University Physics V · topic UPV-15.05
Gamma decay, multipole operators, and isomers
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.
Gamma decay, multipoles and isomers: predict a finite change
A gamma transition is labelled by the angular momentum L the emitted photon carries and by the family of the operator that emits it — electric EL of parity (−1)^L or magnetic ML of parity (−1)^(L+1) — with the two level spins fixing which L are possible and the two parities fixing the family. Use the supplied topic model Y_out = 26.101+1.937*pc-9*log(eg), where Y_out is log₁₀(T½/s) AT L = 4 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. eg is Transition energy, measured in keV. Keep dj (Spin change |Ji − Jf|) = 4 ħ; pc (Parity change: 0 = no, 1 = yes) = 1 fixed. For this exercise use only 204.063 keV ≤ eg ≤ 1345.94 keV 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 eg = 455.275 keV and 615.138 keV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δeg across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
log₁₀(T½/s) AT L = 4 against eg
Predictions from the supplied Gamma decay, multipoles and isomers 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): log₁₀(T½/s) AT L = 4
Horizontal axis (x): eg: Transition energy / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eg: Transition energy / keV | log₁₀(T½/s) AT L = 4 |
|---|---|
| 204.0625 | 7.250131173403489 |
| 489.53125 | 3.8299762119196394 |
| 775 | 2.0342846774432104 |
| 1060.46875 | 0.8085191231317843 |
| 1345.9375 | -0.12322404098424755 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of eg separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At eg = 455.275 keV, substitute into Y_out = 26.101+1.937*pc-9*log(eg): Y₀ = 4.11354.
- At eg = 615.138 keV, the same fixed-parameter model gives Y₁ = 2.93725.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.17629; the input change is Δeg = 159.863 keV.
- Average sensitivity is ΔY_out/Δeg = -0.00735811 1 per keV. 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 = 4.11354, Y₁ = 2.93725; ΔY_out = -1.17629; average sensitivity = -0.00735811 1 per keV.
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.
Gamma decay, multipoles and isomers: infer an input with uncertainty
A gamma transition is labelled by the angular momentum L the emitted photon carries and by the family of the operator that emits it — electric EL of parity (−1)^L or magnetic ML of parity (−1)^(L+1) — with the two level spins fixing which L are possible and the two parities fixing the family. Use the supplied topic model Y_out = 26.101+1.937*pc-9*log(eg), where Y_out is log₁₀(T½/s) AT L = 4 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. eg is Transition energy, measured in keV. Keep dj (Spin change |Ji − Jf|) = 4 ħ; pc (Parity change: 0 = no, 1 = yes) = 1 fixed. For this exercise use only 204.063 keV ≤ eg ≤ 1345.94 keV 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 = 1.4471854, with a bounded reading uncertainty ±0.23556. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown eg by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.114187 keV, 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.
log₁₀(T½/s) AT L = 4 against eg
Predictions from the supplied Gamma decay, multipoles and isomers 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): log₁₀(T½/s) AT L = 4
Horizontal axis (x): eg: Transition energy / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eg: Transition energy / keV | log₁₀(T½/s) AT L = 4 |
|---|---|
| 204.0625 | 7.250131173403489 |
| 489.53125 | 3.8299762119196394 |
| 775 | 2.0342846774432104 |
| 1060.46875 | 0.8085191231317843 |
| 1345.9375 | -0.12322404098424755 |
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(eg) = 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(204.063) = 7.25013 and Y_out(1345.94) = -0.123224. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives eg_line = 204.063 + (1.44719 − 7.25013)(1345.94 − 204.063)/(-0.123224 − 7.25013) = 1102.74 keV. 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 eg = 900.606 keV; substituting back gives Y_out = 1.4471854.
- The interpolation error in the input is |eg_line − eg| = 202.13 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2116254 and 1.6827454. Sort the roots even if the response decreases: 847.933 keV ≤ eg ≤ 956.551 keV.
- The lower and upper input deviations are 52.6731 keV and 55.9451 keV. 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 eg ≈ 900.606 keV; compatible interval [847.933, 956.551] keV. Endpoint interpolation gives 1102.74 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.