(a) Calculate the ratio of probability of finding the 1s electron of hydrogen atom at r = a₀ and at r = 10a₀, where 'r' is the…

(a) The normalized 1s wavefunction of hydrogen is ψ₁ₛ = (1/√π)(1/a₀)^(3/2) e^(−r/a₀). The radial probability density is P(r) = 4πr²|ψ₁ₛ|² = (4/a₀³) r² e^(−2r/a₀). At r = a₀: P(a₀) = (4/a₀³)a₀² e^(−2) = (4/a₀)e^(−2). At r = 10a₀: P(10a₀) = (4/a₀³)(10a₀)² e^(−20) = (400/a₀)e^(−20). Therefore ratio = (4e^(−2))/(400e^(−20)) = e¹⁸/100. Since e¹⁸ ≈ 6.566 × 10⁷, the radial probability density ratio is ≈ 6.566 × 10⁵. If “probability at r” is taken as point probability density |ψ|², the r² factor is omitted and the ratio becomes e¹⁸ ≈ 6.57 × 10⁷. Final: radial probability density ratio = e¹⁸/100 ≈ 6.57 × 10⁵.

(b) Use Hess’s law to construct the Born–Haber cycle for NaCl(s) from its elements in normal states: Na(s) + ½Cl₂(g) → NaCl(s), ΔfH°₂₉₈ = −411 kJ mol⁻¹. The steps are:

• Sublimation of sodium: Na(s) → Na(g), ΔH_sub = +108 kJ mol⁻¹. Energy demanding.
• Ionization of sodium: Na(g) → Na⁺(g) + e⁻, IE = +496 kJ mol⁻¹. Energy demanding.
• Dissociation of chlorine: ½Cl₂(g) → Cl(g), ½D = +121.5 kJ mol⁻¹. Energy demanding.
• Electron gain by chlorine: Cl(g) + e⁻ → Cl⁻(g), EA = −349 kJ mol⁻¹. Energy evolving.
• Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s), ΔH_latt = −787.5 kJ mol⁻¹. Energy evolving. By Hess’s law, ΔfH° = ΔH_sub + IE + ½D + EA + ΔH_latt = 108 + 496 + 121.5 − 349 − 787.5 = −411 kJ mol⁻¹. Energy demanding: sublimation, ionization, chlorine dissociation. Energy evolving: electron gain, lattice formation; overall formation is energy evolving.

(c) Pure germanium and silicon are intrinsic semiconductors. Their small band gaps allow only a few electrons to be thermally excited from the valence band to the conduction band, so conductivity is low. Conductivity can be enhanced greatly by doping, i.e. adding trace amounts of suitable impurities. If a pentavalent element such as P, As or Sb is added to Si or Ge, four valence electrons form covalent bonds, while the fifth is weakly bound at a donor level just below the conduction band. It is easily excited into the conduction band, donating an electron. This gives an n-type semiconductor, with electrons as majority carriers. Example: Si doped with P, or Ge doped with As. If a trivalent element such as B, Al, Ga or In is added, it has only three valence electrons and creates an acceptor level just above the valence band. It accepts an electron from the valence band, leaving a hole. This gives a p-type semiconductor, with holes as majority carriers. Example: Si doped with B, or Ge doped with Ga. Doping at ppm level increases carrier concentration by many orders of magnitude, thereby increasing conductivity.

(d) Use Fourier’s law of heat conduction for steady state: Q/t = kAΔT/L. Here A = 1.50 m², L = 10 cm = 0.10 m, T_hot = 50°C, T_cold = −10°C, so ΔT = 60 K. Given k = 2.4 × 10⁻² J s⁻¹ m⁻¹ K⁻¹. Substituting: Q/t = (2.4 × 10⁻²)(1.50)(60)/(0.10) = 21.6 J s⁻¹ = 21.6 W. Thus the rate of heat transfer by conduction from the warm sheet to the cold sheet is 21.6 W. In steady state, the rate of loss of heat by the warm sheet is also 21.6 W, because the heat it supplies to the air layer is conducted away at the same rate. This assumes air conduction only, steady state, no convection, no radiation and no edge effects. Final: rate of heat transfer = rate of heat loss = 21.6 J s⁻¹ = 21.6 W.

(e) A liquid normally boils when its saturation vapour pressure equals the external pressure. However, boiling requires the formation of vapour bubbles inside the liquid. For a spherical vapour bubble of radius r, surface tension γ produces an excess Laplace pressure 2γ/r, so mechanical equilibrium requires p_v = p_ext + 2γ/r. The Kelvin equation for the curved liquid–vapour interface of a bubble is p_v = p_sat(T) exp(−2γV_m/(rRT)), where V_m is the molar volume of the liquid. Combining these gives p_sat(T) = (p_ext + 2γ/r) exp(2γV_m/(rRT)). Since the exponential factor is greater than 1, p_sat(T) must exceed p_ext. Normal boiling occurs at T_b where p_sat(T_b) = p_ext, so the liquid must be heated above T_b before a small bubble can survive and grow. For very small r, the exponential factor is very large, requiring a large superheat. Nucleation sites such as boiling chips, dust or rough surfaces provide larger effective radii and reduce the required superheat.

