Dt Provided: A formul heet nd tble of phyicl contnt i ttched to thi pper. Ancillry Mteril: DEPARTMENT OF PHYSICS AND ASTRONOMY Autumn 2014 The phyic of mteril 2 hour Intruction: Anwer quetion ONE (Compulory) from Section A nd TWO quetion from Section B. All quetion re mrked out of twenty. The brekdown on the right-hnd ide of the pper i ment guide to the mrk tht cn be obtined from ech prt. Plee clerly indicte the quetion number on which you would like to be exmined on the front cover of your nwer book. Cro through ny work tht you do not wih to be exmined. PHY245 TURN OVER 1
SECTION A COMPULSORY 1. () Nme one electronic nd one mechnicl difference between metl nd cermic. Give brief decription of the phyicl origin of the difference between the two mteril. (b) Given the energy tte digrm below write down the full electronic configurtion of titnium which h tomic number 22. [2] [4] f d p f d p f d d p Energy d p p p 1 2 3 4 5 6 7 Principl quntum number, n (c) Given tht the electronegtivity of zinc nd tellurium re 1.65 nd 2.1 on the Puling cle, clculte the percentge ionic chrcter of the bond formed in ZnTe. [2] PHY245 CONTINUED 2
(d) Redrw the tre-trin plot given below, mrking on it the following: i. The eltic modulu of the mteril. ii. The yield trength t trin offet of 0.02. iii. The tenile trength of the mple. iv. The frcture point. v. The reilience of the mple. [5] Stre Strin Grph Stre Strin (e) I the tre-trin plot from prt (d) of brittle or ductile mteril? Explin your nwer. (f) Wht i the difference between true tre nd engineering tre? [2] (g) For bronze lloy, the tre t which pltic deformtion begin i 275 MP, nd the modulu of elticity i 115 GP. i. Wht i the mximum lod tht my be pplied to pecimen with croectionl re of 325 mm 2 without pltic deformtion? [2] ii. If the originl pecimen length i 115 mm, wht i the mximum length to which it my be tretched without cuing pltic deformtion? [2] [1] PHY245 TURN OVER 3
SECTION B Anwer two quetion 2. () Write down the potentil energy of the Coulomb interction between odium (N + ) ion nd chlorine (Cl - ) ion in vcuum. [2] (b) Clculte the Mdelung contnt of the cubic lttice (FCC) formed by NCl. You hould only ue the firt three ditinct lttice eprtion. Why i the vlue you clculted different from the tndrd vlue of -1.74 for fce centred cubic lttice? [4] (c) Write down the generl form of the Lennrd-Jone potentil indicting the ttrctive nd repulive term. Furthermore explin the phyicl origin of the ditnce dependence of thee term. [6] (d) Uing the Lennrd-Jone potentil bove, derive n expreion for the bond length t equilibrium for two prticle in the potentil. [4] (e) Given tht the modulu of elticity of mteril i given by df E / dr r 0, find the modulu of elticity in term of the ttrctive nd repulive contnt in the Lennrd-Jone potentil. [4] PHY245 CONTINUED 4
3. () Below i tble howing the propertie of 5 mteril (A-E). Mteril Yield Strength (MP) Tenile Strength (MP) Strin t Frcture Frcture Strength (MP) Eltic Modulu (GP) A 310 340 0.23 265 210 B 100 120 0.40 105 150 C 415 550 0.15 500 310 D 700 850 0.14 720 210 E Frcture before yielding 650 250 i. Which mteril experience the gretet percentge reduction in re? Explin your nwer. [4] ii. Which mteril i the tronget one? Explin your nwer. [2] iii. Which mteril i the tiffet one nd why? [2] (b) A tenile tre i pplied to cylindricl br rod which h n initil dimeter of 12.8 mm. Determine the lod required to cue 2.5 10-3 mm chnge in the dimeter of the rod. You my ume tht ll deformtion i eltic nd the Poion rtio for br i 0.34 nd it Young modulu i 97 10 3 MP. [4] (c) The br rod i teted to frcture in tenile meurement, nd it i found to hve frcture trength of of 550 MP. If it cro ectionl dimeter t frcture i 10.7 mm, clculte: i. The ductility in term of percentge re reduction. [2] ii. The true tre t frcture. [2] (d) Drw the tre-trin grph for both brittle mteril nd ductile mteril. Mrk the reilience on both grph. [4] 4. () Wter enter pipe t ditnce Y 1 bove the ground with preure of P 1 nd velocity V 1. It leve the pipe ometime lter with preure of P 2 nd velocity V 2 through n opening tht i Y 2 bove the ground. By equting the net work done nd the mechnicl energy of the ytem derive Bernoulli eqution in it norml form. (b) Ue Bernoulli eqution to explin why wing hped object drgged through wter will move in the direction of the wing ide with the lrger re. You hould ddre ech term in the Bernoulli eqution. [4] (c) A phericl borber of rdiu in n infinite medium borb every prticle reching the urfce. The concentrtion t r = 1 i C 0. i. Rewrite Fick econd lw for ytem with rdil ymmetry. [2] ii. Give n expreion for dorption rte in term of the diffuion contnt, D, nd the concentrtion t infinity. [8] [6] PHY245 TURN OVER 5
