University Physics III · topic UPIII-11.03
Alpha, beta, and gamma decay modes
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, beta, and gamma decay modes: predict a finite change
A decay mode is the route by which an unstable nucleus lowers its mass-energy: α emission removes two protons and two neutrons, β decay turns a neutron into a proton or the reverse at fixed A, and γ emission drops the nucleus to a lower level of the same nuclide. Use the supplied topic model Y_out = 92-2*na+nb-(238-4*na)/(1.98+0.0155*(238-4*na)^(2/3)), where Y_out is OFF THE VALLEY Z − Z₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. na is α decays taken, dimensionless. Keep nb (β⁻ decays taken) = 0 fixed. For this exercise use only 0 ≤ na ≤ 8, at the specified test settings separated by 1. 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 na = 2 and 3. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δna across this interval. State whether the response increases or decreases; explain why a finite difference between these settings does not require interpolation between them.
OFF THE VALLEY Z − Z₀ against na
Predictions from the supplied Alpha, beta, and gamma decay modes 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): OFF THE VALLEY Z − Z₀
Horizontal axis (x): na: α decays taken
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| na: α decays taken | OFF THE VALLEY Z − Z₀ |
|---|---|
| 0 | -0.4173789690291727 |
| 2 | -1.778594498002974 |
| 4 | -3.1173775088579845 |
| 6 | -4.43312682266513 |
| 8 | -5.725208689914027 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of na separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At na = 2, substitute into Y_out = 92-2*na+nb-(238-4*na)/(1.98+0.0155*(238-4*na)^(2/3)): Y₀ = -1.77859.
- At na = 3, the same fixed-parameter model gives Y₁ = -2.45083.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.672232; the input change is Δna = 1.
- Average sensitivity is ΔY_out/Δna = -0.672232 1 per unit input. Its negative sign means that this response decreases over the specified interval.
- This finite difference compares two specified test settings without interpolating between them. A point derivative is not required. Any quantum numbers or particle counts must also obey their physical admissibility conditions.
Answer Y₀ = -1.77859, Y₁ = -2.45083; ΔY_out = -0.672232; average sensitivity = -0.672232 1 per unit input.
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, beta, and gamma decay modes: infer an input with uncertainty
A decay mode is the route by which an unstable nucleus lowers its mass-energy: α emission removes two protons and two neutrons, β decay turns a neutron into a proton or the reverse at fixed A, and γ emission drops the nucleus to a lower level of the same nuclide. Use the supplied topic model Y_out = 92-2*na+nb-(238-4*na)/(1.98+0.0155*(238-4*na)^(2/3)), where Y_out is OFF THE VALLEY Z − Z₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. na is α decays taken, dimensionless. Keep nb (β⁻ decays taken) = 0 fixed. For this exercise use only 0 ≤ na ≤ 8, at the specified test settings separated by 1. 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 = -3.7781702, with a bounded reading uncertainty ±0.42463. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of na consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
OFF THE VALLEY Z − Z₀ against na
Predictions from the supplied Alpha, beta, and gamma decay modes 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): OFF THE VALLEY Z − Z₀
Horizontal axis (x): na: α decays taken
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| na: α decays taken | OFF THE VALLEY Z − Z₀ |
|---|---|
| 0 | -0.4173789690291727 |
| 2 | -1.778594498002974 |
| 4 | -3.1173775088579845 |
| 6 | -4.43312682266513 |
| 8 | -5.725208689914027 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(0) = -0.417379 and Y_out(8) = -5.72521. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives na_line = 0 + (-3.77817 − -0.417379)(8 − 0)/(-5.72521 − -0.417379) = 5.06541. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to na = 5; only the allowed settings 5 lie inside the stated reading bound.
- The interpolation error in the input is |na_line − na| = 0.0654093.
- The compatible allowed values span 5 to 5. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible na: 5. Endpoint interpolation predicts 5.06541; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.