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$\left(|300鉄�鈫�|200鉄�\text{and}|300鉄�鈫�|1\begin{array}{ccc}1& 0& 0\end{array}鉄�\text{violate}螖l=卤1\text{rule.}\right)$

From Eq. 11.76:

$\{\begin{array}{l}\text{if}{m}^{\text{'}}=m,\text{鈥夆赌夆�塼hen}鉄�n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|x\right|n\mathcal{l}m鉄�=鉄�n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|y\right|n\mathcal{l}m鉄�=0\\ \text{if}{m}^{\text{'}}=m卤1,\text{then}鉄�n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|x\right|n\mathcal{l}m鉄�=卤i\left(n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\right|y|n\mathcal{l}m鉄�\\ \text{and}鉄�n^{\text{'}}{\mathcal{l}}^{\text{'}}{m}^{\text{'}}\left|z\right|n\mathcal{l}m鉄�=0\end{array}$(11.76).

$\begin{array}{l}鉄�210\left|r\right|300鉄�=鉄�210\left|z\right|300鉄�\widehat{k}\\ 鉄�21卤1\left|r\right|300鉄�=鉄�21卤1\left|x\right|300鉄�\widehat{i}+鉄�21卤1\left|y\right|300鉄�\widehat{j}\end{array}$

$\begin{array}{l}卤鉄�21卤1\left|x\right|300鉄�=i鉄�21卤1\left|y\right|.\\ Thus|鉄�210|r|300鉄�{|}^2=|鉄�210\left|z\right|300鉄�{|}^2\text{鈥夆赌夆�塧nd鈥夆赌夆��}|鉄�21卤1|r|300鉄�{|}^2=2|鉄�21卤1\left|x\right|300鉄�{|}^2\end{array}$

So there are really just two matrix elements to calculate.

${蠄}_{21m}=R_{21}{Y}_1^m,\text{鈥夆赌夆��}{蠄}_{300}=R_{30}{Y}_0^0.$From Table 4.3:

$\begin{array}{c}鈭�Y_1^0{Y}_0^0\cos 胃\sin 胃d胃d蠒=\sqrt{\frac{3}{4蟺}}\sqrt{\frac{1}{4蟺}}{鈭�}_0^{蟺}{\cos }^2胃\sin 胃d胃{鈭�}_0^{2蟺}d蠒\\ ={\frac{\sqrt{3}}{4蟺}(鈭�\frac{{\cos }^3胃}{3})|}_0^{蟺}\left(2蟺\right)\\ =\frac{\sqrt{3}}{2}\left(\frac{2}{3}\right)\\ =\frac{1}{\sqrt{3}}\end{array}$

$\begin{array}{c}鈭�{\left(Y_1^{卤1}\right)}^{*}{Y}_0^0{\sin }^2胃\cos 蠒d胃d蠒=鈭�\sqrt{\frac{3}{8蟺}}\sqrt{\frac{1}{4蟺}}{鈭�}_0^{蟺}{\sin }^3胃d胃{鈭�}_0^{2蟺}\cos 蠒e^{鈭�i蠒}d蠒\\ =鈭�\frac{1}{4蟺}\sqrt{\frac{3}{2}}\left(\frac{4}{3}\right)[{鈭�}_0^{2蟺}{\cos }^2蠒d蠒鈭�i{鈭�}_0^{2蟺}\cos 蠒\sin 蠒d蠒]\\ =鈭�\frac{1}{蟺\sqrt{6}}(蟺鈭�0)\\ =鈭�\frac{1}{\sqrt{6}}\end{array}$

From Table 4.7:

$\begin{array}{c}K鈮�{鈭�}_0^{\mathrm{鈭�}}{R}_{21}{R}_{30}{r}^{3}dr=\frac{1}{\sqrt{24}{a}^{3/2}}\frac{2}{\sqrt{27}{a}^{3/2}}{鈭�}_0^{\mathrm{鈭�}}\frac{r}{a}{e}^{鈭�r/2a}[1鈭�\frac{2}{3}\frac{r}{a}+\frac{2}{27}{\left(\frac{r}{a}\right)}^2]e^{鈭�r/3a}{r}^{3}dr\\ =\frac{1}{9\sqrt{2}{a}^3}{a}^4{鈭�}_0^{\mathrm{鈭�}}(1鈭�\frac{2}{3}u+\frac{2}{27}{u}^2)u^4{e}^{鈭�5u/6}du\\ =\frac{a}{9\sqrt{2}}[4!{\left(\frac{6}{5}\right)}^5鈭�\frac{2}{3}5!{\left(\frac{6}{5}\right)}^6+\frac{2}{27}6!{\left(\frac{6}{5}\right)}^7]\end{array}$

$\begin{array}{c}=\frac{a}{9\sqrt{2}}\frac{4!6^5}{5^6}(5鈭�\frac{2}{3}6鈰�5+\frac{2}{27}{6}^3)\\ =\frac{a}{9\sqrt{2}}\frac{4!6^5}{5^6}\\ =\frac{2^7{3}^4}{5^6}\sqrt{2}a\end{array}$

So,

$\begin{array}{c}鉄�21卤1\left|x\right|300鉄�=鈭�R_{21}{\left(Y_1^{卤1}\right)}^{*}\left(r\sin 胃\cos 蠒\right)R_{30}{Y}_0^0{r}^2\sin 胃drd胃d蠒\\ =K(鈭�\frac{1}{\sqrt{6}})\end{array}$

$\begin{array}{c}鉄�210\left|z\right|300鉄�=鈭�R_{21}{Y}_1^0\left(r\cos 胃\right)R_{30}{Y}_0^0{r}^2\sin 胃drd胃d蠒\\ =K\left(\frac{1}{\sqrt{3}}\right)\end{array}$

$\begin{array}{c}|鉄�210|r|300鉄�{|}^2=|鉄�210\left|z\right|300鉄�{|}^2\\ =K^2/3;\\ |鉄�21卤1|r|300鉄�{|}^2=2|鉄�21卤1\left|x\right|300鉄�{|}^2\\ =K^2/3\end{array}$

Evidently the three transition rates are equal, and hence 1/3 go by each route.

For each mode,

$\begin{array}{c}A=\frac{{蝇}^3{e}^2|鉄�r鉄�{|}^2}{3蟺{系}_0\mathrm{鈩�}{c}^3}\\ \text{here,}\\ \text{鈥�}蝇=\frac{E_3鈭�E_2}{\mathrm{鈩�}}\\ =\frac{1}{\mathrm{鈩�}}(\frac{E_1}{9}鈭�\frac{E_1}{4})\\ =鈭�\frac{5}{36}\frac{E_1}{\mathrm{鈩�}}\end{array}$,

so the total decay rate is

$\begin{array}{c}R=3{(鈭�\frac{5}{36}\frac{E_1}{\mathrm{鈩�}})}^3\frac{e^2}{3蟺{系}_0\mathrm{鈩�}{c}^3}\frac{1}{3}{\left(\frac{2^7{3}^4}{5^6}\sqrt{2}a\right)}^2\\ =6{\left(\frac{2}{5}\right)}^9{\left(\frac{E_1}{m{c}^2}\right)}^2\left(\frac{c}{a}\right)\end{array}$

$\begin{array}{c}=6{\left(\frac{2}{5}\right)}^9{\left(\frac{13.6}{0.511脳{10}^6}\right)}^2\left(\frac{3脳{10}^8}{0.529脳{10}^{鈭�10}}\right)/s\\ =6.32脳{10}^6/s\\ 蟿=\frac{1}{R}\\ =1.58脳{10}^{鈭�7}s\end{array}$