Quantum chemistry — particle in a box revisited; H-atom solutions: Worked examples

Everything below uses only tools already earned in the parent: the box energy formula, the box wavefunction, the hydrogen energy formula, quantum numbers, and the radial distribution function. Nothing new is assumed.

The scenario matrix

Before solving anything, let us list every case class these topics can throw at you. Each row is a distinct "shape" of question. Every worked example below is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky	Covered by
A	Box: single energy level, plug-and-chug	unit bookkeeping (J vs eV)	Ex 1
B	Box: energy gap $n_{i} \to n_{f}$	uses $n^{2}$ difference, not $n$	Ex 2
C	Box: limiting behaviour ( $L \to \infty$ , $L \to 0$ )	quantization "turns off"/"blows up"	Ex 3
D	Box: probability from wavefunction (integrate $ψ^{2}$ )	node positions, definite integral	Ex 4
E	H-atom: energy level & ionization	negative energies, $r \to \infty$ zero	Ex 5
F	H-atom: transition photon (Rydberg)	difference of $- 1/ n^{2}$ terms, wavelength	Ex 6
G	H-atom: degeneracy & node counting	same- $n$ states, radial vs angular nodes	Ex 7
H	H-atom: most-probable vs expectation radius	the $r^{2}$ shell weight	Ex 8
I	Degenerate / forbidden inputs ( $n = 0$ , $ℓ \geq n$ )	why they are not states	Ex 9
J	Real-world word problem + exam twist	translate physics → formula	Ex 10

Ten examples, ten cells — full coverage.

Worked example Ex 1 — Cell A · Ground-state energy of an electron in a 1 nm box

An electron sits in a 1D box of length $L = 1.00 nm$ . Find its ground-state ( $n = 1$ ) energy in joules and electron-volts.

Forecast: Guess: is it closer to $1 0^{- 20}$ J or $1 0^{- 15}$ J? (Confinement to a nanometre is "gentle" on the atomic scale, so expect the smaller one.)

Write the level formula. $E_{n} = \frac{n ^{2} h ^{2}}{8 m L ^{2}}$ . Why this step? This is the box result from the parent; $n = 1$ is the lowest allowed state.
Insert numbers with $h = 6.626 \times 1 0^{- 34}$ J·s, $m = 9.11 \times 1 0^{- 31}$ kg, $L = 1 0^{- 9}$ m: $E_{1} = \frac{( 6.626 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 1 0 ^{- 9} ) ^{2}} \approx 6.02 \times 1 0^{- 20} J .$ Why this step? Direct substitution; $n^{2} = 1$ .
Convert to eV by dividing by $1.602 \times 1 0^{- 19}$ J/eV: $E_{1} \approx \frac{6.02 \times 1 0 ^{- 20}}{1.602 \times 1 0 ^{- 19}} \approx 0.376 eV .$ Why this step? Chemists compare energies to bond energies (a few eV), so eV is the natural yardstick.

Verify: Units: $\frac{( J\cdots ) ^{2}}{kg \cdot m ^{2}} = \frac{J ^{2} s ^{2}}{kg\cdotm ^{2}}$ . Since $J = kg\cdotm^{2} s^{- 2}$ , one J cancels leaving J. ✓ Magnitude $\sim 1 0^{- 20}$ J matches the forecast (small).

Worked example Ex 3 — Cell C · Limiting behaviour (

$L \to \infty$ and $L \to 0$ ) Explain quantitatively what happens to the box levels and the gap as (a) $L \to \infty$ and (b) $L \to 0$ .

Forecast: A huge box = a free particle. Do you expect the levels to merge or spread apart?

Look at the $L$ -dependence. $E_{n} = \frac{n ^{2} h ^{2}}{8 m L ^{2}} \propto \frac{1}{L ^{2}}$ , and $Δ E_{n \to n + 1} = (2 n + 1) \frac{h ^{2}}{8 m L ^{2}} \propto \frac{1}{L ^{2}}$ . Why this step? All the $L$ info is in the $1/ L^{2}$ ; the case analysis is just its two extremes.
Case $L \to \infty$ : $E_{n} \to 0$ and $Δ E \to 0$ . Levels collapse into a continuum — the free particle. Quantization "turns off." Why this step? No walls → no whole-half-wavelength constraint.
Case $L \to 0$ : $E_{n} \to \infty$ and $Δ E \to \infty$ . Squeezing a particle costs unbounded energy. Why this step? This is the mathematical face of the uncertainty principle: pin down $x$ tightly and kinetic energy explodes.

