Solutons of execrses 1) XPS spectrum analyss The fgure below shows an XPS spectrum measured on the surface of a clean nsoluble homo-polyether. Usng the formulas and tables n ths document, answer the followng questons: - Identfy the peaks ncludng the uger peaks. - Identfy the X-ray source, l or Mg? - Determne the atomc concentratons based on the peak deconvoluton. - Suggest whch polymer from the lst below was measured To dentfy the peaks n the spectrum, we compare the peak postons (bndng energes) wth the bndng energes of carbon and oxygen. We fnd that both elements are present on the surface. Namely at bndng energes of 285eV and 531 ev. 800x10 3 O1s photoelectron counts [a.u.] 600 400 200 C1s 0 800 600 400 200 0 bndng energy [ev] The knetc energes of the Oxygen KLL uger electron s 509 ev. The apparent bndng energy depends on the X-Ray source. For the l-source, the peak has an apparent bndng energy BE=E l -E KLL =1486.6eV-509eV=977.6eV. Ths corresponds to Peak 4 n the peak deconvoluton (table). We conclude that the measurement was done wth an l Ka X- ray source and dentfy the remanng peaks as uger peaks as follows. 1
800x10 3 O1s photoelectron counts [a.u.] 600 400 200 C KLL O KLL C1s 0 1400 1200 1000 800 600 400 200 0 bndng energy [ev] To determne the atomc concentratons, we need to consder the peak areas from the deconvoluton. ccordng to the formula gven for quanttatve XPS, the ntensty depends on mean free path and on photoemsson crossectons. We note already that there are two dfferent carbon peaks (peak 1 and peak 2) and ther relatve ntenstes are n a rato 2:1. We use the relaton C I λσ I λσ I σ I σ to quantfy the frst three peaks 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 I/λσ 23.8E6 47.8E6 23.8E6 λσ 10.55*0.013 10.55*0.013 9.41*0.04 I λσ 23.8E6 For peak 1, we thus have C = = 25% I 95.5E6 λσ 2
I λσ 47.8E6 For peak 2, we thus have C = = 50% I 95.5E6 λσ I λσ 23.8E6 For peak 3, we thus have C = = 25% I 95.5E6 λσ Snce Peaks 1 and 2 belong to the element carbon, we have a total of 75% carbon on the surface and 25% of oxygen, that s a rato C/O=3:1 Lookng at the table, there are two polymers possble wth ths C/O rato: These are PPO and poly(trmethylene oxde). Snce we know from the queston that there are only two dfferent carbon bndng energes n the spectrum, we can exclude PPO. Ths excluson would requre a very good spectral resoluton n a real experment snce the two dfferent ether carbons n PPO have very smlar bndng energes. 3
2) The Lennard-Jones par potental Plot the Lennard-Jones par-potental and the resultng par- force n a graph. t what separaton, r, s the equlbrum dstance (F=0)? Below whch separaton does the repulsve term start to domnate? From chapter 8, we have w(r) = C 1 r n + C 2 r m wth n=6, m=12, C 1=10-77 Jm 6 and C 2 =10-134 Jm 12. The correspondng nteracton force s gven by F(r) = dw(r) dr = 6C 1 + 12C 2 r 7 r 13 4 Lennard-Jones potental and force 40 w(r) F(r) potental, w(r) [J] 2 0 20 0 force, F(r)=-dw(r)/dr [N] -2-20 -4x10-21 -40x10-12 200 300 re 400 500 600 700 800x10-12 dstance, r [m] The equlbrum dstance, r e, s gven by F(r e )=0, so we have: C r 2 2 6 e = 6 2 10 57 3. 55Å C 1 nd the repulsve term starts to domnate at w(r d )=0, so we have n analogy to the above: w(r d ) = 0 r d = C 2 6 6 10 57 3.1Å C 1 We note that the repulsve term s very steep,.e. there s only ~0.4Å dstance between the equlbrum poston and the strongly repulsve wall. 4
