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University Physics V · unit UPV-14

Nuclear 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.

Extra challenging · 01

Nucleons, isospin and the two-nucleon system: infer an input with uncertainty

Isospin is a two-valued internal label that treats the proton and the neutron as one particle, acted on by operators T_i = τ_i/2 obeying the same SU(2) algebra as spin, and approximately conserved by the strong interaction alone. Use the supplied topic model Y_out = rw-tan(rw*sqrt(v/41.472))/sqrt(v/41.472), where Y_out is SCATTERING LENGTH a (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. rw is Well radius R, measured in fm. Keep v (Well depth V₀) = 21.5 MeV fixed. For this exercise use only 1.8425 fm ≤ rw ≤ 2.1575 fm 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 = -11.041413 fm, with a bounded reading uncertainty ±1.0964 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 fm, 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.

Explore the question data

SCATTERING LENGTH a against rw

Predictions from the supplied Nucleons, isospin and the two-nucleon system 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): SCATTERING LENGTH a / fm

SCATTERING LENGTH a against rwPredictions from the supplied Nucleons, isospin and the two-nucleon system 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. Horizontal axis: rw: Well radius R / fm. Vertical axis: SCATTERING LENGTH a / fm. Exact values are available in the data table below.-80.79-60.78-40.78-20.77-0.76851.831.91522.0852.17xy

Horizontal axis (x): rw: Well radius R / fm

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1rw: Well radius R / fm: 1.843SCATTERING LENGTH a / fm: -3.732
Read the exact data as a table
SCATTERING LENGTH a against rw — supplied values
rw: Well radius R / fmSCATTERING LENGTH a / fm
1.8425-3.7321376720742814
1.9212500000000001-5.400315836638095
2-8.560370089302072
2.0787500000000003-16.63898118836597
2.1575-77.8228414842996

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(rw) = 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
  1. Endpoint predictions are Y_out(1.8425) = -3.73214 fm and Y_out(2.1575) = -77.8228 fm. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives rw_line = 1.8425 + (-11.0414 − -3.73214)(2.1575 − 1.8425)/(-77.8228 − -3.73214) = 1.87358 fm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives rw = 2.03465 fm; substituting back gives Y_out = -11.041413 fm.
  4. The interpolation error in the input is |rw_line − rw| = 0.161074 fm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -12.137813 fm and -9.945013 fm. Sort the roots even if the response decreases: 2.02113 fm ≤ rw ≤ 2.04605 fm.
  6. The lower and upper input deviations are 0.0135158 fm and 0.0114026 fm. 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 rw ≈ 2.03465 fm; compatible interval [2.02113, 2.04605] fm. Endpoint interpolation gives 1.87358 fm.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Nuclear radius from the charge form factor: infer an input with uncertainty

The charge form factor F(q) is the normalised Fourier transform of a nucleus's charge density, defined so that the elastic electron cross-section is the point-charge Mott cross-section multiplied by |F(q)|², with F(0) = 1. Use the supplied topic model Y_out = 4.4934/max(c,1), where Y_out is FIRST ZERO q1 (fm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. c is Half-density radius c, measured in fm. Keep s (Gaussian surface width σ) = 0.9 fm; qmax (Data cutoff q_max) = 3 fm⁻¹ fixed. For this exercise use only 2.6375 fm ≤ c ≤ 7.3625 fm 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 = 0.81405861 fm⁻¹, with a bounded reading uncertainty ±0.030562 fm⁻¹. 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 0.0004725 fm, 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.

Explore the question data

FIRST ZERO q1 against c

Predictions from the supplied Nuclear radius from the charge form factor 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): FIRST ZERO q1 / fm⁻¹

FIRST ZERO q1 against cPredictions from the supplied Nuclear radius from the charge form factor 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. Horizontal axis: c: Half-density radius c / fm. Vertical axis: FIRST ZERO q1 / fm⁻¹. Exact values are available in the data table below.0.56660.86181.1571.4521.7472.4493.72456.2767.551xy

Horizontal axis (x): c: Half-density radius c / fm

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1c: Half-density radius c / fm: 2.638FIRST ZERO q1 / fm⁻¹: 1.704
Read the exact data as a table
FIRST ZERO q1 against c — supplied values
c: Half-density radius c / fmFIRST ZERO q1 / fm⁻¹
2.63751.703658767772512
3.818751.176667757774141
50.89868
6.181250.7269403437815976
7.36250.6103089983022072