(f) The ascending order of dipole moments is NF₃ < NH₃ < H₂O. Typical values are NF₃ ≈ 0.23 D, NH₃ ≈ 1.47 D and H₂O ≈ 1.85 D. NH₃ is trigonal pyramidal. Its three N–H bond dipoles point toward nitrogen, and the lone-pair moment on nitrogen reinforces the resultant, giving a sizable net dipole. NF₃ is also pyramidal, but the N–F bond dipoles point from nitrogen to fluorine, opposite in sense to the lone-pair contribution on nitrogen. Therefore they largely cancel, leaving only a very small net dipole moment. H₂O is bent. The two O–H bond dipoles point toward oxygen and their vector sum is large because of the bent geometry; the lone-pair contributions also add to this resultant. Hence water has the largest dipole moment. Final order: NF₃ < NH₃ < H₂O.

(g) Use the benzoic acid combustion to calibrate the calorimeter. Molar mass of benzoic acid, C₇H₆O₂ = 122.12 g mol⁻¹. Moles of benzoic acid = 0.500/122.12 = 4.094 × 10⁻³ mol. Heat released = 4.094 × 10⁻³ × 3227 = 13.21 kJ. Therefore calorimeter constant C_cal = q/ΔT = 13.21/1.236 = 10.69 kJ K⁻¹. For naphthalene, C₁₀H₈, molar mass = 128.17 g mol⁻¹. Moles = 0.300/128.17 = 2.341 × 10⁻³ mol. Heat released by naphthalene = C_cal ΔT = 10.69 × 1.128 = 12.06 kJ. Hence ΔcU(naphthalene) = −q/n = −12.06/(2.341 × 10⁻³) = −5.15 × 10³ kJ mol⁻¹. Since a bomb calorimeter is constant-volume, the value obtained is ΔcU. Final: ΔcU(naphthalene) ≈ −5.15 × 10³ kJ mol⁻¹, i.e. about −5151 kJ mol⁻¹.

(h) For a non-condensing gas at fixed volume, P ∝ T. Neon has critical temperature 44 K, so at 150 K neon remains gaseous. If the sample were pure neon, cooling from 300 K to 150 K would give P = 1.00 × (150/300) = 0.50 atm, not 0.37 atm. Hence it is not pure neon. If the sample were pure ethane, at 150 K its vapour pressure is 0.10 atm. The ideal cooled pressure would be 0.50 atm, which exceeds 0.10 atm, so ethane would condense until the vapour pressure fell to 0.10 atm. The final pressure would then be 0.10 atm, not 0.37 atm. Hence it is not pure ethane. Therefore the sample is a mixture. At 150 K, neon contributes its ideal-gas partial pressure and ethane contributes its vapour pressure 0.10 atm. Let p_Ne(150) = x. Then x + 0.10 = 0.37, so x = 0.27 atm. Back at 300 K, p_Ne(300) = 0.27 × (300/150) = 0.54 atm. Thus initial ethane pressure = 1.00 − 0.54 = 0.46 atm. Composition: mixture of ethane and neon; at 300 K, mole fractions are Ne ≈ 0.54 and C₂H₆ ≈ 0.46.

(i) Use Faraday’s laws of electrolysis. Volume of gold layer = area × thickness = 49.8 cm² × 1 × 10⁻³ cm = 0.0498 cm³. Mass of gold = density × volume = 19.3 g cm⁻³ × 0.0498 cm³ = 0.96114 g. Moles of gold = 0.96114/196.97 = 4.879 × 10⁻³ mol. Gold is in +3 oxidation state, so Au³⁺ + 3e⁻ → Au. Moles of electrons required = 3 × 4.879 × 10⁻³ = 1.464 × 10⁻² mol. Charge Q = n_e F = 1.464 × 10⁻² × 96485 C mol⁻¹ = 1.412 × 10³ C. Given current I = 3.25 A = 3.25 C s⁻¹, time t = Q/I = 1.412 × 10³/3.25 = 434.6 s = 7.24 min. Final: time required ≈ 435 s, i.e. about 7 min 15 s.

(j) The maximum work obtainable from a heat engine is limited by the Carnot efficiency. Convert temperatures to kelvin: T_hot = 400°C + 273.15 = 673.15 K, T_cold = 150°C + 273.15 = 423.15 K. Then η_max = 1 − T_cold/T_hot = 1 − 423.15/673.15 = 250/673.15 = 0.3714. For heat input Q, maximum work W_max = η_max Q = 0.3714 Q = (250/673.15)Q. Thus the turbine can convert at most about 37.1% of the heat input into work. This maximum is achieved only under reversible Carnot conditions: heat is absorbed reversibly at T_hot, rejected reversibly at T_cold, with reversible adiabatic expansion and compression, no friction, no heat leaks, no entropy generation, and infinitesimal temperature differences between the working fluid and the reservoirs. Final: W_max = 0.371 Q, i.e. 37.1% of the heat input.