A{e}^{-br}\right)b}{mr}=-\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}\left(A{e}^{-br}\right) \\ \frac{h^{2}b}{m}=\frac{e^2}{4\pi {\epsilon }_0} \\ b=\frac{m{e}^2}{4\pi {\epsilon }_0{h}^2} \end{gathered}$$

Now, equate the potential energy term to total energy, as the kinetic term is zero.

$$\begin{gathered} -\frac{2h^2\left(A{e}^{-br}\right)b}{mr}=EA{e}^{br} \\ -\frac{h^2\left(A{e}^{-br}\right)b}{mr}=EA{e}^{br} \\ E=-\frac{h^2{b}^2}{2m} \end{gathered}$$

Substitute $\frac{m{e}^2}{4\pi {\epsilon }_0{h}^2}$ for b.

role="math" localid="1659324115735" $$\begin{gathered} E=h^{2}r{\left(\frac{m{e}^2}{4\pi {\epsilon }_0{h}^2}\right)}^2 \\ =-\frac{m{e}^4}{8\pi r_0{h}^2} \end{gathered}$$

Therefore, the angular momentum is zero and the energy is $-\frac{m{e}^4}{8\pi r_0{h}^2}$