University Physics III · unit UPIII-11
Nuclear and Particle Physics
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.
Nuclear constituents, size, and density: infer an input with uncertainty
A nucleus is a bound cluster of Z protons and N neutrons whose radius, measured by scattering, grows as R = R₀A^(1/3) with R₀ ≈ 1.2 fm, so that its volume, not its radius, is proportional to the nucleon number A. Use the supplied topic model Y_out = (2.88*(anum/(1.98+0.0155*anum^(2/3)))*(anum+4)/(kin*anum)), where Y_out is HEAD-ON DISTANCE d (before display rounding) (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kin is Alpha kinetic energy K (lab), measured in MeV. Keep anum (Target nucleon number A) = 197; reach (Reach of the attraction a) = 1.5 fm fixed. For this exercise use only 6.0375 MeV ≤ kin ≤ 35.9625 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 = 9.513941 fm, with a bounded reading uncertainty ±0.46313 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kin by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0029925 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.
HEAD-ON DISTANCE d (before display rounding) against kin
Predictions from the supplied Nuclear constituents, size, and density 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): HEAD-ON DISTANCE d (before display rounding) / fm
Horizontal axis (x): kin: Alpha kinetic energy K (lab) / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kin: Alpha kinetic energy K (lab) / MeV | HEAD-ON DISTANCE d (before display rounding) / fm |
|---|---|
| 6.0375 | 38.27913481910075 |
| 13.518749999999999 | 17.095535938627517 |
| 21 | 11.005251260491463 |
| 28.481249999999996 | 8.114470975477579 |
| 35.9625 | 6.426424093717643 |
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(kin) = 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(6.0375) = 38.2791 fm and Y_out(35.9625) = 6.42642 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives kin_line = 6.0375 + (9.51394 − 38.2791)(35.9625 − 6.0375)/(6.42642 − 38.2791) = 33.0618 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 kin = 24.2918 MeV; substituting back gives Y_out = 9.513941 fm.
- The interpolation error in the input is |kin_line − kin| = 8.77009 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 9.050811 fm and 9.977071 fm. Sort the roots even if the response decreases: 23.1641 MeV ≤ kin ≤ 25.5348 MeV.
- The lower and upper input deviations are 1.12761 MeV and 1.24301 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 kin ≈ 24.2918 MeV; compatible interval [23.1641, 25.5348] MeV. Endpoint interpolation gives 33.0618 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Mass defect and binding energy per nucleon: infer an input with uncertainty
The binding energy B of a nucleus is the energy needed to separate it into free protons and neutrons at rest; it equals the mass deficit Δm, the free nucleons' mass minus the nucleus's, times c², and B/A is its average over the A nucleons. Use the supplied topic model Y_out = 19.708+(74.736-31.537*(a1^(2/3)+a2^(2/3))-0.30026*(a1^(5/3)+a2^(5/3))+0.018209*(a1^2+a2^2))/(a1+a2), where Y_out is FRAGMENTS' MEAN B/A (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a1 is Fragment mass number A₁, dimensionless. Keep a2 (Fragment mass number A₂) = 118 fixed. For this exercise use only 72 ≤ a1 ≤ 92, at the specified test settings separated by 1. 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 = 8.6041641 MeV, with a bounded reading uncertainty ±0.0015387 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of a1 consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
FRAGMENTS' MEAN B/A against a1
Predictions from the supplied Mass defect and binding energy per nucleon 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): FRAGMENTS' MEAN B/A / MeV
Horizontal axis (x): a1: Fragment mass number A₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a1: Fragment mass number A₁ | FRAGMENTS' MEAN B/A / MeV |
|---|---|
| 72 | 8.611249173442125 |
| 77 | 8.61004086165772 |
| 82 | 8.60631837853937 |
| 87 | 8.600253890894114 |
| 92 | 8.592015323288495 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(72) = 8.61125 MeV and Y_out(92) = 8.59202 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a1_line = 72 + (8.60416 − 8.61125)(92 − 72)/(8.59202 − 8.61125) = 79.3673. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to a1 = 84; only the allowed settings 83, 84, 85 lie inside the stated reading bound.
