Chapter 8, Problem 8A.2ST

Interpretation Introduction

Interpretation:

The mean radius of a $3\text{s}$ orbital has to be evaluated by integration.

Concept introduction:

Hydrogen is single electron species.  The mean radius of a $3\text{s}$ orbital is the distance between the nucleus and the electron present in $3\text{s}$ orbital.  The mean radius of $3\text{s}$ orbital is calculated by the following formula.

    $\langle r\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{R}_{3,0}^2}\text{d}r$

Answer to Problem 8A.2ST

The mean radius of a $3\text{s}$ orbital is $\frac{27a_{\text{ο}}}{2Z}$.

Explanation of Solution

The radial distribution function of $3\text{s}$ orbital is shown below.

    $R_{3,0}=\frac{1}{{243}^{1/2}}{\left(\frac{Z}{a_{\text{ο}}}\right)}^{3/2}\left(6-6\rho +{\rho }^2\right){\text{e}}^{-\rho /2}$                                                          (1)

The value of $\rho$ is calculated as shown below.

    $\rho =\frac{2Zr}{n{a}_{\text{ο}}}$                                                                                                         (2)

The value of $n$ for $3\text{s}$ orbital is $3$.

Substitute the value of $n$ in equation (2).

    $\rho =\frac{2Zr}{3a_{\text{ο}}}$

Substitute the value of $\rho$ is calculated as shown below.

    $\begin{array}{c}{R}_{3,0}=\frac{1}{{243}^{1/2}}{\left(\frac{Z}{a_{\text{ο}}}\right)}^{3/2}\left(6-\frac{6\times 2Zr}{3a_{\text{ο}}}+{\left(\frac{2Zr}{3a_{\text{ο}}}\right)}^2\right){\text{e}}^{-2Zr/2\times 3a_{\text{ο}}}\\ =\frac{1}{{243}^{1/2}}{\left(\frac{Z}{a_{\text{ο}}}\right)}^{3/2}\left(6-\frac{4Zr}{a_{\text{ο}}}+\frac{4Z^2{r}^2}{9a_{\text{ο}}^2}\right){\text{e}}^{-Zr/3a_{\text{ο}}}\end{array}$

Where,

• $a_{\text{ο}}$ is Bohr radius.
• $r$ is the radius of hydrogen atom.

The mean radius of a $3\text{s}$ orbital is calculated by the following formula.

    $\langle r\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{R}_{3,0}^2}\text{d}r$                                                                                               (3)

The value of $R_{3,0}^2$ is calculated as shown below.

    $\begin{array}{c}{R}_{3,0}^2={\left(\frac{1}{{243}^{1/2}}{\left(\frac{Z}{a_{\text{ο}}}\right)}^{3/2}\left(6-\frac{4Zr}{a_{\text{ο}}}+\frac{4Z^2{r}^2}{9a_{\text{ο}}^2}\right){\text{e}}^{-Zr/3a_{\text{ο}}}\right)}^2\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3{\left(6-\frac{4Zr}{a_{\text{ο}}}+\frac{4Z^2{r}^2}{9a_{\text{ο}}^2}\right)}^2{\text{e}}^{-2Zr/3a_{\text{ο}}}\left(\because {\left(a+b+c\right)}^2=\left(\begin{array}{l}{a}^2+b^2+c^2\\ +2ab+2bc+2ca\end{array}\right)\right)\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(36+\frac{16Z^2{r}^2}{a_{\text{ο}}^2}+\frac{16Z^4{r}^4}{81a_{\text{ο}}^4}-\frac{48Zr}{a_{\text{ο}}}-\frac{32Z^3{r}^3}{9a_{\text{ο}}^3}+\frac{48Z^2{r}^2}{9a_{\text{ο}}^2}\right){\text{e}}^{-2Zr/3a_{\text{ο}}}\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(36+\frac{192Z^2{r}^2}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^4}{81a_{\text{ο}}^4}-\frac{48Zr}{a_{\text{ο}}}-\frac{32Z^3{r}^3}{9a_{\text{ο}}^3}\right){\text{e}}^{-2Zr/3a_{\text{ο}}}\end{array}$

Substitute the value of $R_{3,0}^2$ in equation (3).

    $\begin{array}{c}\langle r\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(36+\frac{192Z^2{r}^2}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^4}{81a_{\text{ο}}^4}-\frac{48Zr}{a_{\text{ο}}}-\frac{32Z^3{r}^3}{9a_{\text{ο}}^3}\right){\text{e}}^{-2Zr/3a_{\text{ο}}}}\text{d}r\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(36r^3+\frac{192Z^2{r}^5}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^7}{81a_{\text{ο}}^4}-\frac{48Z{r}^4}{a_{\text{ο}}}-\frac{32Z^3{r}^6}{9a_{\text{ο}}^3}\right){\text{e}}^{-2Zr/3a_{\text{ο}}}}\text{d}r\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(36r^3+\frac{192Z^2{r}^5}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^7}{81a_{\text{ο}}^4}-\frac{48Z{r}^4}{a_{\text{ο}}}-\frac{32Z^3{r}^6}{9a_{\text{ο}}^3}\right){\text{e}}^{-2Zr/3a_{\text{ο}}}}\text{d}r\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\begin{array}{l}36r^3{\text{e}}^{-2Zr/3a_{\text{ο}}}+\frac{192Z^2{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}}{81a_{\text{ο}}^4}\\ -\frac{48Z{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}}{a_{\text{ο}}}-\frac{32Z^3{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^3}\end{array}\right)}\text{d}r\end{array}$

Separate the integrals as shown below.

