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

The vector model pictures $\mathbf{L}$ as a cone of fixed length precessing around the $z$-axis, with $L_z$ as its fixed projection. Getting the minimum tilt angle just needs the two extreme lengths of that picture, the full vector length and its largest possible projection.

1. Find $l$ from the given length: $|\mathbf{L}| = \sqrt{l(l+1)}\hbar = \sqrt{12}\hbar$ means $l(l+1)=12$. Since $l$ must be a non-negative integer, checking small values ($l=2$ gives 6, $l=3$ gives 12) shows $l=3$.
2. Longest possible projection: $L_z=m_l\hbar$ can only take integer values from $-l$ to $l$, so the largest projection on the $z$-axis is at $m_l=l=3$, giving $L_z=3\hbar$.
3. Picture the right triangle: the vector $\mathbf{L}$ (length $\sqrt{12}\hbar$), its projection $L_z$ (length $3\hbar$), and the angle $\theta$ between them form a right triangle where $L_z$ is the adjacent side, so $\cos\theta = L_z/|\mathbf{L}|$. A longer projection means a smaller tilt angle, so the biggest $L_z$ (that is, $m_l=+l$) gives the smallest possible $\theta$.
4. Evaluate: $\cos\theta_{min} = \dfrac{3\hbar}{\sqrt{12}\hbar} = \dfrac{3}{2\sqrt3} = \dfrac{\sqrt3}{2}$. This is a standard trigonometric value: $\cos30^\circ = \sqrt3/2$.

Let's summarize:

• The quantum number $l$ comes straight from matching $\sqrt{l(l+1)}$ to the given coefficient of $\hbar$.
• The minimum tilt of $\mathbf{L}$ away from the $z$-axis always happens at the extreme value $m_l=+l$, never at $m_l=0$.

So the minimum angle, rounded to one decimal place, is $\boxed{30.0^\circ}$.