University Physics IV · topic UPIV-09.04
Orbital magnetic moments and the normal Zeeman effect
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.
Orbital moments and the normal Zeeman effect: predict a finite change
An electron's orbital motion gives it a magnetic moment μ = −(e/2mₑ)L, so a field B shifts each sublevel by m_l μ_B B and splits every level into 2l+1 equally spaced rungs — the normal Zeeman effect is the three-line spectrum those rungs produce in singlet atoms. Use the supplied topic model Y_out = 13.996*b, where Y_out is π–σ SPACING μ_B B/h (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Magnetic field B, measured in T. Keep gu (Upper-level g factor) = 1 fixed. For this exercise use only 0.2125 T ≤ b ≤ 1.7875 T 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 b = 0.559 T and 0.7795 T. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
π–σ SPACING μ_B B/h against b
Predictions from the supplied Orbital moments and the normal Zeeman effect 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): π–σ SPACING μ_B B/h / GHz
Horizontal axis (x): b: Magnetic field B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / T | π–σ SPACING μ_B B/h / GHz |
|---|---|
| 0.2125 | 2.97415 |
| 0.60625 | 8.485075 |
| 1 | 13.996 |
| 1.3937499999999998 | 19.506925 |
| 1.7874999999999999 | 25.01785 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 0.559 T, substitute into Y_out = 13.996*b: Y₀ = 7.82376 GHz.
- At b = 0.7795 T, the same fixed-parameter model gives Y₁ = 10.9099 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = 3.08612 GHz; the input change is Δb = 0.2205 T.
- Average sensitivity is ΔY_out/Δb = 13.996 GHz per T. Its positive sign means that this response increases 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 = 7.82376 GHz, Y₁ = 10.9099 GHz; ΔY_out = 3.08612 GHz; average sensitivity = 13.996 GHz per T.
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.
Orbital moments and the normal Zeeman effect: infer an input with uncertainty
An electron's orbital motion gives it a magnetic moment μ = −(e/2mₑ)L, so a field B shifts each sublevel by m_l μ_B B and splits every level into 2l+1 equally spaced rungs — the normal Zeeman effect is the three-line spectrum those rungs produce in singlet atoms. Use the supplied topic model Y_out = 13.996*b, where Y_out is π–σ SPACING μ_B B/h (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Magnetic field B, measured in T. Keep gu (Upper-level g factor) = 1 fixed. For this exercise use only 0.2125 T ≤ b ≤ 1.7875 T 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 = 16.420807 GHz, with a bounded reading uncertainty ±1.2896 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 T, 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.
π–σ SPACING μ_B B/h against b
Predictions from the supplied Orbital moments and the normal Zeeman effect 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): π–σ SPACING μ_B B/h / GHz
Horizontal axis (x): b: Magnetic field B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / T | π–σ SPACING μ_B B/h / GHz |
|---|---|
| 0.2125 | 2.97415 |
| 0.60625 | 8.485075 |
| 1 | 13.996 |
| 1.3937499999999998 | 19.506925 |
| 1.7874999999999999 | 25.01785 |
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(b) = 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(0.2125) = 2.97415 GHz and Y_out(1.7875) = 25.0178 GHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.2125 + (16.4208 − 2.97415)(1.7875 − 0.2125)/(25.0178 − 2.97415) = 1.17325 T. 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 b = 1.17325 T; substituting back gives Y_out = 16.420807 GHz.
- The interpolation error in the input is |b_line − b| = 0 T. This is within the requested numerical tolerance.
- Invert both limits Y_out = 15.131207 GHz and 17.710407 GHz. Sort the roots even if the response decreases: 1.08111 T ≤ b ≤ 1.26539 T.
- The lower and upper input deviations are 0.0921406 T and 0.0921406 T. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer b ≈ 1.17325 T; compatible interval [1.08111, 1.26539] T. Endpoint interpolation gives 1.17325 T.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.