University Physics III · topic UPIII-09.04
Matter waves and electron diffraction
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.
Matter waves and electron diffraction: predict a finite change
A matter wave is the wave associated with any particle of momentum p, with de Broglie wavelength λ = h/p; it diffracts from a periodic structure exactly as light does, and visibly so only when λ is comparable to the structure's spacing. Use the supplied topic model Y_out = (1.2264/d)^2, where Y_out is CUTOFF VOLTAGE, λ = d (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Surface row spacing d, measured in nm. Keep v (Accelerating voltage V) = 54 V fixed. For this exercise use only 0.131875 nm ≤ d ≤ 0.368125 nm 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 d = 0.18385 nm and 0.216925 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CUTOFF VOLTAGE, λ = d against d
Predictions from the supplied Matter waves and 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): CUTOFF VOLTAGE, λ = d / V
Horizontal axis (x): d: Surface row spacing d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Surface row spacing d / nm | CUTOFF VOLTAGE, λ = d / V |
|---|---|
| 0.13187500000000002 | 86.4847109813346 |
| 0.19093750000000004 | 41.25549666480051 |
| 0.25 | 24.064911359999996 |
| 0.3090625 | 15.746051122918328 |
| 0.36812500000000004 | 11.098739533207844 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.18385 nm, substitute into Y_out = (1.2264/d)^2: Y₀ = 44.4976 V.
- At d = 0.216925 nm, the same fixed-parameter model gives Y₁ = 31.9628 V.
- Signed change: ΔY_out = Y₁ − Y₀ = -12.5348 V; the input change is Δd = 0.033075 nm.
- Average sensitivity is ΔY_out/Δd = -378.982 V per nm. 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 = 44.4976 V, Y₁ = 31.9628 V; ΔY_out = -12.5348 V; average sensitivity = -378.982 V per nm.
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.
Matter waves and electron diffraction: infer an input with uncertainty
A matter wave is the wave associated with any particle of momentum p, with de Broglie wavelength λ = h/p; it diffracts from a periodic structure exactly as light does, and visibly so only when λ is comparable to the structure's spacing. Use the supplied topic model Y_out = (1.2264/d)^2, where Y_out is CUTOFF VOLTAGE, λ = d (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Surface row spacing d, measured in nm. Keep v (Accelerating voltage V) = 54 V fixed. For this exercise use only 0.131875 nm ≤ d ≤ 0.368125 nm 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 = 19.746288 V, with a bounded reading uncertainty ±1.2971 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000023625 nm, 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.
CUTOFF VOLTAGE, λ = d against d
Predictions from the supplied Matter waves and 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): CUTOFF VOLTAGE, λ = d / V
Horizontal axis (x): d: Surface row spacing d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Surface row spacing d / nm | CUTOFF VOLTAGE, λ = d / V |
|---|---|
| 0.13187500000000002 | 86.4847109813346 |
| 0.19093750000000004 | 41.25549666480051 |
| 0.25 | 24.064911359999996 |
| 0.3090625 | 15.746051122918328 |
| 0.36812500000000004 | 11.098739533207844 |
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(d) = 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(0.131875) = 86.4847 V and Y_out(0.368125) = 11.0987 V. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.131875 + (19.7463 − 86.4847)(0.368125 − 0.131875)/(11.0987 − 86.4847) = 0.341025 nm. 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 d = 0.275987 nm; substituting back gives Y_out = 19.746288 V.
- The interpolation error in the input is |d_line − d| = 0.0650372 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 18.449188 V and 21.043388 V. Sort the roots even if the response decreases: 0.267346 nm ≤ d ≤ 0.285525 nm.
- The lower and upper input deviations are 0.00864112 nm and 0.00953709 nm. 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 d ≈ 0.275987 nm; compatible interval [0.267346, 0.285525] nm. Endpoint interpolation gives 0.341025 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.