University Physics IV · unit UPIV-11
Atomic 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.
Identical particles and exchange symmetry: infer an input with uncertainty
Exchange symmetry is the requirement that the state of two identical particles either keeps its sign or reverses it when the two particle labels are swapped — symmetric for bosons, antisymmetric for fermions — because the labels themselves are not observable. Use the supplied topic model Y_out = 2*sind(180*a)*sind(360*a), where Y_out is OVERLAP ψ₁(x₁)ψ₂(x₁)L (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. a is Electron 1 position x₁/L, measured in L. Keep t (Exchange mixing angle θ) = 45 ° fixed. For this exercise use only 0.145625 L ≤ a ≤ 0.287375 L 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 = 1.3240317, with a bounded reading uncertainty ±0.030453. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000014175 L, 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.
OVERLAP ψ₁(x₁)ψ₂(x₁)L against a
Predictions from the supplied Identical particles and exchange symmetry 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): OVERLAP ψ₁(x₁)ψ₂(x₁)L
Horizontal axis (x): a: Electron 1 position x₁/L / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Electron 1 position x₁/L / L | OVERLAP ψ₁(x₁)ψ₂(x₁)L |
|---|---|
| 0.145625 | 0.7001463145805644 |
| 0.18106250000000002 | 0.9777963144139875 |
| 0.2165 | 1.2300657167501865 |
| 0.2519375 | 1.4226899703782185 |
| 0.287375 | 1.5270516166473813 |
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(a) = 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(0.145625) = 0.700146 and Y_out(0.287375) = 1.52705. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.145625 + (1.32403 − 0.700146)(0.287375 − 0.145625)/(1.52705 − 0.700146) = 0.252573 L. 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 a = 0.232092 L; substituting back gives Y_out = 1.3240317.
- The interpolation error in the input is |a_line − a| = 0.0204804 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2935787 and 1.3544847. Sort the roots even if the response decreases: 0.226795 L ≤ a ≤ 0.237703 L.
- The lower and upper input deviations are 0.00529758 L and 0.00561049 L. 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 a ≈ 0.232092 L; compatible interval [0.226795, 0.237703] L. Endpoint interpolation gives 0.252573 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Pauli principle and Slater determinants: infer an input with uncertainty
A Slater determinant is the normalised, fully antisymmetric N-electron state built from N one-electron spin-orbitals: rows label the electrons, columns label the spin-orbitals, and the determinant supplies every sign that indistinguishability demands. Use the supplied topic model Y_out = (sin(pi*p)*sin(n*pi/2)+s*sin(n*pi*p))^2/(1+s*s), where Y_out is DENSITY AT MIDPOINT (× 4/L²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. p is Electron 1 held at x₁/L, dimensionless. Keep n (Second orbital index n) = 2; s (Exchange sign s) = -1 fixed. For this exercise use only 0.145625 ≤ p ≤ 0.2165 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 = 0.42976916 × 4/L², with a bounded reading uncertainty ±0.0072606 × 4/L². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000070875, 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.
DENSITY AT MIDPOINT against p
Predictions from the supplied Pauli principle and Slater determinants 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): DENSITY AT MIDPOINT / × 4/L²
Horizontal axis (x): p: Electron 1 held at x₁/L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Electron 1 held at x₁/L | DENSITY AT MIDPOINT / × 4/L² |
|---|---|
| 0.145625 | 0.3140723432832492 |
| 0.16334375 | 0.3658530087318895 |
| 0.1810625 | 0.41191364572185263 |
| 0.19878124999999996 | 0.44998009532285915 |
| 0.21649999999999997 | 0.4781728968508755 |
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(p) = 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(0.145625) = 0.314072 × 4/L² and Y_out(0.2165) = 0.478173 × 4/L². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.145625 + (0.429769 − 0.314072)(0.2165 − 0.145625)/(0.478173 − 0.314072) = 0.195594. 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 p = 0.188859; substituting back gives Y_out = 0.42976916 × 4/L².
