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Física Quântica - Solucionario Eisberg - Resnick, Notas de estudo de Física

Física Quântica

Tipologia: Notas de estudo

2013

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Baixe Física Quântica - Solucionario Eisberg - Resnick e outras Notas de estudo em PDF para Física, somente na Docsity! Solutions Supplement to Accompany | QUANTUM PHYSICS OF ATOMS, MOLECULES, SOLIDS, NUCLEI, AND PARTICLES Second Edition [Walxyalra) = Wolxa)Wstxa)] 1 A Robert Eisberg Robert Resnick Prepared by Edward Derringh This supplement contains solutions to most cf the more-invelved problems in the QUANTUM PHYSICS text; with one exception, solutions to problems ir the Appendices are not included, The supplement is directed toward instructors, and this has influenced the presentation. Not every algebraio step is exhibited, The units have not been displayed explicity in every equation. (SI units are adopted in the supplement, mainly because they are briefer than the text notation.) Rules with regard to significant figures have not been strictly observed, although there should bo no cutlandish Tt is a pleasure to thank Prof, Richard Shurtlo£f Institute cf Tecinology) for preparing the solutions to the problems in Chapter 18. Preparation of the supplement, including eholes of problema, was left to the undersigned, ho vas also his own typist and illustrator. Be would appreciato a note, of up to moderate asperity, from those who detect an error and/or mistake, December 24, 1984 Edward Derringh dl Montocmery Drive Plymouth, MA 02360 nn 2 2 P «férias = Ppfuldu = BA loç) Bi 1 8 vw = cf, = ESBEX IO. 5.4509 x 1014 mz, 5.50 x 107 BZ ip = me, =c = a “ 2 s.51x107 Therefore, Vap = My + v9) = 5.46 x 1034 may do = vo = 9.9 x 10h ne, Eince ent) = Ci ta 2 yr, mumerically; entmd/o? = 1.006 x 1933, hu = 4.37, EaE 3 = 78.04, ôriv) = (1.006 x 1072) (78.04) = 2,289 x 1978, The area of the hole is Au mm (5 x 10)? = 7.054 x 10º mê. Hence, finally, P = k(7.854 x 109) (2.598 x 109) (1.289 x 10 15) (9.9 x 2015, Pe 7,51 W. 15 0 L= amrêgrÊ = qui? x 208)205.67 x 1078) (570099, L=3.685 x 102 u, Le Gm? 2, 26 dn. L 3 de = 5 - dao = 8.094 x 10º kg/s. (3x 10) (bl The mass lost in one year is oem DEE a (4.094 x 10) (3.156 x 107) = 1.252 x 1077 xg. The desired fraction is, then, 3 £ = 8 2 1.292 x 101 A. = 6.5 = 10 2.0 x ra lo ta) The solar constant S Ls defined by L s=“E, am? EF = Earth-sun distance, Lan * rate of energy cutput of the sun. Let R = radius of the earth; the rate P at which energy impinges on the earth is L p= ES aê = mês. The average rate, per m, CÍ arrival Of energy at the earth's Surface is Ed = - ME rs, ER amR 238 Wmê É 401352 W/mê) = 338 wjm?. FE 338 = qu! = (5.67 x 10 Syrº, T = 280 X. 1-19 =E «mr É Rpfa) = É opta) SCE with x = he/AkT. At A = A o x = 4.965, by Problem 18. Thus, Brad = 42.4030 005 nte). Now find x such that Rola) = 0,2860400): mor aê Cr tr = (0.2)42.403m EE nto &-a nº 5 E = 4.203, -l xm= 1.882, x = 10.136. Mmericaliy, - B ie tEda 2 rm (1.38 x 102) (3) E 1 = 4.88 x 1072, so that À = 4.758 x 1073, 1.882 = 2.55 mm, à; = 4.798 x 102/10.136 = 0.473 mm. zo If x= Def ade then, by Problem 18, =. x x x rim RE: 1-21 Ey Problem DO, so that the wavelengths souçht must satisfy 5 ma = k-1707 ED, É a -1 qhey 4 Again let x = hefakr. In terms cf x, the preceding equation becomes É — = à, a Solutions are x = 2.736; a, = 8.050. Since, for À or x = 4.565, these solutions give Me 1.815 ut da = OGIA ru Let à" = 200 nm, A” = 400 mm; then, É A = (3.82) É A he/r"kT " Sep = a.saio?. 1 e - Mumerically, a 3, + = testa poa sso x 2) À asso, tá x 107) (1.38 x 10 9) so that 35587/T [5 = 5 SIA = (3.82) (4)” = 0.1194. Let x = 0588, enon, q = 0.194 = 1, x = 7.375; SBT 2 9.398, T = os = 18,000 E. CRAPTER THO Ei] The photoclectric equation is he = VA + WoA. With V, o” 1.85 V for À = 300 rm, and V5 = 0.82 V for à = 400 mm, he =8.891 x 108 + 3x 10upa 26 pe =s.255 x 102 +4 x 107%. Hence, 8.891 x 1026 4 3 x 1077u, = 5.255 x 1028 4 4 x 1074, (bi wo = 3.636 x 109 3 = 2,27 ev. Therefore, he = 8.891 x 1028 + (3 x 1077) (3.636 x 101), he = 19.799 x 10 É Jem, tab É 26 pe n= BB xD 6.604 x 10 35. 2.998 x tel “g" her 3.636 x 102? = 19.799 x 107287, dp = 5.445 x 107 m = 544.5 1 x In a magmetio field E = mu/0B. 2-33 The number of particles stopped/scattered betwom distances x and x+dx is dl(x) = clixpãx. Hence, for a very thick siab that ultimately etops/scatters all the incident partícles, the average distance a particle travels is x = SE - ea. E o se ay op o the limits cn all x integrais being x = 0 tox== CHAPTER THREE st É «pis = Ta Do ão ele Tux =31 cá - 03 x 10 0 ka 1.813 x 107! = a , a = 1.675 x 107 rg; wvidentiy, the particle is a nevtron. mm ta) E = plo + Ei (6 + E)? = pº 4 ER, er poda? + axegt o Da + ot, Bit E = evandE, = mc, so that 2 2teu) (mc) 1 ES 1 le UE, j= 2 Ji = ( = y = “Uma, and EE, = evfome?, Thezefore, 1 = = great da (5) Monrezsesviseio Limit: eu cemc?; set 2 + aU/2mpe? = 1 to get x = /(mçen 5 = /tomgo = ar B M 8 jo ML aço, te pe = vips, pagão imp ) ivfe) Numerical ly, =34 po = (6.626 x 107º pos) (2,996 x 20º m/0), (1.602 x 10 JfMev) (O * mir) he = 1,240 x 10? mem; bence, sa 3 . à trm) = ni a Modo, 219 ta) =34 p = Bo Eifé x ds (2990 x Jg no). 0,2400100, *oqUme « e 1,602 x 10 3/mev) Pepe sr, == (0.120)? + (0.51)? + E = 0.525 mv; E =E-E,=0.5258- 6.510 = 0.014 MV = 14.8 key, Cb p = Silo me Hd. E 126 Ke. These are garmma-rays, or hard x-rays. lc) The electron microscope is preferable: the quema-rays are difficult to focus, and shieldins would be required, 228 ta) set ax = 10 h 6.626 x 1078 p=ip-qlo- Lu e p = S228 x 17 pams 2.998 x MÓ m/s ( D.S868 ev e 1.602 x 101 pay 8 E = (p%e2+ Ed! = (10.566)? + (5112)% = 511.00095 keys K = E - Eg = 511.00085 - SI] = 0.95 ev. Atoric binding erergies are cn the order cÉ a few electron volts so that this result is consistent with finding electrons inside ators. (bj 4x = 10 m; hence, p = 9.868 Mev/c, from (a). pie? + ed) = (9.8682 + 0.512) = 9.B812 Mey; X=E- E, = 9.812 - 0.510 = 9.37 Me. This is approximately the average binding energy per nucleon, so electrons will tend to escape frem nuclei. (c) For a neutron or proton, p = 9.868 MeV/c, from (b). Using 938 MeV as a rest eneray, E = (pie end" (5.868? + 9362) = 0.052 Mev. = 538.052 Mev; E =E- E = 938.052 - 338 This last result is much less than the average binding energy per nucleon; thus the uncertainty principle is consistent with finding these particles confined inside nuclei. dz ta) Since p, > sp, and x 2 dx, for the smallest E use p= Ap, and x= bx tocbtain « EiiçãE 2 E FE ltP) + AOlax) With ” tp dx = = the minimum energy becomes 2 1,h,2 2 2 E = etquio + SClém) so + ac(um?. 32mºmlhx) (b) Set the derivative equal to zero; 2 E A + 2 k SDS * O Tendo au) CO nO + (0 = qto Substituting this into the expression for E above gives Fato TOS = time 3=34 la) Let the crack be of sero width and 4x = horizontal aíming error (i.e., drop point not exactly above crack). This implies am inítial horizontal speed v, given by Bence, the total horizontal distance X from crack to impact point is x=mx+xe dx + ÉS L Mm 9º mw To minimize, set Jan = ES BA 0h) ZEH = 10m B/9= 1, m = 0.001 kg, then x = (1023035 = 3 x 1916 m, aperesdimate ly. Therefore, 17 = 3 Put an electron behind each - Re siit and cbserve any recoil Ant due to its collísion with a