Physical Chemistry (Advanced)

Q1. (4 marks) Wavefunction and energy: $\psi_n(x) = \sqrt{\tfrac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right),\qquad E_n = \frac{n^2 h^2}{8 m L^2}.$ (1 mark each for correct $\psi_n$ and $E_n$.)

Transition energy $n=1\to2$: $\Delta E = (2^2 - 1^2)\frac{h^2}{8mL^2} = \frac{3h^2}{8 m_e L^2}.$ Numerics: $h^2 = (6.626\times10^{-34})^2 = 4.390\times10^{-67}$; denominator $8 m_e L^2 = 8(9.109\times10^{-31})(1.0\times10^{-9})^2 = 7.287\times10^{-48}$. $\Delta E = \frac{3(4.390\times10^{-67})}{7.287\times10^{-48}} = 1.81\times10^{-18}\ \text{J}.$ (1 mark setup, 1 mark answer $\approx 1.8\times10^{-18}$ J.) Why: energy scales as $n^2$; the transition uses the difference of squares.

Q2. (4 marks)

• Principle: For any normalized trial wavefunction $\phi$, $E_{\text{trial}} = \langle\phi|\hat H|\phi\rangle \ge E_0$, the true ground-state energy; equality holds only if $\phi$ is the true ground state. (2)
• Usefulness: adjustable parameters in $\phi$ can be varied to minimize $E_{\text{trial}}$, giving a rigorous upper bound and an optimized approximate wavefunction. (2)

Q3. (5 marks) (a) Selection rule: $\Delta J = \pm 1$ (also molecule must have permanent dipole). (1) (b) Line for $J\to J+1$: $\tilde\nu_{J\to J+1} = B[(J+1)(J+2) - J(J+1)] = 2B(J+1).$ Spacing between successive lines: $\tilde\nu_{J+1} - \tilde\nu_J = 2B(J+2) - 2B(J+1) = 2B$. (2) (c) $J=0\to1$: $\tilde\nu = 2B(0+1) = 2B = 2(1.93) = 3.86\ \text{cm}^{-1}$. (2) Why: term-value differences telescope, giving equal $2B$ spacing.

Q4. (4 marks)

• Hartree–Fock: each electron moves in the average (mean) field of the others; the central variable is the many-electron wavefunction (a single Slater determinant). (1.5)
• DFT: the ground-state energy is a functional of the electron density $\rho(\mathbf r)$ (Hohenberg–Kohn); central variable is the density. (1.5)
• HF limitation: it neglects electron correlation (Coulomb correlation beyond exchange), so it overestimates energies / misses dispersion. (1)

Q5. (5 marks) Exponent: $x = \dfrac{hc\tilde\nu}{k_B T}$. $hc\tilde\nu = (6.626\times10^{-34})(3.00\times10^{10})(500) = 9.939\times10^{-21}\ \text{J}$. $k_B T = (1.381\times10^{-23})(300) = 4.143\times10^{-21}\ \text{J}$. $x = 9.939/4.143 = 2.399$. (2) $q_{\text{vib}} = \frac{1}{1 - e^{-2.399}} = \frac{1}{1 - 0.0908} = \frac{1}{0.9092} = 1.10.$ (3 marks: exponent 2, evaluation 1.) Why: geometric-series sum of Boltzmann factors over equally spaced levels.

Q6. (4 marks) Langmuir isotherm: $\theta = \frac{K p}{1 + K p}. \quad (2)$

• Low $p$ ($Kp \ll 1$): $\theta \approx K p$ — coverage rises linearly with pressure. (1)
• High $p$ ($Kp \gg 1$): $\theta \to 1$ — saturation (monolayer complete). (1)

Q7. (4 marks)

• CMC: the surfactant concentration above which micelles begin to form spontaneously; below it surfactant exists as free monomers. (2)
• Two properties (any two, 1 each): surface tension (decreases then levels off / becomes ~constant above CMC); molar conductivity (drops in slope above CMC); turbidity/light scattering (rises); osmotic pressure (levels off). (2)

Q8. (4 marks) (a) Overpotential $\eta = E_{\text{applied}} - E_{\text{eq}}$: the extra potential beyond equilibrium needed to drive a net current at a given rate. (1) (b) Slope $b$ (Tafel slope) gives the transfer coefficient $\alpha$ ($b = 2.303RT/\alpha F$), i.e. reaction mechanism/kinetics; the intercept $a$ gives the exchange current density $j_0$ (electrode activity). (2) (c) At low $\eta$: linear regime, $j \approx j_0\dfrac{F\eta}{RT}$ (current proportional to $\eta$). (1)

Q9. (3 marks)

• Stark–Einstein law: each molecule that reacts in the primary photochemical step absorbs one photon (one quantum) of the light causing the reaction. (1.5)
• Primary quantum yield: $\phi = \dfrac{\text{number of molecules undergoing the (primary) process}}{\text{number of photons absorbed}}$. (1.5)

Q10. (3 marks)

• Conductor: valence and conduction bands overlap (or partially filled band), $E_g \approx 0$ → free conduction. (1)
• Insulator: large band gap ($E_g$ large, typically $>3$–4 eV), full valence band, empty conduction band. (1)
• Intrinsic semiconductor: small band gap (~$\le 1$–2 eV) so thermal excitation promotes some electrons to the conduction band, leaving holes. (1)

[
  {"claim":"Q1 particle-in-box n=1->2 transition energy ~1.81e-18 J",
   "code":"h=6.626e-34; me=9.109e-31; L=1.0e-9; dE=3*h**2/(8*me*L**2); result = abs(dE-1.81e-18) < 0.05e-18"},
  {"claim":"Q3c CO J=0->1 wavenumber = 3.86 cm^-1",
   "code":"B=1.93; nu=2*B*1; result = abs(nu-3.86) < 1e-9"},
  {"claim":"Q5 vibrational partition function q_vib ~1.10 at 300 K, 500 cm^-1",
   "code":"h=6.626e-34; c=3.00e10; kB=1.381e-23; T=300; nu=500; x=h*c*nu/(kB*T); q=1/(1-exp(-x)); result = abs(float(q)-1.10) < 0.02"},
  {"claim":"Q5 exponent value ~2.399",
   "code":"h=6.626e-34; c=3.00e10; kB=1.381e-23; T=300; nu=500; x=h*c*nu/(kB*T); result = abs(float(x)-2.399) < 0.01"}
]