1) XPS Spectrum analysis: The figure below shows an XPS spectrum measured on the surface of a clean insoluble homo-polyether. Using the formulas and tables in this document, answer the following questions: - Identify the peaks including the uger peaks. - Identify the X-ray source, l or Mg? - Determine the atomic concentrations based on the peak deconvolution. - Suggest which polymer from the list below was measured (multiple solutions?) simulated XPS spectrum (survey) photoelectron counts [a.u.] 600x10 3 400 200 0 1400 1200 1000 800 600 400 200 0 binding energy [ev] 1
600x10 3 simulated XPS spectrum (detail) photoelectron counts [a.u.] 500 400 300 200 100 0 292 290 288 286 284 282 280 binding energy [ev] photoelectron counts [a.u.] 800x10 3 600 400 200 simulated XPS spectrum (detail) 0 536 534 532 530 528 526 binding energy [ev] Peak Deconvolution: Peak 1 Peak 2 Peak 3 Peak 4 Peak 5 BE 285 287 531 977.6 1226.6 counts 3.27E6 6.55E6 8.98E6 4.77E5 1.19E5 2
2) Plot the Lennard-Jones pair-potential and the resulting pair- force in a graph. t what separation, r, is the equilibrium distance (F=0)? Below which separation does the repulsive term start to dominate? 3) To directly measure Lennard-Jones-type forces between macroscopic bodies, does one need the entire bodies to be present? What is the critical exponent, n, in a potential of type w(r)=-/r n, below which the total energy of interaction starts to depend on the entire body? 4) Compare the maximal Keesom energy to kt. Below what intermolecular distance are dipolar molecules like water prevented from rotating? The dipole moment of water is µ=1.854d (Debye), where 1D=3.33564*10-30 Cm, the relative permittivity of water is ε=78.54 and the permittivity of vacuum is ε 0 =8.854*10 12 s/vm, the Boltzmann constant is k=1.38*10-23 J/K. 5) The DLVO theory is known to fail at high salt concentrations and/or small distances. What could be the reason(s) for this, and, in which sense do you expect the real forces to deviate from the theory? 6) Using rchimedes law, show that two bodies of mass m 1 =V 1 *ρ 1 and m 2 =V 2 *ρ 2 are experiencing a repulsive gravitational force if immersed in a medium of density ρ 3 when either one of the conditions ρ 1 <ρ 3 <ρ 2 or ρ 2 <ρ 3 <ρ 1 applies. 3
Collection of useful equations and constants: The nano cube The macro cube 7x7x7=343 atoms 10 24 atoms 218 surface atoms (63.5%) 6. 10 16 surface atoms (0.000006%) The vacuum monolayer time 3.2 10 τ = s mbar p 6 The XPS energy equation Eb = hν Ekin Θ hν: E b : E kin : X-ray energy (l Kα = 1486.6 ev, Mg Kα = 1253.6 ev) binding energy of photoelectron kinetic energy of photoelectron 4
Binding Energies of C 1s and O 1s Oxygen, O 1s photoelectron energies Carbon, C 1s photoelectron energies Nitrogen, N 1s photoelectron energies 5
uger electron energies Examples of uger energies: Carbon KLL = 260eV; Nitrogen KLL = 379eV; Oxygen KLL = 509eV, 6
Electron mean free path λ = 2 Ekin + B ae kin B E kin B = 0.087 for organics B = 0.096 for inorganics a monolayer can be taken as 3.5Å Examples: For l Kα : For Mg Kα : λ C1s =10.55Å, λ O1s =9.41Å λ C1s =9.48Å, λ O1s =8.19Å Quantitative XPS Measured XPS intensity of one element: I I X rax σ Ω 0 z λ sinα C ( z) e dz I X-ray = Intensity of incident X-ray σ = photoionization crossection Ω = instrumental and unit conversion constant, generally depends on E kin C = atomic concentration of element λ = electron mean free path α = photoelectron emission angle (angular resolved XPS) Example: XPS intensity at normal emission (α = 90 ) and homogeneous atomic concentration X rax 0 z λ I I σ ΩC e dz = I σ ΩC λ X rax Therefore, the atomic concentrations can be obtained from the peak areas, I i, of a measured spectrum using the following simplified relation: I I λσ σ C Ii Ii λσ σ i i i i i Some atomic subshell photoionization crosssections Photoelectron Binding energy Mg Kα l Kα C 1s 285 ev 0.022 0.013 O 1s 531 ev 0.063 0.040 N 1s 400 ev 0.039 0.024 7
Mie potential 10 1 10 0 10-1 y=10-10n /x n 10-2, B: constants r: intermolecular distance w(r) = C 1 + C 2 r n r m Special case Lennard-Jones potential with the exponents n=6 and m=12 and the interaction constants are C 1 =10-77 Jm 6 and C 2 =10-134 Jm 12. Lennard-Jones potential (molecule-surface) W ms (D) = π C ρ 6D 3 pair potential [a.u.] 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 n=6 n=4 n=5 n=3 n=2 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 10 0 distance [m] n=1 C: interaction constant r: intermolecular distance Electrostatic interaction Q w Cb (r) = 1 Q 2 4π ε 0 ε r = z 1 z 2 e 2 4π ε 0 ε r ε: relative permittivity (ε 0 =8.854187 10-12 sv -1 m -1 ) Q 1 and Q 2,: point charges [C] z 1 and z 2 : ionic valency e: elementary charge (e=1.602. 10-19 C) Keesom interaction energy dipoles at a fixed angle: w(r,θ 1,Θ 2,Θ 3 ) = μ 1 μ 2 4πε 0 ε r 3 (2cosΘ 1 cosθ 2 sinθ 1 sinθ 2 cosθ 3 ) Maximum interaction (Θ 1 =Θ 2 =0): w(r,0,0,θ 3 ) = 2μ 1 μ 2 4πε 0 ε r 3 Θ 3 Θ 1 Θ 2 r µ 1 µ2 Freely rotating dipoles: w(r) μ 1 2 μ 2 2 3(4πε 0 ε) 2 kt r 6 for kt > μ 1 μ 2 4πε 0 ε r 3 8
Debye interaction energy dipole field E(r) = μ 1 + 3cos2 Θ 4πε 0 ε r 3 µ Θ r α interaction energy w(r,θ) = 1 2 α E2 (r) = μ 2 α (1+ 3cos 2 Θ) 2(4πε 0 ε) 2 r 6 α: polarizability <cos 2 > = 1 / 3, therefore Debye interaction energy w(r) = μ 2 α (4πε 0 ε ) 2 r 6 London dispersion energy α r α London dispersion interaction w(r) = 3hυα 2 4(4πε 0 ε) 2 r 6 h: is the Planck's constant a: the polarizability, ε: relative permittivity (ε 0 =8.854187 10-12 sv -1 m -1 ) ν: characteristic frequency related to the first ionization potential of the atom or molecule. typical value of hν~2 10-18 9
Van der Waals interactions Collection of different contributions 10