d Ay << qe Doe to the collísicn, the photon's mementum changes. Tn order not to destroy the interference pattem, [o Ren e, «Pol, m the order cÉ the fringe. By conservation cf mmentum, this is aiso the uncertainty in the electron's momentim. Hence, for the electron, it is required that, in order not to destroy the pattern, (tvi (ap) << tê a = z 4-18 The periods of revolution of electron and proton are equal: 2rr 2 4. The motior is about the center cf mass 0F the electror-proton system, so that e mt * Ee Therefore, “pr ves Ge vie É sir rod m/s. “2 (a) Frequency of the fixst Lines w Thy = Rg - pop . Frequency of the series limit: v = en = est - 0). m Therefore, eu uy" ria 6) 2 My e, /0+1) = =. pe agf 51)* + ta) Euu2 TA, = TE = 26.6 =; Ê 5 E = 26.6 - 10.2 = 16.4 ev. o) Emi * 13.6 + 16.4 = 30.0 ev. + (a) By mementum conservation, Dam. c umbinins this with energy conservation gives 2 AB e gm + MZ ay + AM? a ta 4 DT Uh) hua) ——.. Atol (AE) by = ty =hy a * om Tomé vo = va + dim, ve vt Ea. tb) Since% = c/h vg = C/Agy nego dia. 1.4 E = (13.65 - =) = 12.089 ev. Po Naglacting recoil: do "Eos = 102.6 mm. 24 With reccil: -18 at. 2009) 062x108) ox 10?, do me (a q.e73 x 1027) (2.998 x 109)? A = 661 mm. 4-4 The kinetic energy of the electron is K = (0.511 Mov) ds - D. 2 12 x 197 2.998 x 10º = 0.0400, 2 2 1.622 (3.6 Mag 2 A AR 544 av. Hence, En = 54.4 + 409.3 = 463,7 ev, à m 280 = 2.674 mM, 4-35 (a) Eiydrogen Ba: ah “Rã 3. Helim, 2 = 2: A) ad “ROS mp If Ag Agr them Zi . = . nç=4 3engdo o n=6. 4h) Now take into acoxemt the reduced mass us 2.4 24 1,2 MyfDie 1,2 Mel) e mo - J i = à = (Rd. “a “meo amo Me “o amiite Pg mm m md) F dy ço O Mel qa Pe "TR" Fell TO Therefore, Mge > gr so that ds Ee == ada A da Hence, compared to the hydrogen Ho line, the helium 6+4 line savelength is a littie shorter. (b) Since à « pb (the factor 2? 45 combined with 1/r7 = L/nê to alive equal values for E and Be), AA = (4.084 x 1079) (656.3 ne) = 0.268 mem. 442 The momentum associated with the angle 6 is L = Iy. The total emergy E is 2 E=Kegul =. L is indopendent of & for a freely rotating cbject. Hence, by 26 the Eilson-Somerfelá rule, É LO = nh, LS d0 = L(23) = (28) (2n) = nbs UE) = nh, E = RÉ a CHAPTER FIVE va ln) The time-deperdent part of the wavefunction is etitA Cm) À ABA (ritmo Timrefore, 16% = 2m * vo! hj Since E = ho = Zehho, E= ma". (1) The Limiting x can be fomd frm ta According to Example 5-6, the normalizing integral is 4 3 ã (ZE/C) 2mb dx 22 Atl x 1-2" E 2 o E sin FE LRE/O - 2 (2E/c) I, 1a sêmS + pêo tcmid, ta Problem 5-3tc) provides the limits on x; the wavefunction is 1/8 2 7 = tom SA ot tri) 27 32 E Cm), | 2? 5 - tm dio E ax = HE) Ee du, e É Temo! T= te, (b) This seme relation, U = T= 4E, is obeyed by the classical oscillator also. 5-15 Vse the notation be), = pede te cane E am Rs da lz =" Clearly ta bp); = meo de + Tdx= (mply — dk, implying that (xp) and (xp), cannot both be real. Also, by integrating by e mo +. bel; - ate [o Eae) = ap TE ax. Thus, tel, = pdf. TE (xp), is real, this last relation says that Ge), 15 real also, which contradicts the first Finding above. Bence Up), às complex and therefore so is Ge). Moe try == =. 2,2 , =» sobre dç + Bor eu — = = Hb), + teplsh = Sit), + Obg, *p = Re(xp); so that this new Xp 1s real, as desired. al With V = 0, the energy cf the photon is E= po. Replacing the energy E anê momentum p by their operators gives Mm ira dio e Nou set Fix,t) = plbdTIt) and divide the equation by 7 to get niZ-elã-r where K is independent of x and t. Write E = kjc and the tuo equations directly above become EE = suor + qe qt, É = my + peelx Hence, for the photon, Ta ele te-et) E 22 (a), (b) The curvature of 4 15 proporticnal to |V - E]: where fu = El is 1arge the function cecillates rapidly in x, and where |v - E| às small it oscillates less rapidiy (hence, nodes are close together in the former case, farther apart in the latter). In the first state, |V - E| is just large encush to tum 4 over: no nodes. The 10th state will have 10-1 = 9 nodes, Losing to an odd function since V 1s symmetrical about the origin. The wavefunction decays exponential ly wherever VE, the classicalIy forbidden region. For further discussion, see Eemple 5-12, which treats the símilar símpie harmente cacillator potential. 35 tc) Classícaliy, the probability density function P is given by P=BêN, pê che nomalization constant. Energy conservaticn gives vi E = iê tc, the upper sign for 10. Using this, PPBio). To determine B, use the nomalization condition (remo afeto sê =1, mat e veDorE=V= hx. Evalusting the integrais cives P-dEZEIS: Particular values are: PIO) = C/4E; PE/C) = P(-E/C) = =; Soo) = tc2/er?, td) The graph of the classical density function resembles that fer the símple harmnic oscillator, the lack of a horizontal tangent at the origin being the main difference cn a rough sketch. +P due co ca a ns e a 2 0 e ai 36 5-24 See Problem 5-25. 5=25 With no burp, the wavefuncrion will be sinusoldal inside the elassical region of motion and a decaying exponential outside, The lowest energy wavefunction will contain no nodes. In the present situation, in the region of the bp the curvature of the wavefunction will be less than outsido bump, since the curvature-is proporticnal to E - V. This will upset the good behavior of tha mavefunction at larse x associated with the valve E, corresponding to the first bound state without the bump. To Tcorpensate for this reduced curvature in the region of the bamp, a larger curvature (as compared with the no-bump case) is needed outside the bamp. Here V = O so that the curvature is proporticnal to E. Hence, a larger E is required: that is, the first elgenvalue with bump is greater than the first eigenvalue without the Dump. 5=25 By essumption, EE + Uma, Vi = burp potential enerar, Y = wavefunction with no bap in potential. The integral is the area under & curve cf PT vs E Now W = O except mhere it 15 equal to Vo/10, Clearly the area will be larger 1f tho burp is located where Y*Y 1s relatively laroe (i.e., in the center for Ty) than 1£ the bp is placed vhere y*y is small, 1.6., at the edge in this case. Evidently then, E, 1s larger for the centered bump. ..... ado e eme DT, al E | pe + ny x S=27 Schrodinger's equation is E Be-vy-o. In the region in question, V = V5 = constant, E < V,, so that ?