University Physics V · topic UPV-14.02
Nuclear radius and density from 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 radius from the charge form factor: predict a finite change
The charge form factor F(q) is the normalised Fourier transform of a nucleus's charge density, defined so that the elastic electron cross-section is the point-charge Mott cross-section multiplied by |F(q)|², with F(0) = 1. Use the supplied topic model Y_out = 4.4934/max(c,1), where Y_out is FIRST ZERO q1 (fm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. c is Half-density radius c, measured in fm. Keep s (Gaussian surface width σ) = 0.9 fm; qmax (Data cutoff q_max) = 3 fm⁻¹ 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.
FIRST ZERO q1 against c
Predictions from the supplied Nuclear radius from 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): FIRST ZERO q1 / fm⁻¹
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 | FIRST ZERO q1 / fm⁻¹ |
|---|---|
| 2.6375 | 1.703658767772512 |
| 3.81875 | 1.176667757774141 |
| 5 | 0.89868 |
| 6.18125 | 0.7269403437815976 |
| 7.3625 | 0.6103089983022072 |
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 = 4.4934/max(c,1): Y₀ = 1.22203 fm⁻¹.
- At c = 4.3385 fm, the same fixed-parameter model gives Y₁ = 1.0357 fm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.186325 fm⁻¹; the input change is Δc = 0.6615 fm.
- Average sensitivity is ΔY_out/Δc = -0.281671 fm⁻¹ 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 = 1.22203 fm⁻¹, Y₁ = 1.0357 fm⁻¹; ΔY_out = -0.186325 fm⁻¹; average sensitivity = -0.281671 fm⁻¹ 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 radius from the charge form factor: infer an input with uncertainty
The charge form factor F(q) is the normalised Fourier transform of a nucleus's charge density, defined so that the elastic electron cross-section is the point-charge Mott cross-section multiplied by |F(q)|², with F(0) = 1. Use the supplied topic model Y_out = 4.4934/max(c,1), where Y_out is FIRST ZERO q1 (fm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. c is Half-density radius c, measured in fm. Keep s (Gaussian surface width σ) = 0.9 fm; qmax (Data cutoff q_max) = 3 fm⁻¹ 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 = 0.81405861 fm⁻¹, with a bounded reading uncertainty ±0.030562 fm⁻¹. 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.
FIRST ZERO q1 against c
Predictions from the supplied Nuclear radius from 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): FIRST ZERO q1 / fm⁻¹
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 | FIRST ZERO q1 / fm⁻¹ |
|---|---|
| 2.6375 | 1.703658767772512 |
| 3.81875 | 1.176667757774141 |
| 5 | 0.89868 |
| 6.18125 | 0.7269403437815976 |
| 7.3625 | 0.6103089983022072 |
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) = 1.70366 fm⁻¹ and Y_out(7.3625) = 0.610309 fm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives c_line = 2.6375 + (0.814059 − 1.70366)(7.3625 − 2.6375)/(0.610309 − 1.70366) = 6.48198 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 = 0.81405861 fm⁻¹.
- The interpolation error in the input is |c_line − c| = 0.962229 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.78349661 fm⁻¹ and 0.84462061 fm⁻¹. Sort the roots even if the response decreases: 5.32002 fm ≤ c ≤ 5.73506 fm.
- The lower and upper input deviations are 0.199728 fm and 0.21531 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 [5.32002, 5.73506] fm. Endpoint interpolation gives 6.48198 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.