University Physics V · topic UPV-15.03
Alpha decay as a WKB barrier integral
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 as a WKB barrier integral: predict a finite change
The WKB barrier integral reduces α decay to one number: the exponent 2G that κ = √(2μ(V − Q))/ħ accumulates between the nuclear surface and the radius where the Coulomb barrier falls back to Q. Use the supplied topic model Y_out = 2.87992*z/q, where Y_out is 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 proton number Z_d) = 90; rw (Nuclear surface R) = 9.3 fm fixed. For this exercise use only 3.6375 MeV ≤ q ≤ 8.3625 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 = 4.677 MeV and 5.3385 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.
TURNING POINT b against q
Predictions from the supplied Alpha decay as a WKB barrier integral 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): 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 | TURNING POINT b / fm |
|---|---|
| 3.6375 | 71.25575257731958 |
| 4.8187500000000005 | 53.78838910505836 |
| 6 | 43.1988 |
| 7.18125 | 36.09299216710182 |
| 8.3625 | 30.994654708520173 |
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 = 4.677 MeV, substitute into Y_out = 2.87992*z/q: Y₀ = 55.4186 fm.
- At q = 5.3385 MeV, the same fixed-parameter model gives Y₁ = 48.5516 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -6.86699 fm; the input change is Δq = 0.6615 MeV.
- Average sensitivity is ΔY_out/Δq = -10.3809 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 = 55.4186 fm, Y₁ = 48.5516 fm; ΔY_out = -6.86699 fm; average sensitivity = -10.3809 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 as a WKB barrier integral: infer an input with uncertainty
The WKB barrier integral reduces α decay to one number: the exponent 2G that κ = √(2μ(V − Q))/ħ accumulates between the nuclear surface and the radius where the Coulomb barrier falls back to Q. Use the supplied topic model Y_out = 2.87992*z/q, where Y_out is 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 proton number Z_d) = 90; rw (Nuclear surface R) = 9.3 fm fixed. For this exercise use only 3.6375 MeV ≤ q ≤ 8.3625 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 = 39.755021 fm, with a bounded reading uncertainty ±1.3141 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 0.0004725 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.
TURNING POINT b against q
Predictions from the supplied Alpha decay as a WKB barrier integral 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): 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 | TURNING POINT b / fm |
|---|---|
| 3.6375 | 71.25575257731958 |
| 4.8187500000000005 | 53.78838910505836 |
| 6 | 43.1988 |
| 7.18125 | 36.09299216710182 |
| 8.3625 | 30.994654708520173 |
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(3.6375) = 71.2558 fm and Y_out(8.3625) = 30.9947 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 3.6375 + (39.755 − 71.2558)(8.3625 − 3.6375)/(30.9947 − 71.2558) = 7.33439 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 = 6.51975 MeV; substituting back gives Y_out = 39.755021 fm.
- The interpolation error in the input is |q_line − q| = 0.814643 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 38.440921 fm and 41.069121 fm. Sort the roots even if the response decreases: 6.31114 MeV ≤ q ≤ 6.74263 MeV.
- The lower and upper input deviations are 0.208614 MeV and 0.222877 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 ≈ 6.51975 MeV; compatible interval [6.31114, 6.74263] MeV. Endpoint interpolation gives 7.33439 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.