=-Bm->0. Hence, v= 2%, pe, e the general solution, However, pls) = 0, requiring E = 0, vm no, as the wavefunction. Em | Bince | is real, the probability density P is Pevipo qê = Ae DE, Recalling that x is measured from the center of the binding region, the suggested criterion for D gives neah) 1 12 Ba (ta) EP Pao b «ge, 2 aa, - E) sa Use the scheme suggested in Problem 5-26: E= E +JriVIda, ma elgenvalue and eigenfunction of the lowest energy of the infinite, flat, square well potential. From cegiaa 3 (a) Assumina a wave incident from the left: region Li = AI, poli, 4 = (2mE) Ay region 2 p= ME, gil = (mtv, - eia: region 3: qa coltXs po ÍEiX put De O since thero exists enly a wave moving to the richt in this region. Contânvity of the wavefunction at x = 0 and x = a require that: A+B=P+G, “o reta + quê = qolhia, (11) Contiruity cf dy/êx at these seme points yields IA -IkD= RP +k6, (us) ka o] ik,a =kype + cd = 1k Coffia, tv) tb) From (1), A+B=Gm=F From (11), ta+B-ge Fi, gola. qolhia, peted , potua , c(eta = a, = qelhia, tiia) Prom (141), Aik, - Elk, = «KA +B-G) + kG, Alik, + k,) + Bl, — dk,) = 26k,. titia) From (iv), ge RIA 4 BG) + a a tk colia, 43 -Aetaa - epa 4 Gk teta ses a csxpoltio, (iva) Now work with (is), (iiia), (iva). From (lida), =. - [e Fe 1A Oo + 1kp) + BO, = 4x,)). Subgtituting this into (Lia) gives AL + UA + (4 = ax, fe ba) tab) + Bits - dx, /2ç e + (ã + dk 2 Je ME) a colhia, and (iva) becomes A(O + Dm jo dy — 4/3 poi (ve) x + Bilk = 1x,/2k,pebeo + ik 42 ,je ia - e ph ela, Solve for E in (iib) and substitute into (ivb); i£ q, qº are defined by q=1+iky Gel, the result may be written + espiao pude qt +qe k + Siqufza — que ia, = 10 qholkia. k How solve for C/A, using the definitions of q, q*, such as = tê = atos etc., to obtain aiik ) grikia “is aa Im “4 Hence, the transmission coefficient is wc ate asi fee lia - - , VMA quêo as - qressa qua 4 grhsa! T T=160k,/%,) Zig rara 2 hem? qq) qu, T = 160 DI + ESA E? 4 1606, 2, 2,22 (+ K/e) qa 14 CIDA as taça, 16 (ky/ko) Finally, v 2 2 md õ 5 “me Dao so that vêrtv, - E? km qts? T= LATE Ee Vr a ksa 2 T=(1+ eis = é er sda D o o E-5 , À Z£ ka >> 1, then €42 »» e *28 and the transmission coefficien becomes, under these circumstances, Za Now O < E/v, < 1 and therefore E És sEn-Bica Yo the upper limit cocuring at E/4, = 1/2. Hence, 1€ 248 > 4, aZra 160 -É) o “a > 1. Since, in fact, it is assumed that ka >>» 1, Desa >> 1, 16-23 o o and therefore, under these conditions, E E, clksa T=16qll-ge a Wo“ s FRPTE A E % regioes à regioes 2 region 3 e m——| emma mem mol — E = Region 1: q = palbiX , po bx region 2: v = FoloX, co Dux, region 3: v= celkix In these equations, = es - e k = (EIA, k = (mê Yo) e (a) Continaity of the wavefunction at x = 0 and x = a qives A+BRF+G, polkia , co kia « golkia Continuity 0£ db/dx at x = 0 and x = a gives 1k,A - 1x,B = IP - 16, Fix eloa as 1 gertksa ai ix yoeltia, (b) These are the ssme as the corresponding expressions in Probjem 6-5, if in the latter ky 15 replaced with -iks. Making this alteration in the expression for T in Problem 6-5 yíclds for the nes transmission coefficient, 2 2 no = ind E Eno 163) doa qa), 16(k,/k9) Using the expressions for k,, k, given above reduces this to A aiksa,2 0 = ade » P 1 EtÉ - Vo Yo &s (a) The opacity of a barrier is proportional to 2w,a2/Hº and therefore the lower masa particle (proton) has the higher probability cf getting thzcugh. Cb) with vg = 10 Mev, E = 3 Mey, a 0a, Lt folic that E 16 Veil = o) = 3.36. The required massos are m = 1.67) x 1027 xg, m, « 29. For the proton ka = 5.603 and, using the approximate forma, = 3.360 205.800 3,06 x 10º. Since mg Mp, as noted above, KA = 215.803) = 8.207. Bence, for the deuteron, Tg = 3.360 28207) 2 2,5 x 2077, s =19,2 (a) we B- (9x 10º) (Usado o 0 ter 2» 107 =13 vo = Sela gls (432 me. 1.6 x 10 3 (b) E=10kT=(10)(1.38x 10200”) = 1.38 x 193 - 8.625 x 1072 Mey = 0.024. tc) Mumerically, a=2Zr" -r'=2x 135 m; also, “to En) 16 at = o - 0.032; ka = y a = 0.91. Te(l+ Sasha 2, cr D.0073. td) The actual barrier can be V considered as a series cf barriers, coach of constant height but the heights decreasina with r; hence Vo-E diminishes with r and the probability of Im penetration is greater than for E ah equal width barrier cf o constant height Vos eus E . d—- - . 1 mal oj ngnã DO 52 In terms of C, the transmitted wave amplitude, the solutions are 5 asl=z c, 6 selicão,. p= da, 5 c=êle. The desired transmission probability is VC" yomc E. o -2E€ VaRPA x DAMA RA e ec Eat = RE AA. 10 - 2º! 10 - =! qo - =**) (10 — 29) Bow, at; = qikaçãa qo = =*ºy (10 = 2º) = 100 - Io(e la 4 qSlka, 41, (10 = 2*É) (10 = 2Ê) = 101 = Z0cosska. Tr —— BR = ZOcostka” &-20 ta) In the lowest energy state n = 1, y has no nodes, Bence &y must correspond to n=2, by ton=3. Since E cn dE = RA Ery/Er = 3222; Egy = 9 8% lb) Ey the same analysis, 2/8 = 2/2, m=1ev. | E Ee (o) The energy in question Ls mn EE, and therefore the eneroy of the adjscent level is Erin + nº dE, E -E 2 2 Sn mtupon $a sk, Es nm mn (b) Tn the classical limit n + e; but un gl = tin B$do o, es En = n meaning that the energy leveis get so close together as to be Andistinguishable. Hence, quantum effects are not arparent. ta The eigenfunctiens fer odd n are W = Bo cos lnme/a) . Fer nocralization, fz “1 = fax = Eljcos (nme/ajáx = tm [a du, aa o 1 = 282 (a/ne) (nn/6) = 5 E Es for all odd n and, therefore, for n = 3. À 54 cs By virtue cf Problem 6-24, the normalized eigenfuncticns are tg = “É costrex/a). = À x cos? trurxfa) dx =0. t E = Êlccs inuox/a) (1H Êrcce (nr /a) Je - Zoom elru du = 0 ja em (8) x2(n=2) = 0.07782, This changes little with n since the size of the box is fixed, ter - 2 p= fumo (124º Ecos (rms) Jó, Lia ax krm p = - Pta Sa = 222, = Pê) = as.ga0h)2, 55 This incrsases sharpiy with n since E = nº, E « pê, the particle moving faster at higher energies. &-26 (a) Usina the results of the previous problem, 2 h ax = vê = Aa -—+p nm == nrtb. Hence, for n = 3, exp = sta - Ep ar) = 2.624. E (b) The other results are n=L dep = 05H, n=2, bépe 6H. The increase with n is due mainly to the uncertainty in p: ses Problem 625. (c) From (a), the limits as n + — are ax + da hp + e. [im 1 am ha ta [no = drmenêgia = fem - cos (ÊB) ja, — 4a = (hs = Fel (coszu - cosujdu = 0, — = the integrand being an even function ci us 56 2 The Schrodinger equation in three rectanquiar coordinates is De Bay. Me-weco. Inside the cubical region sbere V = 0, yo, mo, ax? avê así w wlxyrz) = KlxbYiy)Ztad. Then, if * denotes the derivative cf a function with respect to its independent variable, CT. E o aenê-o. e This gives E = aê, X = Asin(k 0 + Boos(k 2d, ke = real constant. Sinilarly * = Csin(ky) +Dmos(kyl: & = Esin(k 2) + Foos(k 2). Also, from (4), KE + n2 +12 = men, Since V = « outside the cubical region, p = O at the bondary: D=X(0) =Y(0) = 2(0) + B=D=P=0; 0 =X) = Ya) =zla) + Kan kann ka mn with Meg, = 12,3, ee» « Henoe v= (CAB) sin (nax/2) inn xy/a) sin in vz /a) , És e - Ev? no) Eloi ong erd. a (a) Let M = mass of wing. The zero-point eneroy is By = (0 + Mu = hhu = h/27, T = pericã of oscillation. The actual energy of oscillation is E = aê = us dAÊ = De MAPS. Thus, the value of Mat which E = E, is ht 16.626 x 1074 Ne SEo = “ma dm (10) This is less than the mass cf an electron. Hence E >» E, and tha cbserved vibration is not the zero-point motion. (b) Clearly then, n >> 1 and therefore E=nhv= 2 Mr +» n= rm. As an exemple, take M = 2000 kg: = 1.68 x 109 pg. Therefore, a - - E = 501.055 x 168) — ata .6 x 10181, 4.1 x 10 E = 0.051 ev. 7, ao EE od E 1) “ao Zap (ame) 2), For the limits on 1,0,4 ses Problem 7-8. Now, fora =6, - o? 210 “Tea > For the states withm, = lr af (are EE 1 z" ZE,» tora = E Paço rata mA, and therefore - 2 “ag —— + 5 e T/RacirPpacdado, Sina, (img) ag 7 am. ato + seas This is the same as (*) above, Hence, regardless of the value of Mr Va = E (b) In the case of L = 0, Sooo artes 3020 - efagpet/do, giving = poe À reg soft (2 — /ag)fe "dart sangandaco, s º 2 2 2 domo» ee” dx = — Tanlêneas (2) = 2,» (c) These results are expoctod since, with V « 1), the average potential energy seen by the electron 1s the same for all n=2 states, regardiess Of à. Thus, the expectation value Of an energy will be the same for these states. LI R(r) must satisfy E9.7-17: +2E. Br -vr=2a + DÊ. É Ê Apr maR=, E ada, ak a EM pr, Substituting these into the radial equation gives ais np? der! -vty Ze + prl2, Now E is a constant independent of r, and V = k/r; thus the to tems in () are proportional to 2º, xº2, As r approaches zero, dE *, rel, hence, () + 0, and the equation is satisfied. 