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Level 3 — ProductionPhysical Chemistry (Advanced)

Physical Chemistry (Advanced)

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level 3 — Production (from-scratch derivations, explain-out-loud) Time limit: 45 minutes Total marks: 60

Constants: h=6.626×10−34 J sh = 6.626\times10^{-34}\,\text{J s}, kB=1.381×10−23 J K−1k_B = 1.381\times10^{-23}\,\text{J K}^{-1}, c=3.00×108 m s−1c = 3.00\times10^{8}\,\text{m s}^{-1}, NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\text{mol}^{-1}, F=96485 C mol−1F = 96485\,\text{C mol}^{-1}, R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}.

Q1. Particle in a box — full derivation. (10 marks) Starting from the time-independent Schrödinger equation for a 1D box of length LL with infinite walls: (a) Write the equation inside the box and derive the general solution. (3) (b) Apply boundary conditions to obtain the quantised energies EnE_n. (3) (c) Normalise the wavefunction. (2) (d) For an electron in a box of length L=1.0 nmL = 1.0\,\text{nm}, compute the n=1→2n=1\to2 transition energy in joules. (2)

Q2. Variational principle. (10 marks) (a) State the variational theorem and explain in words why the trial-function energy is an upper bound to the true ground-state energy. (4) (b) For the H-atom (in atomic units, H^=−12∇2−1/r\hat H = -\tfrac12\nabla^2 - 1/r), use the trial function ψ=e−αr\psi = e^{-\alpha r}. Given ⟨T⟩=α2/2\langle T\rangle = \alpha^2/2 and ⟨V⟩=−α\langle V\rangle = -\alpha (both for normalised ψ\psi), minimise E(α)E(\alpha) with respect to α\alpha and find the optimal α\alpha and energy. (4) (c) State whether the exact ground-state energy is recovered and why. (2)

Q3. Rotational–vibrational spectroscopy. (10 marks) (a) Derive the rigid-rotor energy levels EJ=ℏ22IJ(J+1)E_J = \tfrac{\hbar^2}{2I}J(J+1) from H^=L^2/2I\hat H = \hat L^2/2I and give the allowed transition frequencies (in wavenumbers) with rotational constant BB. (4) (b) For 1^1H35^{35}Cl the bond length is 127.5 pm127.5\,\text{pm}; the reduced mass is μ=1.627×10−27 kg\mu = 1.627\times10^{-27}\,\text{kg}. Compute BB in cm−1^{-1}. (4) (c) Explain qualitatively why alternate lines in the P and R branches are not equally spaced (rotational–vibrational coupling). (2)

Q4. Statistical thermodynamics — partition functions. (10 marks) (a) Derive the translational partition function qtransq_{\text{trans}} for a particle in a 1D box in the classical (integral) limit, then state the 3D result. (4) (b) Show how the mean energy ⟨E⟩=kBT2 (∂ln⁡q/∂T)\langle E\rangle = k_B T^2\,(\partial \ln q/\partial T) leads to ⟨E⟩=32kBT\langle E\rangle = \tfrac32 k_B T for 3D translation. (3) (c) For the vibrational mode with qvib=(1−e−θv/T)−1q_{\text{vib}} = (1 - e^{-\theta_v/T})^{-1}, θv=hcν~/kB\theta_v = h c\tilde\nu / k_B, compute qvibq_{\text{vib}} at T=300 KT = 300\,\text{K} for ν~=500 cm−1\tilde\nu = 500\,\text{cm}^{-1}. (3)

Q5. Butler–Volmer & Tafel. (10 marks) (a) Write the Butler–Volmer equation and identify each symbol. (3) (b) Derive the Tafel equation from the Butler–Volmer equation in the high-overpotential anodic limit, and give the Tafel slope. (4) (c) A cell has exchange current density j0=1.0×10−3 A cm−2j_0 = 1.0\times10^{-3}\,\text{A cm}^{-2}, anodic transfer coefficient αa=0.5\alpha_a = 0.5, at T=298 KT = 298\,\text{K}. For an anodic overpotential η=0.20 V\eta = 0.20\,\text{V}, compute the current density jj (Tafel limit). (3)

Q6. Explain-out-loud + Langmuir. (10 marks) (a) Derive the Langmuir isotherm θ=Kp1+Kp\theta = \dfrac{Kp}{1+Kp} from adsorption/desorption rate balance. (4) (b) Explain in words, as if lecturing, the difference between fluorescence and phosphorescence using a Jablonski diagram (spin states, timescales, allowed/forbidden). (4) (c) State the Stark–Einstein law and define primary quantum yield. (2)

Answer keyMark scheme & solutions

Q1 (10)

