University Physics V · unit UPV-15
Radioactive Decay
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
The decay law, Γ, and where it fails: infer an input with uncertainty
The decay law says an unstable state's survival probability falls as e^(−λt) with a rate λ = Γ/ħ fixed by the full width of its energy line, so the expected number of survivors in an ensemble is N₀e^(−λt). Use the supplied topic model Y_out = exp(-0.0015193*gam*t), where Y_out is SURVIVAL S(t) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Elapsed time t, measured in ns. Keep gam (Line width Γ (FWHM)) = 4.7 neV fixed. For this exercise use only 74.375 ns ≤ t ≤ 625.625 ns 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 = 0.053277387, with a bounded reading uncertainty ±0.0062701. 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 0.055125 ns, 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.
SURVIVAL S(t) against t
Predictions from the supplied The decay law, Γ, and where it fails 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): SURVIVAL S(t)
Horizontal axis (x): t: Elapsed time t / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Elapsed time t / ns | SURVIVAL S(t) |
|---|---|
| 74.375 | 0.5879635597313115 |
| 212.1875 | 0.21977095180813358 |
| 350 | 0.08214670868501589 |
| 487.8125 | 0.0307050667627455 |
| 625.625 | 0.011477040772500615 |
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(74.375) = 0.587964 and Y_out(625.625) = 0.011477. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 74.375 + (0.0532774 − 0.587964)(625.625 − 74.375)/(0.011477 − 0.587964) = 585.655 ns. 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 = 410.637 ns; substituting back gives Y_out = 0.053277387.
- The interpolation error in the input is |t_line − t| = 175.017 ns. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.047007287 and 0.059547487. Sort the roots even if the response decreases: 395.056 ns ≤ t ≤ 428.172 ns.
- The lower and upper input deviations are 15.5814 ns and 17.5346 ns. 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 ≈ 410.637 ns; compatible interval [395.056, 428.172] ns. Endpoint interpolation gives 585.655 ns.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Half-life, mean life, width and branching: infer an input with uncertainty
The decay constant λ is an unstable state's hazard rate — the probability per unit time that it decays — equal to Γ/ħ, to 1/τ, and to ln 2/T½, so a width, a mean life and a half-life are three ways of writing one number. Use the supplied topic model Y_out = 12.673*(t1+t2)/(t1*t2), where Y_out is TOTAL WIDTH Γ (×10⁻²⁰ eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t2 is Partial half-life, channel 2, measured in h. Keep t1 (Partial half-life, channel 1) = 10 h fixed. For this exercise use only 4.975 h ≤ t2 ≤ 27.025 h 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 = 1.9550968 ×10⁻²⁰ eV, with a bounded reading uncertainty ±0.032829 ×10⁻²⁰ eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.002205 h, 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.
TOTAL WIDTH Γ against t2
Predictions from the supplied Half-life, mean life, width and branching 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): TOTAL WIDTH Γ / ×10⁻²⁰ eV
Horizontal axis (x): t2: Partial half-life, channel 2 / h
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t2: Partial half-life, channel 2 / h | TOTAL WIDTH Γ / ×10⁻²⁰ eV |
|---|---|
| 4.975 | 3.8146366834170857 |
| 10.487499999999999 | 2.4756909415971395 |
| 15.999999999999998 | 2.0593625 |
| 21.512499999999996 | 1.856399360836723 |
| 27.025 | 1.736236170212766 |
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(t2) = 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(4.975) = 3.81464 ×10⁻²⁰ eV and Y_out(27.025) = 1.73624 ×10⁻²⁰ eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t2_line = 4.975 + (1.9551 − 3.81464)(27.025 − 4.975)/(1.73624 − 3.81464) = 24.7031 h. 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 t2 = 18.4255 h; substituting back gives Y_out = 1.9550968 ×10⁻²⁰ eV.
- The interpolation error in the input is |t2_line − t2| = 6.27758 h. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.9222678 ×10⁻²⁰ eV and 1.9879258 ×10⁻²⁰ eV. Sort the roots even if the response decreases: 17.5861 h ≤ t2 ≤ 19.349 h.