7=11 (a) To avoid infinities, integrate radially to à finite limit R: a = 41 - cose), “quad de “ qued Ea P(Z3.5%) = 4. 1474. db) Por this state: as =5/2 =r/2a Vono = a7tsão o Pre É coss, Since this is already normalized, p= sêar te "en, o P = Mt = costy), PM23.5*) = 11.444. 112 (a) See sketch, following page. () P = (J0s*6-D? 8 = 54.7º, 125.3º. te) 3 henoe Po = O at 0088 = 11//3, giving Pa 0 (0 = 0"), Ca * 1, (3oost6 =D? = 1, 3c0628 = 1 =b, 6=35.3", 90º, 144.7". ru Let (3,2,-1) represent y(n=3,L=2,mp=-1); (2,0,0)* represent Wetn=2,1=0,m,-0) etc. Then it is required to ahos that Ult, = 5(13,0,0)*(3,0,0) + (3,1,0)*(3,1,0) + (3,1,=1)*(3,1,=1) + (3,1,1)*(3,1,1) + (3,2,0)*(3,2,0) + (3,2,0*(3,2,1) + (3,2,-1)8(3,2,-1) + (3,2,2)*(3,2,2) + (3,2,=2)2(3,2,-2)) is independent cÉ 6,4. Now (3,2,-20)*(3,2,-2) = (3,2,2)*(3,2,2), (3,20 "(3,2,-D = (3,2,0º(3,2,0, (3, L=1)*(3,L,=1) = (3,1, 0*(3,1,0, 66 and therefore Vs = 843,0,0)*(3,0,0) + (3,1,0)4(3,1,0) + 2(3,1,1)*(3,1,1) + 13,2,01*(3,2,0) + 203,2, 1*(3,2,1) + 203,2,2)903,2,2)). Nos substitute the specific expressions for the various wavefunctions appearing in the above. L=2 terms: 2((3,2, 0*(3,2,1) + (3,2,2)9(3,2,2)) = —s ut 28/280 (1 + 2co02e - 30060) 218) nas, (3,2,0)*(3,2,0) ai pte2r/das a0osdo - 12. Hence, the sum cf these terma is Z n8erdr/3as Za e 3480) mag, Independent of 6,4j. E=1tems: 213,1, 0*(3,1,1) + (3,1,0)*(3,1,0) = 2 E2, 1,2. -2r/3as as No independent of 6,0. The L = O terms depend on r only. Thus, all terms in V5b, have been acocunted for and their sum found to be independent of direction, so that VãU, 1s spherically syemetric since it depends om Fr only. J-16 ta) Leop = Mlsing ds + coticoss da, tores = Te Pre Eai dO, Ma au Jo = target; ooo = hay a- Therefore, Lesopbaj- * Nlicotosing + cotêcosd) by...» Le cpa Hootolcose + isinpdy,,, = Hootõe!ty, (bl This result cannot be put into the form Eesoptai- = Mano with C independent of r,6,4. Eu The cperator is question is given by 2 1 2. sinta aq? 2 me 2. 2, 12, = tod Esteio do + By Bq.7-13 this may be written 2, = qo - du Do), But E E = og E = ouro + DR- êste - Re), E? Mb =1G + Dy- ue -wrk, by Bg.7-17.- Echrodinger's equation is vê = 2lty - ey. ré Substituting these last tvo results into the expression for LÊ gives e Eat =uL+ no. Ez (a) Let the arca of the cllipse be A; then, mriA-TA T the pericã cf revolution. The angular mementum is L = nx/d8/0ty also, da = tras, so that Lema mi, since dA/dt equals a constant in classical mechanies 1f the force is central. Therefore, L.d le nsid ds (b) This result is identical to Bj.8-5, derived assming a circular orbit. Ba º Em The first apparatus produces two bears, one with spin parallel (in quantum mechanical teres) to the direction of the field (42), the other with spin antiparallei. This latter besm is blocked by the first diaphragm. Hence, a “polarized" besm of atoms enter the second apraratos, field direction +2!. This second ragnet produces a new space quantization along 7". In analogy with the passing of polarized light throuch polaroid (except that the angle for no transmissicn is 90* in the optical case, 180º in the atomic), the second magnet allows only the projection of the entering spins along +2' (not -z"') to pass. Thus, 3f I' is the intensity of the bem entering the second apparatus ani 1 the intensity of the unpolarized bes entering the first, 1=SI'(1 + com) = KIA(1 + cosa). EF! 3 25 The deflectina force is F = u,88,/d2, here Ve 7 Sygr gince £ = O. If D is the deflection and F is constant, 2 2 D = hatí = 40F/m) (Lv) L = length cf maqet and v = speed of the atoms. Thus, 2 dB 2 z D= dl fm) (dB, /dz) My: ge” E meg eg For atoms enitted from the oven, mw? = 2K7 with T = 1233 K. Herce, -23 E. (AD ( 8038 x 1) (1203) 19-0005) . 25 jm. = (0.5)2(2) 19.27 x 102% (4) gts Bs (a) The orbit and spin eneraies are (aumpB and taum. Hence, with respect to B = O, E= SpumçE + ER = + Aa -m+ my E. (bj Fora = 2, E 0,1 qiving the result: k m Tg o o DE (units oÊ mB) -1 o + 74 Thus the energy levei dingram appears as follows: tuo-fold deg. E=0Q a 1 asi mo à NE tuo-fold deg. | -1 twofold deg. lc) The maximum separation is rag TMB = 10.2 eU, a(9.27 x 102% = (10,2) (1.6 x 10715), B=4.4x 10º, = Since 4,) > 0 and & = & the relation “30 + uhz IML +) - ate + IU+D ame += A+ DD. ti) E = 0. In this case (A) reduces to j6+0 > 3/4. But for 2 =0,5=5 (the cnly possibility), so that the relation in question is satisfied, 75 (14) RO. (a) $=2 + 4. Potting this into (A) aives L > [3(t + 0), which clearly is satisfied for all 4 > O, bl j=L-An,n=1, 3,5, Putting this into (a) yielês -Mm=-3 > (6 +). 18) o<(m-DWE+D, which is satisfied for t > 4 (i,e., j> 0), so the relation is obeyed here also. n=3 m this event, (B) gives qe Ii +. Evidently this is not satisfied for £ > &, E < O, but is for D £z <&Botthenj=l-3/2<0, whichis impossíble. a =5 (B) now reduces to o<-zmmem-s. 5 satisfied for some £, e.g. L = 1, bot then j = 8 - 5/2 In fact, put 2 =) + 5/2 to que 0 < 2232 - 77) - 64, a which does not hold for j > 0. Results similar to the last apply to n = 7,9, .«.. eto. Bence, since j > O the inequality is restricted to the values cf 3 given in the problem. 8-19 4 1 (a) Largest j =4+4 = 9/2: 8 largest. = 9/2. The 1 mag! of the vectors are: N 3= 7/0 + DM = (W/2, SE L= LU + A = (20)K, + = Pl E E = vista + 1))JK = 3N/2. o 76 Apply the law of cosines to the L,5,J triangle: 32 = LÊ + 8? - 218008 (180º - 0), 32 = 20 + 24 2/(2008) Joosô, 6 = cos tie) = S8.81º. ft) Sinos 1, às antiparaliel to À and bg is antiparaliel to 8, the angie betueen foi, = S8.91º. lei 3, op == 6 = 25.24". Eu Define the relativistic energy as Eq KM een enftri e É = v/c. The relativistic momento p is p=zv=mesti - 65, Bow To express K in termas of p note that = “a -s0-89%, n-pãl=1+p'ê, so that E= meta +pBi- ne mea + mp. E + ce.) = 1h, 2 4 Ke mo (ap? = dor) E ne 7 Hence, 2 4 Egck+Vve EE su, rel Es Bmje? and therefore 2 Z 4 P ey “Eres Ea o Eta t+. Po emo = If p= Er then 4432 Ep =P /êmçe Now using classical expressions, ín the spirit of the apprecimation, ca Esv=E pl = imite - 2, so that E - - Emp, With E = constant and V = e2/4re,, the above yields the final quoted remalt directly. 