University Physics IV · topic UPIV-07.09
WKB tunnelling: alpha decay, the STM, and field emission
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.
WKB tunnelling and the Gamow exponent: predict a finite change
The WKB estimate replaces a smooth barrier by one dimensionless number — the Gamow exponent 2γ, twice the integral of the decay constant κ between the classical turning points — and predicts a transmission probability exp(−2γ). Use the supplied topic model Y_out = 2.87993*zd/max(ea,0.5), 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; max(a,b) selects the larger value. ea is Alpha energy E, measured in MeV. Keep zd (Daughter proton number Z′) = 82; rc (Contact radius R) = 9 fm fixed. For this exercise use only 4.53125 MeV ≤ ea ≤ 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 ea = 5.3975 MeV and 5.94875 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δea 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 ea
Predictions from the supplied WKB tunnelling and the Gamow exponent 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): ea: Alpha energy E / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ea: Alpha energy E / MeV | OUTER TURNING POINT b / fm |
|---|---|
| 4.53125 | 52.11680220689655 |
| 5.515625 | 42.815503229461754 |
| 6.5 | 36.33142461538461 |
| 7.484375 | 31.552970020876828 |
| 8.46875 | 27.885373874538743 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ea separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ea = 5.3975 MeV, substitute into Y_out = 2.87993*zd/max(ea,0.5): Y₀ = 43.7525 fm.
- At ea = 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 Δea = 0.55125 MeV.
- Average sensitivity is ΔY_out/Δea = -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.
WKB tunnelling and the Gamow exponent: infer an input with uncertainty
The WKB estimate replaces a smooth barrier by one dimensionless number — the Gamow exponent 2γ, twice the integral of the decay constant κ between the classical turning points — and predicts a transmission probability exp(−2γ). Use the supplied topic model Y_out = 2.87993*zd/max(ea,0.5), 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; max(a,b) selects the larger value. ea is Alpha energy E, measured in MeV. Keep zd (Daughter proton number Z′) = 82; rc (Contact radius R) = 9 fm fixed. For this exercise use only 4.53125 MeV ≤ ea ≤ 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 = 34.061734 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 ea 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 ea
Predictions from the supplied WKB tunnelling and the Gamow exponent 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): ea: Alpha energy E / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ea: Alpha energy E / MeV | OUTER TURNING POINT b / fm |
|---|---|
| 4.53125 | 52.11680220689655 |
| 5.515625 | 42.815503229461754 |
| 6.5 | 36.33142461538461 |
| 7.484375 | 31.552970020876828 |
| 8.46875 | 27.885373874538743 |
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(ea) = 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 ea_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 ea = 6.93313 MeV; substituting back gives Y_out = 34.061734 fm.
- The interpolation error in the input is |ea_line − ea| = 0.531994 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 33.135284 fm and 34.988184 fm. Sort the roots even if the response decreases: 6.74954 MeV ≤ ea ≤ 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 ea ≈ 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.