5. () Wht i the min difference between n lloy nd compound, explin thi in term of tructurl mke up nd phyicl propertie. [4] (b) A copper-nickel lloy of compoition 70 wt% Ni nd 30 wt% Cu i lowly heted from temperture of 1300 C. Uing the phe digrm for copper-nickel lloy given below, nwer the following quetion. Compoition (t% Ni) 0 20 40 60 80 100 1600 2800 1500 Liquid 1453 C 1400 2600 Temperture ( C) 1300 Liquidu line B + L Solidu line 2400 Temperture ( F) 1200 2200 1100 1085 C A 2000 1000 0 20 40 60 80 100 (Cu) Compoition (wt% Ni) ( ) (Ni) i. At wht temperture doe the firt liquid phe form? [2] ii. Wht i the compoition of thi liquid phe? [2] iii. At wht temperture doe complete melting of the lloy occur? [4] iv. Wht i the compoition of the lt olid remining prior to complete melting? [4] (c) I it poible to hve copper-nickel lloy tht, t equilibrium, conit of n phe of compoition 37 wt% Ni 63 wt% Cu, nd lo liquid phe of compoition 20 wt% Ni 80 wt% Cu? If o, wht will be the pproximte temperture of the lloy? If thi i not poible, explin why. [4] END OF EXAMINATION PAPER 6
PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Phyicl Contnt electron chrge e = 1.60 10 19 C electron m m e = 9.11 10 31 kg = 0.511 MeV c 2 proton m m p = 1.673 10 27 kg = 938.3 MeV c 2 neutron m m n = 1.675 10 27 kg = 939.6 MeV c 2 Plnck contnt h = 6.63 10 34 J Dirc contnt ( = h/2π) = 1.05 10 34 J Boltzmnn contnt k B = 1.38 10 23 J K 1 = 8.62 10 5 ev K 1 peed of light in free pce c = 299 792 458 m 1 3.00 10 8 m 1 permittivity of free pce ε 0 = 8.85 10 12 F m 1 permebility of free pce µ 0 = 4π 10 7 H m 1 Avogdro contnt N A = 6.02 10 23 mol 1 g contnt R = 8.314 J mol 1 K 1 idel g volume (STP) V 0 = 22.4 l mol 1 grvittionl contnt G = 6.67 10 11 N m 2 kg 2 Rydberg contnt R = 1.10 10 7 m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdiu 0 = 0.529 10 10 m Bohr mgneton µ B = 9.27 10 24 J T 1 fine tructure contnt α 1/137 Wien diplcement lw contnt b = 2.898 10 3 m K Stefn contnt σ = 5.67 10 8 W m 2 K 4 rdition denity contnt = 7.55 10 16 J m 3 K 4 m of the Sun M = 1.99 10 30 kg rdiu of the Sun R = 6.96 10 8 m luminoity of the Sun L = 3.85 10 26 W m of the Erth M = 6.0 10 24 kg rdiu of the Erth R = 6.4 10 6 m Converion Fctor 1 u (tomic m unit) = 1.66 10 27 kg = 931.5 MeV c 2 1 Å (ngtrom) = 10 10 m 1 tronomicl unit = 1.50 10 11 m 1 g (grvity) = 9.81 m 2 1 ev = 1.60 10 19 J 1 prec = 3.08 10 16 m 1 tmophere = 1.01 10 5 P 1 yer = 3.16 10 7
Polr Coordinte x = r co θ y = r in θ da = r dr dθ 2 = 1 ( r ) + 1r 2 r r r 2 θ 2 Sphericl Coordinte Clculu x = r in θ co φ y = r in θ in φ z = r co θ dv = r 2 in θ dr dθ dφ 2 = 1 ( r 2 ) + 1 r 2 r r r 2 in θ ( in θ ) + θ θ 1 r 2 in 2 θ 2 φ 2 f(x) f (x) f(x) f (x) x n nx n 1 tn x ec 2 x e x e x in ( ) 1 x ln x = log e x 1 x co 1 ( x in x co x tn ( 1 x co x in x inh ( ) 1 x coh x inh x coh ( ) 1 x inh x coh x tnh ( ) 1 x ) ) 1 2 x 2 1 2 x 2 2 +x 2 1 x 2 + 2 1 x 2 2 2 x 2 coec x coec x cot x uv u v + uv ec x ec x tn x u/v u v uv v 2 Definite Integrl 0 + + 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 Prt: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx
Serie Expnion (x ) Tylor erie: f(x) = f() + f () + 1! n Binomil expnion: (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 +, in x = x 3! 3! + x5 x2 nd co x = 1 5! 2! + x4 4! ln(1 + x) = log e (1 + x) = x x2 2 + x3 3 n Geometric erie: r k = 1 rn+1 1 r k=0 ( x < 1) Stirling formul: log e N! = N log e N N or ln N! = N ln N N Trigonometry in( ± b) = in co b ± co in b co( ± b) = co co b in in b tn ± tn b tn( ± b) = 1 tn tn b in 2 = 2 in co co 2 = co 2 in 2 = 2 co 2 1 = 1 2 in 2 in + in b = 2 in 1( + b) co 1 ( b) 2 2 in in b = 2 co 1( + b) in 1 ( b) 2 2 co + co b = 2 co 1( + b) co 1 ( b) 2 2 co co b = 2 in 1( + b) in 1 ( b) 2 2 e iθ = co θ + i in θ co θ = 1 ( e iθ + e iθ) 2 nd in θ = 1 ( e iθ e iθ) 2i coh θ = 1 ( e θ + e θ) 2 nd inh θ = 1 ( e θ e θ) 2 Sphericl geometry: in in A = in b in B = in c in C nd co = co b co c+in b in c co A
Vector Clculu 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