University Physics IV · topic UPIV-13.03
The nuclear force and the shell model
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.
Nuclear force and the shell model: predict a finite change
The nuclear shell model treats each nucleon as moving independently in a smooth average well built by all the others — a Woods-Saxon potential plus an inverted spin-orbit term — whose level ordering gives the magic numbers and the ground-state spin and parity. Use the supplied topic model Y_out = -c*(l+1)/2-d*l*(l+1), where Y_out is j = l − 1/2 LEVEL (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Flat-bottom term d, lowers high l, measured in MeV. Keep l (Orbital angular momentum l) = 2; c (Spin-orbit c, negative is nuclear) = -2 MeV fixed. For this exercise use only 0.0371875 MeV ≤ d ≤ 0.312812 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 = 0.097825 MeV and 0.136412 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.
j = l − 1/2 LEVEL against d
Predictions from the supplied Nuclear force and the shell model 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 = l − 1/2 LEVEL / MeV
Horizontal axis (x): d: Flat-bottom term d, lowers high l / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Flat-bottom term d, lowers high l / MeV | j = l − 1/2 LEVEL / MeV |
|---|---|
| 0.0371875 | 2.776875 |
| 0.10609374999999999 | 2.3634375 |
| 0.175 | 1.9500000000000002 |
| 0.24390624999999994 | 1.5365625000000005 |
| 0.31281249999999994 | 1.1231250000000004 |
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 = 0.097825 MeV, substitute into Y_out = -c*(l+1)/2-d*l*(l+1): Y₀ = 2.41305 MeV.
- At d = 0.136412 MeV, the same fixed-parameter model gives Y₁ = 2.18153 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.231525 MeV; the input change is Δd = 0.0385875 MeV.
- Average sensitivity is ΔY_out/Δd = -6 MeV 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 = 2.41305 MeV, Y₁ = 2.18153 MeV; ΔY_out = -0.231525 MeV; average sensitivity = -6 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.
Nuclear force and the shell model: infer an input with uncertainty
The nuclear shell model treats each nucleon as moving independently in a smooth average well built by all the others — a Woods-Saxon potential plus an inverted spin-orbit term — whose level ordering gives the magic numbers and the ground-state spin and parity. Use the supplied topic model Y_out = -c*(l+1)/2-d*l*(l+1), where Y_out is j = l − 1/2 LEVEL (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Flat-bottom term d, lowers high l, measured in MeV. Keep l (Orbital angular momentum l) = 2; c (Spin-orbit c, negative is nuclear) = -2 MeV fixed. For this exercise use only 0.0371875 MeV ≤ d ≤ 0.312812 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 = 1.7680875 MeV, with a bounded reading uncertainty ±0.096744 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 0.0000275625 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.
j = l − 1/2 LEVEL against d
Predictions from the supplied Nuclear force and the shell model 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 = l − 1/2 LEVEL / MeV
Horizontal axis (x): d: Flat-bottom term d, lowers high l / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Flat-bottom term d, lowers high l / MeV | j = l − 1/2 LEVEL / MeV |
|---|---|
| 0.0371875 | 2.776875 |
| 0.10609374999999999 | 2.3634375 |
| 0.175 | 1.9500000000000002 |
| 0.24390624999999994 | 1.5365625000000005 |
| 0.31281249999999994 | 1.1231250000000004 |
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(0.0371875) = 2.77687 MeV and Y_out(0.312812) = 1.12313 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.0371875 + (1.76809 − 2.77687)(0.312812 − 0.0371875)/(1.12313 − 2.77687) = 0.205319 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 = 0.205319 MeV; substituting back gives Y_out = 1.7680875 MeV.
- The interpolation error in the input is |d_line − d| = 0 MeV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.6713435 MeV and 1.8648315 MeV. Sort the roots even if the response decreases: 0.189195 MeV ≤ d ≤ 0.221443 MeV.
- The lower and upper input deviations are 0.016124 MeV and 0.016124 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 ≈ 0.205319 MeV; compatible interval [0.189195, 0.221443] MeV. Endpoint interpolation gives 0.205319 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.