91Ó°ÊÓ

The magnitude of the spin angular momentumis given by

$\mathbf{S}\mathbf{=}\sqrt{\mathbf{s}\left(s+1\right)}\sout{\mathbf{h}}$ .....(1)

The total spin quantum number${\mathrm{S}}_{\mathrm{T}}$ in this case is given by

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{T}}& =& {\mathrm{s}}_1+{\mathrm{s}}_2\\ & =& \frac{1}{2}+\frac{1}{2}\\ {\mathrm{S}}_{\mathrm{T}}& =& 1\end{array}$

since the spins are aligned.

By using Eq. (I) we find the corresponding S_T as

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{T}}& =& \sqrt{1.\left(1+1\right)}\sout{\mathrm{h}}\\ {\mathrm{S}}_{\mathrm{T}}& =& \sqrt{2}\sout{\mathrm{h}}\end{array}$

The individual spins S₁ and S₂ are the same and given by Eq. (1)

$\begin{array}{rcl}{\mathrm{S}}_1& =& {\mathrm{S}}_2=\sqrt{\left(\frac{1}{2}\right).\left(\frac{1}{2}+1\right)}\sout{\mathrm{h}}\\ & =& \frac{\sqrt{3}}{2}\sout{\mathrm{h}}\end{array}$

By squaring the total spin$\stackrel{→}{{\mathrm{S}}_{\mathrm{T}}}=\stackrel{→}{{\mathrm{S}}_1}+\stackrel{→}{{\mathrm{S}}_2}$we get

${\mathrm{S}}_{\mathrm{T}}^2={\mathrm{S}}_1^2+{\mathrm{S}}_2^2+2{\mathrm{S}}_1{\mathrm{S}}_2\mathrm{³¦´Ç²õθ}$ …(2)

Where $\mathrm{θ}$is the angle between the spins.

We can rewrite Eq. (2) as

$\begin{array}{rcl}\cos \mathrm{θ}& =& \frac{{\mathrm{S}}_{\mathrm{T}}^2-{\mathrm{S}}_1^2-{\mathrm{S}}_2^2}{2{\mathrm{S}}_1{\mathrm{S}}_2}\\ \cos \mathrm{θ}& =& \frac{2^2-2×{\displaystyle \frac{3}{4}}}{2×{\displaystyle \frac{3}{4}}}\\ \cos \mathrm{θ}& =& \frac{1}{3}\\ & & \end{array}$

…(3)

From Eq. (3) we find$\mathrm{θ}$ as

$\begin{array}{rcl}\mathrm{θ}& =& \arccos \frac{1}{3}\\ \mathrm{θ}& =& 70.5^{\mathrm{o}}\end{array}$

Hence the angle is, 70.5⁰.