2) Measurement of surface forces To drectly measure Lennard-Jones-type forces between macroscopc bodes, does one need the entre bodes to be present? What s the crtcal exponent, n, n a potental of type w(r)=-/r n, below whch the total energy of nteracton starts to depend on the entre body? We start wth the attractve part of a Me potental: w(r) = C r n We assume addtvty and calculate the total nteracton energy between a sphere of radus, R, wth a sem-nfnte flat plane, whch s made of the same knd of molecules. We wll execute the ntegraton n two steps: Frst, we ntegrate over rngs of equal dstance n the plane. The molecule be stuated at z=d. x r = (z2+x2) 1/2 z=0 z=d dz dx x z D dv f ρ s the number densty of molecules n the plane we obtan the total nteracton energy, w ms (D) between a molecule and a flat surface: z = x= z = x w ms (D) = 2πCρ dz z dx = D x = 2πCρ dz z 2 + x 2 =0 x 2 + z 2 z = D ( ) n 2 n > 2 2πCρ z = dz (2 n) z z 2 n n > 3 2πCρ z 3 n (n 2) (3 n) = D whch s fnte only for n>3. z = z =D ( ) 1 n 2 1 (2 n) 2πCρ = (n 2)(n 3) D3 n x = x =0 5
Now, we wll agan ntegrate the molecule-surface energy to obtan the energy between a sphere and a flat surface usng sectons of equal dstance, (D+z), n the sphere as shown below. x dz z=-d R-z x R z=r z=2r z D (x,z) <=> x 2 = R 2 - (R-z) 2 = (2R -z) z dv = π x 2 dz = π (2R-z) z dz We can thus wrte: 2πCρ W(D) = (n 2)(n 3) 2π 2 Cρ 2 = (n 2)(n 3) z =2 R z= 0 ρdv (dr) dz (D + z) = 2πCρ n 3 (n 2)(n 3) z =2 R z = 0 (2R z)z dz (D + z) n 3 z =2 R dz z= 0 ρπ(2r z)z (D + z) n 3 Ths ntegral depends on the exponent, n, of the nteracton power law, the radus, R, of the sphere as well as on the smallest separaton, D, between the sphere and the flat surface. To fnd out how much of a sphere contrbutes to the measurement of surface forces, we wll nsert values typcally encountered n the Surface Forces pparatus: D=10-10 m, R=1.5. 10-2 (2R z)z m. The followng graphs plot the ntegrand between z=d n 3 (D + z) and z=2*r for dfferent n=1...6. The surface under the plot corresponds then to the value of the ntegral. 6
We readly see that for n=6, only parts of the sphere closer than some 10nm contrbute to the nteracton. We also note that nteractons wth underlyng par potental of n<5 result n body-forces nstead of surface forces. 4x10-6 80x10-9 n=1 n=2 Integrand (2R-z)z/(D+z)^(n-3) 60 40 20 0 200x10-6 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] n=3 Integrand (2R-z)z/(D+z)^(n-3) 3 2 1 0 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] 30x10-3 25 n=4 Integrand (2R-z)z/(D+z)^(n-3) 150 100 50 0 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] Integrand (2R-z)z/(D+z)^(n-3) 20 15 10 5 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] Integrand (2R-z)z/(D+z)^(n-3) 60x10 6 40 20 0 n=5 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] Integrand (2R-z)z/(D+z)^(n-3) 300x10 15 200 100 0 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 z[m] Fgures showng the ntegrand of the total nteracton energy between a sphere and a flat surface. n=6 7
