91Ó°ÊÓ

The z-component of the orbital angular momentum in a hydrogen atom is given by

${\mathit{L}}_{\mathbf{z}}\mathbf{=}{\mathit{m}}_{\mathbf{l}}\sout{\mathbf{h}}$

Where is the magnetic quantum number, it can have values from –l to +l.

The angle the vector L makes with the z-axis is given by

$\mathbf{θ}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\frac{{\mathbf{L}}_{\mathbf{z}}}{\mathbf{L}}$

We know that the magnetic quantum number can have values from -l +l .

For the 5g state, the orbital quantum number is l=4. So the possible values of are$0,Â\pm 1,Â\pm 2,Â\pm 3,Â\pm 4$

Thus in total, there are 9 different states.

We know that the angle made by the orbital angular momentum vector is given by

$\mathrm{θ}={\cos }^{-1}\frac{{\mathrm{L}}_{\mathrm{z}}}{\mathrm{L}}$

So, the largest value of the angle corresponds to the smallest value of

${\left({\mathrm{L}}_{\mathrm{z}}\right)}_{\min }=4\sout{\mathrm{h}}$

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\frac{-4\sout{\mathrm{h}}}{\sqrt{4\left(4+1\right)\sout{\mathrm{h}}}} \\ \mathrm{θ}={\cos }^{-1}\frac{-4}{2\sqrt{5}} \\ \mathrm{θ}=153.4° \end{gathered}$$

The smallest value of the angle corresponds to the largest value of ${\mathrm{L}}_{\mathrm{z}}$

${\left({\mathrm{L}}_{\mathrm{z}}\right)}_{\max }=4\sout{\mathrm{h}}$

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\frac{-4\sout{\mathrm{h}}}{\sqrt{4\left(4+1\right)\sout{\mathrm{h}}}} \\ \mathrm{θ}={\cos }^{-1}\frac{-4}{2\sqrt{5}} \\ \mathrm{θ}=26.57° \end{gathered}$$

Hence, The largest angle corresponds to the states with ${\mathrm{m}}_{\mathrm{l}}=-4$and the angle is ${\mathrm{θ}}_{\min }=153.4°$The smallest angle corresponds to the states with${\mathrm{m}}_{\mathrm{l}}=-4$ and the angle is${\mathrm{θ}}_{\min }=26.57°$