Answer
$54.74^{\circ}$
Work Step by Step
Electrons are spin $\frac{1}{2}$ particle.
The magnitude of spin angular momentum of an electron is given by
$S=\sqrt {s(s+1)}\hbar$
or, $S=\sqrt {\frac{1}{2}(\frac{1}{2}+1)}\hbar$
or, $S=\frac{\sqrt 3}{2}\hbar$
If magnetic field is along $z$-axis, the the $z$ component of spin angular momentum is given by
$S_z=m_s\hbar$
$\implies S_z=\pm\frac{\hbar}{2}\;\;(\because m_s=\pm\frac{1}{2}$)
Let $\theta$ be the angle between angle between the electron spin angular momentum vector and the magnetic field in a Stern–Gerlach experiment. Therefore,
$S_z=S\cos\theta$
$\cos\theta=\frac{S_z}{S}$
Now, the he smaller angle appears at $S_z=+\frac{\hbar}{2}$
Therefore,
$\cos\theta=\frac{\frac{\hbar}{2}}{\frac{\sqrt 3}{2}\hbar}$
$\theta =\cos^{-1}\Big (\frac{1}{\sqrt 3}\Big)\approx54.74^{\circ}$
Therefore, the he smaller angle is $54.74^{\circ}$
