What angles can the spinS of an electron make with thez^_axis?

The magnitude of angular momentum is given as

\[{\bf{L = }}\left( {\sqrt {{\bf{l}}\left( {{\bf{l + 1}}} \right)} } \right)\left( {\frac{{\bf{h}}}{{{\bf{2\pi }}}}} \right)\]

Here,

L = Angular momentum

l = Angular momentum quantum

h = Planck's constant

The magnitude of electron's spin angular momentum is given as

\[{\bf{S = }}\left( {\sqrt {{\bf{s}}\left( {{\bf{s + 1}}} \right)} } \right)\left( {\frac{{\bf{h}}}{{{\bf{2\pi }}}}} \right)\]

Here,

S = Electron's spin angular momentum

s = Spin quantum number

h = Planck's Constant

Consider the formula for the electron spin quantum number in the z axis is:

\({{\bf{S}}_{\bf{z}}}{\bf{ = S}}\frac{{\bf{h}}}{{{\bf{2\pi }}}}\)

Rewrite the formula for the angular momentum in the z axis as:

\[\begin{array}{c}{S_z}{\rm{ }} = {m_s}\frac{h}{{2\pi }}\\ = S\cos (\theta )\\ = \sqrt {s(s + 1)} \frac{h}{{2\pi }}\cos (\theta )\end{array}\]

Rearranging and then solving for the value of θ as:

\[\begin{array}{c}\cos \theta {\rm{ }} = \frac{{{m_s}\frac{h}{{2\pi }}}}{{\sqrt {s\left( {s + 1} \right)} \frac{h}{{2\pi }}}}\\ = \frac{{{m_s}}}{{\sqrt {s\left( {s + 1} \right)} }}\end{array}\]

Derive further as:

\[\begin{array}{c}\theta {\rm{ }} = {\cos ^{ - 1}}\left( {\frac{{{m_s}}}{{\sqrt {s(s + 1)} }}} \right)\\ = {\cos ^{ - 1}}\left( {\frac{{ \pm \frac{1}{2}}}{{\sqrt {\frac{1}{2}\left( {\frac{1}{2} + 1} \right)} }}} \right)\\ = {\cos ^{ - 1}}\left( {\frac{{ \pm \frac{1}{2}}}{{\sqrt 3 }}} \right)\end{array}\]

The first angle is:

\[\begin{array}{c}\theta = {\cos ^{ - 1}}\left( {\frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right)\\ = 54.74^\circ \end{array}\]

The second angle is:

\[\begin{array}{c}\theta = {\cos ^{ - 1}}\left( {\frac{{ - \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right)\\ = 125.26^\circ \end{array}\]

Therefore, the two angles obtained are:54.74^o and 125.26^o.