    $\begin{array}{c}\langle r\rangle =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\begin{array}{l}36r^3{\text{e}}^{-2Zr/3a_{\text{ο}}}+\frac{192Z^2{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^2}+\frac{16Z^4{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}}{81a_{\text{ο}}^4}\\ -\frac{48Z{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}}{a_{\text{ο}}}-\frac{32Z^3{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^3}\end{array}\right)}\text{d}r\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}36r^3{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}+{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{192Z^2{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^2}\text{d}r}+{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{16Z^4{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}}{81a_{\text{ο}}^4}\text{d}r}-\\ {\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{48Z{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}}{a_{\text{ο}}}\text{d}r}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{32Z^3{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}}{9a_{\text{ο}}^3}\text{d}r}\end{array}\right)\\ =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(\begin{array}{l}36{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}+\frac{192Z^2}{9a_{\text{ο}}^2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}+\frac{16Z^4}{81a_{\text{ο}}^4}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}-\\ \frac{48Z}{a_{\text{ο}}}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}-\frac{32Z^3}{9a_{\text{ο}}^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}\end{array}\right)\end{array}$

All the above separated integrals are solved by the identity shown below.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{\text{e}}^{-ax}\text{d}x}=\frac{n!}{a^{n+1}}$                                                                                            (3)

The value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}$ is calculated by using equation (3) as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}=\frac{3!}{{\left(2Z/3a_{\text{ο}}\right)}^4}\\ =\frac{6a_{\text{ο}}^4\times 3^4}{2^4{Z}^4}\end{array}$

The value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}$ is calculated by using equation (3) as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^5{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}=\frac{5!}{{\left(2Z/3a_{\text{ο}}\right)}^6}\\ =\frac{120\times 3^6\times a_{\text{ο}}^6}{2^6{Z}^6}\end{array}$

The value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}$ is calculated by using equation (3) as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^7{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}=\frac{7!}{{\left(2Z/3a_{\text{ο}}\right)}^8}\\ =\frac{5040a_{\text{ο}}^8\times 3^8}{2^8{Z}^8}\end{array}$

The value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}$ is calculated by using equation (3) as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}=\frac{4!}{{\left(2Z/3a_{\text{ο}}\right)}^5}\\ =\frac{24\times a_{\text{ο}}^5\times 3^5}{2^5{Z}^5}\end{array}$

The value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}$ is calculated by using equation (3) as shown below.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^6{\text{e}}^{-2Zr/3a_{\text{ο}}}\text{d}r}=\frac{6!}{{\left(2Z/3a_{\text{ο}}\right)}^7}\\ =\frac{720\times a_{\text{ο}}^7\times 3^7}{2^7{Z}^7}\end{array}$

Substitute all the values of separated integrals in the mean radius of $3\text{s}$ orbital.

    $\begin{array}{c}\langle r\rangle =\frac{1}{243}{\left(\frac{Z}{a_{\text{ο}}}\right)}^3\left(\begin{array}{l}36\times \frac{6a_{\text{ο}}^4\times 3^4}{2^4{Z}^4}+\frac{192Z^2}{9a_{\text{ο}}^2}\times \frac{120\times 3^6\times a_{\text{ο}}^6}{2^6{Z}^6}\\ +\frac{16Z^4}{81a_{\text{ο}}^4}\times \frac{5040a_{\text{ο}}^8\times 3^8}{2^8{Z}^8}-\\ \frac{48Z}{a_{\text{ο}}}\times \frac{24\times a_{\text{ο}}^5\times 3^5}{2^5{Z}^5}-\frac{32Z^3}{9a_{\text{ο}}^3}\times \frac{720\times a_{\text{ο}}^7\times 3^7}{2^7{Z}^7}\end{array}\right)\\ =\left(\begin{array}{l}36\times \frac{6a_{\text{ο}}^4\times 3^4}{2^4{Z}^4}\times {\left(\frac{Z}{a_{\text{ο}}}\right)}^3\times \frac{1}{3^5}+\frac{192Z^2}{9a_{\text{ο}}^2}\times \frac{120\times 3^6\times a_{\text{ο}}^6}{2^6{Z}^6}\times {\left(\frac{Z}{a_{\text{ο}}}\right)}^3\times \frac{1}{3^5}\\ +\frac{16Z^4}{81a_{\text{ο}}^4}\times \frac{5040a_{\text{ο}}^8\times 3^8}{2^8{Z}^8}\times \frac{1}{3^5}\times {\left(\frac{Z}{a_{\text{ο}}}\right)}^3-\\ \frac{48Z}{a_{\text{ο}}}\times \frac{24\times a_{\text{ο}}^5\times 3^5}{2^5{Z}^5}\times \frac{1}{3^5}\times {\left(\frac{Z}{a_{\text{ο}}}\right)}^3-\frac{32Z^3}{9a_{\text{ο}}^3}\times \frac{1}{3^5}\times \frac{720\times a_{\text{ο}}^7\times 3^7}{2^7{Z}^7}\times {\left(\frac{Z}{a_{\text{ο}}}\right)}^3\end{array}\right)\\ =\left(\frac{9a_{\text{ο}}}{2Z}+\frac{240a_{\text{ο}}}{2Z}+\frac{210a_{\text{ο}}}{2Z}-\frac{72a_{\text{ο}}}{2Z}-\frac{360a_{\text{ο}}}{2Z}\right)\\ =\frac{27a_{\text{ο}}}{2Z}\end{array}$

Hence, the mean radius of a $3\text{s}$ orbital is $\frac{27a_{\text{ο}}}{2Z}$.