University Physics IV · topic UPIV-13.06
Gamma decay, multipolarity, and nuclear lifetimes
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, multipolarity and lifetimes: predict a finite change
Gamma decay is the emission of a single photon by a nucleus dropping between two of its own energy levels, with the photon's multipole order L fixed by the spin change and its electric or magnetic character fixed by the parity change. Use the supplied topic model Y_out = -14.15917 - (2/3)*log(am) - 3*log(eg), where Y_out is LOG₁₀ T½ FOR E1 (log₁₀ s). 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 E, measured in MeV. Keep am (Mass number A) = 137 fixed. For this exercise use only 0.257188 MeV ≤ eg ≤ 1.79281 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 eg = 0.595025 MeV and 0.810013 MeV. 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 FOR E1 against eg
Predictions from the supplied Gamma decay, multipolarity and lifetimes 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 FOR E1 / log₁₀ s
Horizontal axis (x): eg: Transition energy E / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eg: Transition energy E / MeV | LOG₁₀ T½ FOR E1 / log₁₀ s |
|---|---|
| 0.2571875 | -13.814399948781361 |
| 0.64109375 | -15.004415007053346 |
| 1.025 | -15.615821974279589 |
| 1.4089062500000002 | -16.030296665177705 |
| 1.7928125000000001 | -16.344254992791875 |
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 = 0.595025 MeV, substitute into Y_out = -14.15917 - (2/3)*log(am) - 3*log(eg): Y₀ = -14.9073 log₁₀ s.
- At eg = 0.810013 MeV, the same fixed-parameter model gives Y₁ = -15.3091 log₁₀ s.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.40187 log₁₀ s; the input change is Δeg = 0.214987 MeV.
- Average sensitivity is ΔY_out/Δeg = -1.86927 log₁₀ s 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 = -14.9073 log₁₀ s, Y₁ = -15.3091 log₁₀ s; ΔY_out = -0.40187 log₁₀ s; average sensitivity = -1.86927 log₁₀ s 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.
Gamma decay, multipolarity and lifetimes: infer an input with uncertainty
Gamma decay is the emission of a single photon by a nucleus dropping between two of its own energy levels, with the photon's multipole order L fixed by the spin change and its electric or magnetic character fixed by the parity change. Use the supplied topic model Y_out = -14.15917 - (2/3)*log(am) - 3*log(eg), where Y_out is LOG₁₀ T½ FOR E1 (log₁₀ s). 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 E, measured in MeV. Keep am (Mass number A) = 137 fixed. For this exercise use only 0.257188 MeV ≤ eg ≤ 1.79281 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 = -15.814575 log₁₀ s, with a bounded reading uncertainty ±0.079452 log₁₀ s. 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 0.000153562 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.
LOG₁₀ T½ FOR E1 against eg
Predictions from the supplied Gamma decay, multipolarity and lifetimes 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 FOR E1 / log₁₀ s
Horizontal axis (x): eg: Transition energy E / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eg: Transition energy E / MeV | LOG₁₀ T½ FOR E1 / log₁₀ s |
|---|---|
| 0.2571875 | -13.814399948781361 |
| 0.64109375 | -15.004415007053346 |
| 1.025 | -15.615821974279589 |
| 1.4089062500000002 | -16.030296665177705 |
| 1.7928125000000001 | -16.344254992791875 |
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(0.257188) = -13.8144 log₁₀ s and Y_out(1.79281) = -16.3443 log₁₀ s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives eg_line = 0.257188 + (-15.8146 − -13.8144)(1.79281 − 0.257188)/(-16.3443 − -13.8144) = 1.4713 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 eg = 1.19392 MeV; substituting back gives Y_out = -15.814575 log₁₀ s.
- The interpolation error in the input is |eg_line − eg| = 0.277377 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -15.894027 log₁₀ s and -15.735123 log₁₀ s. Sort the roots even if the response decreases: 1.12329 MeV ≤ eg ≤ 1.26899 MeV.
- The lower and upper input deviations are 0.0706317 MeV and 0.0750729 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 eg ≈ 1.19392 MeV; compatible interval [1.12329, 1.26899] MeV. Endpoint interpolation gives 1.4713 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.