Verify (numeric anchor): Doubling $L$ from $1$ nm to $2$ nm should divide $E_{1}$ by $4$ : $6.02 \times 1 0^{- 20} /4 = 1.50 \times 1 0^{- 20}$ J. Re-plugging $L = 2 \times 1 0^{- 9}$ into the formula gives exactly that. ✓ See the trend curve.

Worked example Ex 4 — Cell D · Probability of finding the particle in the left third

For the ground state ( $n = 1$ ) in a box $[0, L]$ , what is the probability that the particle lies in $0 \leq x \leq L /3$ ?

Forecast: If probability were spread evenly it would be $1/3 \approx 0.333$ . But $ψ_{1}$ bulges in the middle, so the left third holds less density near $x = 0$ but rising toward $L /3$ . Guess: a bit under $1/3$ ? Or over?

Set up the integral. $P = \int_{0}^{L /3} (\frac{2}{L} sin \frac{π x}{L})^{2} d x = \frac{2}{L} \int_{0}^{L /3} sin^{2} \frac{π x}{L} d x$ . Why this step? $∣ ψ ∣^{2}$ is the probability density; area under it over the region is the probability.
Use $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ . $P = \frac{1}{L} \int_{0}^{L /3} (1 - cos \frac{2 π x}{L}) d x = \frac{1}{L} [x - \frac{L}{2 π} sin \frac{2 π x}{L}]_{0}^{L /3} .$ Why this step? $sin^{2}$ has no elementary antiderivative on its own; the double-angle identity linearises it.
Evaluate at $x = L /3$ : $\frac{2 π}{L} \cdot \frac{L}{3} = \frac{2 π}{3}$ , and $sin \frac{2 π}{3} = \frac{3}{2}$ . $P = \frac{1}{L} (\frac{L}{3} - \frac{L}{2 π} \cdot \frac{3}{2}) = \frac{1}{3} - \frac{3}{4 π} \approx 0.333 - 0.138 = 0.196.$ Why this step? Plugging the limit; the $L$ 's cancel (as they must — probability is dimensionless).

Verify: By symmetry the right third also gives $0.196$ , so the middle third holds $1 - 2 (0.196) = 0.608$ — more than half, consistent with $ψ_{1}$ peaking at the centre. Forecast confirmed: left third $< 1/3$ . ✓

Worked example Ex 5 — Cell E · Energy and ionization of hydrogen from

$n = 1$ Give the energy of the hydrogen ground state and the energy needed to ionize it (remove the electron to $r \to \infty$ ).

Forecast: Ionization energy of hydrogen is a famous number — do you recall it in eV?

Read the level formula. $E_{n} = - \frac{13.6}{n ^{2}} eV$ , so $E_{1} = - 13.6$ eV. Why this step? Negative because the electron is bound below the $r \to \infty$ reference of $0$ .
Ionization = climb to $E_{\infty} = 0$ . $IE = E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6$ eV. Why this step? Ionization means reaching the free state; energy required is the depth of the well.
Convert to J if needed: $13.6 \times 1.602 \times 1 0^{- 19} \approx 2.18 \times 1 0^{- 18}$ J. Why this step? Cross-check against tabulated hydrogen ionization energy.

Verify: Tabulated hydrogen ionization energy is $1312$ kJ/mol. Per atom: $1.312 \times 1 0^{6} /6.022 \times 1 0^{23} = 2.18 \times 1 0^{- 18}$ J. ✓ Matches step 3 exactly.

Worked example Ex 6 — Cell F · Transition photon

$n = 3 \to 2$ (a Balmer line) An electron falls from $n = 3$ to $n = 2$ in hydrogen. Find the photon's energy (eV) and wavelength (nm).

Forecast: $n = 3 \to 2$ is the first Balmer line (H- $α$ ), famously red. So expect $λ$ around $650$ nm.

Take the energy difference. $Δ E = E_{2} - E_{3} = - 13.6 (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = - 13.6 (\frac{1}{4} - \frac{1}{9})$ . Why this step? The photon carries the drop in electron energy; sign tells emission vs absorption.
Compute: $\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$ , so $∣Δ E ∣ = 13.6 \cdot \frac{5}{36} \approx 1.889$ eV (emitted). Why this step? Positive magnitude = energy released as a photon.
Wavelength via $λ = \frac{h c}{∣Δ E ∣}$ , with $h c = 1240$ eV·nm: $λ = \frac{1240}{1.889} \approx 656 nm .$ Why this step? $h c$ in eV·nm turns an eV energy straight into nanometres — cleanest for spectroscopy.

