University Physics V · topic UPV-14.05
The semi-empirical mass formula as a least-squares fit
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.
The mass formula as a least-squares fit: predict a finite change
The semi-empirical mass formula is a five-term liquid-drop expression for nuclear binding energy whose coefficients are not derived from the nuclear force but fitted, by linear least squares, to the measured masses themselves. Use the supplied topic model Y_out = sqrt(((611.562-v*56+s*14.637+a*.2857)^2+(1443.99-v*132+s*25.925+a*7.7576)^2+(2432.7-v*208+s*35.106+a*9.3077)^2+(2761.49-v*238+s*38.405+a*12.252)^2)/4), where Y_out is RMS OF THE FOUR (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Asymmetry a_A, measured in MeV. Keep v (Volume a_V) = 15.75 MeV; s (Surface a_S) = 17.8 MeV fixed. For this exercise use only 23.3062 MeV ≤ a ≤ 23.8837 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 a = 23.4333 MeV and 23.5141 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RMS OF THE FOUR against a
Predictions from the supplied The mass formula as a least-squares fit 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): RMS OF THE FOUR / MeV
Horizontal axis (x): a: Asymmetry a_A / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Asymmetry a_A / MeV | RMS OF THE FOUR / MeV |
|---|---|
| 23.30625 | 9.798934309992083 |
| 23.450625 | 9.207342554922418 |
| 23.595 | 8.753663554739584 |
| 23.739375 | 8.460113458412534 |
| 23.88375 | 8.343610682587958 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 23.4333 MeV, substitute into Y_out = sqrt(((611.562-v*56+s*14.637+a*.2857)^2+(1443.99-v*132+s*25.925+a*7.7576)^2+(2432.7-v*208+s*35.106+a*9.3077)^2+(2761.49-v*238+s*38.405+a*12.252)^2)/4): Y₀ = 9.27152 MeV.
- At a = 23.5141 MeV, the same fixed-parameter model gives Y₁ = 8.98935 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.282163 MeV; the input change is Δa = 0.08085 MeV.
- Average sensitivity is ΔY_out/Δa = -3.48996 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 = 9.27152 MeV, Y₁ = 8.98935 MeV; ΔY_out = -0.282163 MeV; average sensitivity = -3.48996 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.
The mass formula as a least-squares fit: infer an input with uncertainty
The semi-empirical mass formula is a five-term liquid-drop expression for nuclear binding energy whose coefficients are not derived from the nuclear force but fitted, by linear least squares, to the measured masses themselves. Use the supplied topic model Y_out = sqrt(((611.562-v*56+s*14.637+a*.2857)^2+(1443.99-v*132+s*25.925+a*7.7576)^2+(2432.7-v*208+s*35.106+a*9.3077)^2+(2761.49-v*238+s*38.405+a*12.252)^2)/4), where Y_out is RMS OF THE FOUR (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Asymmetry a_A, measured in MeV. Keep v (Volume a_V) = 15.75 MeV; s (Surface a_S) = 17.8 MeV fixed. For this exercise use only 23.3062 MeV ≤ a ≤ 23.8837 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 = 8.6035982 MeV, with a bounded reading uncertainty ±0.038998 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00005775 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.
RMS OF THE FOUR against a
Predictions from the supplied The mass formula as a least-squares fit 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): RMS OF THE FOUR / MeV
Horizontal axis (x): a: Asymmetry a_A / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Asymmetry a_A / MeV | RMS OF THE FOUR / MeV |
|---|---|
| 23.30625 | 9.798934309992083 |
| 23.450625 | 9.207342554922418 |
| 23.595 | 8.753663554739584 |
| 23.739375 | 8.460113458412534 |
| 23.88375 | 8.343610682587958 |
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(a) = 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(23.3062) = 9.79893 MeV and Y_out(23.8837) = 8.34361 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 23.3062 + (8.6036 − 9.79893)(23.8837 − 23.3062)/(8.34361 − 9.79893) = 23.7806 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 a = 23.6585 MeV; substituting back gives Y_out = 8.6035982 MeV.
- The interpolation error in the input is |a_line − a| = 0.122057 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.5646002 MeV and 8.6425962 MeV. Sort the roots even if the response decreases: 23.6406 MeV ≤ a ≤ 23.6777 MeV.
- The lower and upper input deviations are 0.0178808 MeV and 0.019206 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 a ≈ 23.6585 MeV; compatible interval [23.6406, 23.6777] MeV. Endpoint interpolation gives 23.7806 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.