Solutions for Photoelectric, White Dwarf, Fluorescence, Hydrogen Atom, and Nitrogen Vibrat, Exams of Physical Chemistry

Download Solutions for Photoelectric, White Dwarf, Fluorescence, Hydrogen Atom, and Nitrogen Vibrat and more Exams Physical Chemistry in PDF only on Docsity! Chemistry 331 Physical Chemistry Practice Midterm Section _______ Name ________________________________ Given: g = 9.81 m/s2 R = 8.314 J mol-1 K-1 = 0.08206 L atm mol-1 K-1 1 a.m.u. = 1.67 x 10-27 kg Electron charge = 1.62 x 10-19 C 1 Debye = 3.33 x 10-30 Cm 1 atm = 1.0133 x 105 Nm-2 = 760 Torr P = P0exp{-Mgh/RT} ω = kµ , B = h 8π2cµR2 , µ = m1m2 m1 + m2 , λ = hp , T2 T1 = V1 V2 R/CV 1. A researcher has a laser that has a 250 nm wavelength. Which of the metals in the table below can be used to generate electrons? What is the highest energy electron that can be generated? Metal Electron Work Function Al 4.28 eV Au 5.10 eV Cu 4.65 eV In 4.12 eV Mo 4.60 eV Ni 5.15 eV Pt 5.65 eV Ti 4.33 eV W 4.55 eV Zn 4.33 eV Solution: The energy of a 250 nm photon is: E = hc/λ = (6.626 x 10-34 Js)(2.99 x 108 m s-1)/(250 x 10-9 m) = 7.924 x 10-19 J = 4.946 eV The work function of the metal must be greater than 4.946 eV. The metals that satisfy this criterion are Zn, W, Ti, Mo, In, Cu and Al. The highest energy electron will be ejected by the metal with the smallest work function. In this list it is Indium with a work function of 4.12 eV. 1/2mv2 = hν - Φ = 4.946 eV - 4.12 eV = 0.826 eV Energy of a 250 nm photon (4 points) = __________________________________. Appropriate metals that satisy criterion (4 points) = _________________________. Highest energy electron that can be generated (2 points) = ____________________. 2. A white dwarf collapses and then explodes into a supernova. The peak of the blackbody emission is observed to shift from 400 nm to 300 nm. A. and B. What is the change in temperature of the surface of star as it becomes a supernova? C. What is the relative increase in radiative emitted power? Solution: part A. T = 2.88 x 106 nm-K/λmax = 2.88 x 106 nm-K/400 K = 7200 K part B. T = 2.88 x 106 nm-K/λmax = 2.88 x 106 nm-K/300 K = 9600 K part C. ρ = σT4 , ρhot/ρcold = Thot4/Tcold4 = 96004/72004 = 3.16 A. T at 300 nm (3 points) = _____________________________________. B. T at 400 nm (3 points) = _____________________________________. C. Relative increase in radiative power (4 points) = ________________________. 3. A. The molecule Integrafor has a fluorescence quantum yield of 0.75 and an observed decay lifetime of 50 ns. Please determine the fluorescence rate constant (kf) and the internal conversion (or non-radiative decay) rate constant (kIC). Solution: kobs = 1/20 ns = 5 x 107 s-1 kf = Φfkobs = 4 x 107 s-1 kIC = kobs - kf = 107 s-1. kf = ________________________ (5 points) kIC = ___________________ (5 points) B. A scientist synthesizes the molecule bromo-Integrafor with a fluorescence quantum yield of 0.1. The bromine causes a spin transition (singlet à triplet) because of the heavy atom effect. Assuming that kf and kIC are unchanged and that the origin of this effect is a third process is a rate constant for intersystem crossing kISC. Calculate the expected observed fluorescence lifetime for bromo-Integrafor. Please calculate the phosphorescence quantum yield (i.e. the quantum yield for