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

Constants you will reuse (kept in one place so no symbol is ever unexplained):

L1 — Recognition

Problem 1.1

State (a) the allowed quantum number set $n$ for the 1D box, and (b) how many interior nodes the wavefunction $ψ_{4}$ has.

Recall Solution 1.1

(a) The boundary conditions $ψ (0) = ψ (L) = 0$ force $k L = nπ$ with $n$ a positive whole number, so $n = 1, 2, 3, \dots$ . Zero is excluded (it kills the wavefunction). (b) Interior nodes $= n - 1$ . For $n = 4$ : $4 - 1 = 3$ nodes.

Problem 1.2

For a hydrogen orbital with $n = 3$ , list the allowed values of $ℓ$ and, for each, the allowed values of $m_{ℓ}$ .

Recall Solution 1.2

$ℓ$ runs $0, 1, \dots, n - 1$ , so $ℓ = 0, 1, 2$ .

$ℓ = 0$ : $m_{ℓ} = 0$ (one value).
$ℓ = 1$ : $m_{ℓ} = - 1, 0, + 1$ (three values).
$ℓ = 2$ : $m_{ℓ} = - 2, - 1, 0, + 1, + 2$ (five values).

Total states $= 1 + 3 + 5 = 9 = n^{2}$ . That $n^{2}$ is the reason each shell holds $2 n^{2}$ electrons after spin.

L2 — Application

Problem 2.1

An electron sits in a 1D box of length $L = 0.50 nm$ . Find the ground-state energy $E_{1}$ in joules and in eV.

Recall Solution 2.1

WHAT/WHY: ground state means $n = 1$ ; plug into $E_{n} = n^{2} h^{2} /8 m L^{2}$ . $E_{1} = \frac{( 6.626 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 0.50 \times 1 0 ^{- 9} ) ^{2}}$ Numerator: $(6.626 \times 1 0^{- 34})^{2} = 4.39 \times 1 0^{- 67}$ . Denominator: $8 \times 9.11 \times 1 0^{- 31} \times 2.5 \times 1 0^{- 19} = 1.822 \times 1 0^{- 48}$ . $E_{1} = \frac{4.39 \times 1 0 ^{- 67}}{1.822 \times 1 0 ^{- 48}} = 2.41 \times 1 0^{- 19} J .$ In eV: $2.41 \times 1 0^{- 19} /1.602 \times 1 0^{- 19} = 1.50 eV$ .

Problem 2.2

For the same box, what is the energy of the photon emitted in the $n = 3 \to n = 2$ transition (in eV)?

Recall Solution 2.2

WHY: photon energy = level gap = $E_{3} - E_{2} = (9 - 4) E_{1} = 5 E_{1}$ . From Problem 2.1, $E_{1} = 1.50 eV$ . $Δ E = 5 \times 1.50 = 7.52 eV .$ (Using the more precise $E_{1} = 1.504 eV$ : $5 \times 1.504 = 7.52 eV$ .)

L3 — Analysis

Problem 3.1

Show, by explicit integration, that $ψ_{2} (x) = \frac{2}{L} sin \frac{2 π x}{L}$ is normalized, i.e. $\int_{0}^{L} ∣ ψ_{2} ∣^{2} d x = 1$ .

Recall Solution 3.1

WHY normalization: $∣ ψ ∣^{2}$ is a probability density; the total probability of finding the electron somewhere in the box must be exactly $1$ . $\int_{0}^{L} \frac{2}{L} sin^{2} \frac{2 π x}{L} d x .$ Use $sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ : $= \frac{2}{L} \int_{0}^{L} \frac{1}{2} (1 - cos \frac{4 π x}{L}) d x = \frac{1}{L} [x - \frac{L}{4 π} sin \frac{4 π x}{L}]_{0}^{L} .$ The $sin$ term vanishes at both ends (argument is $0$ and $4 π$ , both give $sin = 0$ ): $= \frac{1}{L} [L - 0] = 1 . ✓$ The $2/ L$ prefactor is precisely what makes this come out to $1$ .

Problem 3.2

For the $1 s$ hydrogen orbital, the radial distribution function is $P (r) = 4 π r^{2} ∣ ψ_{1 s} ∣^{2}$ . Using $ψ_{1 s} = \frac{1}{π a _{0}^{3}} e^{- r / a_{0}}$ , find the most probable radius and confirm it is $a_{0}$ , not $0$ . See the figure.

Figure — Quantum chemistry — particle in a box revisited; H-atom solutions

Recall Solution 3.2

WHY the $r^{2}$ : a thin spherical shell at radius $r$ has volume $4 π r^{2} d r$ . Near $r = 0$ the density $∣ ψ ∣^{2}$ is largest, but the shell has almost no volume — so almost no electron lives there. The competition between shrinking density and growing shell volume creates a peak. $P (r) = 4 π r^{2} \cdot \frac{1}{π a _{0}^{3}} e^{- 2 r / a_{0}} = \frac{4}{a _{0}^{3}} r^{2} e^{- 2 r / a_{0}} .$ Maximize: set $\frac{d}{d r} (r^{2} e^{- 2 r / a_{0}}) = 0$ . Product rule: $2 r e^{- 2 r / a_{0}} + r^{2} (- \frac{2}{a _{0}}) e^{- 2 r / a_{0}} = 0.$ Factor $2 r e^{- 2 r / a_{0}}$ (nonzero for $r > 0$ ): $1 - \frac{r}{a _{0}} = 0 \Rightarrow r = a_{0} .$ Look at the figure: the black $∣ ψ ∣^{2}$ curve is highest at $r = 0$ , yet the red $P (r)$ curve peaks out at $r = a_{0}$ — geometry wins.

