Chapter 28, Problem 41E

To determine

Whether the wave function for $2p$ state is normalized or not.

Answer to Problem 41E

The wave function for $2p$ state is normalized.

Explanation of Solution

Write the  expression of the wave function for $2p$ state of hydrogen atom.

    $\Psi \left(r\right)=\frac{1}{{\pi }^{1/2}{\left(2a_0\right)}^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}Cos\theta$        (I)

Here, $a_0$ is first Bohr’s radius, $r$ is the atomic radius and $\theta$ is the azimuthal angle.

Write the expression for the condition of normalization of the wave function.

    ${\displaystyle \int {\left|\Psi \left(r\right)\right|}^2{d}^{3}r}=1$        (II)

Conclusion:

Substitute $\frac{1}{{\pi }^{1/2}{\left(2a_0\right)}^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}Cos\theta$ for $\Psi \left(r\right)$ on the left-hand side of expression (II).

    ${\displaystyle \int {\left|\Psi \left(r\right)\right|}^2{d}^{3}r}={{\displaystyle \underset{0}{\overset{\infty }{\int }}\left|\frac{1}{{\pi }^{1/2}{\left(2a_0\right)}^{3/2}}\frac{r}{a_0}{e}^{-r/2a_0}Cos\theta \right|}}^2{r}^{2}dr{\displaystyle \underset{0}{\overset{\pi }{\int }}Sin\theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }$

Simplify the above expression.

  ${\displaystyle \int {\left|\Psi \left(r\right)\right|}^2{d}^{3}r}=\frac{1}{32\pi a_0^5}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-r/a_0}dr{\displaystyle \underset{0}{\overset{\pi }{\int }}Co{s}^2\theta Sin\theta d\theta }{\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }}$

Evaluate the last integral on the right-hand side of above expression.

    ${\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }=2\pi$

Evaluate the second integral on the right-hand side of expression.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\pi }{\int }}Co{s}^2\theta Sin\theta d\theta }=-{\displaystyle \underset{0}{\overset{\pi }{\int }}Co{s}^2\theta d\left(Cos\theta \right)}\\ =\text{}-\frac{1}{3}{\left[Co{s}^3\theta \right]}_0^{\pi }\\ =-\frac{1}{3}\left[-1-1\right]\\ =\frac{2}{3}\end{array}$

Substitute $x$ for $r/a_0$ in the first integral on the right-hand side of expression.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-r/a_0}dr}={\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(a_{0}x\right)}^4{e}^{-x}\left(a_{0}dx\right)}\\ =\left(a_0^5\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^4{e}^{-x}dx}\end{array}$

Use the identity given below to evaluate the integral on the right-hand side of above expression.

    ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-x}dx}=\Gamma \left(n\right)$

Simplify the expression.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-r/a_0}dr}=\left(a_0^5\right)\Gamma \left(4\right)\\ =24a_0^5\end{array}$

Substitute $24a_0^5$ for ${\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-r/a_0}dr}$, $\frac{2}{3}$ for ${\displaystyle \underset{0}{\overset{\pi }{\int }}Co{s}^2\theta Sin\theta d\theta }$ and $2\pi$ for ${\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }$ in the wave function expression.

    $\begin{array}{c}{\displaystyle \int {\left|\Psi \left(r\right)\right|}^2{d}^{3}r}=\frac{1}{32\pi a_0^5}\left(24a_0^5\right)\left(\frac{2}{3}\right)\left(2\pi \right)\\ =1\end{array}$

Thus, the wave function for $2p$ state is normalized.