- The interpolation error in the input is |p_line − p| = 0.00673569. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.42250856 × 4/L² and 0.43702976 × 4/L². Sort the roots even if the response decreases: 0.185601 ≤ p ≤ 0.192261.
- The lower and upper input deviations are 0.00325761 and 0.00340195. 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 p ≈ 0.188859; compatible interval [0.185601, 0.192261]. Endpoint interpolation gives 0.195594.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Central field and screening: infer an input with uncertainty
The central-field approximation replaces the pairwise repulsion between electrons with a single spherically symmetric potential V(r) for each electron — the angular average of what all the others do — so the many-electron problem separates into one-electron problems again. Use the supplied topic model Y_out = 1+(z-1)*exp(-rv/max(d,0.1)), where Y_out is Z_eff AT THE MARKER (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. rv is Marker radius r, measured in a₀. Keep z (Nuclear charge Z) = 11; d (Core screening radius d) = 0.7 a₀ fixed. For this exercise use only 1.11813 a₀ ≤ rv ≤ 7.18187 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 = 1.0102666, with a bounded reading uncertainty ±0.0014875. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000606375 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.
Z_eff AT THE MARKER against rv
Predictions from the supplied Central field and screening 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): Z_eff AT THE MARKER
Horizontal axis (x): rv: Marker radius r / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rv: Marker radius r / a₀ | Z_eff AT THE MARKER |
|---|---|
| 1.118125 | 3.0243803716440016 |
| 2.6340624999999998 | 1.232152374660293 |
| 4.15 | 1.0266228253421787 |
| 5.6659375 | 1.003053058708692 |
| 7.181875 | 1.0003501193941258 |
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(rv) = 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(1.11813) = 3.02438 and Y_out(7.18187) = 1.00035. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rv_line = 1.11813 + (1.01027 − 3.02438)(7.18187 − 1.11813)/(1.00035 − 3.02438) = 7.15217 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 rv = 4.81701 a₀; substituting back gives Y_out = 1.0102666.
- The interpolation error in the input is |rv_line − rv| = 2.33516 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0087791 and 1.0117541. Sort the roots even if the response decreases: 4.7223 a₀ ≤ rv ≤ 4.92658 a₀.
- The lower and upper input deviations are 0.0947143 a₀ and 0.109565 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 rv ≈ 4.81701 a₀; compatible interval [4.7223, 4.92658] a₀. Endpoint interpolation gives 7.15217 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Configurations, aufbau and Hund's rules: infer an input with uncertainty
The aufbau order is the empirical sequence in which subshells fill — lowest n + l first, and the lower n when n + l ties — and Hund's rules then select, from the states that configuration allows, the term and level that lie lowest. Use the supplied topic model Y_out = abs(abs(3*min(q,5)-min(q,5)*(min(q,5)+1)/2+3*max(q-5,0)-max(q-5,0)*(max(q-5,0)+1)/2)-(min(q,5)-max(q-5,0)+s*(2-s))/2*sign(5.5-q)), where Y_out is HUND GROUND J (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; min(a,b) selects the smaller value; max(a,b) selects the larger value. q is 3d electrons q, dimensionless. Keep s (4s electrons s) = 1 fixed. For this exercise use only 4 ≤ q ≤ 7, at the specified test settings separated by 1. 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 = 4.5, with a bounded reading uncertainty ±0.36. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of q consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
HUND GROUND J against q
Predictions from the supplied Configurations, aufbau and Hund's rules model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): HUND GROUND J
Horizontal axis (x): q: 3d electrons q
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: 3d electrons q | HUND GROUND J |
|---|---|
| 4 | 0.5 |
| 5 | 3 |
| 6 | 4.5 |
| 7 | 5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(4) = 0.5 and Y_out(7) = 5. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 4 + (4.5 − 0.5)(7 − 4)/(5 − 0.5) = 6.66667. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to q = 6; only the allowed settings 6 lie inside the stated reading bound.
- The interpolation error in the input is |q_line − q| = 0.666667.
- The compatible allowed values span 6 to 6. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible q: 6. Endpoint interpolation predicts 6.66667; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.