91影视

The separation of positive and negative charges in a system is referred to as an electric dipole moment.

Initial state is $\left(n,l,m\right)\mathbf{=}\left(2,1,+1\right)$ and the final state is $\left(1,0,0\right)$.

Where, n is the principal quantum number, l is the azimuthal quantum number, ${\mathbf{m}}_{\mathbf{l}}$ is the magnetic quantum number.

Here, the electric dipole moment is given by,

$\mathbf{p}\mathbf{=}\mathbf{-}\mathbf{eRe}\left(鈭�\stackrel{炉}{r}{唯}_{1,0,0}*\left(\stackrel{炉}{r}\right){唯}_{2,1,1}\left(\stackrel{炉}{r}\right)r^2\mathrm{蝉颈苍胃诲谤诲胃诲蠒}\right)$ 鈥�.. (1)

Where, r is the radius, $\mathbf{胃}$ is the colatitude, role="math" localid="1659863355058" $\mathbf{蠒}$ is the azimuth, and $\mathbf{唯}$ is the wave function.

Wave functions can be calculated by,

${\mathrm{唯}}_{1,0,0}*\left(\stackrel{炉}{\mathrm{r}}\right){\mathrm{唯}}_{2,1,1}\left(\stackrel{炉}{\mathrm{r}}\right)=\left(\frac{1}{{\left({\mathrm{a}}_0\right)}^{3/2}}2{\mathrm{e}}^{-\mathrm{r}/\left({\mathrm{a}}_0\right)}\sqrt{\frac{1}{4\mathrm{蟺}}}\right)\left(\frac{1}{{\left(2{\mathrm{a}}_0\right)}^{3/2}}\frac{\mathrm{r}}{\sqrt{3{\mathrm{a}}_0}}{\mathrm{e}}^{-\mathrm{r}/\left(2{\mathrm{a}}_0\right)}\sqrt{\frac{3}{8\mathrm{蟺}}}{\mathrm{e}}^{+\mathrm{颈蠒}}\right)$

Where, ${\mathrm{a}}_0$is the radius of hydrogen atom.

Considering the first integration over $\mathrm{蠒}$, the second term has ${\mathrm{e}}^{+\mathrm{颈蠒}}$, which will cause the z-component to integrate to zero.

The x-component will have,

$$\begin{gathered} {鈭�}_0^{2\mathrm{蟺}}\mathrm{肠辞蝉蠒}\left(\mathrm{肠辞蝉蠒}+\mathrm{颈蝉颈苍蠒}\right)\mathrm{诲蠒}={鈭�}_0^{2\mathrm{蟺}}{\cos }^2\mathrm{蠒诲蠒}+\mathrm{i}{鈭�}_0^{2\mathrm{蟺}}\mathrm{肠辞蝉蠒sin蠒诲蠒} \\ ={\left[\frac{1}{2}\mathrm{蠒}+\frac{1}{4}\sin 2\mathrm{蠒}\right]}_0^{2\mathrm{蟺}}+{\left[-\frac{1}{2}{\cos }^2\mathrm{蠒}\right]}_0^{2\mathrm{蟺}} \\ =\left[\frac{2\mathrm{蟺}}{2}+\frac{1}{4}\sin 4\mathrm{蟺}-0\right]+\mathrm{i}\left[-\frac{1}{2}{\cos }^2\left(2\mathrm{蟺}\right)+\frac{1}{2}{\cos }^2\left(2\mathrm{蟺}\right)\right] \\ =\mathrm{蟺} \end{gathered}$$

You have, y-component as,

$$\begin{gathered} {鈭�}_0^{2\mathrm{蟺}}\mathrm{肠辞蝉蠒}\left(\mathrm{肠辞蝉蠒}+\mathrm{颈蝉颈苍蠒}\right)\mathrm{诲蠒}={鈭�}_0^{2\mathrm{蟺}}{\cos }^2\mathrm{蠒诲蠒}+\mathrm{i}{鈭�}_0^{2\mathrm{蟺}}\mathrm{肠辞蝉蠒sin蠒诲蠒} \\ =0+\mathrm{颈蟺} \end{gathered}$$

Similarly in x and y terms, the role="math" localid="1659863936500" $胃$integration is

${鈭�}_0^{\mathrm{蟺}}{\sin }^3\mathrm{胃诲胃}=\frac{4}{3}$

Now, by putting everything in eq. 1, you get,

$$\begin{gathered} \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}鈭�\mathrm{E}/\mathrm{h}}\frac{\mathrm{蟺}\hat{\mathrm{x}}+\mathrm{颈蟺}\hat{\mathrm{y}}}{8{{\mathrm{蟺补}}^4}_0}\frac{4}{3}鈭�{\mathrm{r}}^2{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\left(2{\mathrm{a}}_0\right)}\mathrm{dr}\right) \\ \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}鈭�\mathrm{E}/\mathrm{h}}\frac{\hat{\mathrm{x}}+\mathrm{i}\hat{\mathrm{y}}}{8{{\mathrm{蟺补}}^4}_0}\frac{4}{3}\frac{4!}{{\left(3/2{\mathrm{a}}_0\right)}^5}\right) \\ \mathrm{p}=-\mathrm{eRe}\left({\mathrm{e}}^{+\mathrm{it}鈭�\mathrm{E}/\mathrm{h}}\left(\hat{\mathrm{x}}+\mathrm{i}\hat{\mathrm{y}}\right){\mathrm{a}}_0\frac{2^7}{3^5}\right) \end{gathered}$$

The amplitude of this vector is

${\mathrm{ea}}_0\frac{2^7}{3^5}=0.53{\mathrm{ea}}_0$

Hence,

$$\begin{gathered} \mathrm{p}=0.53\left(1.6脳{10}^{-19}\mathrm{C}\right)\left(0.0529脳{10}^{-9}\mathrm{m}\right) \\ =4.5脳{10}^{-30}\mathrm{Cm} \end{gathered}$$

The frequency is same as in Example 7.11, hence the transition time will be

$$\begin{gathered} \mathrm{Transition}\mathrm{time}鈮�\frac{12\left(8.85脳{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right){\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right)}^3\left(1.055脳{10}^{-34}\mathrm{J}.\mathrm{s}\right)}{{\left(4.5脳{10}^{-30}\mathrm{Cm}\right)}^2{\left(1.55脳{10}^{16}{\mathrm{s}}^{-1}\right)}^3} \\ 鈮�4\mathrm{ns} \end{gathered}$$

The character of charge oscillation is different, but the estimated transition time is approximately the same as in the example.