91影视

The radial equation for hydrogen is given by:

$\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\left(-\frac{{\mathrm{e}}^2}{4{\mathrm{蟺翱}}_0}\frac{1}{\mathrm{r}}+\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\frac{\mathrm{I}\left(\mathrm{I}+1\right)}{{\mathrm{r}}^2}\right)\mathrm{u}=\mathrm{Eu}$

Where u(r)=rR(r) and the effective potential is the terms inside the square brackets, that is:

$\mathrm{V}\left(\mathrm{r}\right)=-\frac{{\mathrm{e}}^2}{4{\mathrm{蟺翱}}_0}\frac{1}{\mathrm{r}}+\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{\mathrm{I}\left(\mathrm{I}+1\right)}{{\mathrm{r}}^2}$

In this case we have two turning points,$r_1$and$r_2$at both sides of the well, these two points can be determined by setting the potential equal to the energy, so the two turning points are the roots of:

$\mathrm{E}=-\frac{{\mathrm{e}}^2}{4{\mathrm{蟺翱}}_0}\frac{1}{\mathrm{r}}+\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{\mathrm{I}\left(\mathrm{I}+1\right)}{{\mathrm{r}}^2}$ 鈥︹赌︹赌︹赌︹赌︹赌�.(1)

The WKB equations satisfy the condition:

${鈭�}_{{\mathrm{r}}_1}^{{\mathrm{T}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{蟺丑}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌�..(2)

Now we need to find the integral in (2), where ,$\mathrm{p}\left(\mathrm{r}\right)=\sqrt{2\mathrm{m}\left(\mathrm{E}-\mathrm{V}\right)}$, so we have:

$$\begin{gathered} {鈭�}_{r_1}^{f_2}p\left(r\right)dr=\sqrt{2m}{鈭�}_{r_1}^{f_2}\sqrt{E+\frac{e^2}{4蟺{慰}_0}\frac{1}{r}-\frac{h^2}{2m}\frac{I\left(I+1\right)}{r^2}}dr \\ {鈭�}_{{\mathrm{r}}_1}^{{\mathrm{f}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}{鈭�}_{{\mathrm{r}}_1}^{{\mathrm{f}}_2}\frac{1}{\mathrm{r}}\sqrt{-{\mathrm{r}}^2-\frac{{\mathrm{e}}^2}{4{\mathrm{蟺慰}}_0\mathrm{E}}\mathrm{r}+\frac{{\mathrm{h}}^2}{2\mathrm{mE}}\mathrm{I}\left(\mathrm{I}+1\right)}\mathrm{dr} \end{gathered}$$

In the second line we factored out -E . The equation under the square root is identical to equation (1), so the solutions of the equation are $r_1$and $r_2$. To simplify, we let:

$$\begin{gathered} \mathrm{B}=-\frac{{\mathrm{e}}^2}{4{\mathrm{蟺慰}}_0\mathrm{E}} \\ \mathrm{C}鈮�-\frac{{\mathrm{h}}^2}{2\mathrm{mE}}\mathrm{I}\left(\mathrm{I}+1\right) \end{gathered}$$

Thus,

${鈭�}_{{\mathrm{蟿}}_1}^{{\mathrm{r}}^2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}{鈭�}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\frac{1}{\mathrm{r}}\sqrt{-{\mathrm{r}}^2+\mathrm{Br}-\mathrm{Cdr}}$

As we said before, the solutions of the equation under the square root are the two turning points $r_1$and $r_2$, which are given by the general law, as:

$$\begin{gathered} {\mathrm{r}}_1=\frac{\mathrm{B}-\sqrt{{\mathrm{B}}^2-4\mathrm{C}}}{2} \\ {\mathrm{r}}_2=\frac{\sqrt{{\mathrm{B}}^2-4\mathrm{C}}}{2} \end{gathered}$$

So we can write:

$-{\mathrm{r}}^2+\mathrm{Br}-\mathrm{C}=\left(\mathrm{r}-{\mathrm{r}}_1\right)\left({\mathrm{r}}_2-\mathrm{r}\right)$

Thus,

$$\begin{gathered} {鈭�}_{r_1}^{r_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}{鈭�}_{{\mathrm{r}}_1}^{{\mathrm{f}}_2}\sqrt{\left(\mathrm{r}-{\mathrm{r}}_1\right)\left({\mathrm{r}}_2-\mathrm{r}\right)}\mathrm{dr} \\ {鈭�}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}\frac{\mathrm{蟺}}{2}{\left(\sqrt{{\mathrm{r}}^2}-\sqrt{{\mathrm{r}}_1}\right)}^2 \\ {鈭�}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}\frac{\mathrm{蟺}}{2}\left({\mathrm{r}}_1+{\mathrm{r}}_2-2\sqrt{{\mathrm{r}}_1{\mathrm{r}}_2}\right) \\ {鈭�}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{-2\mathrm{mE}}\frac{\mathrm{蟺}}{2}\left(\mathrm{B}-2\sqrt{\mathrm{C}}\right) \end{gathered}$$

$$\begin{gathered} {鈭�}_{r_1}^{r_2}p\left(r\right)dr=\sqrt{-2mE}\frac{蟺}{2}\left(-\frac{e^2}{4蟺{慰}_{0}E}-2\sqrt{-\frac{h^2}{2mE}I\left(I+1\right)}\right) \\ {鈭�}_{r_1}^{r_2}p\left(r\right)dr=\frac{蟺}{2}\left(-\frac{e^2\sqrt{2m}}{4蟺{慰}_0\sqrt{-E}}-2h\sqrt{I\left(I+1\right)}\right) \end{gathered}$$

Where in the second line, I used the given integral in the problem.

Substitute into (2) to get:

$\frac{\mathrm{蟺}}{2}\left(-\frac{{\mathrm{e}}^2\sqrt{2\mathrm{m}}}{4{\mathrm{蟺慰}}_0\sqrt{-\mathrm{E}}}-2\mathrm{h}\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\right)=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{蟺丑}$

Solve for E as:

$$\begin{gathered} -\frac{{\mathrm{e}}^2\sqrt{2\mathrm{m}}}{4{\mathrm{蟺慰}}_0\sqrt{-\mathrm{E}}}-2\mathrm{h}\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}=\left(2\mathrm{n}-1\right)\mathrm{h} \\ -\frac{{\mathrm{e}}^2\sqrt{2\mathrm{m}}}{4{\mathrm{蟺慰}}_0\sqrt{-\mathrm{E}}}=\left(2\mathrm{n}-1\right)\mathrm{h}+2\mathrm{h}\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)} \\ \mathrm{E}=\left(\frac{{\mathrm{e}}^2}{4{\mathrm{蟺慰}}_0}\right)\frac{\mathrm{m}}{2{\mathrm{h}}^2{\left(\mathrm{n}-{\displaystyle \frac{1}{2}}+\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\right)}^2} \end{gathered}$$

Thus,

$\mathrm{E}=\frac{{\mathrm{E}}_0}{{\left(\mathrm{n}-{\displaystyle \frac{1}{2}}+\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\right)}^2}$

Where is the ground state energy of the hydrogen atom, and it is given by the Bohr formula:

$$\begin{gathered} {\mathrm{E}}_0=-{\left(\frac{{\mathrm{e}}^2}{4{\mathrm{蟺慰}}_0}\right)}^2\frac{\mathrm{m}}{2{\mathrm{h}}^2} \\ {\mathrm{E}}_0=13.6\mathrm{eV} \end{gathered}$$

Thus,

$\mathrm{E}=\frac{-13.6\mathrm{eV}}{{\left(\mathrm{n}-{\displaystyle \frac{1}{2}}+\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\right)}^2}$.