(a) Inside box V=0V=0: −ℏ22md2ψdx2=Eψ-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}=E\psi ⇒ ψ′′+k2ψ=0\psi''+k^2\psi=0, k2=2mE/ℏ2k^2=2mE/\hbar^2. General solution ψ=Asin⁡kx+Bcos⁡kx\psi=A\sin kx + B\cos kx. (3) (b) BC ψ(0)=0⇒B=0\psi(0)=0\Rightarrow B=0; ψ(L)=0⇒sin⁡kL=0⇒kL=nπ\psi(L)=0\Rightarrow \sin kL=0\Rightarrow kL=n\pi. Hence En=n2π2ℏ22mL2=n2h28mL2E_n=\dfrac{n^2\pi^2\hbar^2}{2mL^2}=\dfrac{n^2h^2}{8mL^2}. (3) (c) ∫0LA2sin⁡2(nπx/L)dx=A2L/2=1⇒A=2/L\int_0^L A^2\sin^2(n\pi x/L)dx = A^2 L/2 =1 \Rightarrow A=\sqrt{2/L}. (2) (d) ΔE=(4−1)h28mL2=3h28mL2\Delta E=(4-1)\dfrac{h^2}{8mL^2}=\dfrac{3h^2}{8mL^2}. =3(6.626×10−34)28(9.109×10−31)(1.0×10−9)2=\dfrac{3(6.626\times10^{-34})^2}{8(9.109\times10^{-31})(1.0\times10^{-9})^2} =1.807×10−19 J=1.807\times10^{-19}\,\text{J}. (2)

Q2 (10)

(a) Theorem: for any normalised trial ψ\psi, ⟨ψ∣H^∣ψ⟩≥E0\langle\psi|\hat H|\psi\rangle \ge E_0. Reason: expand ψ=∑cnϕn\psi=\sum c_n\phi_n in true eigenstates; ⟨E⟩=∑∣cn∣2En≥E0∑∣cn∣2=E0\langle E\rangle=\sum|c_n|^2 E_n \ge E_0\sum|c_n|^2 = E_0 since every En≥E0E_n\ge E_0. (4) (b) E(α)=α2/2−αE(\alpha)=\alpha^2/2-\alpha. dE/dα=α−1=0⇒α=1dE/d\alpha=\alpha-1=0\Rightarrow \alpha=1; E=1/2−1=−1/2E=1/2-1=-1/2 hartree. (4) (c) Yes — exact ground state −0.5-0.5 hartree is recovered because the trial form e−αre^{-\alpha r} is exactly the true 1s functional form; the variational family contains the exact solution. (2)

Q3 (10)

(a) H^=L^2/2I\hat H=\hat L^2/2I; L^2YJM=ℏ2J(J+1)YJM\hat L^2 Y_{JM}=\hbar^2 J(J+1)Y_{JM} ⇒ EJ=ℏ22IJ(J+1)E_J=\dfrac{\hbar^2}{2I}J(J+1). With B=ℏ4πcI=h8π2cIB=\dfrac{\hbar}{4\pi c I}=\dfrac{h}{8\pi^2 cI}, terms F(J)=BJ(J+1)F(J)=BJ(J+1); Δν~=2B(J+1)\Delta\tilde\nu=2B(J+1) for ΔJ=+1\Delta J=+1. (4) (b) I=μr2=(1.627×10−27)(1.275×10−10)2=2.645×10−47 kg m2I=\mu r^2=(1.627\times10^{-27})(1.275\times10^{-10})^2=2.645\times10^{-47}\,\text{kg m}^2. B=h8π2cI=6.626×10−348π2(3.00×1010 cm/s)(2.645×10−47)B=\dfrac{h}{8\pi^2 c I}=\dfrac{6.626\times10^{-34}}{8\pi^2(3.00\times10^{10}\,\text{cm/s})(2.645\times10^{-47})} =10.58 cm−1=10.58\,\text{cm}^{-1}. (4) (c) In real molecules BB depends on vibrational state (Bv=Be−αe(v+12)B_v = B_e-\alpha_e(v+\tfrac12)); higher-vv has larger bond length (anharmonicity), smaller BB, so line spacing changes across the branch — R-branch lines converge, P-branch diverges. (2)

Q4 (10)

