University Physics V · topic UPV-02.06
Electron diffraction, from Davisson-Germer to molecules
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.
Electron diffraction: predict a finite change
Electron diffraction is elastic scattering of a beam prepared near a momentum eigenstate |k⟩, with intensity emerging only along directions where the momentum transfer ħ(k′ − k) equals ħG for a reciprocal-lattice vector G of the target. Use the supplied topic model Y_out = 135*1.226/(0.123*sqrt(1000*v)), where Y_out is RING r, d = 0.123 nm (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. v is Accelerating voltage V, measured in kV. Keep d (Trial lattice spacing d) = 0.16 nm fixed. For this exercise use only 1.87187 kV ≤ v ≤ 4.62812 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 = 2.47825 kV and 2.86412 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.
RING r, d = 0.123 nm against v
Predictions from the supplied Electron diffraction 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): RING r, d = 0.123 nm / mm
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 | RING r, d = 0.123 nm / mm |
|---|---|
| 1.871875 | 31.101454883479033 |
| 2.5609374999999996 | 26.590079600504616 |
| 3.25 | 23.60355654886283 |
| 3.9390624999999995 | 21.439896840646146 |
| 4.628125 | 19.779558211503495 |
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 = 2.47825 kV, substitute into Y_out = 135*1.226/(0.123*sqrt(1000*v)): Y₀ = 27.03 mm.
- At v = 2.86412 kV, the same fixed-parameter model gives Y₁ = 25.1434 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.88668 mm; the input change is Δv = 0.385875 kV.
- Average sensitivity is ΔY_out/Δv = -4.88936 mm 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 = 27.03 mm, Y₁ = 25.1434 mm; ΔY_out = -1.88668 mm; average sensitivity = -4.88936 mm 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.
Electron diffraction: infer an input with uncertainty
Electron diffraction is elastic scattering of a beam prepared near a momentum eigenstate |k⟩, with intensity emerging only along directions where the momentum transfer ħ(k′ − k) equals ħG for a reciprocal-lattice vector G of the target. Use the supplied topic model Y_out = 135*1.226/(0.123*sqrt(1000*v)), where Y_out is RING r, d = 0.123 nm (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. v is Accelerating voltage V, measured in kV. Keep d (Trial lattice spacing d) = 0.16 nm fixed. For this exercise use only 1.87187 kV ≤ v ≤ 4.62812 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 = 22.57408 mm, with a bounded reading uncertainty ±0.41918 mm. 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 0.000275625 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.
RING r, d = 0.123 nm against v
Predictions from the supplied Electron diffraction 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): RING r, d = 0.123 nm / mm
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 | RING r, d = 0.123 nm / mm |
|---|---|
| 1.871875 | 31.101454883479033 |
| 2.5609374999999996 | 26.590079600504616 |
| 3.25 | 23.60355654886283 |
| 3.9390624999999995 | 21.439896840646146 |
| 4.628125 | 19.779558211503495 |
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(1.87187) = 31.1015 mm and Y_out(4.62812) = 19.7796 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 1.87187 + (22.5741 − 31.1015)(4.62812 − 1.87187)/(19.7796 − 31.1015) = 3.94781 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 = 3.55319 kV; substituting back gives Y_out = 22.57408 mm.
- The interpolation error in the input is |v_line − v| = 0.394627 kV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 22.1549 mm and 22.99326 mm. Sort the roots even if the response decreases: 3.42482 kV ≤ v ≤ 3.68892 kV.
- The lower and upper input deviations are 0.128372 kV and 0.135728 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 ≈ 3.55319 kV; compatible interval [3.42482, 3.68892] kV. Endpoint interpolation gives 3.94781 kV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.