An electron is in anI = 3state of the hydrogen atom, what possible angles might the angular momentum vector make with the z-axis.

The azimuthal quantum number defines the orbital angular momentum and the shape of the orbital. While magnetic quantum number determines the orientation of the orbital in space.

Consider the given data as below.

I = 3

As you know, angular momentum = L = $\mathrm{L}=\sqrt{\mathrm{I}(\mathrm{I}+\mathrm{I})}\times \mathrm{h}$

Where, is the Azimuthal Quantum number and Plank’s constant that is $1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}.$.

Here, if I = 3 the angular momentum is,

role="math" localid="1659324698388" $$\begin{gathered} \mathrm{L}=\sqrt{3\left(3+1\right)}\times \mathrm{h} \\ =\sqrt{12}\mathrm{h} \\ \mathrm{L}=3.46\mathrm{h}......... \\ \left(1\right) \end{gathered}$$

If : I = 3

Thenthe magnetic quantum number is,

$m_l=0,\pm 1\pm 2\pm 3$

Hence, possible z-components are,

${\mathrm{L}}_{\mathrm{z}}=0,\pm \mathrm{h},\pm 2\mathrm{h},\pm 3\mathrm{h}$

You will be finding all the possible angles using the below equation

${\mathrm{L}}_{\mathrm{z}}=\mathrm{L}\mathrm{cos\theta }$ ….. (3)

Where, $\mathrm{\theta }$is the angle of angular momentum vector from z- axis.

Now, by using equations (1) and (2) in equation (3), you get the following.

Substitute 3.46h for L and 0 for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

localid="1659325137879" $$\begin{gathered} 0=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=0 \\ \mathrm{\theta }={\cos }^{-1}\left(0\right) \\ \mathrm{\theta }=90° \end{gathered}$$

Substitute 3.46 for L and h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

localid="1659325202540" $$\begin{gathered} \mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=0.289 \\ \mathrm{\theta }={\cos }^{-1}(0.289) \\ \mathrm{\theta }=73.2° \end{gathered}$$

Substitute 3.46h for L and -h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

$$\begin{gathered} -\mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=-0.289 \\ \mathrm{\theta }={\cos }^{-1}(-0.289) \\ \mathrm{\theta }=106.8° \end{gathered}$$

Substitute 3.46h for L and 2h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

$$\begin{gathered} 2\mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=0.578 \\ \mathrm{\theta }={\cos }^{-1}(0.578) \\ \mathrm{\theta }=54.7° \end{gathered}$$

Substitute 3.46h for L and -2h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

$$\begin{gathered} 2\mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=0.578 \\ \mathrm{\theta }={\cos }^{-1}(0.578) \\ \mathrm{\theta }=125.3° \end{gathered}$$

Substitute 3.46h for L and 3h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

$$\begin{gathered} 3\mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=-0.867 \\ \mathrm{\theta }={\cos }^{-1}(0.867) \\ \mathrm{\theta }=30° \end{gathered}$$

Substitute 3.46h for L and -3h for ${\mathrm{L}}_{\mathrm{z}}$into equation (3), so the angle is,

$$\begin{gathered} -3\mathrm{h}=3.46\times \mathrm{h}\times \cos \mathrm{\theta } \\ \cos \mathrm{\theta }=-0.867 \\ \mathrm{\theta }={\cos }^{-1}(-0.867) \\ \mathrm{\theta }=150° \end{gathered}$$

Hence, the possible angles that the angular momentum vector makes with the z-axis are $\mathrm{\theta }=150°,125.3°,106.8°,90°,73.2°,54.7°,30°.$.