University Physics IV · topic UPIV-13.01
Nuclear size, density, and the charge form factor
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 size and the charge form factor: predict a finite change
The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1. Use the supplied topic model Y_out = n/(4.18879*c^3*(1+9.8696*d*d/(c*c))), where Y_out is INTERIOR DENSITY (fm⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Half-density radius c, measured in fm. Keep n (Mass number A) = 208; d (Surface diffuseness a) = 0.55 fm fixed. For this exercise use only 2.6375 fm ≤ c ≤ 7.3625 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 c = 3.677 fm and 4.3385 fm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
INTERIOR DENSITY against c
Predictions from the supplied Nuclear size and the charge form factor 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): INTERIOR DENSITY / fm⁻³
Horizontal axis (x): c: Half-density radius c / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Half-density radius c / fm | INTERIOR DENSITY / fm⁻³ |
|---|---|
| 2.6375 | 1.8936959753717497 |
| 3.81875 | 0.7401524325931002 |
| 5 | 0.3548712644554244 |
| 6.18125 | 0.19501622271244365 |
| 7.3625 | 0.11792732758963474 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of c separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At c = 3.677 fm, substitute into Y_out = n/(4.18879*c^3*(1+9.8696*d*d/(c*c))): Y₀ = 0.818168 fm⁻³.
- At c = 4.3385 fm, the same fixed-parameter model gives Y₁ = 0.524828 fm⁻³.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.29334 fm⁻³; the input change is Δc = 0.6615 fm.
- Average sensitivity is ΔY_out/Δc = -0.443447 fm⁻³ 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 = 0.818168 fm⁻³, Y₁ = 0.524828 fm⁻³; ΔY_out = -0.29334 fm⁻³; average sensitivity = -0.443447 fm⁻³ 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.
Nuclear size and the charge form factor: infer an input with uncertainty
The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1. Use the supplied topic model Y_out = n/(4.18879*c^3*(1+9.8696*d*d/(c*c))), where Y_out is INTERIOR DENSITY (fm⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Half-density radius c, measured in fm. Keep n (Mass number A) = 208; d (Surface diffuseness a) = 0.55 fm fixed. For this exercise use only 2.6375 fm ≤ c ≤ 7.3625 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 = 0.26891668 fm⁻³, with a bounded reading uncertainty ±0.022648 fm⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown c by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 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.
INTERIOR DENSITY against c
Predictions from the supplied Nuclear size and the charge form factor 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): INTERIOR DENSITY / fm⁻³
Horizontal axis (x): c: Half-density radius c / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Half-density radius c / fm | INTERIOR DENSITY / fm⁻³ |
|---|---|
| 2.6375 | 1.8936959753717497 |
| 3.81875 | 0.7401524325931002 |
| 5 | 0.3548712644554244 |
| 6.18125 | 0.19501622271244365 |
| 7.3625 | 0.11792732758963474 |
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(c) = 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(2.6375) = 1.8937 fm⁻³ and Y_out(7.3625) = 0.117927 fm⁻³. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives c_line = 2.6375 + (0.268917 − 1.8937)(7.3625 − 2.6375)/(0.117927 − 1.8937) = 6.96074 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 c = 5.51975 fm; substituting back gives Y_out = 0.26891668 fm⁻³.
- The interpolation error in the input is |c_line − c| = 1.44099 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.24626868 fm⁻³ and 0.29156468 fm⁻³. Sort the roots even if the response decreases: 5.36355 fm ≤ c ≤ 5.69426 fm.
- The lower and upper input deviations are 0.156201 fm and 0.174512 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 c ≈ 5.51975 fm; compatible interval [5.36355, 5.69426] fm. Endpoint interpolation gives 6.96074 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.