The Stern-Gerlach Experiment — Question 2

Section 40-4 · Physics pages 6634–6635

Question 2 A drift region amplifies the separation acquired inside the magnet Atoms of mass M and horizontal speed v traverse a magnet of length L and gradient G, followed by a field-free drift D. Their vertical moments are ±µ, and initial vertical position and velocity are zero. Use the paraxial approximation. G magnet and detector L D Schematic; labels, not drawing lengths, specify the data. (a) Derive the displacement of either component at the detector. (b) Find the total separation for µ = 9.27 × 10−24 J/T, M = 1.80 × 10−25 kg, G = 100 T/m, L = .100 m, D = .400 m, and v = 500 m/s.
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Question 2 - Solution A drift region amplifies the separation acquired inside the magnet G magnet and detector L D (a) Method and calculation tL = L/v, az = ±µG/M, zL = 1 2azt2 L, vz = aztL. z± = ±µGL Mv2 (D + L/2) . (b) Interpretation and further calculation ∆z = z+ −z−= 2µGL(D + L/2) Mv2 . ∆z = 1.854 × 10−3 m = 1.854 mm . Physical check The force acts only in the magnet, but transverse velocity persists in the drift. Ignoring the magnet-exit velocity substantially underestimates the detector separation.

Question 2

A drift region amplifies the separation acquired inside the magnet

Atoms of mass M and horizontal speed v traverse a magnet of length L and gradient G, followed by a field-free drift D. Their vertical moments are ±μ\pm\mu, and initial vertical position and velocity are zero. Use the paraxial approximation.

Diagram shown in the original SVG worksheet.

  1. Derive the displacement of either component at the detector.

  2. Find the total separation for μ=9.27×10−24J/T\mu=9.27\times10^{-24}\,\mathrm{J/T}, M=1.80×10−25kgM=1.80\times10^{-25}\,\mathrm{kg}, G=100T/mG=100\,\mathrm{T/m}, L=.100mL=.100\,\mathrm m, D=.400mD=.400\,\mathrm m, and v=500m/sv=500\,\mathrm{m/s}.

Question 2 - Solution

A drift region amplifies the separation acquired inside the magnet

Diagram shown in the original SVG worksheet.

(a) Method and calculation

tL=L/v,az=±μG/M,zL=12aztL2,vz=aztL.t_L=L/v,\quad a_z=\pm\mu G/M,\quad z_L=\tfrac12a_zt_L^2,\quad v_z=a_zt_L.z±=±μGLMv2(D+L/2).\boxed{z_\pm=\pm\frac{\mu GL}{Mv^2}(D+L/2)}.

(b) Interpretation and further calculation

Δz=z+−z−=2μGL(D+L/2)Mv2.\Delta z=z_+-z_-=\frac{2\mu GL(D+L/2)}{Mv^2}.Δz=1.854×10−3m=1.854mm.\boxed{\Delta z=1.854\times10^{-3}\,\mathrm m=1.854\,\mathrm{mm}}.

Physical check

The force acts only in the magnet, but transverse velocity persists in the drift. Ignoring the magnet-exit velocity substantially underestimates the detector separation.