- The interpolation error in the input is |a1_line − a1| = 4.6327.
- The compatible allowed values span 83 to 85. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible a1: 83, 84, 85. Endpoint interpolation predicts 79.3673; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Alpha, beta, and gamma decay modes: infer an input with uncertainty
A decay mode is the route by which an unstable nucleus lowers its mass-energy: α emission removes two protons and two neutrons, β decay turns a neutron into a proton or the reverse at fixed A, and γ emission drops the nucleus to a lower level of the same nuclide. Use the supplied topic model Y_out = 92-2*na+nb-(238-4*na)/(1.98+0.0155*(238-4*na)^(2/3)), where Y_out is OFF THE VALLEY Z − Z₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. na is α decays taken, dimensionless. Keep nb (β⁻ decays taken) = 0 fixed. For this exercise use only 0 ≤ na ≤ 8, at the specified test settings separated by 1. 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 = -3.7781702, with a bounded reading uncertainty ±0.42463. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of na consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
OFF THE VALLEY Z − Z₀ against na
Predictions from the supplied Alpha, beta, and gamma decay modes 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): OFF THE VALLEY Z − Z₀
Horizontal axis (x): na: α decays taken
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| na: α decays taken | OFF THE VALLEY Z − Z₀ |
|---|---|
| 0 | -0.4173789690291727 |
| 2 | -1.778594498002974 |
| 4 | -3.1173775088579845 |
| 6 | -4.43312682266513 |
| 8 | -5.725208689914027 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(0) = -0.417379 and Y_out(8) = -5.72521. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives na_line = 0 + (-3.77817 − -0.417379)(8 − 0)/(-5.72521 − -0.417379) = 5.06541. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to na = 5; only the allowed settings 5 lie inside the stated reading bound.
- The interpolation error in the input is |na_line − na| = 0.0654093.
- The compatible allowed values span 5 to 5. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible na: 5. Endpoint interpolation predicts 5.06541; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Q values, two-body kinematics, and conservation ledgers: infer an input with uncertainty
The Q value of a decay or reaction is the rest energy that disappears, Q = (Σm_initial − Σm_final)c²: the kinetic energy released when Q is positive, and the least energy the collision must supply in its centre-of-mass frame when Q is negative. Use the supplied topic model Y_out = q*(ma+mx)/mx, where Y_out is Kₜₕ NONRELATIVISTIC (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. mx is Target mass m_X, measured in u. Keep q (Energy deficit −Q) = 1.2 MeV; ma (Projectile mass mₐ) = 4 u fixed. For this exercise use only 9.95 u ≤ mx ≤ 54.05 u 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 = 1.3302543 MeV, with a bounded reading uncertainty ±0.0062171 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mx by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00441 u, 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.
Kₜₕ NONRELATIVISTIC against mx
Predictions from the supplied Q values, two-body kinematics, and conservation ledgers 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): Kₜₕ NONRELATIVISTIC / MeV
Horizontal axis (x): mx: Target mass m_X / u
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mx: Target mass m_X / u | Kₜₕ NONRELATIVISTIC / MeV |
|---|---|
| 9.95 | 1.6824120603015076 |
| 20.974999999999998 | 1.4288438617401669 |
| 31.999999999999996 | 1.35 |
| 43.02499999999999 | 1.3115630447414293 |
| 54.05 | 1.2888066604995374 |
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(mx) = 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(9.95) = 1.68241 MeV and Y_out(54.05) = 1.28881 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mx_line = 9.95 + (1.33025 − 1.68241)(54.05 − 9.95)/(1.28881 − 1.68241) = 49.4062 u. 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 mx = 36.851 u; substituting back gives Y_out = 1.3302543 MeV.
- The interpolation error in the input is |mx_line − mx| = 12.5552 u. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3240372 MeV and 1.3364714 MeV. Sort the roots even if the response decreases: 35.1722 u ≤ mx ≤ 38.6981 u.
- The lower and upper input deviations are 1.67879 u and 1.84708 u. 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 mx ≈ 36.851 u; compatible interval [35.1722, 38.6981] u. Endpoint interpolation gives 49.4062 u.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.