15 (a) The integrals to examine are tupteBução Seg leBivçãr. Since both w; dp are single electron eigenfunctions, each has the form (n,t,m,). Hence each be written Yntmg em] integral may eftm Lama) * (E) (nam dr. Nou thê parity Of É is odd: P(É) = «E; the parity cf (nst,m,) 18 (=1)º so that the parity of tnetamp tn temp) às (092 and therefore is even regardiess of whether L is odd o even. Thus the parity cÉ the integrand above is odd, and the integral over all space vanishes, (b) Electric dipole moments constant ir time do not exist, since the governing integral above is zero. Only integrals Bz again since the integrand is odd, (iii) Transitions between an odd and an ever-n levels Pe [esneesmama = 2)usin (mu) cos (mu) du, ka : p « Açfioin(wo) , sinbntnem), pyoosin(rm) , costu(ntm),, (rem) Crema) evaluating the final integral. Cleariy the above is not zero for n even, m odd, Bence, transítions are permitted only betasen levels such that cre is cf odd n, the other even n; i.e., the selection rule is me, 33, À5, ces a B-20 The eigenfunctions are 0) = age! m=0,1, 2 e. Pg = ren. Put Roosg = =,, Ring = =. Dropping the absolute value signs, L- sf copio = Efe sero tom, tee Efeito tia M=m, Mp since cos = elf + el). Nom 21 2m Zz [ee = |costngjdo + ferem = Msinzm - icos2m + 1). o Fence, ifn=1, 2, 3 «e [E =0. On the other hand, 1f n = O, then Zu Zn its «o = Thus, 1, É O only if either (1) Mm +1=0, cr (ii) m=1=0: that Ls, only df êm = É. Since sinp = que - ty, ea Aom+do — Albm-l4 ,-E AMO ingão = efe e Jão. The integrand is similar to the one for 1. Therefore, the selection rule is im = 21. f-2 By Eq. 8=43, Bo o Poa vao But F,, depends only on âm; since gm =m, - mg = +1 for both transitions, Po" mg Hence, E] The probability densities ape L I II vEbs = lug tLvg (2)v vç (2) + VE VÊ vg (DV 62) AA III Iv = va CLDAA (Ze (OD eg (2) + 45 (Dea ya (Do, (2). Making the switch interchanging 1 and 2, temes 1 ad Iv interchançe, as do II and III. Thus, uso = Mr tir É + Tv) + HIV É II É TED) = dio AA AA 9-4 From Example 9-2: Va e Arlo DM (4,03) + tg Me (DO) + 4 (DE (209 (0) = 4 (vg (2403) = Vig (Dy, (Dp 63) — Va DA (2 (DD As an example, interchange particles 1 and 3 to qet 1 "a = Tt vg (203) + Va UM (DD vg 3) + Va Dag te 0) - Va (Uva (3) - V Dvd (3) - Vig DD y 024 0, ne The same result is achieved if, instead, particles 1 amd 2, cr 2 and 3 are interchanged, B5 96 The antisymmetric function for three particles is 1 . tw FaTtta vg (20, 03 + Va (Dy 12D (03 + + lh (2) vg 13) - 4 (Uva (vç (3) - “a Ca (2) 63) a a LDA (hp 63 Upon formina Jiabçdr there appears the following terms: (1) Six terms, each the square cf those above; for exmple, SOU Vg E CI (DM (24 arg, = UVA CM, (Ddr, HW (2) vg (2) des) UN (DO, (BD), = QHIMI) = 1, assumina that each save function 1s normalized. Hence, these terms add to 6. (ii) Cross tezms; for exmple, DG g a Bm 4 Cy tm ta 3 ddr, = Va OD (Ddr, HUVg (2a (Dr, MUNDO (DD = (o)tuo) = 0, doe to the orthogonal ty of the eigenfunctiona. All of these cross terms vanish. Tras, the total integral equals 6 and, since 31 = 6 also, “a is normalized as originally written. = Py asmumptico ta tepacetenin! Look first at the symetric space functions; since both of the electrons are In the se (gromd) state, 8 1 “pace = tro Uyog 1) + go! ng th Ea ao 2a, va 2 33/2Rra/asy, Y, must be antisysmetric since the space function was chosen to be symmetrio [electrons in the same level). The coulcrb energy is pias, Ameg Fiz vhece r;, é rj — 7 is the distance beiweun the electrons. Thus V = Wgmindr dação, in which 9,0, are the spin variables. Now V is independent of the spin of the electrons, so that 1f the spin wave funcrion is normalized, then V= WicaceVspace Hp Putting in the wave function qives is escoa veda, riar,sing sino, 108,84 )84,+ "5% in heh E5 = 85000290 0ye8gehyedo)- Now suppose that the antisyemetric space function had been chosen. With both electrons in the ground state, this will be 1 - Vepace = 75!Vo0 Méygg 12) = Pyoo!2 og UU) = 0. Tt may be concluded then that with both electrons in the ground state, the electron spin must be ín the antisymmetric (singlet) etate, coulcmb interaction, being positive, will increase the ground state energy over that calculated ky imering it. 2 (c) If the electron is bond, vir) < O. Clearly 12/20? >0 (recall that L is a constant for central forces). For small encugh Fr. 12/om? >> |vitr)| and v' > 0, indicating a repulsive core in the cne-dimensicnal fommalism, Oy df va r”,n>2 will this core disappear (unless L= 0. Iê (a) The potential in question is V'r vw va L/amê, How, in electron volta, E, rp É tem sold, drag (r/ag) Clearly, for E = 0, V"' = VW; ses Problem 9-3, For L = 1, Vo) ay s3 =ilike = + =v+ = =(27.2, tr/a)? ca] tb) E, = 220 ev. ic) The classical region ef motion is shown on the sketch: it falis within the range for which PY tr), Fig.9-10, is large = (d) For L = O see Fig.9-13. The classically permitted region there falls within r=0.Za, a bit smaller than for £ = 1. This result also corresponds roughly to Fig.9-10, where Po is large at r « 0.54. There ís qualitative agreement betueer classical and quantum results. + (27.2) (rag)? E (a) From Fig.9-15, the lonization energy for the first electror is ZM ev. In the ground state the energy of the , from Fig.9-6, is -78 eV, Thus the enerqy after the First electron is removed is -78 + 24 = -54 ev. The eneroy with both electrons removed is zero; thus the enercy needed to remove the remaining electron is 54 ev. lb) With the first electron gore, the helium atom resembles a hydroqen atom with Z = 2. For such an atom the qround state energy will be = 22(E,p) = 22(-13.6) = -54.4 ev, & 1H and therefore 54.4 ev are required to ionize it. Agreement with ta) is excellent. os o grelalad, eo 0 SRH qe L ——— A — E<o "Ne misioa 1 hola in hshell k + ph hai Kel 4 (c) In the x-ray diagram the levels are inverted, the ground state is taken as Zero energy, and the energy is plotted on a logaritimic scale. IE Es (6) When the hole is in an inner shell the energy differences for likely transitions are large; thus the x-ray diagram, plotting logE, is easier to handle than the standard diagree. (e) Mhen the hole is in an outer shell, the transitions are more likely to be optical, and tho associatod energy differences are small. Hence, the standard diagram is adequate, 9-24 The photom energies are E=HE, Equev) = éto, Use of the last expression gives the following: K, Go): A =0.020mm, E = 59.0 key; Ko ObR): À =0.0184 1, E = 67.4 ke; K MK): A = 0.0179 mm, E = 69.3 key. For the absorption edge, E = 1.2400/0.0178 = 69.7 key = energy needed to ionize the atom by removing an electron from tha K shell. Hence, E = O (ground state) + 69.7 = 68.7 ke = ensray cf the atem vith a hole in the E shell. Then E = 69.7 - 59.0 = 10,7 key; similarly, E, = 69.7 - 67.4 = 2.3 key and E = 69.7 - 69,3 = D.d key, 95 617 k E (Mu) m7 tr E e, a N G=25 ta) The line is emitted abén a hole jurps from then = 2 to anm = 3 level. The eneray required for this is approximately the energy needed to ionize the atom by removins an n= 2 electron. Using the cne-electron fórmula with =: - 10, z 5 - 203.6) - E O,2(33,6) = -870 ev. Thus the required voltage is about AJ0 V. 4b) The wavelength is cbtained from em =E,- E, 2 -E, = 870 ou; À =L4mm. 39-26 (a) The empirical formula is vi=cm-a?;xt=cêa- a). Trus a plot of XY us. £ Is a straight line with a Z-intercept GE 8 = à emAid slope OEM. Poa Bia-ó-1h, the Z-intercept = a = 1.7. Also slope = A «1002093; C=8,6x 19é mi, (6) Fora: a = 1,7 Ca Rm 1x 108 ml, 2 la) The E absorption edge [n = 1) should be aiven by Erê * 3.6 ev (3, mn), edoe cotatts By = (13.6 ELA? = 6.5 key; zon: Ey= (13.6) EB? 7.83 ev. For the inner electrons, the wave functions are essentialLly tydrocente, with an appropriate effective 3. Por the E, line, domo tino = Fi te/ag) Reel, with the selection rule & = £l. The matrix element is 4 De = ietuçoel = 557 à Since É = rsingcoegl + rsinesing] + roostk, the Z-dependence folloss from nte 32R/ 2a ar = 2º — Ls, (82/29) The lifetime becomes ” T-]= A » de £ VePEL,Pb By Exemple 9-8, mo = By = Cage Therefore, 7 Fb 1.41 = Tg larg) =24M x10,, Tas 10º s + Ta * 2.44 x 10716 5. 10z 10-7 For the configuration 4834, £, = 0, 2, =2,8,=8, =. Hence, &'=Zonly, s'=LO. With&'=2,s!=1,3!'