Data Provided: A formula sheet and table of physical constants is attached to this paper. THE PHYSICS OF SOFT CONDENSED MATTER

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1 Dt Provided: A formul sheet nd tble of physicl constnts is ttched to this pper. Ancillry Mteril: Grph pper (liner) DEPARTMENT OF PHYSICS AND ASTRONOMY Autumn Semester (2015) THE PHYSICS OF SOFT CONDENSED MATTER 2 HOURS Instructions: Answer ny 3 questions of your choice out of the 5 questions offered. There is no compulsory question. Questions re mrked out of twenty. The brekdown on the right- hnd side of the pper is ment s guide to the mrks tht cn be obtined from ech prt. Plese clerly indicte the question numbers on which you would like to be exmined on the front cover of the nswer book. Cross through ny work you do not wish to be exmined. 1 TURN OVER

2 1. () (b) The ltent het of vporiztion for wter t mbient pressure is 40 kj/mol. This is remrkbly high for such smll molecule, explin why. Estimte the energy required to remove one wter molecule from within the bulk of wter to vcuum, nd the ctivtion energy for the viscous flow of wter, both in ev. [4] Use the Willims- Lndel- Ferry time- temperture superposition principle, expressed in terms of T, which is the rtio of relxtion time t given temperture T, nd relxtion time t the glss trnsition, T g : log T C 1 T T g C 2 (T T g ) to support the polymer engineer s rule of thumb tht ner the glss trnsition, chnge of temperture of pproximtely 3 o C chnges relxtion time tenfold. (At T g, C 1 = 17.4, C 2 = 51.6 K). [4] (c) Explin wht thermotropic nd lyotropic men in the context of liquid crystls (LCs). [4] (d) The tble shows the melting points (T M ) of semicrystlline polymer s function of its prior crystlliztion temperture, T C. Determine grphiclly the equilibrium melting temperture, T M ( ). [4] T C [K] T M [K] (e) An mphiphile is grdully dded to wter. Sketch grph with the totl volume frction of mphiphile on the horizontl xis, nd the volume frctions occupied by mphiphile monomers (dissolved mphiphile molecules), nd by mphiphile in micelles, on the verticl xis. (You my ssume sphericl micelles). Annotte nd nme the key feture of your digrm. [4] 2 CONTINUED

3 2. The potentil energy of two molecules (1, 2) intercting by Vn der Wls interctions only is given by 12 = A 1 2 / r 6 where A is positive constnt with the dimensions of energy, 1 ( 2 ) is the polrisbility of molecule 1 (molecule 2), which is positive quntity with the dimensions of volume, nd r is the distnce between the molecules. () Show tht the interction prmeter between two molecules s defined in the frmework of lttice theory, = (z / 2k B T) ( ) with positive but unspecified number z, is lwys positive for two molecules tht interct vi Vn der Wls interctions only. Assume 1 2. [8] (b) From the result in 2 (), support the lchemist s proverb simili similibus solvuntur (they spoke in Ltin, it mens like dissolves like ). Wht hs to be like / similr for solubility, i.e. miscibility? [4] At mbient temperture, n-butnol/wter mixtures my phse-seprte into two coexisting frctions. One of the coexisting frctions is mde up of volume frction 1 = 6% = 0.06 butnol, nd 94% wter. The other is mde up of 2 = 72.5% = n- butnol, nd 27.5% wter. (c) (d) Cn the n-butnol/wter system be described by n interction prmeter tht depends on temperture only, (T)? Justify your nswer. [2] How (if t ll) will 50% / 50% (by volume) mix of n- butnol / wter frctionte into coexisting frctions, 1 nd 2? How (if t ll) will 80% / 20% (by volume) mix of n- butnol / wter frctionte into 1 nd 2? [6] 3 TURN OVER

