University Physics IV · topic UPIV-12.02
Ionic, metallic, and van der Waals bonding in crystals
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.
Ionic, metallic and van der Waals bonding: predict a finite change
Ionic, metallic and van der Waals bonding are the three non-covalent ways a solid holds together: transferred charge summed over a whole lattice, valence electrons delocalised across the crystal, and fluctuating dipoles attracting as R⁻⁶. Use the supplied topic model Y_out = a*14.4/r0, where Y_out is COULOMB αe²/4πε₀R₀ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r0 is Nearest-neighbour distance R₀, measured in Å. Keep a (Madelung constant α) = 1.75; n (Born exponent n) = 8 fixed. For this exercise use only 2.34875 Å ≤ r0 ≤ 3.45125 Å 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 r0 = 2.5913 Å and 2.74565 Å. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
COULOMB αe²/4πε₀R₀ against r0
Predictions from the supplied Ionic, metallic and van der Waals bonding 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): COULOMB αe²/4πε₀R₀ / eV
Horizontal axis (x): r0: Nearest-neighbour distance R₀ / Å
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r0: Nearest-neighbour distance R₀ / Å | COULOMB αe²/4πε₀R₀ / eV |
|---|---|
| 2.3487500000000003 | 10.729111229377326 |
| 2.624375 | 9.602286258633008 |
| 2.9000000000000004 | 8.689655172413792 |
| 3.175625 | 7.93544577839008 |
| 3.45125 | 7.301702281781963 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r0 = 2.5913 Å, substitute into Y_out = a*14.4/r0: Y₀ = 9.72485 eV.
- At r0 = 2.74565 Å, the same fixed-parameter model gives Y₁ = 9.17815 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.546694 eV; the input change is Δr0 = 0.15435 Å.
- Average sensitivity is ΔY_out/Δr0 = -3.54191 eV per Å. 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 = 9.72485 eV, Y₁ = 9.17815 eV; ΔY_out = -0.546694 eV; average sensitivity = -3.54191 eV per Å.
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.
Ionic, metallic and van der Waals bonding: infer an input with uncertainty
Ionic, metallic and van der Waals bonding are the three non-covalent ways a solid holds together: transferred charge summed over a whole lattice, valence electrons delocalised across the crystal, and fluctuating dipoles attracting as R⁻⁶. Use the supplied topic model Y_out = a*14.4/r0, where Y_out is COULOMB αe²/4πε₀R₀ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r0 is Nearest-neighbour distance R₀, measured in Å. Keep a (Madelung constant α) = 1.75; n (Born exponent n) = 8 fixed. For this exercise use only 2.34875 Å ≤ r0 ≤ 3.45125 Å 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 = 8.3408495 eV, with a bounded reading uncertainty ±0.15587 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00011025 Å, 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.
COULOMB αe²/4πε₀R₀ against r0
Predictions from the supplied Ionic, metallic and van der Waals bonding 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): COULOMB αe²/4πε₀R₀ / eV
Horizontal axis (x): r0: Nearest-neighbour distance R₀ / Å
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r0: Nearest-neighbour distance R₀ / Å | COULOMB αe²/4πε₀R₀ / eV |
|---|---|
| 2.3487500000000003 | 10.729111229377326 |
| 2.624375 | 9.602286258633008 |
| 2.9000000000000004 | 8.689655172413792 |
| 3.175625 | 7.93544577839008 |
| 3.45125 | 7.301702281781963 |
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(r0) = 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(2.34875) = 10.7291 eV and Y_out(3.45125) = 7.3017 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r0_line = 2.34875 + (8.34085 − 10.7291)(3.45125 − 2.34875)/(7.3017 − 10.7291) = 3.11699 Å. 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 r0 = 3.02127 Å; substituting back gives Y_out = 8.3408495 eV.
- The interpolation error in the input is |r0_line − r0| = 0.0957109 Å. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.1849795 eV and 8.4967195 eV. Sort the roots even if the response decreases: 2.96585 Å ≤ r0 ≤ 3.07881 Å.
- The lower and upper input deviations are 0.0554245 Å and 0.0575354 Å. 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 r0 ≈ 3.02127 Å; compatible interval [2.96585, 3.07881] Å. Endpoint interpolation gives 3.11699 Å.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.