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University Quantum Mechanics I · unit UQMI-15

Introduction to the Hydrogen Atom

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.

Extra challenging · 01

The Coulomb potential and the one-electron model of hydrogen: infer an input with uncertainty

The one-electron model of hydrogen is a single spinless, non-relativistic electron moving in the Coulomb potential energy V(r) = −e²/(4πε₀r) of a fixed point proton, so the whole atom is H = −(ħ²/2mₑ)∇² − e²/(4πε₀r). Use the supplied topic model Y_out = 13.606/(mu*b^2), where Y_out is KINETIC ⟨T⟩ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Cloud size b, measured in a₀. Keep mu (Reduced mass μ) = 1 mₑ fixed. For this exercise use only 1.05062 a₀ ≤ b ≤ 3.64937 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 = 1.9583259 eV, with a bounded reading uncertainty ±0.1405 eV. 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 0.000259875 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.

Explore the question data

KINETIC ⟨T⟩ against b

Predictions from the supplied The Coulomb potential and the one-electron model of hydrogen 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): KINETIC ⟨T⟩ / eV

KINETIC ⟨T⟩ against bPredictions from the supplied The Coulomb potential and the one-electron model of hydrogen 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. Horizontal axis: b: Cloud size b / a₀. Vertical axis: KINETIC ⟨T⟩ / eV. Exact values are available in the data table below.0.56943.6226.6749.72612.780.94671.6482.353.0523.753xy

Horizontal axis (x): b: Cloud size b / a₀

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1b: Cloud size b / a₀: 1.051KINETIC ⟨T⟩ / eV: 12.33
Read the exact data as a table
KINETIC ⟨T⟩ against b — supplied values
b: Cloud size b / a₀KINETIC ⟨T⟩ / eV
1.05062512.326364473145468
1.70031249999999974.706228087565529
2.34999999999999962.463739248528747
2.99968749999999941.5120927806998343
3.6493749999999991.0216296330363415

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
  1. Endpoint predictions are Y_out(1.05062) = 12.3264 eV and Y_out(3.64937) = 1.02163 eV. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives b_line = 1.05062 + (1.95833 − 12.3264)(3.64937 − 1.05062)/(1.02163 − 12.3264) = 3.43405 a₀. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives b = 2.63586 a₀; substituting back gives Y_out = 1.9583259 eV.
  4. The interpolation error in the input is |b_line − b| = 0.798183 a₀. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.8178259 eV and 2.0988259 eV. Sort the roots even if the response decreases: 2.54611 a₀ ≤ b ≤ 2.73583 a₀.
  6. The lower and upper input deviations are 0.0897533 a₀ and 0.0999674 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 b ≈ 2.63586 a₀; compatible interval [2.54611, 2.73583] a₀. Endpoint interpolation gives 3.43405 a₀.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Separating the radial and angular equations: infer an input with uncertainty

Separating the hydrogen equation means substituting ψ = R(r)Y(θ, φ) so that one equation in three variables becomes a radial equation holding the Coulomb potential and E, and an angular equation holding neither, joined by a single constant λ = l(l + 1). Use the supplied topic model Y_out = 52.9*27.2*sqrt(max(0,(1+en*(l*(l+1))/13.6)))/(-en), where Y_out is ALLOWED REGION r₊ − r₋ (pm). 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. en is Trial energy E, measured in eV. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only -6.735 eV ≤ en ≤ -1.065 eV 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 = 316.14471 pm, with a bounded reading uncertainty ±44.289 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000567 eV, 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.

Explore the question data

ALLOWED REGION r₊ − r₋ against en

Predictions from the supplied Separating the radial and angular equations 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): ALLOWED REGION r₊ − r₋ / pm

ALLOWED REGION r₊ − r₋ against enPredictions from the supplied Separating the radial and angular equations 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. Horizontal axis: en: Trial energy E / eV. Vertical axis: ALLOWED REGION r₊ − r₋ / pm. Exact values are available in the data table below.-27.91301.5630.8960.21290-6.962-5.431-3.9-2.369-0.8382xy

Horizontal axis (x): en: Trial energy E / eV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1en: Trial energy E / eV: -6.735ALLOWED REGION r₊ − r₋ / pm: 20.89
Read the exact data as a table
ALLOWED REGION r₊ − r₋ against en — supplied values
en: Trial energy E / eVALLOWED REGION r₊ − r₋ / pm
-6.73520.88763154462731
-5.317500000000001126.34555636046645
-3.9000000000000004240.9375620843347
-2.482500000000001461.84583949784235
-1.06500000000000041240.7583970979767