- The lower and upper input deviations are 0.839396 h and 0.923543 h. 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 t2 ≈ 18.4255 h; compatible interval [17.5861, 19.349] h. Endpoint interpolation gives 24.7031 h.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Alpha decay as a WKB barrier integral: infer an input with uncertainty
The WKB barrier integral reduces α decay to one number: the exponent 2G that κ = √(2μ(V − Q))/ħ accumulates between the nuclear surface and the radius where the Coulomb barrier falls back to Q. Use the supplied topic model Y_out = 2.87992*z/q, where Y_out is TURNING POINT b (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Decay energy Q, measured in MeV. Keep z (Daughter proton number Z_d) = 90; rw (Nuclear surface R) = 9.3 fm fixed. For this exercise use only 3.6375 MeV ≤ q ≤ 8.3625 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 = 39.755021 fm, with a bounded reading uncertainty ±1.3141 fm. 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 0.0004725 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.
TURNING POINT b against q
Predictions from the supplied Alpha decay as a WKB barrier integral 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): TURNING POINT b / fm
Horizontal axis (x): q: Decay energy Q / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Decay energy Q / MeV | TURNING POINT b / fm |
|---|---|
| 3.6375 | 71.25575257731958 |
| 4.8187500000000005 | 53.78838910505836 |
| 6 | 43.1988 |
| 7.18125 | 36.09299216710182 |
| 8.3625 | 30.994654708520173 |
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(3.6375) = 71.2558 fm and Y_out(8.3625) = 30.9947 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 3.6375 + (39.755 − 71.2558)(8.3625 − 3.6375)/(30.9947 − 71.2558) = 7.33439 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 = 6.51975 MeV; substituting back gives Y_out = 39.755021 fm.
- The interpolation error in the input is |q_line − q| = 0.814643 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 38.440921 fm and 41.069121 fm. Sort the roots even if the response decreases: 6.31114 MeV ≤ q ≤ 6.74263 MeV.
- The lower and upper input deviations are 0.208614 MeV and 0.222877 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 ≈ 6.51975 MeV; compatible interval [6.31114, 6.74263] MeV. Endpoint interpolation gives 7.33439 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Beta decay, phase space and the Kurie plot: infer an input with uncertainty
A beta spectrum is continuous because the released energy is shared with a neutrino, and its shape is a count of how many ways that sharing can happen — phase space, scaled by the Coulomb pull of the daughter nucleus on the departing lepton. Use the supplied topic model Y_out = (2*((q+511)/511)^4-9*((q+511)/511)^2-8)*sqrt(((q+511)/511)^2-1)/60+((q+511)/511)*ln(((q+511)/511)+sqrt(((q+511)/511)^2-1))/4, where Y_out is PHASE-SPACE f (Z = 0) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm. q is Endpoint Q, measured in keV. Keep z (Daughter charge Z_d (− for β⁺)) = 20; m (Neutrino mass m_νc² (exaggerated)) = 0 keV fixed. For this exercise use only 395.625 keV ≤ q ≤ 1104.38 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 = 2.0482579, with a bounded reading uncertainty ±0.28953. 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 0.070875 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.
PHASE-SPACE f (Z = 0) against q
Predictions from the supplied Beta decay, phase space and the Kurie plot 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): PHASE-SPACE f (Z = 0)
Horizontal axis (x): q: Endpoint Q / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Endpoint Q / keV | PHASE-SPACE f (Z = 0) |
|---|---|
| 395.625 | 0.11806077930312059 |
| 572.8125 | 0.48411307376739054 |
| 750 | 1.384977517195499 |
| 927.1875 | 3.218141706606861 |
| 1104.375 | 6.5243925123654165 |
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(395.625) = 0.118061 and Y_out(1104.38) = 6.52439. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 395.625 + (2.04826 − 0.118061)(1104.38 − 395.625)/(6.52439 − 0.118061) = 609.168 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 q = 827.963 keV; substituting back gives Y_out = 2.0482579.
- The interpolation error in the input is |q_line − q| = 218.795 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.7587279 and 2.3377879. Sort the roots even if the response decreases: 796.755 keV ≤ q ≤ 855.941 keV.
- The lower and upper input deviations are 31.2077 keV and 27.9788 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 q ≈ 827.963 keV; compatible interval [796.755, 855.941] keV. Endpoint interpolation gives 609.168 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.