L4 — Synthesis

Problem 4.1

A 1D box models a linear conjugated dye. It has $N$ $π$ -electrons filling the lowest $N /2$ levels (2 per level, spin-paired). The lowest absorption promotes an electron from the highest filled level ( $n = N /2$ ) to the lowest empty one ( $n = N /2 + 1$ ). Derive a formula for the transition energy $Δ E$ in terms of $N$ and $L$ , then compute the wavelength for $N = 6$ , $L = 0.70 nm$ .

Recall Solution 4.1

WHAT/WHY: the highest filled level is $n_{top} = N /2$ ; the transition is $N /2 \to N /2 + 1$ . $Δ E = [(N /2 + 1)^{2} - (N /2)^{2}] \frac{h ^{2}}{8 m _{e} L ^{2}} .$ Expand the bracket: $(N /2 + 1)^{2} - (N /2)^{2} = N + 1$ . So $Δ E = (N + 1) \frac{h ^{2}}{8 m _{e} L ^{2}} .$ For $N = 6$ : factor $= 7$ . First compute the box unit: $\frac{h ^{2}}{8 m _{e} L ^{2}} = \frac{4.39 \times 1 0 ^{- 67}}{8 ( 9.11 \times 1 0 ^{- 31} ) ( 0.70 \times 1 0 ^{- 9} ) ^{2}} = \frac{4.39 \times 1 0 ^{- 67}}{3.572 \times 1 0 ^{- 48}} = 1.229 \times 1 0^{- 19} J .$ $Δ E = 7 \times 1.229 \times 1 0^{- 19} = 8.60 \times 1 0^{- 19} J .$ Wavelength from $λ = h c /Δ E$ (with $c = 3.00 \times 1 0^{8} m/s$ ): $λ = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.00 \times 1 0 ^{8} )}{8.60 \times 1 0 ^{- 19}} = 2.31 \times 1 0^{- 7} m = 231 nm .$

L5 — Mastery

Problem 5.1

Prove that the two lowest box states $ψ_{1}$ and $ψ_{2}$ are orthogonal: $\int_{0}^{L} ψ_{1} ψ_{2} d x = 0$ . Explain what orthogonality means physically and why it must hold for eigenstates of $\hat{H}$ .

Recall Solution 5.1

WHAT: orthogonality means the overlap integral is zero. WHY it must hold: $\hat{H}$ is a Hermitian operator (Operators and Eigenvalues); eigenstates with different eigenvalues ( $E_{1} \neq = E_{2}$ ) are guaranteed orthogonal. Physically it means measuring energy and getting $E_{1}$ leaves no chance of the state secretly being $ψ_{2}$ — the states are perfectly distinguishable. $\int_{0}^{L} \frac{2}{L} sin \frac{π x}{L} sin \frac{2 π x}{L} d x .$ Product-to-sum: $sin A sin B = \frac{1}{2} [cos (A - B) - cos (A + B)]$ , with $A - B = - π x / L$ , $A + B = 3 π x / L$ : $= \frac{1}{L} \int_{0}^{L} [cos \frac{π x}{L} - cos \frac{3 π x}{L}] d x .$ Integrate: $= \frac{1}{L} [\frac{L}{π} sin \frac{π x}{L} - \frac{L}{3 π} sin \frac{3 π x}{L}]_{0}^{L} .$ At $x = L$ : $sin π = 0$ and $sin 3 π = 0$ . At $x = 0$ : both $sin 0 = 0$ . $= \frac{1}{L} [0 - 0] = 0 . ✓$

Problem 5.2

Using $E_{n} = - 13.6/ n^{2} eV$ for hydrogen, show that the series limit (electron falling from $n = \infty$ to $n = 1$ ) equals the ionization energy, and compute the longest-wavelength line of the Lyman series ( $n = 2 \to 1$ ).

Recall Solution 5.2

WHY $E < 0$ and converging: bound states have negative energy relative to the free electron at $r \to \infty$ ( $E = 0$ ). As $n \to \infty$ , $E_{n} \to 0^{-}$ , so the levels crowd toward zero — a ceiling, unlike the box whose levels rise forever. This ceiling is the ionization threshold. Series limit / ionization energy: $Δ E_{\infty \to 1} = E_{\infty} - E_{1} = 0 - (- 13.6) = 13.6 eV .$ Longest-wavelength Lyman line = smallest gap in the series = $n = 2 \to 1$ : $Δ E = E_{2} - E_{1} = - \frac{13.6}{4} - (- 13.6) = 13.6 (1 - \frac{1}{4}) = 10.2 eV .$ Convert: $10.2 eV = 1.634 \times 1 0^{- 18} J$ . $λ = \frac{h c}{Δ E} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.00 \times 1 0 ^{8} )}{1.634 \times 1 0 ^{- 18}} = 1.216 \times 1 0^{- 7} m = 121.6 nm .$ (This is the famous Lyman- $α$ line, deep UV — compare Bohr Model vs Quantum Mechanics, which gives the same number.)

Recall wrap-up

Recall Cover-and-check summary

Box ground energy for $L$ ? ::: $E_{1} = h^{2} /8 m L^{2}$ ; scales as $1/ L^{2}$ .
Interior nodes of $ψ_{n}$ ? ::: $n - 1$ .
Most probable $1 s$ radius vs peak of $∣ ψ ∣^{2}$ ? ::: $r = a_{0}$ vs $r = 0$ ; the $4 π r^{2}$ shell factor shifts the peak.
Why are box eigenstates orthogonal? ::: $\hat{H}$ is Hermitian; different-energy eigenstates overlap to zero.
Ionization energy of H? ::: $13.6 eV$ (the $n = \infty$ series limit).
Lyman- $α$ wavelength? ::: $121.6 nm$ .