=3,2,1qivino 2,1 levels. For &' = 2, 8'=0, )' = 2 only, resultina in a lo, estate, By the Lande interval rule, the separation is do, -?o3/00, - o) = 3/2. The eneray shifts themselves are SE = R(ING! + DD) (a! + 1) — stds* + 1)), do: 4 =0; Jo, ME = 4Ks giving JD dE = = *o,: as =. The latter three shifts chey the Lande rule. EE uau oasis jt=ã 3a, E aà tm! =a -— doc anl>-—— —— =-— O =] 3, ta “a a = Gplitiing seculo Rd 3 "om ras + 103 10-8 The magnitudes of the vectors are: = IM L'= vd 8! = vd. Apelyina the cosine Law: gts! Lê. 29'L'cosã, 2 = 12 +6 -272c088, 6 = 19.47º. Again resorting tô the cosine rules 1? es? 4 Lê 4 25'L'coso, 12=2+6 + 2/1204, p= 5.7. Turning to the mecnetic momentos ne = Um “= a = “eu vê = uy? + ujó + 2ujuicoss, RR az + a/Izubcoesa.14" + qu = 6.69035 . ve = ul +2- Zepucosa, A = 6 + 21,599 - 2/5(4.6503)c0s0 + a = 29.50º, so that F0,=5") = 29.50º = 19.67* = 10.03". 104 os (a) On Fig.9-13, the colums reveal the last shell being filled, the row the numbers of electrons in that shall. Therefore, hs o 182, 282, 2º, 382, at: 182, 282, 20º, 382, 3p!s Mas. 152, 252, 2º, 382, 32. (b) Lg; the configuration represents a filled shell, and tus aii the angular romenta are zero, leading to 7 Di: there is a síncio valence electron (s = 8! = 4); thus às' +1=2; &º = 1 giving a P-stai 7 3º = 3/2, 1/2 with the smaller j* 1ying Jouer, leading to . : here there are tão £t = 1 electrens; s' = 1,0 and t' =2,1, O. For the lover enerqy pick the larger 5'. This gives as possibilities: = 3t= 32 D31 W=1; 3" =2,1,0: Pã 3, 5. 4 =0 5º =1; 1 The ?D, and 25, states are prohibited by the Exctuslon princípio. DE te Des “states, the smallest )' los lowest; hence the ground state configuration should be a 10-11 For a single multiplet s* and &* have the ssme valve for each level, By the interval mile, is Ep = Ke E, i E = Mg E $ E Rs 3 E A As. '2 195 Therefore Eme do dd Ut Up 3405 Since j=1, j,=d j=3% d)=2 3)m 1 5)" 0 Bt jt=Lta!, Est-], ces |L'-s' |, so that plrst=4, 4t-at=0 Lt=8'=2, and hence the results are Qt=st=2; 3! =4,3,2,1,0. Jg-1s tal The g-factor is ita) + sta! g=1+ + ” (1) For q > 2, tie) est (ste) — Et(uta). te) I£ jt= 14º +s*, this becomes Dema s dtst), which is impossible. But 1f j' = L' - s!' the relation gives 5" > E", so that j'=s* - L*, a contradíiction. So try )' =8s' = 4" in (*), which will now reduce to 2 <s', which is acceptable, For example, s' = 2, A'=1,)j'=2-1= 1, civing q = 5/2 > 2, as required, (11) For the case q < 1, the fommila for q requires that 3!) + s! fatal) < EMMA. qua) If 39! =84º +8*, (4) becomes 106 posts + L's!, which is impossible. But 1£ )' = 2º - =", (*4) reduces to 2º > 9", For exmple, 1º =2,8'=1,5j'=2-1=1,9=4<l. (bl Consider this second caser L' = 2, s'=1,j'=1, g=i Then, Lee ovM St = 26, Tm /2m, ROD,ÊO = 150º. E = 2, dra Lj equal in lensth to É', js ce ireçes É; de ae aoia Ti rs foda indicatins that q = 6. AEseemassentaas za 10-17 (a) Ip, yielês three levels, the JP; five, 2P) three, and the “Pp qives one; thus the total mwber is 12. Wb) For the 35 electron £ = 0; 22441) = 2; For the 3p electron £ = 1; 2(21+1) = 6. Clesrly, with (6) (2) = 12, the field has removed the deceneracy completely. E 18 For a sinalet s = (, so there is only orbital angular rementum to consider, The potential eneray of orlentation is me- qi. 107 For orbital anular rements q = 1 so that La O. cut? Lºm ue 1 É às in the z-direction, H Mm ue B=-,LB=-YLA=-mb = um, giving rise to 22+1 levels. Since s = O, 4 = jj mking this substitution, and inserting a factor q = 1, leads to “E = uam, which agrecs with Eg.10-22 in the case g = 1. 10-18 In the classical model, picture the memetic field buildina vp to a valve B from zero in time T. Faraday's law requires am induced electric field E, pose This imparts an additional velocity 4v to the electrons, which circulate either clockwise or counterciockwise, Hence, E asa Er E erB er Tempra The new spoeds aro era vv yieldirg frequencies v given by “2 Ter O MO E-hy-hy ifÊB E = + corresponding to Eg.10-22 with q = 1, &mj = 1. CHAPTEP ELEVEN nd =? The Debye specific heat is, wlth y = 6/T, ta) Ba. 11-25 is - ma? = . 13 7 E = By 1 E-1 * Wir Ey definition, For y << 1, the second tem becomes hu/kT =. du, 2, Y = 1 Na ( - né 1 Qevegimred) = PENELA nes? fe) Let x = k7/hv; T+ 0 implies x + O also, Then, in tens oÊ x | implying that fx , e =2 Lim = 1. vd + . poe -1 Ae x+0, 0!* > 1; therotoro q, + 381/72, tence, foz mali The first tem is y x, á - ame ço a |-E-a. G = E faço, | = Nem, D IÉ y 16 small then, over the range of integration, so 15 X« Pet Pu rl+ sos extdegee. Expandi ng the integrand, ” x 2ué qué y Hence E E E é alzo ia r 4|-Eas + 4y E Lim xe * = «, A .2 restos 1 + 3 x+ 0 o and Therefore Ema -m Lim RÉ - 1) = 3h coa nei O " For small T, the Dulono-Petit result. op o ne O qr per 2, 11-10 (a) Let the sample be a square of sides a, so that A = a?, and oriented as shosm, p.ll4. If standing waves are set up in the 112 material, then, na udjacent undrs Elx,t) = Enpsin(2mx/A )stndrvt, Elyt) = EgySin(2ny/A, ) sinZave. There must be nodes at the houndary: hence, ina/A = na Zme/a, =n Bol, = 012, ss If the wave makes an anqle a with the x-axis, & with the praxis, ard has wavelenath À, then A = A/cosa; = A/cosg, 2,.24 ng rm BB? (costa + cos?B) = 2, Ea tn) “ E-Ememp! us Each frequency is represented by 3 e a Ro ig z£ (mê + np! = 2av/v, then the points nen, represent v. It follows that all points on a circle of radius 2au/v stand for frequency v. The density cf the no is TS Thus with == (notas 23%, the muber of o ce ds ntxjdr = Rar « iordr, and this is the allowed mmber of frequencies batwomn v and tdos z ptuidy = 38) Paul = 2a = Biivoy, (b) There are still gy modas par role, £o that ) m (rias my EB vão, 0 where À 1s chosen so that there are N atoms in the samples = 22,2); qm ã 3 À tu) Ye viIN mA) a The average energy of each oscillator is E E Er so that the total enerey of each mole will be E= Entao» fre Ea. IE x = bu/kT, é RAS Sai o But A= mv tmvês using this gives 2 x e =-1 “m o Finally, observe that x = hy /k7 is dimensionlesa; bence, hy/k has the dimensions of a temperature; let walk = 0h, giving Hm E= See, fe Pr corpo os 6 Timm). (e) The specific heat per mole is Hr 8 copie tado If T/o <<, 6/7 >> IT; tut Lin É p= 0, être TT 4 7 and s/T selo [e ste peter a -=1-— = a Einite . B/Des |e* - 1 e-1 o o Thus, 2 Limic) = T. im Al=il (a) Since the atoms are distinquishahle, use Eoltzmann statístics: pp per dai EMT 4 n=aA; n=a E since E, = 0, E, = E. With NM = total rumber of atoms, nm +n;=M n=al ref E) ae 1+e The total energy is E=mã + mão = n8 = 68 = MEO 1l+e (b) As usual, the specific heat is mc tÊyZe E/S = n= = rs VT gre (ci Let x = E/kT; then, =% nn. use Por x << 1, T large, e «1 and c,/Mk = ia?. On the other hand, with x >> 1 Tamall, CX = 0, amd c/8k = x2e *, A aketch oÉ the specific heat vs. T is on the next paso. 122 11-24 The eneray of a Bose system 19, with x = E/kT, 2 E- [proa = UE cano) * oem E am. hn o ec -1 0 As in Problem 11-23, put “ID. + Pica se, RT et. 1 witha = e? tho energy becomes E = ELtand Mer 2a (2 (e 4 pa 2 4 le 4 Jam H But, k e” 2 Kas = verba = nr) = n 2d Therefore, 3/2 = E- etglnpenaa + Ta 4 VE 4 nd). 1=25 “The average eneroy per particle in a Fermi gas is, approximately, Since the average energy in a classical cas is 3kT/2, quantum degeneracy occurs when the second tem in the above is not regligible compared to the first; 1.e., certainly when 193 soa v 3, rã Now 1£ KT << E, then : : É = Ecomyn2 Do Eat? Substitutirs this into the inequality above yields Hmm, which clearly is satisfied. Thus, when E >> KT, quantum effects must be considered. 