Calculate the probability that the electron in a hydrogen atom would be found within 30 degrees of the xy-plane, irrespective of radius, for (a) I=0 ,${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$; (b) role="math" localid="1660014331933" $\mathbf{I}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{m}}_{\mathbf{I}}\mathbf{=}\mathbf{\pm }\mathbf{1}$and (c) $\mathbf{I}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{m}}_{\mathbf{I}}\mathbf{=}\mathbf{\pm }\mathbf{2}$. (d) As angular momentum increases, what happens to the orbits whose z-components of angular momentum are the maximum allowed?

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 - ${\left|\Psi \right|}^{\mathbf{2}}$at that point.

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

$\mathit{P}\mathbf{=}{\mathbf{\int }}_{\mathbf{\pi }\mathbf{/}\mathbf{3}}^{\mathbf{2}\mathbf{\pi }\mathbf{/}\mathbf{3}}{\left({⊝}_{\mathcal{l},m\mathcal{l}}\left(\theta \right)\right)}^{\mathbf{2}}\mathbf{}\mathbf{2}\mathbf{\pi sin\theta d\theta }$

Where, is the Orbital wave function and $\mathbf{\theta }$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,

$$\begin{gathered} P_{0,0}={\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}{\left(\sqrt{\frac{1}{4\mathrm{\pi }}}\right)}^{2}2\mathrm{\pi sin\theta d\theta } \\ ={\left|\frac{1}{2}\left(-\mathrm{cos\theta }\right)\right|}_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3} \\ =\frac{1}{2}\left(-\cos \left(\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\frac{\mathrm{\pi }}{3}\right)\right) \end{gathered}$$

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

$$\begin{gathered} P_{1,1}={\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}{\left(\sqrt{\frac{3}{8\mathrm{\pi }}}\sin \theta \right)}^{2}2\mathrm{\pi sin\theta d\theta } \\ =\frac{3}{4}{\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}\left(1-{\cos }^2\mathrm{\theta }\right)\mathrm{sin\theta d\theta } \\ =\frac{3}{4}{\left|\left(-\mathrm{cos\theta }+\frac{{\cos }^3\mathrm{\theta }}{3}\right)\right|}_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3} \\ =\frac{3}{4}\left(-\cos \left(\frac{2\mathrm{\pi }}{3}\right)+\frac{{\cos }^3\left({\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{3}+\cos \left(\frac{\mathrm{\pi }}{3}\right)-\frac{{\cos }^3\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{3}\right) \end{gathered}$$

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

$$\begin{gathered} {\mathrm{P}}_{2,2}={\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}{\left(\sqrt{\frac{15}{32\mathrm{\pi }}}{\sin }^2\mathrm{\theta }\right)}^{2}2\mathrm{\pi sin\theta d\theta } \\ =\frac{15}{16}{\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}{\sin }^4\mathrm{\theta sin\theta d\theta } \\ =\frac{15}{16}{\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}{\left(1-{\cos }^2\mathrm{\theta }\right)}^2\mathrm{sin\theta d\theta } \\ =\frac{15}{16}{\int }_{\mathrm{\pi }/3}^{2\mathrm{\pi }/3}\left(1-2{\cos }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }\right)\mathrm{sin\theta d\theta } \end{gathered}$$

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

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.