3) Keesom energy and thermal actvaton Compare the maxmal Keesom energy to kt. Below what ntermolecular dstance are dpolar molecules lke water prevented from rotatng? The dpole moment of water s µ=1.854d (Debye), where 1D=3.33564*10-30 Cm, the relatve permttvty of water s ε=78.54 and the permttvty of vacuum s ε 0 =8.854*10 12 s/vm, the Boltzmann constant s k=1.38*10-23 J/K. We know the Keesom energy to be the dpole-dpole nteracton energy. If the relatve angles between the dpoles are fxed, the nteracton s gven by: Θ 1 Θ 3 µ 1 r µ2 Θ2 w(r,θ 1,Θ 2,Θ 3 ) = µ 1 µ 2 4πε 0 ε r 3 (2cosΘ 1 cosθ 2 snθ 1 snθ 2 cosθ 3 ) Maxmum nteracton wll occur when the two dpoles are lyng n a lne (Θ 1 =Θ 2 =0) and one has: w(r,0,0,θ 3 ) = 2µ 1 µ 2 4πε 0 ε r 3 the radus, r th, whch corresponds to the energy kt (~4.14. 10-21 J) s thus gven by: r th = 2 (1.854Debye 3.33564 10 30 Cm 2µ 1 µ 2 Debye ) 2 3 3 4πε 0 ε kt s 1.28Å, 12.5664 8.854 10 12 Vm 78.54 4.14 10 21 J whch s a very small dstance. We conclude that, at ambent temperatures, the Keesom energy s not enough to freeze the rotaton of water molecules. 4) Breakdown of DLVO theory The DLVO theory s known to fal at hgh salt concentratons and/or small dstances. What could be the reason(s) for ths, and, n whch sense do you expect the real forces to devate from the theory? 8
DLVO theory s a contnuum theory, t thus breaks down at small separatons and/or hgh salt concentratons. Somewhat aganst common ntuton, there seems to be a strong attracton at small separatons, whch mght be due to formaton of salt crystal structure. If that s true, then there mght also be more than one secondary mnmum n realty. The detals of the known DLVO breakdown are stll hghly debated. The Van der Waals attracton between ons of even the same charge seems to play an mportant role. For example, negatve ons of a mnmum sze can ndeed adsorb to negatvely charged surfaces. 5) rchmedes and Lfshtz Usng rchmedes law, show that two bodes of mass m 1 =V 1 *ρ 1 and m 2 =V 2 *ρ 2 are experencng a repulsve gravtatonal force f mmersed n a medum of densty ρ 3 when ether one of the condtons ρ 1 <ρ 3 <ρ 2 or ρ 2 <ρ 3 <ρ 1 apples. Van der Waals forces and gravtatonal forces have very dfferent dstance-dependence (see ntegraton n exercse 2). In fact, ntegraton of the total gravtatonal energy of any body wth an nfnte medum would yeld an nfnte energy. We can however show that the Lfshtz theory s analogue to the rchmedes flotaton law for two specal cases: rchmedes flotaton law states that a body of volume, V 2, and mass densty, ρ 2, floatng n a medum of densty, ρ 3, experences a flotaton force, F =V 2 *ρ 3 *g, where g~9.81m/s 2 s earth's gravtatonal acceleraton. F =V 2 *ρ 3 *g ρ 2 ρ 3 F g =V 2 *ρ 2 *g The rchmedes flotaton force acts n opposte drecton to the gravtatonal force, F g =V 2 *ρ 2 *g, so that the total force on the submerged body s, F=F -F g. In rchmedes model, body 1 s represented by earth and has a mass densty ρ 1, underneath the beaker. So, we have a smple soluton: F = F F g = V 2 g(ρ 3 ρ 2 ), whch s ndeed repulsve for ρ 1 >ρ 3 >ρ 2. 9