Verify: $656$ nm is exactly the measured H- $α$ line — deep red. ✓ Forecast confirmed.

Worked example Ex 7 — Cell G · Degeneracy and node counting at

$n = 3$ For hydrogen, list the $ℓ$ values allowed at $n = 3$ , count states (the degeneracy), and give radial/angular nodes of the $3 d$ orbital.

Forecast: How many distinct orbitals share $n = 3$ ? (Hint: think $n^{2}$ .)

Allowed $ℓ$ : $ℓ = 0, 1, 2$ (i.e. $3 s, 3 p, 3 d$ ), since $ℓ$ runs $0 \dots n - 1$ . Why this step? $ℓ < n$ comes from the radial finiteness condition in the parent.
Count $m_{ℓ}$ per $ℓ$ : $2 ℓ + 1$ values each → $1 + 3 + 5 = 9$ orbitals. In pure hydrogen all 9 share $E_{3} = - 13.6/9 \approx - 1.51$ eV. Why this step? Energy depends only on $n$ (the accidental $1/ r$ degeneracy), so same $n$ = same energy.
Nodes of $3 d$ ( $n = 3, ℓ = 2$ ): radial $= n - ℓ - 1 = 0$ ; angular $= ℓ = 2$ ; total $= n - 1 = 2$ . Why this step? These node rules were derived from separation of variables; $3 d$ has zero radial nodes — all its structure is angular.

Verify: Sum check: $\sum_{ℓ = 0}^{2} (2 ℓ + 1) = n^{2} = 9$ . ✓ And radial+angular $= 0 + 2 = 2 = n - 1$ . ✓

Worked example Ex 8 — Cell H · Most-probable vs mean radius of the

$1 s$ electron For hydrogen $1 s$ , find the most-probable radius $r_{mp}$ and the mean (expectation) radius $⟨ r ⟩$ , in units of $a_{0}$ .

Forecast: Both are "typical" radii, yet they differ. Which is larger — the peak of the distribution or its average? (A long tail to large $r$ pulls the average one way…)

Most probable maximizes $P (r) = 4 π r^{2} ∣ ψ_{1 s} ∣^{2} \propto r^{2} e^{- 2 r / a_{0}}$ : $\frac{d}{d r} (r^{2} e^{- 2 r / a_{0}}) = (2 r - \frac{2 r ^{2}}{a _{0}}) e^{- 2 r / a_{0}} = 0 \Rightarrow r_{mp} = a_{0} .$ Why this step? The $r^{2}$ shell-volume weight beats the density's peak at $0$ ; setting the derivative to $0$ finds the balance point.
Mean radius is $⟨ r ⟩ = \int_{0}^{\infty} r P (r) d r$ with normalized $P$ . For $1 s$ the standard integral gives $⟨ r ⟩ = \frac{\int _{0}^{\infty} r ^{3} e ^{- 2 r / a_{0}} d r}{\int _{0}^{\infty} r ^{2} e ^{- 2 r / a_{0}} d r} = \frac{3 ! / ( 2/ a _{0} ) ^{4}}{2 ! / ( 2/ a _{0} ) ^{3}} = \frac{3}{2} a_{0} .$ Why this step? $⟨ r ⟩$ weights each radius by probability; the $\int r^{n} e^{- α r} = n! / α^{n + 1}$ formula does the integrals.
Compare: $⟨ r ⟩ = 1.5 a_{0} > r_{mp} = a_{0}$ . Why this step? The distribution is right-skewed (long tail out to large $r$ ), so the mean sits beyond the peak.

Verify: Numerically $r_{mp} = a_{0} \approx 52.9$ pm; $⟨ r ⟩ = 1.5 a_{0} \approx 79.4$ pm. The forecast — mean $>$ peak — is confirmed by the skew. ✓

Worked example Ex 9 — Cell I · Why

$n = 0$ and $ℓ = n$ are NOT states Show explicitly that (a) $n = 0$ in the box gives no particle, and (b) $ℓ = n$ is forbidden for the atom.

Forecast: These look like harmless "extra" values. What breaks?

