Chapter 8, Problem 8A.2P

(a)

Interpretation Introduction

Interpretation:

The hydrogenic $1\text{s}$ and $2\text{s}$ orbitals are mutually orthogonal has to be shown by explicit integration.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 8A.2P

The hydrogenic $1\text{s}$ and $2\text{s}$ orbitals are mutually orthogonal.

Explanation of Solution

The wavefunction of $1\text{s}$ orbital is given by the expression shown below.

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

Where,

• $Z$ is the atomic number of the atom.
• $a_0$ is the Bohr’s radius.
• $r$ is the distance electron from nucleus.

The wavefunction of $2\text{s}$ orbital is given by the expression shown below.

  ${\psi }_{\text{1s}}={\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{1}{4\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{3/2}\left(2-\frac{Zr}{a_0}\right){\text{e}}^{-Zr/2a_0}$                                                 (2)

The orthogonality of the two wave functions can be shown as follows.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\Psi }_{2s}^{*}{\Psi }_{1s}}}}{r}^2\sin \theta drd\theta d\varphi =0$                                                                    (3)

Substitute the value in the above function as follows.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\Psi }_{2s}^{*}{\Psi }_{1s}}}}{r}^2\sin \theta drd\theta d\varphi ={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\begin{array}{l}\left(2{\left(\frac{1}{4\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{3/2}{\text{e}}^{-Zr/a_0}\right)\times \\ \left(\begin{array}{l}{\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{1}{4\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{3/2}\times \\ \left(2-\frac{Zr}{a_0}\right){\text{e}}^{-Zr/2a_0}\end{array}\right)\end{array}\right)}}}{r}^2\sin \theta drd\theta d\varphi \\ =2\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left(\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2-\frac{Zr}{a_0}\right)r^{2}dr}\times \\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\end{array}\right)\end{array}$

Substitute the value of left hand side of the above equation equal to zero for a orthogonal function.

  $\begin{array}{c}0=2\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2-\frac{Zr}{a_0}\right)r^{2}dr}{\left[-\cos \theta \right]}_0^{\pi }{\left[\varphi \right]}_0^{2\pi }\right)\\ =2\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2r^2-\frac{Z{r}^3}{a_0}\right)dr}\left(-\left[\cos \pi -\cos 0\right]\left[2\pi -0\right]\right)\right)\end{array}$

Substitute the values of $\cos \pi =-1$ and $\cos 0=1$ in the above equation.

  $\begin{array}{c}0=-2\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2r^2-\frac{Z{r}^3}{a_0}\right)dr}\left(\left[-1-1\right]\left[2\pi -0\right]\right)\right)\\ =-2\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2r^2-\frac{Z{r}^3}{a_0}\right)dr}\left(\left[-2\right]\left[2\pi -0\right]\right)\right)\\ =4\left(2\pi \right)\left(\frac{1}{4\pi }\right){\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2r^2-\frac{Z{r}^3}{a_0}\right)dr}\right)\\ =2{\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-3Zr/2a_0}\right)\left(2r^2-\frac{Z{r}^3}{a_0}\right)dr}\right)\end{array}$

The above equation can be simplified as shown below.

  $0=2{\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left(2r^2\left({\text{e}}^{-3Zr/2a_0}\right)-\frac{Z{r}^3\left({\text{e}}^{-3Zr/2a_0}\right)}{a_0}\right)dr}\right)$                             (4)

The defiant integration for product of linear function and exponential function is shown below.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{\text{e}}^{-bx}dx}=\frac{n!}{b^{n+1}}$

The equation (4) is expanding similarly as the above relation. The value of $n$ for first integration term is $2$ and the value of $n$ for second integration term is $3$. The value of $b$ for both integration terms is $3Z/2a_0$.