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
  1. Endpoint predictions are Y_out(2.6375) = 1.70366 fm⁻¹ and Y_out(7.3625) = 0.610309 fm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives c_line = 2.6375 + (0.814059 − 1.70366)(7.3625 − 2.6375)/(0.610309 − 1.70366) = 6.48198 fm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives c = 5.51975 fm; substituting back gives Y_out = 0.81405861 fm⁻¹.
  4. The interpolation error in the input is |c_line − c| = 0.962229 fm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.78349661 fm⁻¹ and 0.84462061 fm⁻¹. Sort the roots even if the response decreases: 5.32002 fm ≤ c ≤ 5.73506 fm.
  6. The lower and upper input deviations are 0.199728 fm and 0.21531 fm. 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 ≈ 5.51975 fm; compatible interval [5.32002, 5.73506] fm. Endpoint interpolation gives 6.48198 fm.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

The nuclide chart and mass tables: infer an input with uncertainty

The mass excess of a nuclide is the energy equivalent of its neutral-atom mass minus its mass number times the atomic mass unit, Δ = (M_at − A u)c², and it is the quantity nuclear mass tables actually list. Use the supplied topic model Y_out = 238-4*a, where Y_out is MASS NUMBER A (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Alpha decays n_α, dimensionless. Keep b (Beta-minus decays n_β) = 2 fixed. For this exercise use only 0 ≤ a ≤ 9, 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 = 218, with a bounded reading uncertainty ±2.88. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of a 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.

Explore the question data

MASS NUMBER A against a

Predictions from the supplied The nuclide chart and mass tables 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): MASS NUMBER A

MASS NUMBER A against aPredictions from the supplied The nuclide chart and mass tables 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. Horizontal axis: a: Alpha decays n_α. Vertical axis: MASS NUMBER A. Exact values are available in the data table below.200.6210.3220229.7239.4-0.362.074.56.939.36xy

Horizontal axis (x): a: Alpha decays n_α

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1a: Alpha decays n_α: 0MASS NUMBER A: 238
Read the exact data as a table
MASS NUMBER A against a — supplied values
a: Alpha decays n_αMASS NUMBER A
0238
2230
5218
7210
9202

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
  1. Endpoint predictions are Y_out(0) = 238 and Y_out(9) = 202. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives a_line = 0 + (218 − 238)(9 − 0)/(202 − 238) = 5. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to a = 5; only the allowed settings 5 lie inside the stated reading bound.
  4. The interpolation error in the input is |a_line − a| = 0.
  5. 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.
  6. 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 a: 5. Endpoint interpolation predicts 5; check admissibility against the discrete settings.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

Mass defect and separation energies: infer an input with uncertainty

The binding energy of a nucleus is the energy equivalent of its mass defect — how much less the bound nucleus weighs than the free protons and neutrons it contains — and a separation energy is the change in that binding when one nucleon is removed. Use the supplied topic model Y_out = 9-4*s+d, where Y_out is S_n AT EVEN N (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Pairing gap Δ, measured in MeV. Keep s (Smooth fall of S_n per neutron) = 0.5 MeV fixed. For this exercise use only 0.265625 MeV ≤ d ≤ 2.23438 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 = 8.4665625 MeV, with a bounded reading uncertainty ±0.11517 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 0.000196875 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.

Explore the question data

S_n AT EVEN N against d

Predictions from the supplied Mass defect and separation energies 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): S_n AT EVEN N / MeV

S_n AT EVEN N against dPredictions from the supplied Mass defect and separation energies 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. Horizontal axis: d: Pairing gap Δ / MeV. Vertical axis: S_n AT EVEN N / MeV. Exact values are available in the data table below.7.1877.7188.258.7829.3130.18690.71841.251.7822.313xy

Horizontal axis (x): d: Pairing gap Δ / MeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1d: Pairing gap Δ / MeV: 0.2656S_n AT EVEN N / MeV: 7.266
Read the exact data as a table
S_n AT EVEN N against d — supplied values
d: Pairing gap Δ / MeVS_n AT EVEN N / MeV
0.2656257.265625
0.75781257.7578125
1.258.25
1.74218758.7421875
2.2343759.234375

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
  1. Endpoint predictions are Y_out(0.265625) = 7.26563 MeV and Y_out(2.23438) = 9.23438 MeV. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives d_line = 0.265625 + (8.46656 − 7.26563)(2.23438 − 0.265625)/(9.23438 − 7.26563) = 1.46656 MeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d = 1.46656 MeV; substituting back gives Y_out = 8.4665625 MeV.
  4. The interpolation error in the input is |d_line − d| = 8.88178e-16 MeV. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 8.3513925 MeV and 8.5817325 MeV. Sort the roots even if the response decreases: 1.35139 MeV ≤ d ≤ 1.58173 MeV.
  6. The lower and upper input deviations are 0.11517 MeV and 0.11517 MeV. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer d ≈ 1.46656 MeV; compatible interval [1.35139, 1.58173] MeV. Endpoint interpolation gives 1.46656 MeV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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