Chapter 36, Problem 42P

To determine

The ground state hydrogen atom wave function is solution to which equation.

Answer to Problem 42P

The ground state hydrogen atom wave function is solution to Schrodinger’s equation in spherical coordinatesis $\frac{-{\hslash }^2}{2m{r}^2}\left\{\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)+\left[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \psi }{\partial \theta }\right)+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2\psi }{\partial {\varphi }^2}\right]\right\}+U\left(r\right)\psi =E\psi$ .

Explanation of Solution

Given:

The ground state hydrogen atom wave functionis ${\psi }_{100}={\pi }^{-1/2}{\left(Z/a_o\right)}^{3/2}{e}^{-Zr/a_o}$ .

Formula Used:

The expression for Schrodinger’s equation in spherical coordinates is given by,

  $\frac{-{\hslash }^2}{2m{r}^2}\left\{\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)+\left[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \psi }{\partial \theta }\right)+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2\psi }{\partial {\varphi }^2}\right]\right\}+U\left(r\right)\psi =E\psi$

The expression $U\left(r\right)$ for is given by,

  $U\left(r\right)=-\frac{kZ{e}^2}{r}$

The expression for $a_o$ is given by,

  $a_o=\frac{{\hslash }^2}{mk{e}^2}$

Calculation:

The ground state is having spherical symmetry. So,

  $\left[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \psi }{\partial \theta }\right)+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2\psi }{\partial {\varphi }^2}\right]=0$

The value of $\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)$ is calculated as,

  $\begin{array}{c}\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)=\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\left(\frac{{\left(Z/a_o\right)}^{3/2}{e}^{-Zr/a_o}}{{\pi }^{1/2}}\right)\right)\\ =\frac{\partial }{\partial r}\left(r^2\left(\frac{{\left(Z/a_o\right)}^{3/2}{e}^{-Zr/a_o}}{{\pi }^{1/2}}\right)\times \frac{-Z}{a_o}\right)\\ =\frac{{\left(Z/a_o\right)}^{3/2}}{{\pi }^{1/2}}\times \frac{-Z}{a_o}\left[2r{e}^{-Zr/a_o}+r^2{e}^{-Zr/a_o}\times \frac{-Z}{a_o}\right]\\ =\frac{r{e}^{-Zr/a_o}}{{\pi }^{1/2}}{\left(\frac{Z}{a_o}\right)}^{5/2}\left[\frac{rZ}{a_o}-2\right]\end{array}$

The Schrodinger’s equation in spherical coordinates is calculated as,

  $\begin{array}{c}\frac{-{\hslash }^2}{2m{r}^2}\left\{\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)+\left[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \psi }{\partial \theta }\right)+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2\psi }{\partial {\varphi }^2}\right]\right\}+U\left(r\right)\psi =E\psi \\ \frac{-{\hslash }^2}{2m{r}^2}\left\{\frac{r{e}^{-Zr/a_o}}{{\pi }^{1/2}}{\left(\frac{Z}{a_o}\right)}^{5/2}\left[\frac{rZ}{a_o}-2\right]+0\right\}+\left(-\frac{kZ{e}^2}{r}\right)\left(\frac{e^{-Zr/a_o}}{{\pi }^{1/2}}{\left(\frac{Z}{a_o}\right)}^{3/2}\right)=E\frac{e^{-Zr/a_o}}{{\pi }^{1/2}}{\left(\frac{Z}{a_o}\right)}^{3/2}\\ \frac{-{\hslash }^2}{2mr}\left(\frac{Z}{a_o}\right)\left[\frac{rZ}{a_o}-2\right]+\left(-\frac{kZ{e}^2}{r}\right)=E\\ \frac{-{\hslash }^2}{2mr}\left(\frac{Z}{\frac{{\hslash }^2}{mk{e}^2}}\right)\left[\frac{rZ}{\frac{{\hslash }^2}{mk{e}^2}}-2\right]+\left(-\frac{kZ{e}^2}{r}\right)=E\end{array}$

Solve further as,

  $\begin{array}{c}\frac{-{\hslash }^2}{2mr}\left(\frac{Z}{\frac{{\hslash }^2}{mk{e}^2}}\right)\left[\frac{rZ}{\frac{{\hslash }^2}{mk{e}^2}}-2\right]+\left(-\frac{kZ{e}^2}{r}\right)=E\\ E=-\frac{Z^2{k}^2{e}^{4}m}{2{\hslash }^2}\end{array}$

The above expression for $E$ is correct for the ground state energy.

Conclusion:

Therefore, the ground state hydrogen atom wave function ${\psi }_{100}={\pi }^{-1/2}{\left(Z/a_o\right)}^{3/2}{e}^{-Zr/a_o}$ is a solution to Schrodinger’s equation in spherical coordinates is $\frac{-{\hslash }^2}{2m{r}^2}\left\{\frac{\partial }{\partial r}\left(r^2\frac{\partial \psi }{\partial r}\right)+\left[\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \psi }{\partial \theta }\right)+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2\psi }{\partial {\varphi }^2}\right]\right\}+U\left(r\right)\psi =E\psi$ .