4 3. The degree of polymeriztion, N, gives the number of repet units in one polymer chin. When polymer chin dissolves in solvent, it coils up into coil of size R. The concentrtion, c, of polymer solution is mesured in units of mss of dissolved polymer per unit volume of solution. The overll number of repet units per unit volume of solution is proportionl to c. () When R scles with N s R ~ N, how does the volume occupied by one dissolved coil, V 1, scle with N? How does the number of dissolved polymer chins, k, scle with c, nd N? How does the volume frction of solution occupied by the k dissolved polymer coils, k ~ V k = kv 1, scle with c nd N? [8] The trnsition from dilute to semidilute polymer solution is defined s the concentrtion, c*, where dissolved polymer coils begin to overlp. (b) (c) (d) Find the power b for the scling lw c* ~ N b. Hint: Work with k. First express b generlly in terms of, then evlute for good solvent ( = 3 / 5 ) nd solvent ( = ½). Show tht for poor solvent (coil globule trnsition, = 1 / 3 ), c* is independent of N. [8] Explin why c* cn be surprisingly smll for high polymer solution (i.e., solution of polymer with lrge N) in good solvent. [2] Explin why semidilute polymer solutions disply viscoelstic, rther thn purely viscous, mechnicl behviour. [2] 4 CONTINUED

5 4. () (b) (c) (d) (e) (f) Rod- shped molecules my lign prllel to ech other to form nemtic liquid crystlline (LC) phse either due to the nisotropy of their mutul interctions, or due to reduced excluded volume. Which of the two is considered by the Mier- Supe theory of the nemtic LC phse? [2] Nme or describe two types of moleculr shpes tht my led to (thermotropic) liquid crystlline phses. [4] Assume you hve set of different moleculr mterils tht ll disply nemtic LC phse, but t elevted tempertures. Describe resonble ttempt to prepre n mbient temperture nemtic LC phse from these mterils, without synthesis of new mterils. [2] Assume tht in nemtic LC, 80% of molecules re ligned with their long moleculr xis exctly prllel to the director, nd 20% re ligned exctly orthogonl to the director. Give the nemtic order prmeter, S = ½ P 2 (cos ) = ½ 3cos 2 1. [2] Nme function tht the lignment lyer in n LC disply (LCD) cell performs, other thn lignment itself. [2] Why is chirl dopnt dded to the LC in twisted nemtic (TN) LCD cell? [2] The criticl electricl field E Fr for the Frederiks trnsition in n LCD cell is given by E Fr 2 d K 1 1 A 4 (K 3 2K 2 ) 1/ 2 Wherein d is the thickness of the cell, K 1, K 2, K 3 re elstic constnts relting to different deformtions of LC lignment, nd A = 0, wherein is the (dimensionless) difference of dielectric constnts prllel / orthogonl to lignment nd 0 is the vcuum permittivity, 0 = F/m. (g) Wht re the units of mesurement for K 1, K 2, nd K 3? Explin how the given eqution shows tht the criticl or switching voltge V Fr is independent of the thickness of n LCD cell. Why is it nevertheless desirble to use very thin cells in LCD screens for moving imges, e.g. TV screens? [6] 5 TURN OVER

6 5. () Why will the melting temperture T M of crystl lmelle in semicrystlline polymer lwys be lower thn the polymer s equilibrium melting point, T M ( )? How cn we nevertheless determine T M ( )? [4] The ttched tble shows the lmellr thickness, l*, of polyethylene (PE) crystlline lmelle crystllized t vrious tempertures, T C. The equilibrium melting point, T M ( ), of PE is independently determined to be K. T C [K] l* [nm] (b) By plotting n pproprite grph show tht the given dt re consistent with the reltionship l * (T C ) 2 f T M ( ) H M T M ( ) T C [8] (c) Extrct /, nd 2 f / H M, from your grph. Quote them with their units. [4] (d) Use the reltion T M (l * ) T M ( ) 1 2 f H M l * to give the melting point of lmelle crystllized t T C = 401 K. [4] END OF EXAMINATION PAPER 6 CONTINUED

7 PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physicl Constnts electron chrge e = C electron mss m e = kg = MeV c 2 proton mss m p = kg = MeV c 2 neutron mss m n = kg = MeV c 2 Plnck s constnt h = J s Dirc s constnt ( = h/2π) = J s Boltzmnn s constnt k B = J K 1 = ev K 1 speed of light in free spce c = m s m s 1 permittivity of free spce ε 0 = F m 1 permebility of free spce µ 0 = 4π 10 7 H m 1 Avogdro s constnt N A = mol 1 gs constnt R = J mol 1 K 1 idel gs volume (STP) V 0 = 22.4 l mol 1 grvittionl constnt G = N m 2 kg 2 Rydberg constnt R = m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdius 0 = m Bohr mgneton µ B = J T 1 fine structure constnt α 1/137 Wien displcement lw constnt b = m K Stefn s constnt σ = W m 2 K 4 rdition density constnt = J m 3 K 4 mss of the Sun M = kg rdius of the Sun R = m luminosity of the Sun L = W mss of the Erth M = kg rdius of the Erth R = m Conversion Fctors 1 u (tomic mss unit) = kg = MeV c 2 1 Å (ngstrom) = m 1 stronomicl unit = m 1 g (grvity) = 9.81 m s 2 1 ev = J 1 prsec = m 1 tmosphere = P 1 yer = s

