University Physics V · topic UPV-14.08
The shell model: spin-orbit splitting and magic numbers
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 shell model and spin–orbit splitting: predict a finite change
The nuclear spin–orbit term is a one-body operator V_ls(r) l⋅s with V_ls negative, which splits every l > 0 orbital into j = l ± ½ and, because the splitting grows as 2l + 1, drives the largest-l j = l + ½ orbital out of its own oscillator shell. Use the supplied topic model Y_out = 20*d+2*v, where Y_out is 1g9/2 BELOW N = 4 LEVEL (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is l² depression coefficient D, measured in MeV. Keep v (Spin-orbit coefficient |V_ls|) = 1.4 MeV; m (Mass number A) = 92 fixed. For this exercise use only 0.127625 MeV ≤ d ≤ 0.332375 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 d = 0.17267 MeV and 0.201335 MeV. 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.
1g9/2 BELOW N = 4 LEVEL against d
Predictions from the supplied The shell model and spin–orbit splitting 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): 1g9/2 BELOW N = 4 LEVEL / MeV
Horizontal axis (x): d: l² depression coefficient D / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: l² depression coefficient D / MeV | 1g9/2 BELOW N = 4 LEVEL / MeV |
|---|---|
| 0.12762500000000002 | 5.3525 |
| 0.1788125 | 6.376250000000001 |
| 0.22999999999999998 | 7.3999999999999995 |
| 0.2811875 | 8.423749999999998 |
| 0.332375 | 9.447499999999998 |
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 = 0.17267 MeV, substitute into Y_out = 20*d+2*v: Y₀ = 6.2534 MeV.
- At d = 0.201335 MeV, the same fixed-parameter model gives Y₁ = 6.8267 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.5733 MeV; the input change is Δd = 0.028665 MeV.
- Average sensitivity is ΔY_out/Δd = 20 MeV per MeV. 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 = 6.2534 MeV, Y₁ = 6.8267 MeV; ΔY_out = 0.5733 MeV; average sensitivity = 20 MeV 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.
The shell model and spin–orbit splitting: infer an input with uncertainty
The nuclear spin–orbit term is a one-body operator V_ls(r) l⋅s with V_ls negative, which splits every l > 0 orbital into j = l ± ½ and, because the splitting grows as 2l + 1, drives the largest-l j = l + ½ orbital out of its own oscillator shell. Use the supplied topic model Y_out = 20*d+2*v, where Y_out is 1g9/2 BELOW N = 4 LEVEL (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is l² depression coefficient D, measured in MeV. Keep v (Spin-orbit coefficient |V_ls|) = 1.4 MeV; m (Mass number A) = 92 fixed. For this exercise use only 0.127625 MeV ≤ d ≤ 0.332375 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 = 7.85045 MeV, with a bounded reading uncertainty ±0.23956 MeV. 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 0.000020475 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.
1g9/2 BELOW N = 4 LEVEL against d
Predictions from the supplied The shell model and spin–orbit splitting 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): 1g9/2 BELOW N = 4 LEVEL / MeV
Horizontal axis (x): d: l² depression coefficient D / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: l² depression coefficient D / MeV | 1g9/2 BELOW N = 4 LEVEL / MeV |
|---|---|
| 0.12762500000000002 | 5.3525 |
| 0.1788125 | 6.376250000000001 |
| 0.22999999999999998 | 7.3999999999999995 |
| 0.2811875 | 8.423749999999998 |
| 0.332375 | 9.447499999999998 |
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(0.127625) = 5.3525 MeV and Y_out(0.332375) = 9.4475 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 0.127625 + (7.85045 − 5.3525)(0.332375 − 0.127625)/(9.4475 − 5.3525) = 0.252523 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 d = 0.252522 MeV; substituting back gives Y_out = 7.85045 MeV.
- The interpolation error in the input is |d_line − d| = 1.11022e-16 MeV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 7.61089 MeV and 8.09001 MeV. Sort the roots even if the response decreases: 0.240544 MeV ≤ d ≤ 0.2645 MeV.
- The lower and upper input deviations are 0.011978 MeV and 0.011978 MeV. 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 ≈ 0.252522 MeV; compatible interval [0.240544, 0.2645] MeV. Endpoint interpolation gives 0.252523 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.