91影视

As you know that the probability of finding an electron in the given point can be found using the square of the orbital wave function - ${\left|唯\right|}^{\mathbf{2}}$at that point.

Hence, the required probability that the electron in a hydrogen atom would be found withing $\mathbf{30}\mathbf{掳}$of the x-y plane, can be found using the following formula

$\mathit{P}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{蟺}\mathbf{/}\mathbf{3}}^{\mathbf{2}\mathbf{蟺}\mathbf{/}\mathbf{3}}{\left({鈯�}_{\mathcal{l},m\mathcal{l}}\left(胃\right)\right)}^{\mathbf{2}}\mathbf{}\mathbf{2}\mathbf{蟺蝉颈苍胃诲胃}$

Where, is the Orbital wave function and $\mathbf{胃}$Angle from the x-y plane.

Here, I is the azimuthal quantum number and ${\mathrm{m}}_{\mathrm{I}}$is the magnetic quantum number.

The probability is define by,

$$\begin{gathered} P_{0,0}={鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}{\left(\sqrt{\frac{1}{4\mathrm{蟺}}}\right)}^{2}2\mathrm{蟺蝉颈苍胃诲胃} \\ ={\left|\frac{1}{2}\left(-\mathrm{肠辞蝉胃}\right)\right|}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3} \\ =\frac{1}{2}\left(-\cos \left(\frac{2\mathrm{蟺}}{3}\right)+\cos \left(\frac{\mathrm{蟺}}{3}\right)\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{0,0}=\frac{1}{2}\left(-\left(-0.5\right)+0.5\right) \\ =\frac{1}{2} \\ =0.5 \end{gathered}$$

$$\begin{gathered} P_{1,1}={鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}{\left(\sqrt{\frac{3}{8\mathrm{蟺}}}\sin 胃\right)}^{2}2\mathrm{蟺蝉颈苍胃诲胃} \\ =\frac{3}{4}{鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}\left(1-{\cos }^2\mathrm{胃}\right)\mathrm{蝉颈苍胃诲胃} \\ =\frac{3}{4}{\left|\left(-\mathrm{肠辞蝉胃}+\frac{{\cos }^3\mathrm{胃}}{3}\right)\right|}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3} \\ =\frac{3}{4}\left(-\cos \left(\frac{2\mathrm{蟺}}{3}\right)+\frac{{\cos }^3\left({\displaystyle \frac{2\mathrm{蟺}}{3}}\right)}{3}+\cos \left(\frac{\mathrm{蟺}}{3}\right)-\frac{{\cos }^3\left({\displaystyle \frac{\mathrm{蟺}}{3}}\right)}{3}\right) \end{gathered}$$

$$\begin{gathered} P_{1,1}=\frac{3}{4}\left(-\left(-0.5\right)+\frac{{\left(-0.5\right)}^3}{3}+0.5-\frac{{\left(0.5\right)}^3}{3}\right) \\ =\frac{3}{4}\left(0.5-0.04167+0.5-0.04167\right) \\ =0.688 \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{2,2}={鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}{\left(\sqrt{\frac{15}{32\mathrm{蟺}}}{\sin }^2\mathrm{胃}\right)}^{2}2\mathrm{蟺蝉颈苍胃诲胃} \\ =\frac{15}{16}{鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}{\sin }^4\mathrm{胃蝉颈苍胃诲胃} \\ =\frac{15}{16}{鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}{\left(1-{\cos }^2\mathrm{胃}\right)}^2\mathrm{蝉颈苍胃诲胃} \\ =\frac{15}{16}{鈭�}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3}\left(1-2{\cos }^2\mathrm{胃}+{\cos }^4\mathrm{胃}\right)\mathrm{蝉颈苍胃诲胃} \end{gathered}$$

$$\begin{gathered} P_{2,2}=\frac{15}{16}{\left|\left(-\cos 胃-2\frac{{\cos }^3\left({\displaystyle \frac{2\mathrm{蟺}}{3}}\right)}{3}+\frac{{\cos }^5\left({\displaystyle \frac{2\mathrm{蟺}}{3}}\right)}{5}\right)\right|}_{\mathrm{蟺}/3}^{2\mathrm{蟺}/3} \\ =\frac{202}{256} \\ =0.793 \end{gathered}$$

The trend from Step 2 to Step 4, i.e., from 0.5 to 0.793 shows that as the angular momentum increases, the maximum z-component states are more nearly restricted to the xy plane.