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The Schrodinger equation for the one-electron flatrogen atom is

$\mathbf{-}\frac{\mathit{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\right)}\mathit{蠄}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{胃}\mathbf{)}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}}{\mathbf{鈭�}{\mathbf{胃}}^{\mathbf{2}}}\mathit{蠄}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{胃}\mathbf{)}\mathbf{+}\mathit{U}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{蠄}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{胃}\mathbf{)}\mathbf{=}\mathit{E}\mathit{蠄}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{胃}\mathbf{)}$

(a)

Separates the variables of the Schrodinger equation for the one-electron flatrogen atom.

$$\begin{gathered} -\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{鈭�}{鈭�\mathrm{r}}\left(\mathrm{r}\frac{鈭�}{鈭�\mathrm{r}}\right)蠄(\mathrm{r},\mathrm{胃})-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{{鈭�}^2}{鈭�{\mathrm{胃}}^2}蠄(\mathrm{r},\mathrm{胃})+U\left(\mathrm{r}\right)蠄(\mathrm{r},\mathrm{胃})=E蠄(\mathrm{r},\mathrm{胃}) \\ \frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{{鈭�}^2}{鈭�{\mathrm{胃}}^2}蠄(\mathrm{r},\mathrm{胃})=\left[-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{鈭�}{鈭�\mathrm{r}}\left(\mathrm{r}\frac{鈭�}{鈭�\mathrm{r}}\right)+U\left(r\right)-E\right]蠄(\mathrm{r},\mathrm{胃}) \\ \frac{{鈭�}^2}{鈭�{\mathrm{胃}}^2}蠄(\mathrm{r},\mathrm{胃})=\frac{2m{r}^2}{\sout{\mathrm{h}}}\left[-\frac{\sout{\mathrm{h}}}{2\mathrm{m}}\frac{1}{\mathrm{r}}\frac{鈭�}{鈭�\mathrm{r}}\left(\mathrm{r}\frac{鈭�}{鈭�\mathrm{r}}\right)+U\left(r\right)-E\right]蠄(\mathrm{r},\mathrm{胃}) \\ \frac{{鈭�}^2}{鈭�{\mathrm{胃}}^2}蠄(\mathrm{r},\mathrm{胃})=\left[-r\frac{鈭�}{鈭�\mathrm{r}}\left(r\frac{鈭�}{鈭�\mathrm{r}}\right)-\frac{2m{r}^2}{\sout{\mathrm{h}}}\left(E-U\left(r\right)\right)\right]蠄(\mathrm{r},\mathrm{胃}) \end{gathered}$$

Substitute $R\left(r\right)鈯�\left(胃\right)$for $蠄\left(r,胃\right)$into above equation.

$$\begin{gathered} \frac{{鈭�}^2}{鈭�{胃}^2}R\left(r\right)鈯�\left(胃\right)=\left[-r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)\right]R\left(r\right)鈯�\left(胃\right) \\ \frac{{鈭�}^2}{鈭�{胃}^2}R\left(r\right)鈯�\left(胃\right)=-r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)鈯�\left(胃\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)R\left(r\right)鈯�\left(胃\right) \\ R\left(r\right)\frac{{鈭�}^2}{鈭�{胃}^2}鈯�\left(胃\right)=-r鈯�\left(胃\right)\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)R\left(r\right)鈯�\left(胃\right) \end{gathered}$$

Divide the above equation by.

role="math" localid="1660042716100" $$\begin{gathered} \frac{R\left(r\right)}{R\left(r\right)鈯�\left(胃\right)}\frac{{鈭�}^2}{鈭�{胃}^2}鈯�\left(胃\right)=-\frac{r鈯�\left(胃\right)}{R\left(r\right)鈯�\left(胃\right)}\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)\frac{R\left(r\right)鈯�\left(胃\right)}{R\left(r\right)鈯�\left(胃\right)} \\ \frac{1}{鈯�\left(胃\right)}\frac{{鈭�}^2}{鈭�{胃}^2}鈯�\left(胃\right)=-\frac{r}{R\left(r\right)}\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)............\left(i\right) \end{gathered}$$

