University Physics IV · topic UPIV-13.02
Binding energy and the semi-empirical mass formula
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.
Binding energy and the mass formula: predict a finite change
The binding energy of a nucleus is the energy needed to take it apart into free nucleons, and it equals the amount by which the nucleus's mass falls short of the summed masses of those nucleons, multiplied by c². Use the supplied topic model Y_out = (a*(93.582+c/a^(1/3))/(2*(c*a^(2/3)+92.8))), where Y_out is Z AT VALLEY FLOOR (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Coulomb coefficient a_C, measured in MeV. Keep a (Mass number A) = 64; d (Isobar picked: Z − Z at floor) = 0 fixed. For this exercise use only 0.46375 MeV ≤ c ≤ 0.93625 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 c = 0.5677 MeV and 0.63385 MeV. 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.
Z AT VALLEY FLOOR (before display rounding) against c
Predictions from the supplied Binding energy and the mass formula 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): Z AT VALLEY FLOOR (before display rounding)
Horizontal axis (x): c: Coulomb coefficient a_C / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Coulomb coefficient a_C / MeV | Z AT VALLEY FLOOR (before display rounding) |
|---|---|
| 0.46375 | 29.91752145280383 |
| 0.5818749999999999 | 29.373019292919402 |
| 0.7 | 28.84830769230769 |
| 0.818125 | 28.34232694305411 |
| 0.9362499999999999 | 27.854091668213023 |
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 = 0.5677 MeV, substitute into Y_out = (a*(93.582+c/a^(1/3))/(2*(c*a^(2/3)+92.8))): Y₀ = 29.4373.
- At c = 0.63385 MeV, the same fixed-parameter model gives Y₁ = 29.1398.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.29752; the input change is Δc = 0.06615 MeV.
- Average sensitivity is ΔY_out/Δc = -4.49766 1 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 = 29.4373, Y₁ = 29.1398; ΔY_out = -0.29752; average sensitivity = -4.49766 1 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.
Binding energy and the mass formula: infer an input with uncertainty
The binding energy of a nucleus is the energy needed to take it apart into free nucleons, and it equals the amount by which the nucleus's mass falls short of the summed masses of those nucleons, multiplied by c². Use the supplied topic model Y_out = (a*(93.582+c/a^(1/3))/(2*(c*a^(2/3)+92.8))), where Y_out is Z AT VALLEY FLOOR (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Coulomb coefficient a_C, measured in MeV. Keep a (Mass number A) = 64; d (Isobar picked: Z − Z at floor) = 0 fixed. For this exercise use only 0.46375 MeV ≤ c ≤ 0.93625 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 = 28.623428, with a bounded reading uncertainty ±0.1154. 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 0.00004725 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.
Z AT VALLEY FLOOR (before display rounding) against c
Predictions from the supplied Binding energy and the mass formula 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): Z AT VALLEY FLOOR (before display rounding)
Horizontal axis (x): c: Coulomb coefficient a_C / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Coulomb coefficient a_C / MeV | Z AT VALLEY FLOOR (before display rounding) |
|---|---|
| 0.46375 | 29.91752145280383 |
| 0.5818749999999999 | 29.373019292919402 |
| 0.7 | 28.84830769230769 |
| 0.818125 | 28.34232694305411 |
| 0.9362499999999999 | 27.854091668213023 |
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(0.46375) = 29.9175 and Y_out(0.93625) = 27.8541. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives c_line = 0.46375 + (28.6234 − 29.9175)(0.93625 − 0.46375)/(27.8541 − 29.9175) = 0.760081 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 c = 0.751975 MeV; substituting back gives Y_out = 28.623428.
- The interpolation error in the input is |c_line − c| = 0.00810636 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 28.508028 and 28.738828. Sort the roots even if the response decreases: 0.7252 MeV ≤ c ≤ 0.778971 MeV.
- The lower and upper input deviations are 0.0267751 MeV and 0.0269958 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 c ≈ 0.751975 MeV; compatible interval [0.7252, 0.778971] MeV. Endpoint interpolation gives 0.760081 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.