University Quantum Mechanics II · topic UQMII-15.01
Cross-sections and the classical benchmarks
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.
Cross-sections and the classical benchmarks: predict a finite change
The differential cross-section dσ/dΩ is the rate at which particles scatter into unit solid angle about a direction, divided by the incident flux; it is an area per steradian that belongs to the interaction, and its integral σ is the beam area the target removes. Use the supplied topic model Y_out = 32.35/(en*en*sind(th/2)^4), where Y_out is RUTHERFORD dσ/dΩ(θ₀) (b sr⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. en is CM energy E, α on gold, measured in MeV. Keep th (Marker angle θ₀) = 60 °; rs (Hard-sphere radius R) = 10 fm fixed. For this exercise use only 2.4875 MeV ≤ en ≤ 13.5125 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 en = 4.913 MeV and 6.4565 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RUTHERFORD dσ/dΩ(θ₀) against en
Predictions from the supplied Cross-sections and the classical benchmarks 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): RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹
Horizontal axis (x): en: CM energy E, α on gold / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: CM energy E, α on gold / MeV | RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹ |
|---|---|
| 2.4875 | 83.65041286836197 |
| 5.2437499999999995 | 18.823930526294866 |
| 7.999999999999999 | 8.087500000000006 |
| 10.756249999999998 | 4.473758044405494 |
| 13.5125 | 2.8348028044749074 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 4.913 MeV, substitute into Y_out = 32.35/(en*en*sind(th/2)^4): Y₀ = 21.4438 b sr⁻¹.
- At en = 6.4565 MeV, the same fixed-parameter model gives Y₁ = 12.4165 b sr⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -9.02723 b sr⁻¹; the input change is Δen = 1.5435 MeV.
- Average sensitivity is ΔY_out/Δen = -5.84854 b sr⁻¹ 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 = 21.4438 b sr⁻¹, Y₁ = 12.4165 b sr⁻¹; ΔY_out = -9.02723 b sr⁻¹; average sensitivity = -5.84854 b sr⁻¹ 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.
Cross-sections and the classical benchmarks: infer an input with uncertainty
The differential cross-section dσ/dΩ is the rate at which particles scatter into unit solid angle about a direction, divided by the incident flux; it is an area per steradian that belongs to the interaction, and its integral σ is the beam area the target removes. Use the supplied topic model Y_out = 32.35/(en*en*sind(th/2)^4), where Y_out is RUTHERFORD dσ/dΩ(θ₀) (b sr⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. en is CM energy E, α on gold, measured in MeV. Keep th (Marker angle θ₀) = 60 °; rs (Hard-sphere radius R) = 10 fm fixed. For this exercise use only 2.4875 MeV ≤ en ≤ 13.5125 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 = 6.098397 b sr⁻¹, with a bounded reading uncertainty ±0.48954 b sr⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0011025 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.
RUTHERFORD dσ/dΩ(θ₀) against en
Predictions from the supplied Cross-sections and the classical benchmarks 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): RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹
Horizontal axis (x): en: CM energy E, α on gold / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: CM energy E, α on gold / MeV | RUTHERFORD dσ/dΩ(θ₀) / b sr⁻¹ |
|---|---|
| 2.4875 | 83.65041286836197 |
| 5.2437499999999995 | 18.823930526294866 |
| 7.999999999999999 | 8.087500000000006 |
| 10.756249999999998 | 4.473758044405494 |
| 13.5125 | 2.8348028044749074 |
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(en) = 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(2.4875) = 83.6504 b sr⁻¹ and Y_out(13.5125) = 2.8348 b sr⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 2.4875 + (6.0984 − 83.6504)(13.5125 − 2.4875)/(2.8348 − 83.6504) = 13.0673 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 en = 9.21275 MeV; substituting back gives Y_out = 6.098397 b sr⁻¹.
- The interpolation error in the input is |en_line − en| = 3.85453 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.608857 b sr⁻¹ and 6.587937 b sr⁻¹. Sort the roots even if the response decreases: 8.86385 MeV ≤ en ≤ 9.60638 MeV.
- The lower and upper input deviations are 0.3489 MeV and 0.393634 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 en ≈ 9.21275 MeV; compatible interval [8.86385, 9.60638] MeV. Endpoint interpolation gives 13.0673 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.