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(en) = 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
  1. Endpoint predictions are Y_out(-6.735) = 20.8876 pm and Y_out(-1.065) = 1240.76 pm. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives en_line = -6.735 + (316.145 − 20.8876)(-1.065 − -6.735)/(1240.76 − 20.8876) = -5.36264 eV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives en = -3.2763 eV; substituting back gives Y_out = 316.14471 pm.
  4. The interpolation error in the input is |en_line − en| = 2.08634 eV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 271.85571 pm and 360.43371 pm. Sort the roots even if the response decreases: -3.61966 eV ≤ en ≤ -2.98869 eV.
  6. The lower and upper input deviations are 0.343363 eV and 0.287605 eV. 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 en ≈ -3.2763 eV; compatible interval [-3.61966, -2.98869] eV. Endpoint interpolation gives -5.36264 eV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

The angular solutions: the quantum numbers l and m: infer an input with uncertainty

The angular solutions of any central-potential problem are the spherical harmonics Yₗᵐ(θ, φ), labelled by an integer l ≥ 0 that fixes L² = l(l + 1)ħ² and an integer m with |m| ≤ l that fixes L_z = mħ. Use the supplied topic model Y_out = sqrt((l*(l+1)-clamp(m,-l,l)^2)/2), where Y_out is ΔL_x = ΔL_y (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; clamp(x,a,b) confines x to [a,b]. m is Magnetic quantum number m, dimensionless. Keep l (Orbital quantum number l) = 2 fixed. For this exercise use only -2 ≤ m ≤ 0, 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 = 1.5811388 ħ, with a bounded reading uncertainty ±0.058564 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of m 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.

Explore the question data

ΔL_x = ΔL_y against m

Predictions from the supplied The angular solutions: the quantum numbers l and m model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): ΔL_x = ΔL_y / ħ

ΔL_x = ΔL_y against mPredictions from the supplied The angular solutions: the quantum numbers l and m model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: m: Magnetic quantum number m. Vertical axis: ΔL_x = ΔL_y / ħ. Exact values are available in the data table below.0.97071.1681.3661.5641.761-2.08-1.54-1-0.460.08xy

Horizontal axis (x): m: Magnetic quantum number m

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1m: Magnetic quantum number m: -2ΔL_x = ΔL_y / ħ: 1
Read the exact data as a table
ΔL_x = ΔL_y against m — supplied values
m: Magnetic quantum number mΔL_x = ΔL_y / ħ
-21
-11.5811388300841898
01.7320508075688772

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
  1. Endpoint predictions are Y_out(-2) = 1 ħ and Y_out(0) = 1.73205 ħ. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives m_line = -2 + (1.58114 − 1)(0 − -2)/(1.73205 − 1) = -0.412299. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to m = -1; only the allowed settings -1 lie inside the stated reading bound.
  4. The interpolation error in the input is |m_line − m| = 0.587701.
  5. The compatible allowed values span -1 to -1. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
  6. 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 m: -1. Endpoint interpolation predicts -0.412299; check admissibility against the discrete settings.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

The radial equation and its quoted solution: infer an input with uncertainty

The radial equation is the one-dimensional Schrödinger equation obeyed by u(r) = rR(r) on the half-line from r = 0 to ∞, with u(0) = 0 and an effective potential that adds the centrifugal barrier ħ²l(l + 1)/2μr² to the Coulomb well. Use the supplied topic model Y_out = -13.6/(nr+l+1)^2, where Y_out is ENERGY Eₙ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. l is Orbital quantum number l, dimensionless. Keep nr (Radial nodes nᵣ) = 1 fixed. For this exercise use only 0 ≤ l ≤ 2, 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 = -1.5111111 eV, with a bounded reading uncertainty ±0.204 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of l 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.

Explore the question data

ENERGY Eₙ against l

Predictions from the supplied The radial equation and its quoted solution model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): ENERGY Eₙ / eV

ENERGY Eₙ against lPredictions from the supplied The radial equation and its quoted solution model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: l: Orbital quantum number l. Vertical axis: ENERGY Eₙ / eV. Exact values are available in the data table below.-3.502-2.813-2.125-1.437-0.748-0.080.4611.542.08xy

Horizontal axis (x): l: Orbital quantum number l

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1l: Orbital quantum number l: 0ENERGY Eₙ / eV: -3.4
Read the exact data as a table
ENERGY Eₙ against l — supplied values
l: Orbital quantum number lENERGY Eₙ / eV
0-3.4
1-1.511111111111111
2-0.85

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
  1. Endpoint predictions are Y_out(0) = -3.4 eV and Y_out(2) = -0.85 eV. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives l_line = 0 + (-1.51111 − -3.4)(2 − 0)/(-0.85 − -3.4) = 1.48148. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to l = 1; only the allowed settings 1 lie inside the stated reading bound.
  4. The interpolation error in the input is |l_line − l| = 0.481481.
  5. The compatible allowed values span 1 to 1. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
  6. 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 l: 1. Endpoint interpolation predicts 1.48148; check admissibility against the discrete settings.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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