University Physics V · topic UPV-14.06
Isobaric mass parabolas and the line of beta stability
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.
Isobaric parabolas and beta stability: predict a finite change
An isobaric mass parabola is the semi-empirical mass formula evaluated at fixed mass number A, where it reduces to a quadratic in proton number Z, split into two offset branches by the pairing term whenever A is even. Use the supplied topic model Y_out = (0.782+4*aa)/(2*(0.711/ma^(1/3)+4*aa/ma)), where Y_out is MINIMUM Z_min (protons). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. aa is Asymmetry coefficient a_A, measured in MeV. Keep ma (Mass number A) = 100; ap (Pairing coefficient a_P) = 11.2 MeV fixed. For this exercise use only 19.0625 MeV ≤ aa ≤ 26.9375 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 aa = 20.795 MeV and 21.8975 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δaa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MINIMUM Z_min against aa
Predictions from the supplied Isobaric parabolas and beta stability 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): MINIMUM Z_min / protons
Horizontal axis (x): aa: Asymmetry coefficient a_A / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| aa: Asymmetry coefficient a_A / MeV | MINIMUM Z_min / protons |
|---|---|
| 19.0625 | 42.062715260142184 |
| 21.03125 | 42.69127733151359 |
| 23 | 43.22759159918886 |
| 24.96875 | 43.69057721435361 |
| 26.9375 | 44.09431085806354 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of aa separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At aa = 20.795 MeV, substitute into Y_out = (0.782+4*aa)/(2*(0.711/ma^(1/3)+4*aa/ma)): Y₀ = 42.6212 protons.
- At aa = 21.8975 MeV, the same fixed-parameter model gives Y₁ = 42.9374 protons.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.316211 protons; the input change is Δaa = 1.1025 MeV.
- Average sensitivity is ΔY_out/Δaa = 0.286813 protons 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 = 42.6212 protons, Y₁ = 42.9374 protons; ΔY_out = 0.316211 protons; average sensitivity = 0.286813 protons 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.
Isobaric parabolas and beta stability: infer an input with uncertainty
An isobaric mass parabola is the semi-empirical mass formula evaluated at fixed mass number A, where it reduces to a quadratic in proton number Z, split into two offset branches by the pairing term whenever A is even. Use the supplied topic model Y_out = (0.782+4*aa)/(2*(0.711/ma^(1/3)+4*aa/ma)), where Y_out is MINIMUM Z_min (protons). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. aa is Asymmetry coefficient a_A, measured in MeV. Keep ma (Mass number A) = 100; ap (Pairing coefficient a_P) = 11.2 MeV fixed. For this exercise use only 19.0625 MeV ≤ aa ≤ 26.9375 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 = 43.439415 protons, with a bounded reading uncertainty ±0.098234 protons. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown aa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0007875 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.
MINIMUM Z_min against aa
Predictions from the supplied Isobaric parabolas and beta stability 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): MINIMUM Z_min / protons
Horizontal axis (x): aa: Asymmetry coefficient a_A / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| aa: Asymmetry coefficient a_A / MeV | MINIMUM Z_min / protons |
|---|---|
| 19.0625 | 42.062715260142184 |
| 21.03125 | 42.69127733151359 |
| 23 | 43.22759159918886 |
| 24.96875 | 43.69057721435361 |
| 26.9375 | 44.09431085806354 |
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(aa) = 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(19.0625) = 42.0627 protons and Y_out(26.9375) = 44.0943 protons. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives aa_line = 19.0625 + (43.4394 − 42.0627)(26.9375 − 19.0625)/(44.0943 − 42.0627) = 24.399 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 aa = 23.8663 MeV; substituting back gives Y_out = 43.439415 protons.
- The interpolation error in the input is |aa_line − aa| = 0.532699 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 43.341181 protons and 43.537649 protons. Sort the roots even if the response decreases: 23.4577 MeV ≤ aa ≤ 24.2873 MeV.
- The lower and upper input deviations are 0.408581 MeV and 0.421002 MeV. 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 aa ≈ 23.8663 MeV; compatible interval [23.4577, 24.2873] MeV. Endpoint interpolation gives 24.399 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.