8 Polr Coordintes x = r cos θ y = r sin θ da = r dr dθ 2 = 1 ( r ) + 1r 2 r r r 2 θ 2 Sphericl Coordintes Clculus x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r 2 sin θ dr dθ dφ 2 = 1 ( r 2 ) + 1 r 2 r r r 2 sin θ ( sin θ ) + θ θ 1 r 2 sin 2 θ 2 φ 2 f(x) f (x) f(x) f (x) x n nx n 1 tn x sec 2 x e x e x sin ( ) 1 x ln x = log e x 1 x cos 1 ( x sin x cos x tn ( 1 x cos x sin x sinh ( ) 1 x cosh x sinh x cosh ( ) 1 x sinh x cosh x tnh ( ) 1 x ) ) 1 2 x x 2 2 +x 2 1 x x x 2 cosec x cosec x cot x uv u v + uv sec x sec x tn x u/v u v uv v 2 Definite Integrls x n e x dx = n! (n 0 nd > 0) n+1 π e x2 dx = π x 2 e x2 dx = 1 2 Integrtion by Prts: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx

9 Series Expnsions (x ) Tylor series: f(x) = f() + f () + 1! n Binomil expnsion: (x + y) n = (1 + x) n = 1 + nx + k=0 ( ) n x n k y k k n(n 1) x 2 + ( x < 1) 2! (x )2 f () + 2! nd (x )3 f () + 3! ( ) n n! = k (n k)!k! e x = 1+x+ x2 2! + x3 x3 +, sin x = x 3! 3! + x5 x2 nd cos x = 1 5! 2! + x4 4! ln(1 + x) = log e (1 + x) = x x2 2 + x3 3 n Geometric series: r k = 1 rn+1 1 r k=0 ( x < 1) Stirling s formul: log e N! = N log e N N or ln N! = N ln N N Trigonometry sin( ± b) = sin cos b ± cos sin b cos( ± b) = cos cos b sin sin b tn ± tn b tn( ± b) = 1 tn tn b sin 2 = 2 sin cos cos 2 = cos 2 sin 2 = 2 cos 2 1 = 1 2 sin 2 sin + sin b = 2 sin 1( + b) cos 1 ( b) 2 2 sin sin b = 2 cos 1( + b) sin 1 ( b) 2 2 cos + cos b = 2 cos 1( + b) cos 1 ( b) 2 2 cos cos b = 2 sin 1( + b) sin 1 ( b) 2 2 e iθ = cos θ + i sin θ cos θ = 1 ( e iθ + e iθ) 2 nd sin θ = 1 ( e iθ e iθ) 2i cosh θ = 1 ( e θ + e θ) 2 nd sinh θ = 1 ( e θ e θ) 2 Sphericl geometry: sin sin A = sin b sin B = sin c sin C nd cos = cos b cos c+sin b sin c cos A

10 Vector Clculus A B = A x B x + A y B y + A z B z = A j B j A B = (A y B z A z B y ) î + (A zb x A x B z ) ĵ + (A xb y A y B x ) ˆk = ɛ ijk A j B k A (B C) = (A C)B (A B)C A (B C) = B (C A) = C (A B) grd φ = φ = j φ = φ x î + φ y ĵ + φ z ˆk div A = A = j A j = A x x + A y y + A z z ) curl A = A = ɛ ijk j A k = ( Az y A y z φ = 2 φ = 2 φ x + 2 φ 2 y + 2 φ 2 z 2 ( φ) = 0 nd ( A) = 0 ( A) = ( A) 2 A ( Ax î + z A ) ( z Ay ĵ + x x A ) x y ˆk