91Ó°ÊÓ

The orbital angular momentum of the valence electron in the silver atom is zero.

The spin angular momentum of light (SAM) is the component of angular momentum of light associated with the quantum spin and rotation between the photon's polarisation degrees of freedom.

Using the magnitude of the spin angular momentum and its value of the z-component, we can get the two cases of smaller and larger semi-classical angles.

Formulas:

The magnitude of the spin angular momentum in terms of is,

$\stackrel{→}{\mathrm{S}}=\sqrt{s\left(s+1\right)\sout{h}}$ ….. (1)

The z-component of the orbital angular momentum is,

$S_z=m_s\sout{h}$ ….. (2)

Here, the spin momentum is $m_s=Â\pm \frac{1}{2}$.

The semi-classical angle between a vector and its z-component,

$\mathrm{θ}={\cos }^{-1}\left(\frac{a_z}{a}\right)$ ….. (3)

The magnitude of the spin angular momentum can be given using equation (1) and the value $\mathrm{s}=\frac{1}{2}\left(\mathrm{spin}\right)$ as follows:

$$\begin{gathered} s=\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)\sout{h}} \\ =\frac{\sqrt{3\sout{h}}}{2} \end{gathered}$$

Now, the smaller semi-classical angle can be given using equation (2) and the above value in equation (3) as follows: (for ${\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$)

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{{\displaystyle \frac{1}{2}}\sout{h}}{{\displaystyle \frac{\sqrt{3\sout{h}}}{2}}}\right) \\ ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =54.7° \end{gathered}$$

Hence, the value of the angle is $54.7°$.

Now, the smaller semi-classical angle can be given using equation (2) and the above value in equation (3) as follows: (for${\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$ )

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{{\displaystyle -\frac{1}{2}\sout{h}}}{{\displaystyle \frac{\sqrt{3\sout{h}}}{2}}}\right) \\ ={\cos }^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ =125° \end{gathered}$$

Hence, the value of the angle is$125°$ .