(a) q=∑ne−βEn≈∫0∞e−βn2h2/8mL2dn=(2πmkBTh2)1/2Lq=\sum_n e^{-\beta E_n}\approx\int_0^\infty e^{-\beta n^2 h^2/8mL^2}dn = \left(\dfrac{2\pi m k_B T}{h^2}\right)^{1/2}L. 3D: qtrans=(2πmkBTh2)3/2V=V/Λ3q_{\text{trans}}=\left(\dfrac{2\pi m k_B T}{h^2}\right)^{3/2}V=V/\Lambda^3. (4) (b) ln⁡q=32ln⁡T+const\ln q=\tfrac32\ln T + \text{const}; ⟨E⟩=kBT2∂ln⁡q∂T=kBT2⋅32T=32kBT\langle E\rangle=k_BT^2\dfrac{\partial\ln q}{\partial T}=k_BT^2\cdot\dfrac{3}{2T}=\tfrac32k_BT. (3) (c) θv=hcν~kB=(6.626×10−34)(3.00×1010)(500)1.381×10−23=719.4 K\theta_v=\dfrac{hc\tilde\nu}{k_B}=\dfrac{(6.626\times10^{-34})(3.00\times10^{10})(500)}{1.381\times10^{-23}}=719.4\,\text{K}. θv/T=2.398\theta_v/T=2.398; qvib=1/(1−e−2.398)=1/(1−0.0909)=1.100q_{\text{vib}}=1/(1-e^{-2.398})=1/(1-0.0909)=1.100. (3)

Q5 (10)

(a) j=j0[exp⁡ ⁣(αaFηRT)−exp⁡ ⁣(−αcFηRT)]j=j_0\left[\exp\!\left(\dfrac{\alpha_a F\eta}{RT}\right)-\exp\!\left(-\dfrac{\alpha_c F\eta}{RT}\right)\right]; j0j_0 exchange current density, αa,αc\alpha_a,\alpha_c transfer coefficients, η\eta overpotential. (3) (b) For η≫0\eta\gg0 second term negligible: j≈j0eαaFη/RTj\approx j_0 e^{\alpha_a F\eta/RT} ⇒ η=−RTαaFln⁡j0+RTαaFln⁡j\eta=-\dfrac{RT}{\alpha_a F}\ln j_0 + \dfrac{RT}{\alpha_a F}\ln j. Tafel form η=a+blog⁡j\eta=a+b\log j, slope b=2.303RTαaFb=\dfrac{2.303RT}{\alpha_a F} ≈ 0.118 V/decade. (4) (c) αaFηRT=0.5(96485)(0.20)8.314(298)=3.895\dfrac{\alpha_a F\eta}{RT}=\dfrac{0.5(96485)(0.20)}{8.314(298)}=3.895; j=10−3e3.895=10−3(49.1)=4.91×10−2 A cm−2j=10^{-3}e^{3.895}=10^{-3}(49.1)=4.91\times10^{-2}\,\text{A cm}^{-2}. (3)

Q6 (10)

(a) Rate ads =kap(1−θ)=k_a p(1-\theta), rate des =kdθ=k_d\theta. Equilibrium: kap(1−θ)=kdθk_a p(1-\theta)=k_d\theta ⇒ Kp(1−θ)=θKp(1-\theta)=\theta, K=ka/kdK=k_a/k_d ⇒ θ=Kp1+Kp\theta=\dfrac{Kp}{1+Kp}. (4) (b) Fluorescence: emission from excited singlet S1→S0S_1\to S_0, spin-allowed, fast (10−910^{-9}–10−710^{-7} s). Phosphorescence: from triplet T1→S0T_1\to S_0 after intersystem crossing S1⇝T1S_1\rightsquigarrow T_1; spin-forbidden, slow (10−310^{-3} s to s), longer wavelength. Jablonski diagram shows S0,S1,T1S_0,S_1,T_1 with radiative/non-radiative arrows. (4) (c) Stark–Einstein: each molecule absorbs one photon in the primary photochemical act. Primary quantum yield ϕ=molecules reacting (primary process)photons absorbed\phi=\dfrac{\text{molecules reacting (primary process)}}{\text{photons absorbed}}. (2)

[
{"claim":"Q1d PIB transition energy ~1.807e-19 J","code":"h=6.626e-34; m=9.109e-31; L=1.0e-9; dE=3*h**2/(8*m*L**2); result = abs(dE-1.807e-19)<2e-21"},
{"claim":"Q2b optimal alpha=1, E=-0.5","code":"from sympy import symbols,diff,solve; a=symbols('a'); E=a**2/2-a; sol=solve(diff(E,a),a)[0]; result = (sol==1) and (E.subs(a,1)==Rational(-1,2))"},
{"claim":"Q3b B ~10.58 cm-1","code":"h=6.626e-34; c=3.00e10; mu=1.627e-27; r=1.275e-10; I=mu*r**2; B=h/(8*pi**2*c*I); result = abs(float(B)-10.58)<0.2"},
{"claim":"Q4c q_vib ~1.100","code":"h=6.626e-34; c=3.00e10; k=1.381e-23; nu=500; T=300; th=h*c*nu/k; q=1/(1-exp(-th/T)); result = abs(float(q)-1.100)<0.01"},
{"claim":"Q5c j ~4.91e-2 A/cm2","code":"F=96485; R=8.314; T=298; j0=1e-3; x=0.5*F*0.20/(R*T); j=j0*exp(x); result = abs(float(j)-4.91e-2)<2e-3"}
]