University Physics III · topic UPIII-10.06
Electron spin and the Stern-Gerlach experiment
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Electron spin and the Stern-Gerlach experiment: predict a finite change
Electron spin is an intrinsic angular momentum with quantum number s = ½: its component along whatever axis you measure is only ever +ħ/2 or −ħ/2, and it carries a magnetic moment μ = −gₛ(e/2mₑ)S with gₛ ≈ 2.0023. Use the supplied topic model Y_out = 2*grad/v^2*(2.5918*len^2+31.101*len), where Y_out is SPOT SEPARATION (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Atom speed v, measured in m s⁻¹. Keep grad (Field gradient ∂B_z/∂z) = 800 T m⁻¹; len (Magnet length L) = 4 cm fixed. For this exercise use only 463.75 m s⁻¹ ≤ v ≤ 936.25 m s⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at v = 567.7 m s⁻¹ and 633.85 m s⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPOT SEPARATION against v
Predictions from the supplied Electron spin and the Stern-Gerlach experiment model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SPOT SEPARATION / mm
Horizontal axis (x): v: Atom speed v / m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Atom speed v / m s⁻¹ | SPOT SEPARATION / mm |
|---|---|
| 463.75 | 1.2340345333149279 |
| 581.875 | 0.7838550520847153 |
| 700 | 0.541625469387755 |
| 818.125 | 0.39651153925838684 |
| 936.25 | 0.3027690631567501 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 567.7 m s⁻¹, substitute into Y_out = 2*grad/v^2*(2.5918*len^2+31.101*len): Y₀ = 0.823488 mm.
- At v = 633.85 m s⁻¹, the same fixed-parameter model gives Y₁ = 0.660575 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.162913 mm; the input change is Δv = 66.15 m s⁻¹.
- Average sensitivity is ΔY_out/Δv = -0.00246278 mm per m s⁻¹. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.823488 mm, Y₁ = 0.660575 mm; ΔY_out = -0.162913 mm; average sensitivity = -0.00246278 mm per m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Electron spin and the Stern-Gerlach experiment: infer an input with uncertainty
Electron spin is an intrinsic angular momentum with quantum number s = ½: its component along whatever axis you measure is only ever +ħ/2 or −ħ/2, and it carries a magnetic moment μ = −gₛ(e/2mₑ)S with gₛ ≈ 2.0023. Use the supplied topic model Y_out = 2*grad/v^2*(2.5918*len^2+31.101*len), where Y_out is SPOT SEPARATION (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Atom speed v, measured in m s⁻¹. Keep grad (Field gradient ∂B_z/∂z) = 800 T m⁻¹; len (Magnet length L) = 4 cm fixed. For this exercise use only 463.75 m s⁻¹ ≤ v ≤ 936.25 m s⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.46934085 mm, with a bounded reading uncertainty ±0.024986 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.04725 m s⁻¹, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
SPOT SEPARATION against v
Predictions from the supplied Electron spin and the Stern-Gerlach experiment model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SPOT SEPARATION / mm
Horizontal axis (x): v: Atom speed v / m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Atom speed v / m s⁻¹ | SPOT SEPARATION / mm |
|---|---|
| 463.75 | 1.2340345333149279 |
| 581.875 | 0.7838550520847153 |
| 700 | 0.541625469387755 |
| 818.125 | 0.39651153925838684 |
| 936.25 | 0.3027690631567501 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(v) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(463.75) = 1.23403 mm and Y_out(936.25) = 0.302769 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 463.75 + (0.469341 − 1.23403)(936.25 − 463.75)/(0.302769 − 1.23403) = 851.736 m s⁻¹. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives v = 751.975 m s⁻¹; substituting back gives Y_out = 0.46934085 mm.
- The interpolation error in the input is |v_line − v| = 99.7608 m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.44435485 mm and 0.49432685 mm. Sort the roots even if the response decreases: 732.724 m s⁻¹ ≤ v ≤ 772.828 m s⁻¹.
- The lower and upper input deviations are 19.2509 m s⁻¹ and 20.8526 m s⁻¹. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer v ≈ 751.975 m s⁻¹; compatible interval [732.724, 772.828] m s⁻¹. Endpoint interpolation gives 851.736 m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.