Question

Calculate the (a) smaller and (b) larger value of the semi-classical angle between the electron spin angular momentum vector and the magnetic field in a Stern–Gerlach experiment. Bear in mind that the orbital angular momentum of the valence electron in the silver atom is zero.

Short answer

1. The smaller value of the semi-classical angle between the electron spin angular momentum vector and the magnetic field in a Stern-Gerlach experiment is $54.7°$.
2. The larger value of the semi-classical angle between the electron spin angular momentum vector and the magnetic field in a Stern-Gerlach experiment is$125°$ .

Step-by-step solution

The given data:

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

Understanding the concept of spin angular momentum:

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,

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

The z-component of the orbital angular momentum is,

$S_z=m_s\sout{h}$ ….. (2)

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

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

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

(a) Calculation of the smaller semi-classical angle

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 ${\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$)

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

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

(b) Calculation of the smaller semi-classical angle

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

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

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