11-26 ae T=0K, 1 0xE<Es 0 E> Ep- Hence, the total nunber of particles is 25 3,5 r- mea = [ma = frio MEL”. h o Thus, the averace eneroy per fermicn is o E- drama -ifpma. o o 3,5% 3,4 E = 1 fntuçam 1 16 pz iss RE 3 = 387/5, using the expression for N found first. ng) = 124 li-28 The depth of the well às Vo = my + E = 4.8 ev + Bo. Honco, 1t is cnly necessary to compute the Fermi energy. If KT «<< Epe then, à = EA. The electron density is a 19,3 x 10 ) (6. =s9x 108 w*. 151 Using this and m = 9.11 x 101 kg gives E =5.53 e; Vou 4.8+5.53= 10.3 ev. Algo It 1s aivon that 2E<2E nm = Eixm)*; nt) = 0 E> E (ai The mmber of particles is Fe n= urna - | inftamm dar = 4(2m)! be, o so that E = tn /amaê, (bl The average energy is femme - frotoem - 4 em para, o Usirg the expression for N found in (a) gives E = E/3. 11-30 For silver, =5.5eVat Ta 300 E (T << 105 E, above waich the material is classical in behavior). Since KT = 1/40 ev, kT << Ep, and the Ferai distribution is close to the T=0FK distribution. àpprocimate the distribution by n = 1 for enorcies less than &-kT, n=0 for enercies areater than , and in tha transition reg. the straight line kT dio E ks E<E +. (Any reasonable approximation will yield the síme final Desult.) The number $ of partícles with energies greater than the Fermi encray is E A qui PIRES =|NBnindE = 3 (2?) AU sã + E TaE, 3,4 = Am) 4 452,1, K,52. np, EM E nr EE Mto asia E = 2ev tam?) Sermb mo, the total number of particles, the value calculated at T = OK may be used: see Problem 11-26, Hence É = dona = IM ev, 5 no = 973, ev, 8 5.5ev CHAPIER THEIVE Ag=1 The potential energy of E and CI” separated by distance r is =19, v= 1 Ida 1 (eis ,. x , 6 = 10 2 E meters) " im, TF ires r 9 -19, v = (210 041.62 1000) cy o Láá gy tr in mm). E (in mm) x 10 The required dissociatior energy is the necative cf the energy reguired to assenble a KC1 molecule from neutral atoms of K and 1, initially am Infinite distance apart. This latter process irvolves (1) comoving an electron from the E atom; energy nesded = 4,34 EV; (Ly attaching the electron to the Cl atam; eneray required = -3.82 eV li.e., energy is released); (444) moving the neviy created jons from infinity tô their equilibius positions at a separation oÉ 0.279 mm; this requires Lis (o, - Tag = 5161 ev of energy (energy is Liberated). Hence, the energy needed to form the molecule ls 4.34 - 3.82 - 5.16 = -4.64 ev. Thus, to dissociate a KCl molecule requires +4,64 EV Of energy. lEs A bound EBr molecule must have negative total energy. Hence, the possible separstion distances are Eounded by that for which = O. By Problem 12-1, the total energy of a KBr molecule is E” na 3.5- 4,3) e, 126 127 Rg in mma Hence, Pomax * Pº is given by ldd - 0.8 -0; Rº= Bm. 2-5 (a) 1£ a = WÊ/20e7, then np = nptzr + peste), at the desired level, Gn fer = nte (2 -atr +03) =0, «yr ES! re tit-va Et (b) Prom p.426, for HCL, 1 = 2.66 x 1079 gem; alo, -23 xr = Li28l x 10 1 (600). 6,05932 ex; 1.602 x 10 2 q.oss x 102? E- 955 x 10 = 0.00261 ev. (2.66 x 1097) (1.602 x 101º) Therefore, from la), = (OD 0 395 + red. lg The rotational energies are Ep=rir+ Z From Table 12-1, p.429, for E,. KÊ/2 = 7.56 x 10? ev. Thus, F g g g E sible light, visible light photons must be energetic enough to excite the oxygen cr nitrosen EM aa 4 barely sufficient (but close). All other visible light photons are less enerçetic. 12-23 For the laser, À = 694.3 1m; = 8.320 x 1014 ma, using e = 3 x 10º m/s. 2 a E ta) Ep = rlz + DH/2I. mo gr = 2; E = 6/20 = 14.88 x 10 ev. Eg = 0, so My! mB, - By = E, giving vº = 3.593 x 10 a The Raman lines have frequency v=v! = 4.321 x 1019 - 4 x 10 = 0.4317 x 10) Hz, Or Ay = 694.9 nm IE there are enough molécules ín the r = 1 state, the r = 1 to r = 3 transition can be chserved. But E; - E) = 1OK2/21 = 24.8 x 104 ev, giving v' = 5.99 x 10:C pr. Again, the line frequency is v - vº = (0.4321 - 0.0006) x 10! = 0.4315 x 10! Hz, indicating that A) = 695.2 em. () The intensity ratio equals the level poculation ratio. Since the 1, line originates at r = O and 1, with 1 = 1, s aging = (2r + 10 Er/kT, mm * Eno” ana, With Ey = 2(2.48 x 10 ev), kT = 1/40 eV, this intensity ratio |s 2,94. 133 fc) Por vibrations, HE = hu, giving v* = vç. Por Ro, u, = "2360 cale yhich transiates into v! = 7.08 x 1013 Hz. Bence, the line frequencies are Wa ve 10 1) = 0.4321 x 10! gr * 0.0708 x 101º Hz, n =vu- v o1 w = 0.5029 An = 596.5 mm mi yx 100 pm; 10 “= 0.3613 da = 830.3 m. (8) As in QD), Eg/Toy = mig 2 EBD q qtde, Since vw; = 7.08 x 10 Hz, by, = 0.293 eU, At z00m temperature KT = 1/40 ev, and these rubers give = 10.293) (40) - Log = E =8.1x10 . 12-24 The energy of the rotational levels, above r = O, are given by E=r(r+ Ka. For r=: VÊJZI = 0.0005 ev tr =0) =p - ' pp = 0.0004 ev qu" = 1). Thus the energies of all the levels be tha can now assigqed, and are & 134 a FRED = PMS 1 —————— |.opas e arT—— E v.2ms 27 7 1 FE rem 3 n D TF na ' + euersg lo [e] 172 34 SETE 3 n.nnEs 3 2 mao , T L nom à 1 dn 5 These infrared transitions are not seen in molecules with identical ruclei. These molecules display only the Raman spectrum. The Raman spectrum is seen in ron-identical ruclei moecuies also. For these transiticns the selection rules are av =t, dr =0, *2. The resulting transitions are shon above. TE the muclei have identical spin each equal to zero, then one set of the Raman lines will be missing: i,e., those that originate either on even-r or odd-r levels. 135 t E 4a 13-1 (a) Metallie: charge density unifom, like en electron sas. (b] Tonic: altematins positive and negative distribution. (c) Molecular: molecules retain identíties, charce zero between them, (4) Covalent: electrons shared, highest charge density betaser molecules. 13-4 From the text, construct the following table. Transparent er opaque Melting Type Of Solid in visibler Point Mallesbility Qonducttvity molecular low tonic transparent high covalent* most cpacue high metallie reflective high (4) Properties vary, depending on bond eneray. soft hara hard e] Eleet. & Th. poor poor fair excellent Also, by Problem 13-8, for a metal the resistivity increases linsarly with T near roce temperature. From p.497, the conductivity oÉ a semiconductor increases with T. Use of these considerations gives the following results: (a) metal 15) covalent, since the conductivity increases with T; (c) covalent semiconductor; (d) ionic (L£ covalent, conductivity may increase wlth T); te) molecular, by virtue oÉ low melting point. 15 (a) vm SÉ = (98) É. since É.8 = rpcosa, a bok < E=- tá - in. O x 136 r 137 Bence, Et lol pd T - fposo 1 pt, dney 1º e] 1 2 2 = «b52n + scos?o - coco: r “co 2 v=-—E q + 3c0g) E. (árco) É (b) The force is derived from the potential eneray. bebe fito tir do 2 £--—S Pro + 3ecs2s)P + 30in280,) = E, timeç) E since a > O, F-<0, indicating an attractive force. 13-6 From Fig.12-1, E = S.lev, r=0.24 mm. Since E = err, 5.1 = (LE(0.24 x 10%), E=2.1x 101º nc. lrE (a) By Bq-l3-Ltab, on En. DE A E» analogy with the classical Boltamarm das, mar = Der Var 14Z the aversos number cf electrons per atom is ALOj(Z) + (O) 2; E 10 + 90 Liz Maoy "1 Therefore, the formula for the Fermi energy of the alloy aives 2z a E Pe atioy = MPE, cy = (1.066) 6.52) = 6.95 ev. (b) The bend width is 62s2/2ma? and depends solely cn the Internuclear spacino. This, by assumption, is unchanged, end therefore there is no difference in band wiéth between copper and the alloy. 