Answer
\( \psi_{2p_z}^2 \) at \( r = a_0 \) for \( \theta = 0^\circ \) is given by \( \frac{1}{32\pi}\left(\frac{Z}{a_0}\right)^3\sigma^2 e^{-a} \).
\( \psi_{2p_z}^2 \) at \( r = a_0 \) for \( \theta = 90^\circ \) is equal to 0.
Work Step by Step
To calculate the value of \( \psi_{2p_z}^2 \) at \( r = a_0 \) for \( \theta = 0^\circ \) and \( \theta = 90^\circ \), we need to substitute the given values into the wave function expression and square it.
Let's start with \( \theta = 0^\circ \) (z-axis):
\[
\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-a/2}\cos\theta
\]
Substituting \( r = a_0 \) and \( \theta = 0^\circ \) into the equation:
\[
\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-a/2}\cos(0^\circ)
\]
Since \( \cos(0^\circ) = 1 \), the equation simplifies to:
\[
\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-a/2}
\]
Now, let's calculate \( \psi_{2p_z}^2 \) for \( \theta = 0^\circ \):
\[
\psi_{2p_z}^2 = \left(\frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-a/2}\right)^2
\]
Simplifying further:
\[
\psi_{2p_z}^2 = \frac{1}{32\pi}\left(\frac{Z}{a_0}\right)^3\sigma^2 e^{-a}
\]
Now, let's calculate \( \psi_{2p_z}^2 \) for \( \theta = 90^\circ \) (xy-plane):
\[
\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-a/2}\cos(90^\circ)
\]
Since \( \cos(90^\circ) = 0 \), the equation simplifies to:
\[
\psi_{2p_z} = 0
\]
Therefore, \( \psi_{2p_z}^2 \) for \( \theta = 90^\circ \) is also equal to 0.
