Calculate the electric dipole moment p and estimate the transition time for a hydrogen atom electron making an electric dipole transition from the

$\left(n,l,m\right)\mathbf{=}\left(3,2,0\right)$to the $\left(2,1,0\right)$ state.

Initial state is $\left(n,l,m_l\right)\mathbf{=}\left(3,2,0\right)$ and the final state is $\left(2,1,0\right)$.

Where, n is the principal quantum number, l is the azimuthal quantum number, m_l is the magenetic quantum number.

Here, the electric dipole moment is given by,

$\mathbf{p}\mathbf{=}\mathbf{-}\mathbf{eRe}\mathbf{\left(}\mathbf{\int }\overline{\mathbf{r}}{\mathbf{\Psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}\mathbf{*}\mathbf{\left(}\overline{\mathbf{r}}\mathbf{\right)}{\mathbf{\Psi }}_{\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{0}}\mathbf{\left(}\overline{\mathbf{r}}\mathbf{\right)}{\mathbf{r}}^{\mathbf{2}\mathbf{}\mathbf{}}\mathbf{sin\theta drd\theta d\varphi }\mathbf{\right)}$ ….. (1)

Where,$\mathbf{\Psi }$is the wave function, $\mathit{\theta }$ is the colatitude, $\mathbf{\varphi }$ is the azimuth, and r is the radius.

Wave functions can be calculated by,

$$\begin{gathered} {\mathrm{\Psi }}_{2,1,0}*\left(\overline{\mathrm{r}}\right){\mathrm{\Psi }}_{3,2,0}\left(\overline{\mathrm{r}}\right)=\left(\frac{1}{{\left({\mathrm{a}}_0\right)}^{3/2}}\frac{\mathrm{r}}{\sqrt{3{\mathrm{a}}_0}}{\mathrm{e}}^{-\mathrm{r}/\left({\mathrm{a}}_0\right)}\sqrt{\frac{3}{4\mathrm{\pi }}}\mathrm{cos\theta }\right)\left(\frac{1}{{\left(3{\mathrm{a}}_0\right)}^{3/2}}\frac{2\sqrt{2}{r}^2}{27\sqrt{5}{a}_0^2}{\mathrm{e}}^{-\mathrm{r}/\left(2{\mathrm{a}}_0\right)}\sqrt{\frac{5}{16\mathrm{\pi }}}\left(3{\cos }^2\mathrm{\theta }-1\right)\right) \\ =\frac{1}{{a_0}^3}\frac{1}{2\sqrt{2}}\frac{1}{3\sqrt{3}}\frac{2\sqrt{2}{r}^3}{\sqrt{3}{a}_0}{e}^{-5r/\left(6a_0\right)}\frac{1}{4\pi }\sqrt{\frac{3\times 5}{4}}\cos \theta \left(3{\cos }^2\theta -1\right) \\ =\frac{\sqrt{3}}{8\pi 3^5{a}_0^6}{r}^3{e}^{-5r/\left(6a_0\right)}\cos \theta \left(3{\cos }^2\theta -1\right) \end{gathered}$$

$$\begin{gathered} \overline{r}=\frac{1}{8.3^{9/2}{a}_0^6}{\int }_0^{\infty }{r}^6{e}^{-5r/\left(6a_0\right)}dr{\int }_0^{\pi }\left(3{\cos }^4\theta -{\cos }^2\theta \right)\sin \theta dr \\ =\frac{1}{8.3^{9/2}{a}_0^6}\frac{6!}{{\left(5/6a_0\right)}^7}\left(-\frac{3}{5}{\cos }^5\theta +\frac{1}{3}{\cos }^3\theta \right) \\ =\frac{720{\left(6a_0\right)}^7}{8.3^{9/2}{a_0}^6{5}^7}\left(\frac{6}{5}-\frac{2}{3}\right) \end{gathered}$$

$\overline{\mathrm{r}}=1.23{\mathrm{a}}_0$ ….. (2)

From eq. (1) and eq. (2), you get

$p=-\left(1.23a_0\right)\cos \left(t∆E/h\right)$

Youalso know that,

role="math" localid="1659870308402" $$\begin{gathered} ∆\mathrm{E}=(-13.6\mathrm{eV})\left(\frac{1}{3^2}-\frac{1}{2^2}\right) \\ =1.89\mathrm{eV} \end{gathered}$$

Hence,

$$\begin{gathered} \mathrm{p}=-(1.23{\mathrm{a}}_0)\cos \left[\left(\mathrm{t}1.89\mathrm{eV}\right)/\mathrm{h}\right] \\ =\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(1.23\times 0.0529\times {10}^{-9}\mathrm{m}\right) \\ =1.04\times {10}^{-29}\mathrm{Cm} \end{gathered}$$

Define the angular frequency as below.

$\begin{array}{l}\mathrm{\omega }=\frac{\left|{\mathrm{E}}_{\mathrm{i}}-{\mathrm{E}}_{\mathrm{f}}\right|}{\mathrm{h}}\\ =\frac{1.89\mathrm{eV}\times 1.6\mathrm{x}{10}^{-19}\mathrm{J}/\mathrm{eV}}{6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}}\\ =2.86\times {10}^{15}{\mathrm{s}}^{-1}\end{array}$

Hence, the transition time will be given by

$$\begin{gathered} \mathrm{Transition}\mathrm{time}\cong \frac{12\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right){\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^3\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}{{\left(1.04\times {10}^{-29}\mathrm{Cm}\right)}^2{\left(2.86\times {10}^{15}{\mathrm{s}}^{-1}\right)}^3} \\ \cong 1.2\times {10}^{-7}s \end{gathered}$$