Chapter 8, Problem 8A.8P

(a)

Interpretation Introduction

Interpretation:

The expectation value of $1/r$ for a hydrogenic $\text{1s}$ orbital has to be calculated.

Concept introduction:

Expectation value is the average value of the large number of measurements of an observable.  The expectation value of a measurable parameter is calculated by the formula,

  $\langle \Omega \rangle ={\displaystyle \int {\psi }^{*}\widehat{\left(\Omega \right)}\psi \text{d}\tau }$

Where,

• $\psi$ is the wave function.
• $\Omega$ is an observable.

Answer to Problem 8A.8P

The expectation value of $1/r$ for a hydrogenic $\text{1s}$ orbital is $\frac{1}{a_0}$.

Explanation of Solution

The normalized wavefunction for hydrogenic $\text{1s}$ (${\psi }_{\text{1s}}$) orbital is given below.

    ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}$

The expectation value of $1/r$ for a hydrogenic $\text{1s}$ orbital is calculated as shown below.

    $\langle \frac{1}{r}\rangle ={\displaystyle \int {\psi }_{1s}^{*}}\left(\frac{1}{r}\right){\psi }_{1s}\text{d}\tau$                                                                                     (1)

Where,

• $\text{d}\tau$ is the volume element.

The value of $\text{d}\tau$ is given below.

    $\text{d}\tau =r^2\text{d}r\sin \theta \text{d}\theta \text{d}\varphi$

The ${\psi }_{1s}^{*}$ of a hydrogenic $\text{1s}$ orbital is the same as the ${\psi }_{\text{1s}}$.

    ${\psi }_{1s}^{*}=\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}$

Substitute the value of ${\psi }_{1s}^{*}$, $\text{d}\tau$ and ${\psi }_{\text{1s}}$ in (1).

    $\begin{array}{c}\langle \frac{1}{r}\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}\left(\frac{1}{r}\right)\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}{r}^2\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r\times \left(-{\left[\cos \theta \right]}_0^{\pi }\right)\times {\left[\varphi \right]}_0^{2\pi }}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta =-{\left[\cos \theta \right]}_0^{\pi }\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r\times \left(-{\left[-1-1\right]}_0^{\pi }\right)\times \left[2\pi -0\right]}\cos \pi =-1\text{and}\cos 0=1\end{array}$

The above equation is simplified below.

    $\begin{array}{c}\langle \frac{1}{r}\rangle =\frac{4\pi }{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r}\\ =\frac{4}{a_0^3}\left(\frac{1!}{{\left(\frac{2}{a_0}\right)}^2}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{\text{n}}{e}^{-\alpha x}\text{d}r}=\frac{n!}{{\left(\alpha \right)}^{n+1}}\\ =\frac{4}{a_0^3}\times \frac{a_0{}^2}{4}\\ =\frac{1}{a_0}\end{array}$

Therefore, the expectation value of $1/r$ for a hydrogenic $\text{1s}$ orbital is $\frac{1}{a_0}$.

(b)

Interpretation Introduction

Interpretation:

The expectation value of $1/r$ for a hydrogenic $\text{2s}$ orbital has to be calculated.

Concept introduction:

Expectation value is the average value of the large number of measurements of an observable.  The expectation value of a measurable parameter is calculated by the formula,

  $\langle \Omega \rangle ={\displaystyle \int {\psi }^{*}\widehat{\left(\Omega \right)}\psi \text{d}\tau }$

Where,

• $\psi$ is the wave function.
• $\Omega$ is an observable.

Answer to Problem 8A.8P

The expectation value of $1/r$ for a hydrogenic $\text{2s}$ orbital is $\frac{1}{4a_0}$.

Explanation of Solution

The normalized wavefunction for hydrogenic $\text{2s}$ (${\psi }_{\text{2s}}$) orbital is given below.

    ${\psi }_{\text{2s}}=\frac{1}{4\sqrt{2\pi }{a}_0^{3/2}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$

The expectation value of $1/r$ for a hydrogenic $\text{2s}$ orbital is calculated as shown below.

    $\langle \frac{1}{r}\rangle ={\displaystyle \int {\psi }_{\text{2s}}^{*}}\left(\frac{1}{r}\right){\psi }_{\text{2s}}\text{d}\tau$                                                                                   (1)

Where,

• $\text{d}\tau$ is the volume element.

The value of $\text{d}\tau$ is given below.

    $\text{d}\tau =r^2\text{d}r\sin \theta \text{d}\theta \text{d}\varphi$

The ${\psi }_{\text{2s}}^{*}$ of a hydrogenic $\text{2s}$ orbital is the same as the ${\psi }_{\text{2s}}$.

    ${\psi }_{\text{2s}}^{*}=\frac{1}{4\sqrt{2\pi }{a}_0^{3/2}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$

Substitute the value of ${\psi }_{\text{2s}}^{*}$, $\text{d}\tau$ and ${\psi }_{\text{2s}}$ in (1).

