University Physics IV · unit UPIV-13
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.
Nuclear size and the charge form factor: infer an input with uncertainty
The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1. Use the supplied topic model Y_out = n/(4.18879*c^3*(1+9.8696*d*d/(c*c))), where Y_out is INTERIOR DENSITY (fm⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Half-density radius c, measured in fm. Keep n (Mass number A) = 208; d (Surface diffuseness a) = 0.55 fm 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 = 0.26891668 fm⁻³, with a bounded reading uncertainty ±0.022648 fm⁻³. 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.
INTERIOR DENSITY against c
Predictions from the supplied Nuclear size and 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): INTERIOR DENSITY / fm⁻³
Horizontal axis (x): c: Half-density radius c / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Half-density radius c / fm | INTERIOR DENSITY / fm⁻³ |
|---|---|
| 2.6375 | 1.8936959753717497 |
| 3.81875 | 0.7401524325931002 |
| 5 | 0.3548712644554244 |
| 6.18125 | 0.19501622271244365 |
| 7.3625 | 0.11792732758963474 |
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(2.6375) = 1.8937 fm⁻³ and Y_out(7.3625) = 0.117927 fm⁻³. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives c_line = 2.6375 + (0.268917 − 1.8937)(7.3625 − 2.6375)/(0.117927 − 1.8937) = 6.96074 fm. 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 = 5.51975 fm; substituting back gives Y_out = 0.26891668 fm⁻³.
- The interpolation error in the input is |c_line − c| = 1.44099 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.24626868 fm⁻³ and 0.29156468 fm⁻³. Sort the roots even if the response decreases: 5.36355 fm ≤ c ≤ 5.69426 fm.
- The lower and upper input deviations are 0.156201 fm and 0.174512 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 [5.36355, 5.69426] fm. Endpoint interpolation gives 6.96074 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Nuclear force and the shell model: infer an input with uncertainty
The nuclear shell model treats each nucleon as moving independently in a smooth average well built by all the others — a Woods-Saxon potential plus an inverted spin-orbit term — whose level ordering gives the magic numbers and the ground-state spin and parity. Use the supplied topic model Y_out = -c*(l+1)/2-d*l*(l+1), where Y_out is j = l − 1/2 LEVEL (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Flat-bottom term d, lowers high l, measured in MeV. Keep l (Orbital angular momentum l) = 2; c (Spin-orbit c, negative is nuclear) = -2 MeV fixed. For this exercise use only 0.0371875 MeV ≤ d ≤ 0.312812 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 = 1.7680875 MeV, with a bounded reading uncertainty ±0.096744 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 0.0000275625 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.
j = l − 1/2 LEVEL against d
Predictions from the supplied Nuclear force and the shell model 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): j = l − 1/2 LEVEL / MeV
Horizontal axis (x): d: Flat-bottom term d, lowers high l / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Flat-bottom term d, lowers high l / MeV | j = l − 1/2 LEVEL / MeV |
|---|---|
| 0.0371875 | 2.776875 |
| 0.10609374999999999 | 2.3634375 |
| 0.175 | 1.9500000000000002 |
| 0.24390624999999994 | 1.5365625000000005 |
| 0.31281249999999994 | 1.1231250000000004 |
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(0.0371875) = 2.77687 MeV and Y_out(0.312812) = 1.12313 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.0371875 + (1.76809 − 2.77687)(0.312812 − 0.0371875)/(1.12313 − 2.77687) = 0.205319 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 d = 0.205319 MeV; substituting back gives Y_out = 1.7680875 MeV.
- The interpolation error in the input is |d_line − d| = 0 MeV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.6713435 MeV and 1.8648315 MeV. Sort the roots even if the response decreases: 0.189195 MeV ≤ d ≤ 0.221443 MeV.
- The lower and upper input deviations are 0.016124 MeV and 0.016124 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 ≈ 0.205319 MeV; compatible interval [0.189195, 0.221443] MeV. Endpoint interpolation gives 0.205319 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Alpha decay as barrier penetration: infer an input with uncertainty
Alpha decay is a tunnelling problem: an alpha already inside the nucleus escapes through the Coulomb barrier its own charge builds with the daughter's, and its half-life is set almost entirely by the exponential of a WKB integral across the classically forbidden region. Use the supplied topic model Y_out = 2.88*z/q, where Y_out is OUTER TURNING POINT b (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Decay energy Q, measured in MeV. Keep z (Daughter charge Z_d) = 86; rc (Contact radius R) = 9.2 fm fixed. For this exercise use only 4.58437 MeV ≤ q ≤ 8.91562 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 = 34.27415 fm, with a bounded reading uncertainty ±0.97406 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000433125 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.
OUTER TURNING POINT b against q
Predictions from the supplied Alpha decay as barrier penetration 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): OUTER TURNING POINT b / fm
Horizontal axis (x): q: Decay energy Q / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Decay energy Q / MeV | OUTER TURNING POINT b / fm |
|---|---|
| 4.584375 | 54.026993865030676 |
| 5.667187499999999 | 43.70421836228288 |
| 6.749999999999999 | 36.693333333333335 |
| 7.832812499999999 | 31.620825852782765 |
| 8.915624999999999 | 27.78044164037855 |
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(q) = 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(4.58437) = 54.027 fm and Y_out(8.91562) = 27.7804 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 4.58437 + (34.2741 − 54.027)(8.91562 − 4.58437)/(27.7804 − 54.027) = 7.84402 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 q = 7.22644 MeV; substituting back gives Y_out = 34.27415 fm.
- The interpolation error in the input is |q_line − q| = 0.617585 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 33.30009 fm and 35.24821 fm. Sort the roots even if the response decreases: 7.02674 MeV ≤ q ≤ 7.43782 MeV.
- The lower and upper input deviations are 0.199698 MeV and 0.21138 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 q ≈ 7.22644 MeV; compatible interval [7.02674, 7.43782] MeV. Endpoint interpolation gives 7.84402 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.