The Stern-Gerlach Experiment — Question 4

Question 4

A measured spot separation can determine an atomic magnetic moment

A calibrated apparatus produces separation $\Delta z=2\mu GL(D+L/2)/(Mv^2)$. Two neutral species traverse it with the same speed. Species A has mass $M_A$, moment magnitude $\mu_A$ and separation $s_A$. Species B has $M_B=1.50M_A$ and $s_B=.800s_A$.

1. Infer $\mu_B/\mu_A$.

2. If the speeds instead satisfy $v_B=1.10v_A$ with the same observed separations, recalculate the ratio.

Question 4 - Solution

A measured spot separation can determine an atomic magnetic moment

(a) Method and calculation

$$\frac{s_B}{s_A}=\frac{\mu_B}{\mu_A}\frac{M_A}{M_B}\quad(v_A=v_B).$$$$\boxed{\mu_B/\mu_A=.800(1.50)=1.20}.$$

(b) Interpretation and further calculation

$$\frac{s_B}{s_A}=\frac{\mu_B}{\mu_A}\frac{M_A}{M_B}\frac{v_A^2}{v_B^2}.$$$$\boxed{\mu_B/\mu_A=.800(1.50)(1.10)^2=1.452}.$$

Physical check

The symbol $\mu_B$ here labels species B, not the Bohr magneton. Beam speeds and masses must be accounted for before interpreting separation as moment.