From the above equation, variables r and$胃$ is separated on either side of the equation alone, leading deduction that both the sides must equal to constant, which is C ,

$$\begin{gathered} \frac{1}{鈯�\left(胃\right)}\frac{d^2}{d{胃}^2}鈯�\left(胃\right)=C \\ \frac{d^2}{d{胃}^2}鈯�\left(胃\right)=C鈯�\left(胃\right).....\left(ii\right) \end{gathered}$$ 鈥︹��. (ii)

Hence the required equation$\frac{d^2}{d{胃}^2}鈯�\left(胃\right)=C鈯�\left(胃\right)$ is proved.

(b)

The solution of equation (i) depends on the sign of the constant C . If the sign of the constant C is positive then the solution of an equation must be non-periodic and real exponential. But if the sign of the constant is negative, then the solution of the equation must be periodic of the complex exponential. Therefore, the sign of C should be negative.

Hence the sign of C is negative.

(c)

It is proved that the sign of the constant is negative in part (b).

Rewrite the equation (1).

$\frac{d^2}{d{胃}^2}鈯�\left(胃\right)=-\left|C\right|鈯�\left(胃\right)$ 鈥︹�� (iii)

Let assume the angular solution has the complex exponential that therefore,$鈯�\left(胃\right)=e^{i伪胃}$ for the real scalar$伪$ .

$$\begin{gathered} {伪}^2鈯�\left(胃\right)=-\left|C\right|鈯�\left(胃\right) \\ {伪}^2=-\left|C\right| \\ 伪=\sqrt{\left|C\right|} \end{gathered}$$

Hence the complex exponential of the form$e^{i\sqrt{\left|C\right|}胃}$ is acceptable solution for$鈯�\left(胃\right)$ .

(d)

The function$鈯�\left(胃\right)$ is said to be periodic when full rotation about the polar axis brings the same value of$鈯�\left(胃\right)$ .

Therefore, the periodic condition is$鈯�\left(胃\right)=鈯�\left(胃+2\mathrm{蟺}\right)$ .

$$\begin{gathered} e^{i\sqrt{\left|C\right|}\left(胃+2\mathrm{蟺}\right)}=e^{i\sqrt{\left|C\right|}胃} \\ 2\mathrm{蟺}\sqrt{\left|\mathrm{C}\right|}=2\mathrm{蟺尘}\left(\mathrm{m}=0卤1,卤2,.....\right) \end{gathered}$$

From the above, the allowed values of C are$m=0卤1,卤2,....$ that is$C=0,-1,-4,....$ .

Hence, the allowed values of$C$ are $C=0,-1,-4,.....$.

(e)

The z component of the angular momentum is in the cylindrical coordinates is given by,

$L_z=-i\sout{h}\frac{鈭�}{鈭�胃}$

Thus, the solution is given by,

$$\begin{gathered} L_z鈯�\left(胃\right)=-i\sout{h}\frac{鈭�}{鈭�胃}{e}^{i\sqrt{\left|C\right|}胃} \\ {L}_z鈯�\left(胃\right)=-\sout{h}\sqrt{\left|C\right|}鈯�\left(胃\right) \end{gathered}$$

The angular momentum$L_z=\sout{h}\sqrt{\left|C\right|}$, with$\sqrt{\left|C\right|}=0,卤1,卤2,......$ .

The above discussion tells us that property quantized according to the value C is component of angular momentum.

(f)

Determine the radial equation by substituting the left side of the equation (i) to C .

$$\begin{gathered} -\frac{r}{R\left(r\right)}\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)-\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)=C \\ r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)=0........\left(iv\right) \end{gathered}$$

Hence the radial equation is$r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)=0$ .

