University Physics V · topic UPV-13.07
The Zeeman effect: Lande factors to Paschen-Back
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 Zeeman effect and Paschen-Back: predict a finite change
The Zeeman effect is the splitting of atomic levels by H_Z = (μ_B/ħ)(L + g_s S)⋅B, with the pattern set by which basis diagonalises the larger of the field and spin-orbit terms: coupled |j m_j⟩ in weak field, uncoupled |m_l m_s⟩ in strong. Use the supplied topic model Y_out = (b-1/3)/2 + sqrt((1+b*gam/3)^2/4 + 2*b^2*gam^2/9) - 1/3 - b*(1+gam/3)/2, where Y_out is j-MIXING SHIFT (Δ_fs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. gam is Spin anomaly g_s − 1, dimensionless. Keep b (Field μ_B B / Δ_fs) = 2 fixed. For this exercise use only 0.1275 ≤ gam ≤ 1.0725 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 gam = 0.3354 and 0.4677. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δgam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
j-MIXING SHIFT against gam
Predictions from the supplied The Zeeman effect and Paschen-Back 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): j-MIXING SHIFT / Δ_fs
Horizontal axis (x): gam: Spin anomaly g_s − 1
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Spin anomaly g_s − 1 | j-MIXING SHIFT / Δ_fs |
|---|---|
| 0.1275 | 0.013158393259743884 |
| 0.36374999999999996 | 0.08837247885759636 |
| 0.5999999999999999 | 0.20000000000000018 |
| 0.8362499999999999 | 0.3294305189137734 |
| 1.0724999999999998 | 0.46830400135163264 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of gam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At gam = 0.3354, substitute into Y_out = (b-1/3)/2 + sqrt((1+b*gam/3)^2/4 + 2*b^2*gam^2/9) - 1/3 - b*(1+gam/3)/2: Y₀ = 0.0768895 Δ_fs.
- At gam = 0.4677, the same fixed-parameter model gives Y₁ = 0.134444 Δ_fs.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0575543 Δ_fs; the input change is Δgam = 0.1323.
- Average sensitivity is ΔY_out/Δgam = 0.435029 Δ_fs 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 = 0.0768895 Δ_fs, Y₁ = 0.134444 Δ_fs; ΔY_out = 0.0575543 Δ_fs; average sensitivity = 0.435029 Δ_fs 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.
The Zeeman effect and Paschen-Back: infer an input with uncertainty
The Zeeman effect is the splitting of atomic levels by H_Z = (μ_B/ħ)(L + g_s S)⋅B, with the pattern set by which basis diagonalises the larger of the field and spin-orbit terms: coupled |j m_j⟩ in weak field, uncoupled |m_l m_s⟩ in strong. Use the supplied topic model Y_out = (b-1/3)/2 + sqrt((1+b*gam/3)^2/4 + 2*b^2*gam^2/9) - 1/3 - b*(1+gam/3)/2, where Y_out is j-MIXING SHIFT (Δ_fs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. gam is Spin anomaly g_s − 1, dimensionless. Keep b (Field μ_B B / Δ_fs) = 2 fixed. For this exercise use only 0.1275 ≤ gam ≤ 1.0725 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 = 0.25539828 Δ_fs, with a bounded reading uncertainty ±0.031936 Δ_fs. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000945, 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.
j-MIXING SHIFT against gam
Predictions from the supplied The Zeeman effect and Paschen-Back 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): j-MIXING SHIFT / Δ_fs
Horizontal axis (x): gam: Spin anomaly g_s − 1
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Spin anomaly g_s − 1 | j-MIXING SHIFT / Δ_fs |
|---|---|
| 0.1275 | 0.013158393259743884 |
| 0.36374999999999996 | 0.08837247885759636 |
| 0.5999999999999999 | 0.20000000000000018 |
| 0.8362499999999999 | 0.3294305189137734 |
| 1.0724999999999998 | 0.46830400135163264 |
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(gam) = 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(0.1275) = 0.0131584 Δ_fs and Y_out(1.0725) = 0.468304 Δ_fs. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives gam_line = 0.1275 + (0.255398 − 0.0131584)(1.0725 − 0.1275)/(0.468304 − 0.0131584) = 0.630453. 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 gam = 0.70395; substituting back gives Y_out = 0.25539828 Δ_fs.
- The interpolation error in the input is |gam_line − gam| = 0.0734973. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.22346228 Δ_fs and 0.28733428 Δ_fs. Sort the roots even if the response decreases: 0.644686 ≤ gam ≤ 0.76177.
- The lower and upper input deviations are 0.0592644 and 0.0578205. 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 gam ≈ 0.70395; compatible interval [0.644686, 0.76177]. Endpoint interpolation gives 0.630453.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.