University Physics IV · topic UPIV-09.05
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.
The Stern–Gerlach experiment: predict a finite change
The Stern–Gerlach experiment passes a beam of neutral atoms through a strongly inhomogeneous magnetic field, so each atom deflects in proportion to μ_z; the beam lands in 2j + 1 discrete traces rather than the continuous band a classical moment distribution would leave. Use the supplied topic model Y_out = 213.1*g*bp/v^2, where Y_out is SPLIT AT PLATE (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. Keep bp (Gradient ∂B_z/∂z) = 1400 T/m; g (Spin g factor) = 2 fixed. For this exercise use only 363.75 m/s ≤ v ≤ 836.25 m/s 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 = 467.7 m/s and 533.85 m/s. 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.
SPLIT AT PLATE against v
Predictions from the supplied 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): SPLIT AT PLATE / mm
Horizontal axis (x): v: Atom speed v / m/s
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 | SPLIT AT PLATE / mm |
|---|---|
| 363.75 | 4.5095735761268765 |
| 481.875 | 2.5696424035354224 |
| 600 | 1.6574444444444445 |
| 718.125 | 1.1570213929545576 |
| 836.25 | 0.8532360951914935 |
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 = 467.7 m/s, substitute into Y_out = 213.1*g*bp/v^2: Y₀ = 2.72776 mm.
- At v = 533.85 m/s, the same fixed-parameter model gives Y₁ = 2.09364 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.634119 mm; the input change is Δv = 66.15 m/s.
- Average sensitivity is ΔY_out/Δv = -0.00958608 mm per m/s. 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 = 2.72776 mm, Y₁ = 2.09364 mm; ΔY_out = -0.634119 mm; average sensitivity = -0.00958608 mm per m/s.
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.
The Stern–Gerlach experiment: infer an input with uncertainty
The Stern–Gerlach experiment passes a beam of neutral atoms through a strongly inhomogeneous magnetic field, so each atom deflects in proportion to μ_z; the beam lands in 2j + 1 discrete traces rather than the continuous band a classical moment distribution would leave. Use the supplied topic model Y_out = 213.1*g*bp/v^2, where Y_out is SPLIT AT PLATE (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. Keep bp (Gradient ∂B_z/∂z) = 1400 T/m; g (Spin g factor) = 2 fixed. For this exercise use only 363.75 m/s ≤ v ≤ 836.25 m/s 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 = 1.4037171 mm, with a bounded reading uncertainty ±0.082572 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, 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.
SPLIT AT PLATE against v
Predictions from the supplied 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): SPLIT AT PLATE / mm
Horizontal axis (x): v: Atom speed v / m/s
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 | SPLIT AT PLATE / mm |
|---|---|
| 363.75 | 4.5095735761268765 |
| 481.875 | 2.5696424035354224 |
| 600 | 1.6574444444444445 |
| 718.125 | 1.1570213929545576 |
| 836.25 | 0.8532360951914935 |
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(363.75) = 4.50957 mm and Y_out(836.25) = 0.853236 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 363.75 + (1.40372 − 4.50957)(836.25 − 363.75)/(0.853236 − 4.50957) = 765.113 m/s. 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 = 651.975 m/s; substituting back gives Y_out = 1.4037171 mm.
- The interpolation error in the input is |v_line − v| = 113.138 m/s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3211451 mm and 1.4862891 mm. Sort the roots even if the response decreases: 633.606 m/s ≤ v ≤ 672.041 m/s.
- The lower and upper input deviations are 18.3693 m/s and 20.0656 m/s. 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 ≈ 651.975 m/s; compatible interval [633.606, 672.041] m/s. Endpoint interpolation gives 765.113 m/s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.