University Physics IV · topic UPIV-11.07
Spin-orbit coupling, fine structure, and the 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.
Spin-orbit coupling and the Zeeman effect: predict a finite change
Fine structure is the splitting of a gross atomic level by the three corrections of order α² beyond the Coulomb problem, spin-orbit coupling, the relativistic kinetic correction and the Darwin term, whose sum in hydrogen depends only on n and the total angular momentum j. Use the supplied topic model Y_out = 0.0578838*b/max(1.5*z,0.01), where Y_out is RATIO μ_B B / ΔE_fs (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. z is Spin-orbit constant ζ, measured in meV. Keep b (Applied magnetic field B) = 6 T fixed. For this exercise use only 0.39125 meV ≤ z ≤ 1.80875 meV 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 z = 0.7031 meV and 0.90155 meV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RATIO μ_B B / ΔE_fs against z
Predictions from the supplied Spin-orbit coupling and the 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): RATIO μ_B B / ΔE_fs
Horizontal axis (x): z: Spin-orbit constant ζ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Spin-orbit constant ζ / meV | RATIO μ_B B / ΔE_fs |
|---|---|
| 0.39125 | 0.5917832587859425 |
| 0.745625 | 0.3105249958088852 |
| 1.1 | 0.21048654545454545 |
| 1.4543749999999998 | 0.15919910614525143 |
| 1.8087499999999999 | 0.12800840359364205 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of z separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At z = 0.7031 meV, substitute into Y_out = 0.0578838*b/max(1.5*z,0.01): Y₀ = 0.329306.
- At z = 0.90155 meV, the same fixed-parameter model gives Y₁ = 0.256819.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0724872; the input change is Δz = 0.19845 meV.
- Average sensitivity is ΔY_out/Δz = -0.365267 1 per meV. 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 = 0.329306, Y₁ = 0.256819; ΔY_out = -0.0724872; average sensitivity = -0.365267 1 per meV.
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.
Spin-orbit coupling and the Zeeman effect: infer an input with uncertainty
Fine structure is the splitting of a gross atomic level by the three corrections of order α² beyond the Coulomb problem, spin-orbit coupling, the relativistic kinetic correction and the Darwin term, whose sum in hydrogen depends only on n and the total angular momentum j. Use the supplied topic model Y_out = 0.0578838*b/max(1.5*z,0.01), where Y_out is RATIO μ_B B / ΔE_fs (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. z is Spin-orbit constant ζ, measured in meV. Keep b (Applied magnetic field B) = 6 T fixed. For this exercise use only 0.39125 meV ≤ z ≤ 1.80875 meV 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 = 0.18435432, with a bounded reading uncertainty ±0.0084519. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00014175 meV, 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.
RATIO μ_B B / ΔE_fs against z
Predictions from the supplied Spin-orbit coupling and the 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): RATIO μ_B B / ΔE_fs
Horizontal axis (x): z: Spin-orbit constant ζ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Spin-orbit constant ζ / meV | RATIO μ_B B / ΔE_fs |
|---|---|
| 0.39125 | 0.5917832587859425 |
| 0.745625 | 0.3105249958088852 |
| 1.1 | 0.21048654545454545 |
| 1.4543749999999998 | 0.15919910614525143 |
| 1.8087499999999999 | 0.12800840359364205 |
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(z) = 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(0.39125) = 0.591783 and Y_out(1.80875) = 0.128008. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives z_line = 0.39125 + (0.184354 − 0.591783)(1.80875 − 0.39125)/(0.128008 − 0.591783) = 1.63653 meV. 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 z = 1.25593 meV; substituting back gives Y_out = 0.18435432.
- The interpolation error in the input is |z_line − z| = 0.380607 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.17590242 and 0.19280622. Sort the roots even if the response decreases: 1.20087 meV ≤ z ≤ 1.31627 meV.
- The lower and upper input deviations are 0.055055 meV and 0.0603457 meV. 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 z ≈ 1.25593 meV; compatible interval [1.20087, 1.31627] meV. Endpoint interpolation gives 1.63653 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.