Question

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?

Short answer

(a) For I=0,${\mathrm{m}}_{\mathrm{I}}=0$Probability is 0.5 .

(b) For I=1,${\mathrm{m}}_1=\pm 1$Probability is 0.688.

(c) For I=2,${\mathrm{m}}_{\mathrm{I}}=\pm 2$Probability is 0.793.

(d) As the angular momentum increases, the maximum z-component states are more nearly restricted to the xy plane.

Step-by-step solution

Required formula:

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.

(a) Probability for I=0,mI=0:

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

The probability is define by,

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

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

(b) Probability for I=1,mI=±1 :

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

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

(c) Probability for I=2,mI=±2:

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

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

(d) Conclusion:

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.