The opposte case (ρ 1 <ρ 3 <ρ 2 ) s equvalent due to nterchangeablty of ndces 1 and 2. For a generalzed rchmedes law, one would have to ntegrate the nteracton over the mass densty dstrbutons. If the medum has an nfnte extenson, the total nteracton energy wll be nfnte due to the long range of the gravtatonal potental. We can get around ths problem, f we consder the system n terms of a varaton of energy. In vacuum (ρ 3 =0), we know that the gravtatonal energy between the bodes 1 & 2 s gven by: E gravty = G V 1ρ 1 V 2 ρ 2 r 1 ρ 3 ρ 1 ρ 2 r 1 r 2 In the medum 3, the varaton of total energy, resultng from a densty exchange,.e. exchange of body 2 to a larger dstance, r 2,=r 1 + r s gven by: E(r 2 ) = V 1ρ 1 V 2 ρ 2 + V 1ρ 1 V 2 ρ 3 V 1ρ 1 V 2 ρ 2 V 1ρ 1 V 2 ρ 3 r 2 r 1 r 1 r 2 = V V 1 2ρ 1 (ρ 3 ρ 2 ) V V 1 2ρ 1 (ρ 3 ρ 2 ) r 1 r 2 and we thus obtan the nteracton energy, E' 1 of body 1 wth body 2 n medum 3, by exchangng wth a volume, V 2, at nfnte dstance (.e. puttng the zero of the energy scale dfferently): E' = E(r 2 ) = V 1 V 2ρ 1 (ρ 3 ρ 2 ) r 1, 10
whch s the same as the rchmedes flotaton law, but expressed as an nteracton energy wth zero energy at nfnte dstance between the bodes. 11
Collecton of useful equatons and constants: The nano cube The macro cube 7x7x7=343 atoms 10 24 atoms 210 surface atoms (60%) 6. 10 16 surface atoms (0.000006%) The vacuum monolayer tme 3.2 10 τ = s mbar p 6 The XPS energy equaton Eb = hν Ekn Θ hν: E b : E kn : X-ray energy (l Kα = 1486.6 ev, Mg Kα = 1253.6 ev) bndng energy of photoelectron knetc energy of photoelectron 12
Bndng Energes of C 1s and O 1s Oxygen, O 1s photoelectrons Carbon, C 1s photoelectrons 13
uger electron energes 14
Electron mean free path λ = 2 Ekn + B ae kn B = 0.087 for organcs B = 0.096 for norgancs B E kn Quanttatve XPS I = I X rax σ Me potental 10 1 10 0 10-1 y=10-10n /x n 10-2, B: constants r: ntermolecular dstance w(r) = C 1 + C 2 r n r m Specal case Lennard-Jones potental wth the exponents n=6 and m=12 and the nteracton constants are C 1 =10-77 Jm 6 and C 2 =10-134 Jm 12. par potental [a.u.] 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 n=6 n=5 n=4 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 dstance [m] n=1 15
Lennard-Jones potental (molecule-surface) W ms (D) = π C ρ 6D 3 C: nteracton constant r: ntermolecular dstance Electrostatc nteracton Q w Cb (r) = 1 Q 2 4π ε 0 ε r = z z 1 2 e2 4π ε 0 ε r ε: relatve permttvty (ε 0 =8.854187 10-12 sv -1 m -1 ) Q 1 and Q 2,: pont charges [C] z 1 and z 2 : onc valency e: elementary charge (e=1.602. 10-19 C) Keesom nteracton energy dpoles at a fxed angle: Θ 3 Θ 1 Θ 2 r µ 1 µ2 w(r,θ 1,Θ 2,Θ 3 ) = µ 1 µ 2 4πε 0 ε r 3 (2cosΘ 1 cosθ 2 snθ 1 snθ 2 cosθ 3 ) Maxmum nteracton (Θ 1 =Θ 2 =0): Freely rotatng dpoles: w(r,0,0,θ 3 ) = 2µ 1 µ 2 4πε 0 ε r 3 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 Debye nteracton energy dpole feld E(r) = µ 1 + 3cos2 Θ 4πε 0 ε r 3 µ Θ r α 16
nteracton energy w(r,θ) = 1 2 α E2 (r) = µ 2 α (1+ 3cos 2 Θ) 2(4πε 0 ε) 2 r 6 α: polarzablty <cos 2 > = 1 / 3, therefore Debye nteracton energy w(r) = µ 2 α (4πε 0 ε ) 2 r 6 London dsperson energy α r α London dsperson nteracton w(r) = 3hυα 2 4(4πε 0 ε) 2 r 6 h: s the Planck's constant a: the polarzablty, ε: relatve permttvty (ε 0 =8.854187 10-12 sv -1 m -1 ) ν: characterstc frequency related to the frst onzaton potental of the atom or molecule. typcal value of hν~2 10-18 Van der Waals nteractons Collecton of dfferent contrbutons 17
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