University Physics IV · topic UPIV-04.05
The de Broglie hypothesis
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.
The de Broglie hypothesis: predict a finite change
The de Broglie hypothesis assigns every particle a wavelength λ = h/p set by its momentum alone, so the wave relation first established for light applies to matter as well. Use the supplied topic model Y_out = 38.783/sqrt(max(v,1)), where Y_out is λ NON-RELATIVISTIC (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. v is Accelerating voltage V, measured in kV. Keep w (Window width along the beam) = 24 pm fixed. For this exercise use only 21.4688 kV ≤ v ≤ 143.531 kV 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 v = 48.3225 kV and 65.4112 kV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
λ NON-RELATIVISTIC against v
Predictions from the supplied The de Broglie hypothesis 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): λ NON-RELATIVISTIC / pm
Horizontal axis (x): v: Accelerating voltage V / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Accelerating voltage V / kV | λ NON-RELATIVISTIC / pm |
|---|---|
| 21.46875 | 8.370242123105983 |
| 51.984375 | 5.379042637702979 |
| 82.5 | 4.269867770289564 |
| 113.015625 | 3.648145530119555 |
| 143.53125 | 3.2371898251196276 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 48.3225 kV, substitute into Y_out = 38.783/sqrt(max(v,1)): Y₀ = 5.57913 pm.
- At v = 65.4112 kV, the same fixed-parameter model gives Y₁ = 4.79529 pm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.78384 pm; the input change is Δv = 17.0887 kV.
- Average sensitivity is ΔY_out/Δv = -0.0458687 pm per kV. 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 = 5.57913 pm, Y₁ = 4.79529 pm; ΔY_out = -0.78384 pm; average sensitivity = -0.0458687 pm per kV.
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.
The de Broglie hypothesis: infer an input with uncertainty
The de Broglie hypothesis assigns every particle a wavelength λ = h/p set by its momentum alone, so the wave relation first established for light applies to matter as well. Use the supplied topic model Y_out = 38.783/sqrt(max(v,1)), where Y_out is λ NON-RELATIVISTIC (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. v is Accelerating voltage V, measured in kV. Keep w (Window width along the beam) = 24 pm fixed. For this exercise use only 21.4688 kV ≤ v ≤ 143.531 kV 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 = 3.9597818 pm, with a bounded reading uncertainty ±0.10839 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0122063 kV, 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.
λ NON-RELATIVISTIC against v
Predictions from the supplied The de Broglie hypothesis 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): λ NON-RELATIVISTIC / pm
Horizontal axis (x): v: Accelerating voltage V / kV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Accelerating voltage V / kV | λ NON-RELATIVISTIC / pm |
|---|---|
| 21.46875 | 8.370242123105983 |
| 51.984375 | 5.379042637702979 |
| 82.5 | 4.269867770289564 |
| 113.015625 | 3.648145530119555 |
| 143.53125 | 3.2371898251196276 |
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(v) = 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(21.4688) = 8.37024 pm and Y_out(143.531) = 3.23719 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 21.4688 + (3.95978 − 8.37024)(143.531 − 21.4688)/(3.23719 − 8.37024) = 126.348 kV. 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 v = 95.9269 kV; substituting back gives Y_out = 3.9597818 pm.
- The interpolation error in the input is |v_line − v| = 30.4213 kV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.8513918 pm and 4.0681718 pm. Sort the roots even if the response decreases: 90.8833 kV ≤ v ≤ 101.402 kV.
- The lower and upper input deviations are 5.04354 kV and 5.47533 kV. 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 v ≈ 95.9269 kV; compatible interval [90.8833, 101.402] kV. Endpoint interpolation gives 126.348 kV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.