Orbital Angular Momentum Magnitude

The magnitude of the orbital angular momentum ($\vec{L}$) of an electron is given by the formula:

$ L = \sqrt{l(l+1)}\hbar $

Where '$l$' is the azimuthal quantum number and '$\hbar$' is the reduced Planck constant.

We are given $L = \sqrt{12}\hbar$. By comparing this with the formula:

$ \sqrt{l(l+1)}\hbar = \sqrt{12}\hbar $

$ l(l+1) = 12 $

Solving the quadratic equation $l^2 + l - 12 = 0$ yields $(l+4)(l-3)=0$. Since $l$ must be non-negative, we get $l=3$.

Z-Component of Angular Momentum

The z-component of the orbital angular momentum ($L_z$) is given by:

$ L_z = m_l \hbar $

Where '$m_l$' is the magnetic quantum number. For a given '$l$', the possible values of '$m_l$' range from $-l$ to $+l$ in integer steps. In this case, for $l=3$, '$m_l$' can be $-3, -2, -1, 0, 1, 2, 3$.

Angle Calculation

The angle '$\theta$' between the angular momentum vector $\vec{L}$ and the z-axis is related by:

$ \cos \theta = \frac{L_z}{L} = \frac{m_l \hbar}{\sqrt{l(l+1)}\hbar} = \frac{m_l}{\sqrt{12}} $

To find the minimum angle, we need the value of $\cos \theta$ closest to 1 (for $0 \le \theta \le \pi$). This occurs when $L_z$ has the maximum possible magnitude.

• The maximum possible value for $|m_l|$ is 3.
• Therefore, the maximum magnitude for $L_z$ is $3\hbar$.

Substituting the maximum $|m_l|$ value:

$ \cos \theta = \frac{\pm 3}{\sqrt{12}} = \frac{\pm 3}{2\sqrt{3}} = \frac{\pm \sqrt{3}}{2} $

The possible angles are:

• If $\cos \theta = \frac{\sqrt{3}}{2}$, then $\theta = 30^\circ$.
• If $\cos \theta = -\frac{\sqrt{3}}{2}$, then $\theta = 150^\circ$.

The minimum angle between $\vec{L}$ and the z-axis is $30^\circ$. Rounded to one decimal place, this is $30.0^\circ$.