University Physics V · topic UPV-13.05
Fine structure: relativistic, Darwin, and spin-orbit terms
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.
Fine structure: relativistic, Darwin and spin-orbit: predict a finite change
Fine structure is the order-α² splitting of an atomic level produced by three corrections to the Schrödinger Hamiltonian — relativistic kinetic energy, the Darwin contact term, and spin-orbit coupling — whose sum, for a one-electron atom, depends only on n and j. Use the supplied topic model Y_out = 11.3207*(8*d-13*r+7*r/3+8*w/3), where Y_out is 2s − 2p₁/₂ GAP (μeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Darwin term, l = 0 only, dimensionless. Keep r (Relativistic term −p⁴/8m³c²) = 1; w (Spin-orbit term ξ(r)L⋅S, l > 0) = 1 fixed. For this exercise use only 0.10625 ≤ d ≤ 0.89375 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 d = 0.2795 and 0.38975. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
2s − 2p₁/₂ GAP against d
Predictions from the supplied Fine structure: relativistic, Darwin and spin-orbit 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): 2s − 2p₁/₂ GAP / μeV
Horizontal axis (x): d: Darwin term, l = 0 only
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Darwin term, l = 0 only | 2s − 2p₁/₂ GAP / μeV |
|---|---|
| 0.10625 | -80.94300500000001 |
| 0.303125 | -63.1129025 |
| 0.5 | -45.282799999999995 |
| 0.6968749999999999 | -27.452697500000003 |
| 0.8937499999999999 | -9.622595000000006 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.2795, substitute into Y_out = 11.3207*(8*d-13*r+7*r/3+8*w/3): Y₀ = -65.2525 μeV.
- At d = 0.38975, the same fixed-parameter model gives Y₁ = -55.2677 μeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 9.98486 μeV; the input change is Δd = 0.11025.
- Average sensitivity is ΔY_out/Δd = 90.5656 μeV per unit input. 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 = -65.2525 μeV, Y₁ = -55.2677 μeV; ΔY_out = 9.98486 μeV; average sensitivity = 90.5656 μeV per unit input.
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.
Fine structure: relativistic, Darwin and spin-orbit: infer an input with uncertainty
Fine structure is the order-α² splitting of an atomic level produced by three corrections to the Schrödinger Hamiltonian — relativistic kinetic energy, the Darwin contact term, and spin-orbit coupling — whose sum, for a one-electron atom, depends only on n and j. Use the supplied topic model Y_out = 11.3207*(8*d-13*r+7*r/3+8*w/3), where Y_out is 2s − 2p₁/₂ GAP (μeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Darwin term, l = 0 only, dimensionless. Keep r (Relativistic term −p⁴/8m³c²) = 1; w (Spin-orbit term ξ(r)L⋅S, l > 0) = 1 fixed. For this exercise use only 0.10625 ≤ d ≤ 0.89375 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 = -37.437555 μeV, with a bounded reading uncertainty ±4.1722 μeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
2s − 2p₁/₂ GAP against d
Predictions from the supplied Fine structure: relativistic, Darwin and spin-orbit 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): 2s − 2p₁/₂ GAP / μeV
Horizontal axis (x): d: Darwin term, l = 0 only
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Darwin term, l = 0 only | 2s − 2p₁/₂ GAP / μeV |
|---|---|
| 0.10625 | -80.94300500000001 |
| 0.303125 | -63.1129025 |
| 0.5 | -45.282799999999995 |
| 0.6968749999999999 | -27.452697500000003 |
| 0.8937499999999999 | -9.622595000000006 |
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(d) = 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(0.10625) = -80.943 μeV and Y_out(0.89375) = -9.6226 μeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 0.10625 + (-37.4376 − -80.943)(0.89375 − 0.10625)/(-9.6226 − -80.943) = 0.586625. 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 d = 0.586625; substituting back gives Y_out = -37.437555 μeV.
- The interpolation error in the input is |d_line − d| = 0. This is within the requested numerical tolerance.
- Invert both limits Y_out = -41.609755 μeV and -33.265355 μeV. Sort the roots even if the response decreases: 0.540557 ≤ d ≤ 0.632693.
- The lower and upper input deviations are 0.0460683 and 0.0460683. 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 d ≈ 0.586625; compatible interval [0.540557, 0.632693]. Endpoint interpolation gives 0.586625.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.