    $\begin{array}{c}\langle \frac{1}{r}\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{4\sqrt{2\pi }{a}_0^{3/2}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}\left(\frac{1}{r}\right)\frac{1}{4\sqrt{2\pi }{a}_0^{3/2}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}{r}^2\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{16\times 2\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(2-\frac{r}{a_0}\right)}^2{e}^{-r/a_0}r\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{16\times 2\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}r\left(4+\frac{r^2}{a_0^2}-\frac{4r}{a_0}\right)e^{-r/a_0}\text{d}r}\left(-{\left[\cos \theta \right]}_0^{\pi }\right)\times {\left[\varphi \right]}_0^{2\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta =-{\left[\cos \theta \right]}_0^{\pi }\\ =\frac{1}{16\times 2\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}r\left(4+\frac{r^2}{a_0^2}-\frac{4r}{a_0}\right)e^{-r/a_0}\text{d}r}\times \left(-{\left[-1-1\right]}_0^{\pi }\right)\times \left[2\pi -0\right]\cos \pi =-1\text{and}\cos 0=1\end{array}$

The above equation is simplified below.

    $\begin{array}{c}\langle \frac{1}{r}\rangle =\frac{4\pi }{16\times 2\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}r\left(4+\frac{r^2}{a_0^2}-\frac{4r}{a_0}\right)e^{-r/a_0}\text{d}r}\\ =\frac{4\pi }{16\times 2\pi a_0^3}\left(4{\displaystyle \underset{0}{\overset{\infty }{\int }}r{e}^{-r/a_0}\text{dr}}+\frac{1}{a_0^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{dr}}-\frac{4}{a_0}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^2{e}^{-r/a_0}\text{dr}}\right)\\ =\frac{4\pi }{16\times 2\pi a_0^3}\left(4\times \frac{1}{{\left(\frac{1}{a_0}\right)}^2}+\frac{1}{a_0^2}\times \frac{3!}{{\left(\frac{1}{a_0}\right)}^4}-\frac{4}{a_0}\times \frac{2!}{{\left(\frac{1}{a_0}\right)}^3}\right)\\ =\frac{4\pi }{16\times 2\pi a_0^3}\left(4a_0^2+6a_0^2-8a_0^2\right)\end{array}$

The above equation is further solved below.

    $\begin{array}{c}\langle \frac{1}{r}\rangle =\frac{4\pi }{16\times 2\pi a_0^3}\left(4a_0^2+6a_0^2-8a_0^2\right)\\ =\frac{4\pi }{16\times 2\pi a_0^3}\times 2a_0^2\\ =\frac{1}{4a_0}\end{array}$

Therefore, the expectation value of $1/r$ for a hydrogenic $\text{2s}$ orbital is $\frac{1}{4a_0}$.

(c)

Interpretation Introduction

Interpretation:

The expectation value of $1/r$ for a hydrogenic $\text{2p}$ orbital has to be calculated.

Concept introduction:

Expectation value is the average value of the large number of measurements of an observable.  The expectation value of a measurable parameter is calculated by the formula,

  $\langle \Omega \rangle ={\displaystyle \int {\psi }^{*}\widehat{\left(\Omega \right)}\psi \text{d}\tau }$

Where,

• $\psi$ is the wave function.
• $\Omega$ is an observable.

Answer to Problem 8A.8P

The expectation value of $1/r$ for a hydrogenic $\text{2p}$ orbital is $\frac{1}{8a_0}$.

Explanation of Solution

The normalized wavefunction for hydrogenic $\text{2p}$ (${\psi }_{\text{2p}}$) orbital is given below.

    ${\psi }_{\text{2p}}=\frac{1}{8\sqrt{2\pi }{a}_0^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}\sin \theta e^{i\varphi }$

The expectation value of $1/r$ for a hydrogenic $\text{2p}$ orbital is calculated as shown below.

    $\langle \frac{1}{r}\rangle ={\displaystyle \int {\psi }_{\text{2p}}^{*}}\left(\frac{1}{r}\right){\psi }_{\text{2p}}\text{d}\tau$                                                                                    (1)

Where,

• $\text{d}\tau$ is the volume element.

The value of $\text{d}\tau$ is given below.

    $\text{d}\tau =r^2\text{d}r\sin \theta \text{d}\theta \text{d}\varphi$

The ${\psi }_{\text{2p}}^{*}$ of a hydrogenic $\text{2p}$ orbital is given below.

    ${\psi }_{\text{2p}}^{*}=\frac{1}{8\sqrt{2\pi }{a}_0^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}\sin \theta e^{-i\varphi }$

Substitute the value of ${\psi }_{\text{2s}}^{*}$, $\text{d}\tau$ and ${\psi }_{\text{2s}}$ in (1).