emission from the triplet state ΦISC). Solution: The quantum under quenching conditions becomes Φ f = k f k f + k IC + k quench 7. Calculate the DeBroglie wavelength of an electron ejected from sodium metal by a 400 nm photon. The work function of sodium is 1.82 eV. (A useful conversion factor is 1 eV = 8065.5 cm-1). The wavenumber of a 400 nm photon is 107/400 nm = 25,000 cm-1. The workfunction is 1.82 eV(8065.5 cm-1/eV) = 14,670 cm-1. The kinetic energy of an ejected electron is 10,330 cm-1 = 2.05 x 10-19 J. The momentum of the electron is p = √2mE = √2(9.1 x 10-31 kg)(2.05 x 10-19 J) = 6.1 x 10-25 kgm/s. λ = h/p = 6.626 x 10-34 Js/6.11 x 10-25 kgm/s = 1.08 x 10-9 m ≈ 10 Å. 8. A. Calculate the pressure on the top of Mt. Mitchell at 2000 meters elevation assuming 0 meters elevation has a pressure of 1 atm. Solution: Use the barometric pressure formula P = P0{-Mgh/RT} = exp{-(0.029 kg/mol)(9.8 m/s2)(2000 m)/(8.31 J/mol-K)/(298 K)} = 0.794 atm Since P0 = 1 atm I have just left this as a factor of 1 multiplying the exponent. Pressure on the top of Mt. Mitchell = _______0.794 atm_________________. B. Calculate the adiabatic cooling effect on air that rises from sea level to the elevation of Mt. Mitchell. Assume an initial temperature of 310 K. T2 = T1 V1 V2 R/Cv = T1 P2 P1 R/Cv = (310 K) 0.794 1 2/5 = 282 K since P1V1 = P2V2 Air temperature on the top of Mt. Mitchell = _________282 K______________. 9. In 1890 a scientist descends in a diving bell. The volume of the gas inside the bell is 2000 L. What is the volume at a depth of 15 m below the surface of the ocean at 20oC, noting that sea water has a density of 1.03 x 103 gL-1: Solution: at 15 m below sea level the external pressure is: P = ρgh = (1000 kg/m3)(9.8 m/s2)(15 m) = 147000 Pa = 1.47 bar Now the gas is compressible so P1V1 = P2V2 and V2 = V1(P1/P2) = (2000 L)(1/1.47) = 1360 L V(Liters) = ______1360 ___________. 10. An ideal gas is initially at 20.00 atm and 500 K. Its volume is initially 0.5 L and the gas expands isothermally to 120.0 L under the following conditions: (1.) Pexternal = 0 (2.) Pexternal = 0.333 atm (3.) Pexternal = Pgas (reversible expansion) For each of the above conditions calculate ∆U, q, and w, for the gas. ∆U(J) q(J) w(J) (1.) ___0______ ____0______ ____0_______ (2.) ___0______ ___4031____ __-4031_____ (3.) ___0______ ___5693____ __-5693_____ Solution: You can say immediately that ∆U = 0 for all of the conditions since this is an isothermal expansion. Another consequence that q = -w for all. for (1.) no work is done since the expansion occurs against an external pressure of zero. Therefore, the heat must be zero. For (2.) the expansion is against a constant pressure so w = -Pext∆V = -(0.333 atm)(120-0.5 L) = -39.8 L-atm = -4031 J For (3.) w –nRTln(V2/V1) = -(0.25 mol)(8.31 J/mol-K)(500 K)ln(120/0.5) = -5693 J We must find n=PV/RT = (20 atm)(0.5 L)/0.08026L-atm/mol-L/500 K = 0.25 11. Calculate the dipole moment of CO assuming its bond length is 1.23Å and 0.025 units of charge are displaced from the oxygen to the carbon. Solution: µ = ezd = e(0.025)(1.23 Å) = 0.03 eÅ = 0.144 Debye (using the fact that 1 D = 4.8 eÅ) Alternatively you may use MKS units. e = 1.62 x 10-19 C and 1 Å = 10-10 m So 1 e Å = 1.62 x 10-29 Cm and 3.33 x 10-30 Cm = 1 Debye You get the same answer of course.