Box $n = 0$ : $k = \frac{nπ}{L} = 0$ , so $ψ_{0} = \frac{2}{L} sin 0 = 0$ everywhere. Why this step? A wavefunction that is zero everywhere means $\int ∣ ψ ∣^{2} = 0 \neq = 1$ — cannot be normalized, so it describes no particle.
Consequence: the lowest real box state is $n = 1$ , carrying zero-point energy $E_{1} = \frac{h ^{2}}{8 m L ^{2}} > 0$ — see Zero-point Energy. A confined particle can never be perfectly still. Why this step? Establishes that the floor is not zero energy.
Atom $ℓ = n$ : allowed $ℓ$ stops at $n - 1$ . If we forced $ℓ = n$ , the radial nodes $n - ℓ - 1 = - 1$ would be negative — impossible; the radial equation has no finite, decaying solution. Why this step? Node count cannot be negative; that is the algebraic fingerprint of a non-state.

Verify: Node check at each real level is $\geq 0$ : for $n = 2$ , $ℓ$ can be $0$ (radial nodes $1$ ) or $1$ (radial nodes $0$ ) — both non-negative; $ℓ = 2$ would give $- 1$ . ✓ Forbidden confirmed.

Worked example Ex 10 — Cell J · Real-world word problem + exam twist (a quantum dot)

A spherical quantum dot is modelled crudely as a 1D box holding an electron. Its first absorption ( $n = 1 \to 2$ ) appears at $λ = 620$ nm (orange). Estimate the box length $L$ . Exam twist: then predict whether a smaller dot glows redder or bluer.

Forecast: Smaller box = larger gap = higher-energy photon. Higher energy means shorter $λ$ = bluer. Hold that thought.

Photon energy → gap. $Δ E = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{620 nm} = 2.00$ eV $= 3.20 \times 1 0^{- 19}$ J. Why this step? The absorbed photon supplies exactly the level gap.
Relate gap to $L$ . $Δ E = 3 \frac{h ^{2}}{8 m L ^{2}} \Rightarrow L = \frac{3 h ^{2}}{8 m Δ E}$ . Why this step? Invert the box gap formula for $L$ ; the $n = 1 \to 2$ factor is $(4 - 1) = 3$ .
Plug numbers. $L = \frac{3 ( 6.626 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 3.20 \times 1 0 ^{- 19} )} \approx 7.5 \times 1 0^{- 10} m = 0.75 nm .$ Why this step? A sub-nanometre length is exactly the quantum-dot regime — the model self-validates.
Exam twist — direction of colour shift. $Δ E \propto 1/ L^{2}$ : shrink $L$ → bigger gap → shorter $λ$ → bluer emission. Why this step? Ties the size-tunable colour of real quantum dots straight to $E_{n} \propto 1/ L^{2}$ .

Verify: Feed $L = 0.75$ nm back: $Δ E = 3 (6.626 \times 1 0^{- 34})^{2} / [8 (9.11 \times 1 0^{- 31}) (0.75 \times 1 0^{- 9})^{2}] \approx 3.21 \times 1 0^{- 19}$ J $\approx 2.00$ eV → $λ \approx 620$ nm. ✓ Self-consistent, and the blue-shift-for-smaller-dot forecast is confirmed.

Recall

Recall Cover the answers

Box gap $n = 1 \to 2$ scales as which factor of $E_{1}$ ? ::: $3 \times E_{1}$ (because $2^{2} - 1^{2} = 3$ ). As $L \to \infty$ , what happens to box levels? ::: They collapse to a continuum ( $E_{n} \to 0$ , gaps $\to 0$ ) — the free particle. Probability of the $n = 1$ particle in the left third of the box? ::: $\frac{1}{3} - \frac{3}{4 π} \approx 0.196$ . Ionization energy of hydrogen from $n = 1$ ? ::: $13.6$ eV $= 2.18 \times 1 0^{- 18}$ J. Wavelength of the $n = 3 \to 2$ hydrogen line? ::: $656$ nm (red H- $α$ ). Degeneracy of the $n = 3$ shell in pure hydrogen? ::: $n^{2} = 9$ orbitals, all at $- 1.51$ eV. Most-probable vs mean radius of $1 s$ ? ::: $r_{mp} = a_{0}$ ; $⟨ r ⟩ = 1.5 a_{0}$ . Why is $n = 0$ not a box state? ::: $ψ \equiv 0$ — cannot be normalized, no particle. Smaller quantum dot glows…? ::: Bluer (bigger gap, since $Δ E \propto 1/ L^{2}$ ).

See also: Schrödinger Equation · Quantum Numbers · Atomic Orbitals and Shapes · Bohr Model vs Quantum Mechanics · Operators and Eigenvalues.