University Physics V · topic UPV-14.01
Protons, neutrons, and the two-nucleon system
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.
Nucleons, isospin and the two-nucleon system: predict a finite change
Isospin is a two-valued internal label that treats the proton and the neutron as one particle, acted on by operators T_i = τ_i/2 obeying the same SU(2) algebra as spin, and approximately conserved by the strong interaction alone. Use the supplied topic model Y_out = rw-tan(rw*sqrt(v/41.472))/sqrt(v/41.472), where Y_out is SCATTERING LENGTH a (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. rw is Well radius R, measured in fm. Keep v (Well depth V₀) = 21.5 MeV fixed. For this exercise use only 1.8425 fm ≤ rw ≤ 2.1575 fm 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 rw = 1.9118 fm and 1.9559 fm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SCATTERING LENGTH a against rw
Predictions from the supplied Nucleons, isospin and the two-nucleon system 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): SCATTERING LENGTH a / fm
Horizontal axis (x): rw: Well radius R / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rw: Well radius R / fm | SCATTERING LENGTH a / fm |
|---|---|
| 1.8425 | -3.7321376720742814 |
| 1.9212500000000001 | -5.400315836638095 |
| 2 | -8.560370089302072 |
| 2.0787500000000003 | -16.63898118836597 |
| 2.1575 | -77.8228414842996 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rw separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rw = 1.9118 fm, substitute into Y_out = rw-tan(rw*sqrt(v/41.472))/sqrt(v/41.472): Y₀ = -5.14712 fm.
- At rw = 1.9559 fm, the same fixed-parameter model gives Y₁ = -6.51459 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.36747 fm; the input change is Δrw = 0.0441 fm.
- Average sensitivity is ΔY_out/Δrw = -31.0085 fm per fm. 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 = -5.14712 fm, Y₁ = -6.51459 fm; ΔY_out = -1.36747 fm; average sensitivity = -31.0085 fm per fm.
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.
Nucleons, isospin and the two-nucleon system: infer an input with uncertainty
Isospin is a two-valued internal label that treats the proton and the neutron as one particle, acted on by operators T_i = τ_i/2 obeying the same SU(2) algebra as spin, and approximately conserved by the strong interaction alone. Use the supplied topic model Y_out = rw-tan(rw*sqrt(v/41.472))/sqrt(v/41.472), where Y_out is SCATTERING LENGTH a (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. rw is Well radius R, measured in fm. Keep v (Well depth V₀) = 21.5 MeV fixed. For this exercise use only 1.8425 fm ≤ rw ≤ 2.1575 fm 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 = -11.041413 fm, with a bounded reading uncertainty ±1.0964 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 fm, 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.
SCATTERING LENGTH a against rw
Predictions from the supplied Nucleons, isospin and the two-nucleon system 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): SCATTERING LENGTH a / fm
Horizontal axis (x): rw: Well radius R / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rw: Well radius R / fm | SCATTERING LENGTH a / fm |
|---|---|
| 1.8425 | -3.7321376720742814 |
| 1.9212500000000001 | -5.400315836638095 |
| 2 | -8.560370089302072 |
| 2.0787500000000003 | -16.63898118836597 |
| 2.1575 | -77.8228414842996 |
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(rw) = 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(1.8425) = -3.73214 fm and Y_out(2.1575) = -77.8228 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rw_line = 1.8425 + (-11.0414 − -3.73214)(2.1575 − 1.8425)/(-77.8228 − -3.73214) = 1.87358 fm. 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 rw = 2.03465 fm; substituting back gives Y_out = -11.041413 fm.
- The interpolation error in the input is |rw_line − rw| = 0.161074 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -12.137813 fm and -9.945013 fm. Sort the roots even if the response decreases: 2.02113 fm ≤ rw ≤ 2.04605 fm.
- The lower and upper input deviations are 0.0135158 fm and 0.0114026 fm. 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 rw ≈ 2.03465 fm; compatible interval [2.02113, 2.04605] fm. Endpoint interpolation gives 1.87358 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.