    $\begin{array}{c}\langle \frac{1}{r}\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{8\sqrt{2\pi }{a}_0^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}\sin \theta e^{i\varphi }\left(\frac{1}{r}\right)\frac{1}{8\sqrt{2\pi }{a}_0^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}\sin \theta e^{-i\varphi }{r}^2\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{64\times 2\pi a_0^3\times a_0^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^3\theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}{e}^{-i\varphi }{e}^{i\varphi }\text{d}\varphi }}\\ =\frac{1}{64\times 2\pi a_0^3\times a_0^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^3\theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{64\times 2\pi a_0^3\times a_0^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{d}r}\times \frac{4}{3}\times {\left[\varphi \right]}_0^{2\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^3\theta }\text{d}\theta =\frac{4}{3}\end{array}$

The above equation is simplified below.

    $\begin{array}{c}\langle \frac{1}{r}\rangle =\frac{1}{64\times 2\pi a_0^3\times a_0^2}\times \frac{4}{3}\times 2\pi {\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{d}r}\\ =\frac{4}{64\times 3a_0^5}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-r/a_0}\text{d}r}\\ =\frac{4}{64\times 3a_0^5}\left(\frac{3!}{{\left(\frac{1}{a_0}\right)}^4}\right)\\ =\frac{4}{64\times 3a_0^5}\times 6a_0^4\end{array}$

The above equation is further solved below.

    $\langle \frac{1}{r}\rangle =\frac{1}{8a_0}$

Therefore, the expectation value of $1/r$ for a hydrogenic $\text{2p}$ orbital is $\frac{1}{8a_0}$.

(d)

Interpretation Introduction

Interpretation:

Whether the equation, $\langle 1/r\rangle =1/\langle r\rangle$ holds true has to be determined.

Concept introduction:

Expectation value is the average value of the large number of measurements of an observable.  The expectation value of a measurable parameter is calculated by the formula,

  $\langle \Omega \rangle ={\displaystyle \int {\psi }^{*}\widehat{\left(\Omega \right)}\psi \text{d}\tau }$

Where,

• $\psi$ is the wave function.
• $\Omega$ is an observable.

Answer to Problem 8A.8P

The equation, $\langle 1/r\rangle =1/\langle r\rangle$ holds true.

Explanation of Solution

The normalized wavefunction for hydrogenic $\text{1s}$ (${\psi }_{\text{1s}}$) orbital is given below.

    ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}$

The expectation value of $1/\langle r\rangle$ for a hydrogenic $\text{1s}$ orbital is calculated as shown below.

    $\frac{1}{\langle r\rangle }={\displaystyle \int {\psi }_{1s}^{*}}\left(\frac{1}{r}\right){\psi }_{1s}\text{d}\tau$                                                                                     (1)

Where,

• $\text{d}\tau$ is the volume element.

The value of $\text{d}\tau$ is given below.

    $\text{d}\tau =r^2\text{d}r\sin \theta \text{d}\theta \text{d}\varphi$

The ${\psi }_{1s}^{*}$ of a hydrogenic $\text{1s}$ orbital is the same as the ${\psi }_{\text{1s}}$.

    ${\psi }_{1s}^{*}=\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}$

Substitute the value of ${\psi }_{1s}^{*}$, $\text{d}\tau$ and ${\psi }_{\text{1s}}$ in (1).

    $\begin{array}{c}\frac{1}{\langle r\rangle }={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}\left(\frac{1}{r}\right)\frac{1}{\sqrt{\pi }{a}_0^{3/2}}{e}^{-r/a_0}{r}^2\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta {\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{d}\varphi }}\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r\times \left(-{\left[\cos \theta \right]}_0^{\pi }\right)\times {\left[\varphi \right]}_0^{2\pi }}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta }\text{d}\theta =-{\left[\cos \theta \right]}_0^{\pi }\\ =\frac{1}{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r\times \left(-{\left[-1-1\right]}_0^{\pi }\right)\times {\left[\varphi \right]}_0^{2\pi }}\cos \pi =-1\text{and}\cos 0=1\end{array}$

The above equation is simplified below.

    $\begin{array}{c}\frac{1}{\langle r\rangle }=\frac{4\pi }{\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2r/a_0}r\text{d}r}\\ =\frac{4}{a_0^3}\left(\frac{1!}{{\left(\frac{2}{a_0}\right)}^2}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{\text{n}}{e}^{-\alpha x}\text{d}r}=\frac{n!}{{\left(\alpha \right)}^{n+1}}\\ =\frac{4}{a_0^3}\times \frac{a_0{}^2}{4}\\ =\frac{1}{a_0}\end{array}$

Therefore, the value of $\langle 1/r\rangle =1/\langle r\rangle$ is same.

Therefore, the equation, $\langle 1/r\rangle =1/\langle r\rangle$ holds true.