91Ó°ÊÓ

Assume the electron's wave function is an eigenfunction of the total angular momentum operator \(\hat{L}^2\). The eigenvalue of \(\hat{L}^2\) can be expressed as: \[\hat{L}^2 \psi = L(L + 1)\hbar^2 \psi\] where \(L = \ell\) is the azimuthal quantum number, and \(\hbar\) is the reduced Planck constant. For this problem, we are given that \(n = 2\) and \(\ell = 1\). Therefore, we can calculate the magnitude of angular momentum \(L\) by plugging the values of \(\ell\) and \(\hbar\) into the equation: \[L = \sqrt{\ell(\ell + 1)} \hbar\] \[L = \sqrt{1(1 + 1)} \hbar\]

Now, let's find the magnitude of L by simplifying the expression: \[L = \sqrt{2} \hbar\] This is the magnitude of the angular momentum for an electron with \(n = 2\) and \(\ell = 1\).

The allowed values of \(L_z\) are given by the expression: \[m_\ell \hbar\] where \(m_\ell\) ranges from \(-\ell\) to \(\ell\) in integer steps. For our problem, \(\ell = 1\), so the allowed values for \(L_z\) are: \[L_z = -1 \hbar, 0, +1 \hbar\] These are the allowed values of \(L_z.\)

To find the angles \(\theta\) between the positive z-axis and \(\overline{\mathbf{L}}\), we will use the relation: \[L_z = L \cos{\theta}\] For each allowed value of \(L_z\), let's find the corresponding angles: (a) For \(L_z = -1 \hbar\): \[\theta = \cos^{-1} \left(\frac{-\hbar}{\sqrt{2}\hbar}\right)\] (b) For \(L_z=0\): \[\theta = \cos^{-1} \left(\frac{0}{\sqrt{2}\hbar}\right)\] (c) For \(L_z = +1 \hbar\): \[\theta = \cos^{-1} \left(\frac{\hbar}{\sqrt{2}\hbar}\right)\] Now, let's find the angles for each: (a) \(\theta = \cos^{-1} \left(-\frac{1}{\sqrt{2}}\right) \approx 135^{\circ}\) (b) \(\theta = \cos^{-1}(0) \approx 90^{\circ}\) (c) \(\theta = \cos^{-1} \left(\frac{1}{\sqrt{2}}\right) \approx 45^{\circ}\) These are the angles between the positive z-axis and \(\overline{\mathbf{L}}\) for the allowed values of \(L_z\).