University Physics V · topic UPV-06.08
Alpha decay: the Coulomb barrier and the Geiger-Nuttall law
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.
Alpha decay and the Geiger–Nuttall law: predict a finite change
Alpha decay is the tunnelling of a preformed alpha cluster through the Coulomb barrier of its daughter, so the decay constant is an attempt frequency times exp(−2G), where 2G is the WKB action across the forbidden gap between the nuclear surface and the outer turning point. Use the supplied topic model Y_out = 2.879928*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) = 82; rn (Contact radius R) = 9 fm fixed. For this exercise use only 4.53125 MeV ≤ q ≤ 8.46875 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. Calculate Y_out at q = 5.3975 MeV and 5.94875 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δq across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
OUTER TURNING POINT b against q
Predictions from the supplied Alpha decay and the Geiger–Nuttall law 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 |
|---|---|
| 4.53125 | 52.116766013793104 |
| 5.515625 | 42.81547349575071 |
| 6.5 | 36.33139938461539 |
| 7.484375 | 31.5529481085595 |
| 8.46875 | 27.885354509225092 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of q separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At q = 5.3975 MeV, substitute into Y_out = 2.879928*z/q: Y₀ = 43.7525 fm.
- At q = 5.94875 MeV, the same fixed-parameter model gives Y₁ = 39.6981 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.05439 fm; the input change is Δq = 0.55125 MeV.
- Average sensitivity is ΔY_out/Δq = -7.35491 fm per MeV. Its negative sign means that this response decreases 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 = 43.7525 fm, Y₁ = 39.6981 fm; ΔY_out = -4.05439 fm; average sensitivity = -7.35491 fm per MeV.
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.
Alpha decay and the Geiger–Nuttall law: infer an input with uncertainty
Alpha decay is the tunnelling of a preformed alpha cluster through the Coulomb barrier of its daughter, so the decay constant is an attempt frequency times exp(−2G), where 2G is the WKB action across the forbidden gap between the nuclear surface and the outer turning point. Use the supplied topic model Y_out = 2.879928*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) = 82; rn (Contact radius R) = 9 fm fixed. For this exercise use only 4.53125 MeV ≤ q ≤ 8.46875 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 = 34.06171 fm, with a bounded reading uncertainty ±0.92645 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 0.00039375 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 and the Geiger–Nuttall law 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 |
|---|---|
| 4.53125 | 52.116766013793104 |
| 5.515625 | 42.81547349575071 |
| 6.5 | 36.33139938461539 |
| 7.484375 | 31.5529481085595 |
| 8.46875 | 27.885354509225092 |
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(4.53125) = 52.1168 fm and Y_out(8.46875) = 27.8854 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 4.53125 + (34.0617 − 52.1168)(8.46875 − 4.53125)/(27.8854 − 52.1168) = 7.46512 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 = 6.93313 MeV; substituting back gives Y_out = 34.06171 fm.
- The interpolation error in the input is |q_line − q| = 0.531994 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 33.13526 fm and 34.98816 fm. Sort the roots even if the response decreases: 6.74954 MeV ≤ q ≤ 7.12697 MeV.
- The lower and upper input deviations are 0.183582 MeV and 0.193848 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 ≈ 6.93313 MeV; compatible interval [6.74954, 7.12697] MeV. Endpoint interpolation gives 7.46512 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.