University Physics IV · unit UPIV-12
Molecular and Solid-State Physics
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Covalent bonding and molecular orbitals: infer an input with uncertainty
A molecular orbital is a one-electron state that spans every nucleus in the molecule; in the LCAO approximation it is a weighted sum of atomic orbitals, with the weights fixed by minimising the energy. Use the supplied topic model Y_out = 54.422*((z*z-z-(1-exp(-2*z*r)*(1+z*r))/r)*exp(-z*r)*(1+z*r+z*z*r*r/3)-z*(z-2)*exp(-z*r)*(1+z*r))/(1-exp(-2*z*r)*(1+z*r+z*z*r*r/3)^2), where Y_out is σg–σu* GAP (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. r is Nuclear separation R, measured in a₀. Keep z (Orbital exponent ζ) = 1 fixed. For this exercise use only 2.6375 a₀ ≤ r ≤ 7.3625 a₀ 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 = 0.76641231 eV, with a bounded reading uncertainty ±0.090231 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 a₀, 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.
σg–σu* GAP against r
Predictions from the supplied Covalent bonding and molecular orbitals 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): σg–σu* GAP / eV
Horizontal axis (x): r: Nuclear separation R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Nuclear separation R / a₀ | σg–σu* GAP / eV |
|---|---|
| 2.6375 | 6.760930580278907 |
| 3.81875 | 2.8654448163440684 |
| 5 | 1.160076560980908 |
| 6.18125 | 0.4463045184677629 |
| 7.3625 | 0.1648737517547915 |
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(r) = 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(2.6375) = 6.76093 eV and Y_out(7.3625) = 0.164874 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 2.6375 + (0.766412 − 6.76093)(7.3625 − 2.6375)/(0.164874 − 6.76093) = 6.9316 a₀. 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 r = 5.51975 a₀; substituting back gives Y_out = 0.76641231 eV.
- The interpolation error in the input is |r_line − r| = 1.41185 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.67618131 eV and 0.85664331 eV. Sort the roots even if the response decreases: 5.38137 a₀ ≤ r ≤ 5.67452 a₀.
- The lower and upper input deviations are 0.13838 a₀ and 0.154774 a₀. 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 r ≈ 5.51975 a₀; compatible interval [5.38137, 5.67452] a₀. Endpoint interpolation gives 6.9316 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Rotational spectra and the rigid rotor: infer an input with uncertainty
A rigid rotor is a two-atom molecule modelled as two point masses held a fixed distance apart and free to tumble; its allowed energies are E_J = J(J+1)ħ²/2I, where I = μR² is the moment of inertia about the centre of mass. Use the supplied topic model Y_out = 156.81/sqrt(b), where Y_out is BOND LENGTH WITH CO MASSES (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Rotational constant B̃, measured in cm⁻¹. Keep t (Temperature T) = 300 K; q (Distortion D̃/B̃) = 0 ×10⁻⁴ fixed. For this exercise use only 1.58688 cm⁻¹ ≤ b ≤ 3.71312 cm⁻¹ 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 = 92.338879 pm, with a bounded reading uncertainty ±1.6442 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000212625 cm⁻¹, 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.
BOND LENGTH WITH CO MASSES against b
Predictions from the supplied Rotational spectra and the rigid rotor 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): BOND LENGTH WITH CO MASSES / pm
Horizontal axis (x): b: Rotational constant B̃ / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Rotational constant B̃ / cm⁻¹ | BOND LENGTH WITH CO MASSES / pm |
|---|---|
| 1.586875 | 124.48080715887716 |
| 2.1184374999999998 | 107.73726018713388 |
| 2.65 | 96.32761727073189 |
| 3.1815625 | 87.91308599143515 |
| 3.713125 | 81.37742995890225 |
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(b) = 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(1.58688) = 124.481 pm and Y_out(3.71312) = 81.3774 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 1.58688 + (92.3389 − 124.481)(3.71312 − 1.58688)/(81.3774 − 124.481) = 3.17241 cm⁻¹. 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 b = 2.88389 cm⁻¹; substituting back gives Y_out = 92.338879 pm.
- The interpolation error in the input is |b_line − b| = 0.288519 cm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 90.694679 pm and 93.983079 pm. Sort the roots even if the response decreases: 2.78387 cm⁻¹ ≤ b ≤ 2.9894 cm⁻¹.
- The lower and upper input deviations are 0.100022 cm⁻¹ and 0.105512 cm⁻¹. 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 b ≈ 2.88389 cm⁻¹; compatible interval [2.78387, 2.9894] cm⁻¹. Endpoint interpolation gives 3.17241 cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Vibrational and rotation-vibration spectra: infer an input with uncertainty
A vibrational spectrum is the set of transitions between the quantised stretching levels of a bond, E_v = (v+½)ħω; in a gas of Σ-state diatomics such as HCl each one must be accompanied by a rotational change, so it appears as a band of lines rather than a single line. Use the supplied topic model Y_out = (sqrt(max(t,1)/(2*1.4388*max(b0,0.1)))-0.5), where Y_out is BRIGHTEST J″ (before display rounding) (dimensionless). 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. b0 is Rotational constant B₀ (v = 0), measured in cm⁻¹. Keep d (Bond stretch B₀ − B₁) = 0.3 cm⁻¹; t (Temperature T) = 300 K fixed. For this exercise use only 5.74375 cm⁻¹ ≤ b0 ≤ 11.2562 cm⁻¹ 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 = 2.8835504, with a bounded reading uncertainty ±0.051034. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 cm⁻¹, 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.
BRIGHTEST J″ (before display rounding) against b0
Predictions from the supplied Vibrational and rotation-vibration spectra 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): BRIGHTEST J″ (before display rounding)
Horizontal axis (x): b0: Rotational constant B₀ (v = 0) / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b0: Rotational constant B₀ (v = 0) / cm⁻¹ | BRIGHTEST J″ (before display rounding) |
|---|---|
| 5.74375 | 3.760373263861288 |
| 7.121875 | 3.326028859184726 |
| 8.5 | 3.002159747431779 |
| 9.878125 | 2.748689136681243 |
| 11.25625 | 2.5433265154180007 |
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(b0) = 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(5.74375) = 3.76037 and Y_out(11.2562) = 2.54333. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b0_line = 5.74375 + (2.88355 − 3.76037)(11.2562 − 5.74375)/(2.54333 − 3.76037) = 9.71524 cm⁻¹. 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 b0 = 9.10637 cm⁻¹; substituting back gives Y_out = 2.8835504.
- The interpolation error in the input is |b0_line − b0| = 0.608863 cm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.8325164 and 2.9345844. Sort the roots even if the response decreases: 8.83776 cm⁻¹ ≤ b0 ≤ 9.38742 cm⁻¹.
- The lower and upper input deviations are 0.26861 cm⁻¹ and 0.281045 cm⁻¹. 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 b0 ≈ 9.10637 cm⁻¹; compatible interval [8.83776, 9.38742] cm⁻¹. Endpoint interpolation gives 9.71524 cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.