Question

The subatomic omega particle has spin $\mathbf{s}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}$. What angles might its intrinsic angular in momentum vector make with the z-axis?

Short answer

Angles that intrinsic angular momentum vector${39.2}^{{}^{°}},{75}^{{}^{°}},{105}^{{}^{°}}\text{and}{140.8}^{{}^{°}}$

Step-by-step solution

Given information: 

$\mathrm{s}=\frac{3}{2}$

Concept of angular momentum:

Following formula is used to calculate projection of spin angular moment with respect to z-axis,$\mathbf{cos\theta }\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{z}}}{\mathbf{\left|}\mathbf{S}\mathbf{\right|}}$ Angular momentum along z-axis is,${\mathbf{S}}_{\mathbf{z}}\mathbf{=}{\mathbf{m}}_{\mathbf{s}}\mathbf{h}$ spin angular momentum vector's magnitude is$\mathbf{\left|}\mathbf{S}\mathbf{\right|}\mathbf{=}\sqrt{\mathbf{S}\mathbf{(}\mathbf{S}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{h}}^{\mathbf{2}}}$.

Evaluate the spin angular momentum

The omega particle has spin $\frac{3}{2}$. Its components of angular momentum along the $\mathrm{z}$-axis are therefore ${\mathrm{S}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{s}}\mathrm{h}$.

For ${\mathrm{m}}_{\mathrm{z}}=-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$

The magnitude of its spin angular momentum vector is calculated by substituting $\frac{3}{2}$for $\mathrm{S}$ in the equation $\left|\mathrm{S}\right|=\sqrt{\mathrm{S}(\mathrm{S}+1){\mathrm{h}}^2}$as follows.

$\left|\mathrm{S}\right|=\sqrt{\mathrm{S}(\mathrm{S}+1){\mathrm{\hslash }}^2}=\sqrt{\left(\frac{3}{2}\right)\left(\frac{3}{2}+1\right){\mathrm{\hslash }}^2}=\sqrt{\frac{15}{4}}\mathrm{h}$

The possible angles with respect to the z-axis are shown in the diagram.

For${\mathrm{m}}_{\mathrm{z}}=\frac{3}{2}$the angle is equal to

$\begin{array}{c}\mathrm{cos\theta }=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{\frac{3}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=\frac{3}{\sqrt{15}}=0.77460\\ \mathrm{\theta }={39.2}^{°}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=\frac{1}{2}$the angle is equal to

$\begin{array}{c}\mathrm{cos\theta }=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{\frac{1}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=\frac{1}{\sqrt{15}}=0.25820\\ \mathrm{\theta }={75.0}^{°}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=-\frac{1}{2}$the angle is equal to

$\begin{array}{c}\mathrm{cos\theta }=\frac{{\mathrm{S}}_2}{\left|\mathrm{S}\right|}=\frac{-\frac{1}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=-0.25820\\ \mathrm{\theta }={105}^{°}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=-\frac{3}{2}$ the angle is equal to

$\begin{array}{c}\mathrm{cos\theta }=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{-\frac{3}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=-\frac{3}{\sqrt{15}}=-0.77460\\ \mathrm{\theta }={140.8}^{°}\end{array}$

Therefore, Angles that intrinsic angular momentum vector ${39.2}^{{}^{°}},{75}^{{}^{°}},{105}^{{}^{°}}\text{and}{140.8}^{{}^{°}}$.