    $\begin{array}{c}0=-2{\left(\frac{1}{8}\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^3\left(2\left(\frac{2!}{{\left(3Z/2a_0\right)}^3}\right)-\frac{Z}{a_0}\left(\frac{3!}{{\left(3Z/2a_0\right)}^4}\right)\right)\\ 0=\frac{\sqrt{2}}{2a^3}\left[2\left(\frac{2\times 8a_0^3}{27Z^3}\right)-\frac{Z}{a_0}\left(\frac{6\times 16a_0^4}{81Z^4}\right)\right]\\ 0=\frac{\sqrt{2}}{2a^3}\left[\left(\frac{32a_0^3}{27Z^3}\right)-\left(\frac{32a_0^3}{27Z^3}\right)\right]\\ 0=0\end{array}$

The right hand side and left hand side of the above expression is zero. Therefore, the hydrogenic $1\text{s}$ and $2\text{s}$ orbitals are mutually orthogonal.

(b)

Interpretation Introduction

Interpretation:

The hydrogenic $2{\text{p}}_x$ and $2{\text{p}}_y$ orbitals are mutually orthogonal has to be shown by explicit integration.

Concept introduction:

For the orthogonality of the two different wave functions, the product of the wave functions is integrated over the entire limits. It is expressed by the equation as given below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\Psi }_m{\Psi }_n}=0$

Where, ${\Psi }_m$ and ${\Psi }_n$ are two different wave functions. The essential condition for two wave functions to be orthogonal to each other is $m\ne n$.

Answer to Problem 8A.2P

The hydrogenic $2{\text{p}}_x$ and $2{\text{p}}_y$ orbitals are mutually orthogonal.

Explanation of Solution

The wavefunction of $2{\text{p}}_x$ orbital is given by the expression shown below.

  ${\psi }_{{\text{2p}}_x}={\left(\frac{1}{8\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{5/2}r\sin \theta e^{-i\varphi }{\text{e}}^{-Zr/2a_0}$                                                         (1)

Where,

• $Z$ is the atomic number of the atom.
• $a_0$ is the Bohr’s radius.
• $r$ is the distance electron from nucleus.

The wavefunction of $2{\text{p}}_y$ orbital is given by the expression shown below.

  ${\psi }_{{\text{2p}}_y}=-{\left(\frac{1}{8\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{5/2}r\sin \theta e^{+i\varphi }{\text{e}}^{-Zr/2a_0}$                                                      (2)

The orthogonality of the two wave functions can be shown as follows.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\psi }_{\text{1s}}^{*}{\psi }_{\text{2s}}}}}{r}^2\sin \theta drd\theta d\varphi =0$                                                                      (3)

Substitute the value in the above function as follows.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\psi }_{\text{1s}}^{*}{\psi }_{\text{2s}}}}}{r}^2\sin \theta drd\theta d\varphi ={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\begin{array}{l}{\left(\frac{1}{8\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{5/2}\times \\ r\sin \theta e^{-i\varphi }{\text{e}}^{-Zr/2a_0}\times \\ -{\left(\frac{1}{8\pi }\right)}^{1/2}{\left(\frac{Z}{a_0}\right)}^{5/2}\\ \times r\sin \theta e^{-i\varphi }{\text{e}}^{-Zr/2a_0}\end{array}\right)}}}{r}^2\sin \theta drd\theta d\varphi$

The above equation is further simplified to an expression as shown below.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\psi }_{\text{1s}}^{*}{\psi }_{\text{2s}}}}}{r}^2\sin \theta drd\theta d\varphi =\left(\frac{1}{8\pi }\right){\left(\frac{Z}{a_0}\right)}^5\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-Zr/a_0}\right)r^{3}dr}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^2\theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\right)$  (4)

The integration of $\sin \theta$ with a definite limit is shown below.

  ${\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^2\theta d\theta }=0$

Substitute the value of above expression in the equation (4).

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\psi }_{\text{1s}}^{*}{\psi }_{\text{2s}}}}}{r}^2\sin \theta drd\theta d\varphi =\left(\frac{1}{8\pi }\right){\left(\frac{Z}{a_0}\right)}^5\left({\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\text{e}}^{-Zr/a_0}\right)r^{3}dr}\left(0\right){\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\right)\\ =0\end{array}$

The above equation is same as equation (3). Therefore, the hydrogenic $2{\text{p}}_x$ and $2{\text{p}}_y$ orbitals are mutually orthogonal.