University Physics IV · topic UPIV-13.05
Beta decay, neutrinos, and Fermi's theory
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.
Beta decay and Fermi's theory: predict a finite change
Beta decay is the weak conversion of a neutron into a proton or a proton into a neutron inside a nucleus, emitting an electron or a positron together with an antineutrino or neutrino that shares the released energy with it. Use the supplied topic model Y_out = sqrt((q+0.511)^2-0.261121), where Y_out is MAX ELECTRON pc (MeV/c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. q is Endpoint energy Q, measured in MeV. Keep z (Daughter proton number Z) = 16; s (s = +1 for β⁻, −1 for β⁺) = 1 fixed. For this exercise use only 0.57 MeV ≤ q ≤ 1.83 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 = 0.8472 MeV and 1.0236 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.
MAX ELECTRON pc against q
Predictions from the supplied Beta decay and Fermi's theory 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): MAX ELECTRON pc / MeV/c
Horizontal axis (x): q: Endpoint energy Q / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Endpoint energy Q / MeV | MAX ELECTRON pc / MeV/c |
|---|---|
| 0.5700000000000001 | 0.9525964518094743 |
| 0.885 | 1.2991131590435068 |
| 1.2000000000000002 | 1.632911510156016 |
| 1.5150000000000001 | 1.9604986610553963 |
| 1.83 | 2.284548095357154 |
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 = 0.8472 MeV, substitute into Y_out = sqrt((q+0.511)^2-0.261121): Y₀ = 1.25841 MeV/c.
- At q = 1.0236 MeV, the same fixed-parameter model gives Y₁ = 1.44702 MeV/c.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.188617 MeV/c; the input change is Δq = 0.1764 MeV.
- Average sensitivity is ΔY_out/Δq = 1.06926 MeV/c per MeV. Its positive sign means that this response increases 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 = 1.25841 MeV/c, Y₁ = 1.44702 MeV/c; ΔY_out = 0.188617 MeV/c; average sensitivity = 1.06926 MeV/c 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.
Beta decay and Fermi's theory: infer an input with uncertainty
Beta decay is the weak conversion of a neutron into a proton or a proton into a neutron inside a nucleus, emitting an electron or a positron together with an antineutrino or neutrino that shares the released energy with it. Use the supplied topic model Y_out = sqrt((q+0.511)^2-0.261121), where Y_out is MAX ELECTRON pc (MeV/c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. q is Endpoint energy Q, measured in MeV. Keep z (Daughter proton number Z) = 16; s (s = +1 for β⁻, −1 for β⁺) = 1 fixed. For this exercise use only 0.57 MeV ≤ q ≤ 1.83 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 = 1.7776105 MeV/c, with a bounded reading uncertainty ±0.076041 MeV/c. 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 0.000126 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.
MAX ELECTRON pc against q
Predictions from the supplied Beta decay and Fermi's theory 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): MAX ELECTRON pc / MeV/c
Horizontal axis (x): q: Endpoint energy Q / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Endpoint energy Q / MeV | MAX ELECTRON pc / MeV/c |
|---|---|
| 0.5700000000000001 | 0.9525964518094743 |
| 0.885 | 1.2991131590435068 |
| 1.2000000000000002 | 1.632911510156016 |
| 1.5150000000000001 | 1.9604986610553963 |
| 1.83 | 2.284548095357154 |
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(0.57) = 0.952596 MeV/c and Y_out(1.83) = 2.28455 MeV/c. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.57 + (1.77761 − 0.952596)(1.83 − 0.57)/(2.28455 − 0.952596) = 1.35045 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 = 1.3386 MeV; substituting back gives Y_out = 1.7776105 MeV/c.
- The interpolation error in the input is |q_line − q| = 0.011847 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.7015695 MeV/c and 1.8536515 MeV/c. Sort the roots even if the response decreases: 1.26564 MeV ≤ q ≤ 1.4118 MeV.
- The lower and upper input deviations are 0.0729571 MeV and 0.0731961 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 ≈ 1.3386 MeV; compatible interval [1.26564, 1.4118] MeV. Endpoint interpolation gives 1.35045 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.