Question

(a) Construct the spatial wave function $\mathbf{\left(}\mathbf{\psi }\mathbf{\right)}$for hydrogen in the state $\mathbf{n}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{I}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{.}$Express your answer as a function of $\mathbf{r}\mathbf{,}\mathbf{}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{a}$(the Bohr radius) only—no other variables ($\mathbf{p}\mathbf{,}\mathbf{z}\mathbf{}\mathbf{,}$etc.) or functions ($\mathit{p}\mathbf{,}\mathit{v}\mathbf{,}$etc.), or constants ($\mathbf{A}\mathbf{,}{\mathbf{c}}_{\mathbf{0}}\mathbf{,}$etc.), or derivatives, allowed (π is okay, and e, and 2, etc.).

(b) Check that this wave function is properly normalized, by carrying out the appropriate integrals over, $\mathbf{\theta }\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\varphi }\mathbf{.}$

(c) Find the expectation value of ${\mathbf{r}}^{\mathbf{s}}$in this state. For what range of s (positive and negative) is the result finite?

Short answer

$\left(\mathrm{a}\right){\mathrm{\psi }}_{321}={\mathrm{R}}_{32}{\mathrm{Y}}_2^1=-\frac{1}{\sqrt{\mathrm{\pi }}}\frac{1}{81{\mathrm{a}}^{7/2}}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/3\mathrm{a}}{\mathrm{sin\theta cos\theta e}}^{\mathrm{i\varphi }}.$

(b)The wave function is properly normalized by 1

Step-by-step solution

(a) Constructing the spatial wave function for hydrogen.

$$\begin{gathered} {\mathrm{\psi }}_{321}={\mathrm{R}}_{32}{\mathrm{Y}}_2^1=\frac{4}{81\sqrt{30}}\frac{1}{{\mathrm{a}}^{3/2}}{\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}^2{\mathrm{e}}^{-\mathrm{r}/3\mathrm{a}}\left[-\sqrt{\frac{15}{8\mathrm{\pi }}}{\mathrm{sin\theta cos\theta e}}^{\mathrm{i\varphi }}\right] \\ =-\frac{1}{\sqrt{\mathrm{\pi }}}\frac{1}{81{\mathrm{a}}^{7/2}}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/3\mathrm{a}}{\mathrm{sin\theta cos\theta e}}^{\mathrm{i\varphi }}. \end{gathered}$$

 Step2: (b) Checking the wave function is properly normalized.

$$\begin{gathered} \int {\left|\psi \right|}^2{d}^{3}r=\frac{1}{\pi }\frac{1}{\pi {\left(81\right)}^2{a}^7}\int \left(r^4{e}^{-2r/3a}{\sin }^2\theta {\cos }^2\theta \right)r^2\sin \theta drd\theta d\varphi . \\ =\frac{1}{\pi {\left(81\right)}^2{a}^7}2\pi {\int }_0^{\propto }{r}^6{e}^{-2r/3a}dr{\int }_0^{\pi }\left(1-{\cos }^2\theta \right){\cos }^2\theta {\cos }^2\theta d\theta . \\ =\frac{2}{{\left(81\right)}^2{a}^7}\left[6!{\left(\frac{3a}{2}\right)}^7\right]{\overline{)\left[-\frac{{\cos }^3\theta }{3}+\frac{{\cos }^5\theta }{5}\right]}}_0^{\pi }. \\ =\frac{2}{3^8{a}^7}6.5.4.3.2\frac{3^7{a}^7}{2^7}\left[\frac{2}{3}-\frac{2}{5}\right]=\frac{3.5}{4}.\frac{4}{15}=1. \end{gathered}$$

 Step3: (c)Finding the exceptional value of rs