Chapter 28, Problem 52E

22782-28-52E                                                                                    AID: 1825 | 15/4/2020

To determine

The average value of $z^2$ and $r^2$.

Answer to Problem 52E

The average value of $\langle z^2\rangle$ is $z^2$ and the average value of $\langle r^2\rangle$ is $12a_0^2$.

Explanation of Solution

Write the expression for the wave function in $2s$ state.

    $\psi \left(r\right)=\frac{1}{\sqrt{32\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$        (I)

Write the expression the average value of $z^2$.

    $\langle z^2\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{z}^2{\left|\psi \left(r\right)\right|}^{2}dV}$        (II)

Write the expression the average value of $r^2$.

    $\langle r^2\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^2{\left|\psi \left(r\right)\right|}^{2}dV}$        (III)

Conclusion:

Substitute $\frac{1}{\sqrt{32\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$ for $\psi \left(r\right)$ and $4\pi r^{2}dr$ for $dV$ in expression (II).

    $\langle z^2\rangle =\frac{1}{32\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{z}^{2}4\pi r^2{\left(2-\frac{r}{a_0}\right)}^2{e}^{-r/a_0}dr}$

Substitute $2$ for $\left(2-\frac{r}{a_0}\right)$ in the above expression.

  $\langle z^2\rangle =\frac{1}{32\pi a_0^3}\left(16\pi z^2\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^2{e}^{-r/a_0}dr}$

Use the following identity to evaluate the above expression.

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

Rearrange the expression for $\langle z^2\rangle$.

    $\begin{array}{c}\langle z^2\rangle =\frac{1}{32\pi a_0^3}\left(16\pi z^2\right)\left(2a_0^3\right)\\ =z^2\end{array}$

Substitute $\frac{1}{\sqrt{32\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$ for $\psi \left(r\right)$ and $4\pi r^{2}dr$ for $dV$ in expression (III).

    $\langle r^2\rangle =\frac{1}{32\pi a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^{2}4\pi r^2{\left(2-\frac{r}{a_0}\right)}^2{e}^{-r/a_0}dr}$

Substitute $2$ for $\left(2-\frac{r}{a_0}\right)$ in the above expression.

  $\langle r^2\rangle =\frac{1}{32\pi a_0^3}\left(16\pi \right){\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-r/a_0}dr}$

Use the following identity to evaluate the above expression.

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

Rearrange the expression for $\langle r^2\rangle$.

    $\begin{array}{c}\langle r^2\rangle =\frac{1}{32\pi a_0^3}\left(16\pi \right)\left(24a_0^5\right)\\ =12a_0^2\end{array}$

Thus, the average value of $\langle z^2\rangle$ is $z^2$ and the average value of $\langle r^2\rangle$ is $12a_0^2$.