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Chapter 7: Problem 179

The wave function for the 2\(p_{z}\) orbital in the hydrogen atom is $$\psi_{2 p_{z}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta$$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p_{z}}^{2}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis ) and for } \theta=90^{\circ}\right.\) (xy plane).

Short Answer

Expert verified
For \(\theta = 0^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = \frac{1}{32\pi}\mathrm{e}^{-1}\). For \(\theta = 90^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = 0\).

Step by step solution

01

Plug in the values for \(r\) and \(a_0\) into the wave function formula

We are given the wave function for the 2\(p_{z}\) orbital as: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta \] It is also given that \(r = a_0\), and since \(\sigma = Z\left(\frac{r}{a_0}\right)\), we can plug in these values: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} Z \mathrm{e}^{-Z / 2} \cos \theta \] Note that for a hydrogen atom, \(Z = 1\), so the equation becomes: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos \theta \]
02

Calculate the square of the wave function for \(\theta = 0^\circ\)

At \(\theta = 0^\circ\), we have: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos 0^\circ \] Since \(\cos 0^\circ = 1\), the wave function is: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \] Now, we can find the square of the wave function: \[ \psi_{2 p_{z}}^2 = \left(\frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2}\right)^2 = \frac{1}{32\pi}\mathrm{e}^{-1} \]
03

Calculate the square of the wave function for \(\theta = 90^\circ\)

At \(\theta = 90^\circ\), we have: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos 90^\circ \] Since \(\cos 90^\circ = 0\), the wave function is: \[ \psi_{2 p_{z}} = 0 \] Thus, the square of the wave function is: \[ \psi_{2 p_{z}}^2 = (0)^2 = 0 \]
04

Results

For \(\theta = 0^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = \frac{1}{32\pi}\mathrm{e}^{-1}\). For \(\theta = 90^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = 0\).

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