(g)

Given that $U\left(r\right)=-b/r$

Recall the equation (iv),

$r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-U\left(r\right)\right)+C\right)R\left(r\right)\overline{)=0}$

Substitute $-b/r$for $U\left(r\right)$into above equation.

$$\begin{gathered} r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E-\left(-\frac{b}{r}\right)\right)+C\right)R\left(r\right)=0 \\ r\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E+\frac{b}{r}\right)+C\right)R\left(r\right)=0.......\left(v\right) \end{gathered}$$

The term $\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)$can be calculated as,

$$\begin{gathered} \frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)=\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}{e}^{-r/a}\right) \\ \frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)=-\frac{鈭�}{鈭�r}\left(\frac{r}{a}{e}^{-r/a}\right) \\ \frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)=\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right) \end{gathered}$$

Substitute $\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right)$for $\frac{鈭�}{鈭�r}\left(r\frac{鈭�}{鈭�r}\right)R\left(r\right)$into equation (v).

$$\begin{gathered} r\left(-\frac{1}{a}+\frac{r}{a^2}\right)R\left(r\right)+\left(\frac{2m{r}^2}{\sout{h}}\left(E+\frac{b}{r}\right)+C\right)R\left(r\right)=0 \\ \left[\left(\frac{1}{a^2}+\frac{2mE}{\sout{h}}\right)r^2+\left(-\frac{1}{a}+\frac{2mr}{{\sout{h}}^2}\right)r+C\right]R\left(r\right)=0......\left(vi\right) \end{gathered}$$

From the above discussion, R(r) is the solution under some requirement and c = 0 is the one of them.

(h)

Since the C = 0 is the required condition for R(r) to be solution.

Substitute 0 for C into the equation (vi).

$$\begin{gathered} \left[\left(\frac{1}{a^2}+\frac{2mE}{{\sout{h}}^2}\right)r^2+\left(-\frac{1}{a}+\frac{2mb}{{\sout{h}}^2}\right)r+0\right]R\left(r\right)=0 \\ \left[\left(\frac{1}{a^2}+\frac{2mE}{{\sout{h}}^2}\right)r^2+\left(-\frac{1}{a}+\frac{2mb}{{\sout{h}}^2}\right)r\right]R\left(r\right)=0 \end{gathered}$$

The other requirement to satisfy the R(r) into above equation are $\frac{1}{a}=\frac{2mb}{{\sout{h}}^2}$and $-\frac{1}{a^2}=\frac{2mE}{{\sout{h}}^2}$.

Now the value of the is given by,

$$\begin{gathered} \frac{1}{a}=\frac{2mb}{{\sout{h}}^2} \\ a=\frac{{\sout{h}}^2}{2mb}......\left(vii\right) \end{gathered}$$

The allowed energy is given by,

localid="1660046666625" $$\begin{gathered} -\frac{1}{a^2}=\frac{2mE}{{\sout{h}}^2} \\ E=\frac{{\sout{h}}^2}{2m{a}^2} \end{gathered}$$

Substitute localid="1660046671256" $\frac{{\sout{h}}^2}{2mb}$for into above equation.

localid="1660046660072" $$\begin{gathered} E=\frac{{\sout{h}}^2}{2m{\left({\displaystyle \frac{{\sout{h}}^2}{2mb}}\right)}^2} \\ E=\frac{4m^2{b}^2脳{\sout{h}}^2}{2m脳{\sout{h}}^2} \\ E=-\frac{2m{b}^2}{{\sout{h}}^2}........\left(viii\right) \end{gathered}$$

From the equation (vii) and (viii), the required value of C is 0 . It follows the ground state wave function,

localid="1660046678252" $$\begin{gathered} {蠄}_0\left(r,胃\right)=R_0{鈯�}_0\left(胃\right) \\ {蠄}_0\left(r,胃\right)=A{e}^{-2mbr/{\sout{h}}^2} \end{gathered}$$

Here, A is the normalization constant.

Hence the value of a is localid="1660046691263" $\frac{{\sout{h}}^2}{2mb}$, the ground state energy islocalid="1660046684817" $-\frac{2mb}{{\sout{h}}^2}$and eave wave function of flatrogen atom is$A{e}^{-2mbr/{\sout{h}}^2}$.