Question

The orbital angular momentum (\(\mathbf{L}\)) of an electron is \(\sqrt{12}\,\hbar\). The minimum angle between \(\mathbf{L}\) and its \(z\)-component (in degrees) is (rounded off to one decimal place).

Hint:

Find \(l\) from \(\sqrt{l(l+1)}=\sqrt{12}\), then use \(\cos\theta = m_l/\sqrt{l(l+1)}\) with the largest allowed \(m_l=l\) to get the minimum angle.

Answer 1

Correct Answer: 30

Step 1: Identify the azimuthal quantum number \(l\).
The magnitude of the orbital angular momentum vector is given by
\[ |\mathbf{L}| = \sqrt{l(l+1)}\,\hbar \] We are told \(|\mathbf{L}| = \sqrt{12}\,\hbar\), so
\[ l(l+1) = 12 \] Testing integers, \(l=3\) gives \(3\times4=12\), which works. So \(l=3\).

Step 2: Recall the z-component of angular momentum.
The z-component is quantised as \(L_z = m_l\hbar\), where \(m_l\) runs from \(-l\) to \(+l\) in integer steps. For \(l=3\), the allowed \(m_l\) values are \(-3,-2,-1,0,1,2,3\).

Step 3: Find the angle between \(\mathbf{L}\) and the z-axis.
By the vector model, the angle \(\theta\) between \(\mathbf{L}\) and the \(z\)-axis satisfies
\[ \cos\theta = \frac{L_z}{|\mathbf{L}|} = \frac{m_l\hbar}{\sqrt{l(l+1)}\,\hbar} = \frac{m_l}{\sqrt{l(l+1)}} \] Since \(\sqrt{l(l+1)}\) is fixed, \(\theta\) is smallest when \(\cos\theta\) is largest, which happens at the largest allowed \(m_l\), namely \(m_l=+l=3\).

Step 4: Compute the minimum angle.
\[ \cos\theta_{min} = \frac{l}{\sqrt{l(l+1)}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2} = 0.8660 \] \[ \theta_{min} = \cos^{-1}(0.8660) = 30.0^{\circ} \]
Final Answer:
The minimum angle between \(\mathbf{L}\) and its z-component is \[ \boxed{30.0^{\circ}} \]