University Physics IV · topic UPIV-13.07
Decay chains, equilibrium, and counting statistics
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.
Decay chains, equilibrium and counting statistics: predict a finite change
A decay chain is a sequence of nuclides in which each member is created by the decay of the one before it and removed by its own decay, so its population obeys a gain-minus-loss equation rather than a simple exponential. Use the supplied topic model Y_out = b*r/(sign(r-1)*max(abs(r-1),0.05))*(1-exp(-ln(2)*(r-1)*t)), where Y_out is RATIO A₂/A₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; exp(x) means eˣ; abs(x) means |x|; max(a,b) selects the larger value. t is Cursor time t, measured in parent t½. Keep r (Half-life ratio t½ parent / t½ daughter) = 10.8; b (Branching fraction b into daughter) = 0.875 fixed. For this exercise use only 0.6375 parent t½ ≤ t ≤ 3.63 parent t½ 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 t = 1.29585 parent t½ and 1.7148 parent t½. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RATIO A₂/A₁ against t
Predictions from the supplied Decay chains, equilibrium and counting statistics 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): RATIO A₂/A₁
Horizontal axis (x): t: Cursor time t / parent t½
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / parent t½ | RATIO A₂/A₁ |
|---|---|
| 0.6375 | 0.9515939839809691 |
| 1.3856249999999999 | 0.9642069223458558 |
| 2.13375 | 0.9642852251349336 |
| 2.881875 | 0.9642857112490015 |
| 3.63 | 0.9642857142668619 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 1.29585 parent t½, substitute into Y_out = b*r/(sign(r-1)*max(abs(r-1),0.05))*(1-exp(-ln(2)*(r-1)*t)): Y₀ = 0.964141.
- At t = 1.7148 parent t½, the same fixed-parameter model gives Y₁ = 0.964277.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.000136565; the input change is Δt = 0.41895 parent t½.
- Average sensitivity is ΔY_out/Δt = 0.000325969 1 per parent t½. 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 = 0.964141, Y₁ = 0.964277; ΔY_out = 0.000136565; average sensitivity = 0.000325969 1 per parent t½.
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.
Decay chains, equilibrium and counting statistics: infer an input with uncertainty
A decay chain is a sequence of nuclides in which each member is created by the decay of the one before it and removed by its own decay, so its population obeys a gain-minus-loss equation rather than a simple exponential. Use the supplied topic model Y_out = b*r/(sign(r-1)*max(abs(r-1),0.05))*(1-exp(-ln(2)*(r-1)*t)), where Y_out is RATIO A₂/A₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; exp(x) means eˣ; abs(x) means |x|; max(a,b) selects the larger value. t is Cursor time t, measured in parent t½. Keep r (Half-life ratio t½ parent / t½ daughter) = 10.8; b (Branching fraction b into daughter) = 0.875 fixed. For this exercise use only 0.6375 parent t½ ≤ t ≤ 3.63 parent t½ 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 = 0.96428566, with a bounded reading uncertainty ±8.14e-9. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925 parent t½, 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.
RATIO A₂/A₁ against t
Predictions from the supplied Decay chains, equilibrium and counting statistics 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): RATIO A₂/A₁
Horizontal axis (x): t: Cursor time t / parent t½
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / parent t½ | RATIO A₂/A₁ |
|---|---|
| 0.6375 | 0.9515939839809691 |
| 1.3856249999999999 | 0.9642069223458558 |
| 2.13375 | 0.9642852251349336 |
| 2.881875 | 0.9642857112490015 |
| 3.63 | 0.9642857142668619 |
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(t) = 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(0.6375) = 0.951594 and Y_out(3.63) = 0.964286. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.6375 + (0.964286 − 0.951594)(3.63 − 0.6375)/(0.964286 − 0.951594) = 3.62999 parent t½. 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 t = 2.45739 parent t½; substituting back gives Y_out = 0.96428566.
- The interpolation error in the input is |t_line − t| = 1.1726 parent t½. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.96428565 and 0.96428567. Sort the roots even if the response decreases: 2.43682 parent t½ ≤ t ≤ 2.4813 parent t½.
- The lower and upper input deviations are 0.0205681 parent t½ and 0.0239159 parent t½. 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 t ≈ 2.45739 parent t½; compatible interval [2.43682, 2.4813] parent t½. Endpoint interpolation gives 3.62999 parent t½.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.