University Physics V · unit UPV-16
Nuclear Fission and Fusion
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
Induced fission and neutron moderation: infer an input with uncertainty
Induced fission is the decay of a compound nucleus formed by neutron capture, and moderation is the elastic slowing-down that delivers the neutron to the energy where that capture is most likely to end in fission. Use the supplied topic model Y_out = ln(e0*1000000/0.0253)/(1+(((a-1)/(a+1))^2+0.0000001)*ln(((a-1)/(a+1))^2+0.0000001)/(1-((a-1)/(a+1))^2+0.0000001)), where Y_out is COLLISIONS TO 0.025 eV (collisions). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. e0 is Fission neutron energy E₀, measured in MeV. Keep a (Moderator mass number A) = 12; n (Collisions elapsed n) = 20 fixed. For this exercise use only 0.871875 MeV ≤ e0 ≤ 3.62812 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 = 116.81473 collisions, with a bounded reading uncertainty ±0.33407 collisions. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown e0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625 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.
COLLISIONS TO 0.025 eV against e0
Predictions from the supplied Induced fission and neutron moderation 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): COLLISIONS TO 0.025 eV / collisions
Horizontal axis (x): e0: Fission neutron energy E₀ / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| e0: Fission neutron energy E₀ / MeV | COLLISIONS TO 0.025 eV / collisions |
|---|---|
| 0.871875 | 110.004468153047 |
| 1.5609374999999999 | 113.69590242555233 |
| 2.25 | 116.01348323614863 |
| 2.9390624999999995 | 117.70684222819486 |
| 3.628125 | 119.04186138701697 |
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(e0) = 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(0.871875) = 110.004 collisions and Y_out(3.62812) = 119.042 collisions. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives e0_line = 0.871875 + (116.815 − 110.004)(3.62812 − 0.871875)/(119.042 − 110.004) = 2.94889 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 e0 = 2.55319 MeV; substituting back gives Y_out = 116.81473 collisions.
- The interpolation error in the input is |e0_line − e0| = 0.395701 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 116.48066 collisions and 117.1488 collisions. Sort the roots even if the response decreases: 2.4221 MeV ≤ e0 ≤ 2.69137 MeV.
- The lower and upper input deviations are 0.131084 MeV and 0.138178 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 e0 ≈ 2.55319 MeV; compatible interval [2.4221, 2.69137] MeV. Endpoint interpolation gives 2.94889 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Chain reactions and critical size: infer an input with uncertainty
A core is critical when the buckling its composition demands, B_m² = (k∞ − 1)/L², equals the lowest eigenvalue B_g² of −∇² on that core's geometry with the flux vanishing at the extrapolated boundary. Use the supplied topic model Y_out = 10000*(pi/(rcore+1.79))^2, where Y_out is GEOMETRIC BUCKLING B_g² (×10⁻⁴ cm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rcore is Core radius R, measured in cm. Keep kinf (Infinite multiplication k∞) = 1.2; lsq (Diffusion area L²) = 40 cm² fixed. For this exercise use only 30.0938 cm ≤ rcore ≤ 104.906 cm 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 = 16.424011 ×10⁻⁴ cm⁻², with a bounded reading uncertainty ±1.1632 ×10⁻⁴ cm⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rcore by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00748125 cm, 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.
GEOMETRIC BUCKLING B_g² against rcore
Predictions from the supplied Chain reactions and critical size 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): GEOMETRIC BUCKLING B_g² / ×10⁻⁴ cm⁻²
Horizontal axis (x): rcore: Core radius R / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rcore: Core radius R / cm | GEOMETRIC BUCKLING B_g² / ×10⁻⁴ cm⁻² |
|---|---|
| 30.09375 | 97.08697171980977 |
| 48.796875 | 38.56772678455436 |
| 67.5 | 20.556947309451918 |
| 86.203125 | 12.746832008669559 |
| 104.90625 | 8.669647381247731 |
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(rcore) = 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(30.0938) = 97.087 ×10⁻⁴ cm⁻² and Y_out(104.906) = 8.66965 ×10⁻⁴ cm⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rcore_line = 30.0938 + (16.424 − 97.087)(104.906 − 30.0938)/(8.66965 − 97.087) = 98.3451 cm. 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 rcore = 75.7294 cm; substituting back gives Y_out = 16.424011 ×10⁻⁴ cm⁻².
- The interpolation error in the input is |rcore_line − rcore| = 22.6157 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 15.260811 ×10⁻⁴ cm⁻² and 17.587211 ×10⁻⁴ cm⁻². Sort the roots even if the response decreases: 73.122 cm ≤ rcore ≤ 78.6294 cm.
- The lower and upper input deviations are 2.60738 cm and 2.90007 cm. 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 rcore ≈ 75.7294 cm; compatible interval [73.122, 78.6294] cm. Endpoint interpolation gives 98.3451 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Point kinetics, inhour roots and feedback: infer an input with uncertainty
Point kinetics is the approximation that the flux keeps one fixed spatial shape, so a whole core is a scalar amplitude n(t) plus six delayed-precursor concentrations, obeying a linear system whose eigenvalues — the inhour roots — set the reactor period. Use the supplied topic model Y_out = (b-r)/(0.0767*r)+10/r, where Y_out is ONE-GROUP PERIOD (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Step reactivity ρ (pcm), measured in pcm. Keep b (Delayed fraction β_eff (pcm)) = 650 pcm fixed. For this exercise use only 29.125 pcm ≤ r ≤ 170.875 pcm 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 = 60.362941 s, with a bounded reading uncertainty ±3.5621 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.014175 pcm, 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.
ONE-GROUP PERIOD against r
Predictions from the supplied Point kinetics, inhour roots and feedback 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): ONE-GROUP PERIOD / s
Horizontal axis (x): r: Step reactivity ρ (pcm) / pcm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Step reactivity ρ (pcm) / pcm | ONE-GROUP PERIOD / s |
|---|---|
| 29.125 | 278.2781138262334 |
| 64.5625 | 118.37866696284665 |
| 100 | 71.80795306388525 |
| 135.4375 | 49.60788501698766 |
| 170.875 | 36.61589201221949 |
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(r) = 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(29.125) = 278.278 s and Y_out(170.875) = 36.6159 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 29.125 + (60.3629 − 278.278)(170.875 − 29.125)/(36.6159 − 278.278) = 156.946 pcm. 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 r = 115.593 pcm; substituting back gives Y_out = 60.362941 s.
- The interpolation error in the input is |r_line − r| = 41.3534 pcm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 56.800841 s and 63.925041 s. Sort the roots even if the response decreases: 110.242 pcm ≤ r ≤ 121.488 pcm.
- The lower and upper input deviations are 5.35001 pcm and 5.89576 pcm. 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 r ≈ 115.593 pcm; compatible interval [110.242, 121.488] pcm. Endpoint interpolation gives 156.946 pcm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.