What angles might the intrinsic angular momentum vector make with the z-axis for a deuteron? (See Table 8.1)

The spin angular momentum, or intrinsic angular momentum, the possible value of the total spin angular momentum can be found from all the possible orientations of electrons within the atom

The spins of elementary particles are analogous to the spins of macroscopic bodies. In fact, the spin of a planet is the sum of the spins and the orbital angular momenta of all its elementary particles.

The intrinsic angular momentum is given by $S=\sqrt{s\left(s+1\right)h}$. Since deuteron has spin ,

$s=1$this means that $s=\sqrt{2} h$.

We also know that the z-component of spin is given by

$S_z=m_{s}h$

where in general$m_s=-s,...,+s.$,Therefore, for deuteron,$S_z=-h,0,+h$.

Because the angle between S and${\text{S}}_{\text{z}}$is given by

.$\theta ={\cos }^{-1}\left(\frac{S_z}{s}\right)$

Plugging in the values of${\text{S}}_{\text{z}}$for deuteron, we find that the possible angles are

${\theta }_1=135°$

${\theta }_2=90°$

${\theta }_3=45°$

Therefore the angles might the intrinsic angular momentum vector makes with the Z-axis for a deuteron are ${\theta }_1=135°$, or ${\theta }_2=90°$, or${\theta }_3=45°$