University Physics V · topic UPV-16.01
Binding energy, the liquid drop, and the fission barrier
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 fission barrier: predict a finite change
The fission barrier is the extra energy a nucleus must acquire to reach the saddle-point shape beyond which surface tension can no longer hold it together against its own Coulomb repulsion. Use the supplied topic model Y_out = 7.12*am^(2/3)*d*d, where Y_out is SURFACE COST (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Quadrupole deformation α, dimensionless. Keep z (Proton number Z) = 92; am (Mass number A) = 236 fixed. For this exercise use only 0.023375 ≤ d ≤ 0.196625 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 d = 0.06149 and 0.085745. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SURFACE COST against d
Predictions from the supplied Binding energy and the fission barrier 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): SURFACE COST / MeV
Horizontal axis (x): d: Quadrupole deformation α
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Quadrupole deformation α | SURFACE COST / MeV |
|---|---|
| 0.023375 | 0.14856775747545284 |
| 0.0666875 | 1.2092335900402558 |
| 0.10999999999999999 | 3.2900818264460137 |
| 0.1533125 | 6.391112466692729 |
| 0.196625 | 10.512325510780398 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.06149, substitute into Y_out = 7.12*am^(2/3)*d*d: Y₀ = 1.02809 MeV.
- At d = 0.085745, the same fixed-parameter model gives Y₁ = 1.99912 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.971032 MeV; the input change is Δd = 0.024255.
- Average sensitivity is ΔY_out/Δd = 40.0343 MeV per unit input. 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 = 1.02809 MeV, Y₁ = 1.99912 MeV; ΔY_out = 0.971032 MeV; average sensitivity = 40.0343 MeV per unit input.
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 fission barrier: infer an input with uncertainty
The fission barrier is the extra energy a nucleus must acquire to reach the saddle-point shape beyond which surface tension can no longer hold it together against its own Coulomb repulsion. Use the supplied topic model Y_out = 7.12*am^(2/3)*d*d, where Y_out is SURFACE COST (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Quadrupole deformation α, dimensionless. Keep z (Proton number Z) = 92; am (Mass number A) = 236 fixed. For this exercise use only 0.023375 ≤ d ≤ 0.196625 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 = 4.5288488 MeV, with a bounded reading uncertainty ±0.65704 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000017325, 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.
SURFACE COST against d
Predictions from the supplied Binding energy and the fission barrier 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): SURFACE COST / MeV
Horizontal axis (x): d: Quadrupole deformation α
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Quadrupole deformation α | SURFACE COST / MeV |
|---|---|
| 0.023375 | 0.14856775747545284 |
| 0.0666875 | 1.2092335900402558 |
| 0.10999999999999999 | 3.2900818264460137 |
| 0.1533125 | 6.391112466692729 |
| 0.196625 | 10.512325510780398 |
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(d) = 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(0.023375) = 0.148568 MeV and Y_out(0.196625) = 10.5123 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 0.023375 + (4.52885 − 0.148568)(0.196625 − 0.023375)/(10.5123 − 0.148568) = 0.0965998. 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 d = 0.129057; substituting back gives Y_out = 4.5288488 MeV.
- The interpolation error in the input is |d_line − d| = 0.0324577. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.8718088 MeV and 5.1858888 MeV. Sort the roots even if the response decreases: 0.119329 ≤ d ≤ 0.138102.
- The lower and upper input deviations are 0.00972842 and 0.00904481. 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 d ≈ 0.129057; compatible interval [0.119329, 0.138102]. Endpoint interpolation gives 0.0965998.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.