13-17 la) From Table 11-2, p.408, E = 3.1 eV for Na. By definition, 4 Ep = KT: Te” 3.6 x 10 K. (bj Room temperature is about 300 E, much less than the Fermi temperature above. Hence, tô à quod ptT=0K approximation, ânstesd of 300 K and use F-D statistics (classical not applicable since T << To). tc) Ey Bample 13-2, TE 52 = 0.008. 13-18 (a) By definíticn, 4 É vp = (287/m) = 1.20 x 10 m/s, since Ep = 4.72 e. (b) The de Broglie wavelength is a=b. = sit 0.564 mm. ll = 10 1, (2.29 x 109) 143 This is comparable to the interatomic spacirs a, which car be estimated as follows: from Table 13-1, p.d5l, (ngTt = 2.70 x 101 npc, so that nt = (0.70 x 102º) (1.602 x 1922) = 2.723 x 102 mw, a=n 3 0.30 mm, 13-20 The Fermi distribution is emudged cover a region = 2kT wide about the Fermi energy. OF the May electrors (1.€., pl di one valence ejectron per atom; ses Table 13-1), NayT/T their energy increased upon heating (by Example 13-2, Eca Te >> The ipi cem Pic sa z Uz Cy Tp OD +K, where K = the constant energy cf the other electrons. Bence, a Ca = qr = 2a Por the lattice, Sat Re and since k = R8, + c = .21I.2 e =-S = 5 séooo = 0.005, the Fermi temperature from Problem 13-17 (a). L3=22 In mmentm space, tuo electrons cccupy each point, cut to the surface Of a sphere of radius Pp: exterior points are empty. Electrons with »-momentum = Py. en a circle, oriented as shom on the sketch, next page. The number É of electrons with Léa -mementam = A, ig proportional to the area the circle: L.e., + e 2rÊ = 2rtpÉ - pd), = 2mêo - péph, E=1- p/o). 13-; The ratio of the mumber Mof electrons in the conduction band to the total raember 3 is, by Example 13-€, aum = o ii da cu Comparina conditions at the tuo temperatures T, and T;: Blat 28/88, = (7,/7;) epi => With T; = 300 E, the rumber rmtio = 20, E = 0.67 ev, this 0.248 = 15/20" 3884/M4, m=37X 145 la) The Fermi distribution is nto « (ENA gy, and the density of states is 3,4% nem =am-E)% à = SRD h Bence, the number of conduction electrons is dad A. (ro - 2) He FEM 4 iyrlar, Er If the material is a seniconductor or insulator, it is expected that, over the range of integration, (E = Ep) >> KT and tus also (E - Ep) >> kT, so that Ra = fa - Ee EH) x, n refrão ve — 2h !E Fo ag, E ns pet The value cf the integral is /n/2 and therefore, insertíng the expression for A, E 3/2 (Ep) e. E ES (ami) e (b) For the valence band, RG) = AME, — 25. 146 The number n, cf holes is given from eme [rm af, ente art 0 o a [em a Since lt follows that Í m Ê É 5 é Ê 5 : : ô E number of electrons in the conduction band; 1,0., nv = state) 2a Cr, To obtain this result from direct integration, note that 1f (EE) >> KT, then (E-£) >> kT over the range cf integration. 147 Therefore, [a- ne, ya = [x gig - e FDA, a o = [a - ni - compra, nt VD, 0 he first integral is [x -n'w=-2n/2, write the second as [x, - te Da 2 qem Pen, o KT But d SkT >> 1, so it may be asgmed that >» 1; then, replace lower limit with infinity with Little error, and this second integral becomes E] eos E E per 32, o Thus, non eAà GS MMA do, For E use the T = 0 valve: E a-ade- uia = Sn? Therefore, the same expression as obtained above is derived here also: i.e., for nv. By Example 14-1, Te =4.2 E; hence, 2 19,2 «À E = 2.008 x 10))—!L602 x 10 FE reg DE a E (4.2) 341.381 x 10 == 1300 mm. 142 The relevant equations are ds fra. Io E= 53. (a) The first of the equations above implies that dg fra --2-a8 If p= 0, then da/6t = 0 civing FP = constant, not necessarily ofi-a - o. bl) IfE=0, Fut a current exists on the surface and outer 1 oE the superconductor; 1.e., j) É O. Hence p = O unless «di = 0. 152 153 J=3 Consider the total field inside the superconductor as the mm oÊ an external field and an interral field de to the currerts set up in the material as the extemal field changes. Py Lenz's law, if the total field Lg zero and one tries to change the extemai fieid, the induced currents flow so as to set up internal fields cppositely directed, so that the total field remains zero. If the material can do this, the currents must be able to respond precisely to the changing extemal field, 1.€., p = O. Hence, Meissner effect implies p = D. Lonz's law states that an induced current will Flow in a sense so as to oppose the change that produced it, but this carries no implication that the original change will be amsul led completely even 1f p = 0 (above, it was assumed that complete armuliment cocurs). For example, if dt = E(t), the current set up may be independent of t, Ín which case Bp = O (Meissner effect) carmot be satisfied for all time. Ld-4 sermel parasegartiom supercondnciino ê= à 154 148 The isctope effect is that nr. = constant. For naturally occuring varadim, in atomic mess units, M = 0.5576 (50.9440) + D.0024(49.94727) = 50.9416. Hence, k (50.5416) “(5.300) = asse Ta E. = 5.352 E. 16-10 Eg.'s 14-2, 14-3 are B-ufirud R=od. rap a ug = -uçfis He Hed x=-l. From Eq. 14-4, Ã= Tê): ua + nÊê = é. The last cf these with y = =1 gives B = 0, consistent with the above. él (a) Since B=tnç, =6, =5 po lxI]0 03 x10 0 04x 30% nm. (4x x 10) (1) 155 (b) For me ka-mole, n = 6.022 x 1026, then, 2 =24 Mat E = = 16.022 x 10 2.8) (9.27 x 10 Eos 22.4 lá-iz (a) mith x = pB/KI, x .ck M = ny É E € +e e +e x = 13 = 0.549. (b) By Fig. 14-56, at M = Mo B/T = 0.53 T/K. Therefore, x= 12, E 0.549 =— E =s(0,53), 1.381 x 10 v= La x 102377. 14-13 The magnetization is mem ER asmming Curie's law (y = C/T]. Hence, 8; - BT, + T = 0.01 X. 156 Já=14 The mametization is x -K E.e M= Au “mt Es with x = pB/KT. IE uB/KP >> 1d, thenel > 1, e “= 1, 80 that ef >» e. x e Mem my, 4 tha saturation valve. 14-15 antiparalLel After sore antiparallel=spin electrons make the transition to parallel-gpín, as indicated, the excess in cÉ elactrons with parallel spin over those with antiparallel spin is represerted by twice tné shaded area, which may be aporaximated by a rectangle. Thus, E] Bv (am h tn = 24,8) (nt ONCEÇOE = 248) (1146 Ora, Ss in which the factor of 4 occurs since the upper curve stands for, Orly half the electrons. The total rurber cf particles N of a Fermi cas is 157 "= EPs, so that the excess per unit volume is É = ImByze, where n = mumber cÉ electrons per unit voltme. Now E = kT and Vu "SM = (2hu E) = w giving , Mm CGI = SBa/26 = Bye 2 a (b) For copper Ep = Kip = 7.1 eU (Table 11-2) and n = 1.1364 x 1029 ="3 (Table 13-1). Substitution of these into fonmia above for the susceptihility aives y = 1.62 x 105. J4-16 (a) The rumbor of dipoles per unit volume, aligned parallel and antiparallel to the field is ' n, = cnesb/XT, n, = ent, The eneroy associated with parallel aligiment is “se, with antiparallel alicrment +jB. Hence, the total energy is U=vin (8) +n,CuB)) = pace MET qB/T, Now, with x = VB/eT, =X, x n,+n =n= ente +€e cats Also, Put this into the expression for U above to get 163 ' Using this to eliminate E, cives =4.454, rma+ (0.55)in(l + Ze ) = à + O. 0126, = me . 2 hrose 15-9 D=K + » Ko Eu mg area) cose Conservation 0f rass-energy and of momentum (classical form is used) require that 15-11 2 qi Ent) +90 = 22, 089 Ma = PpYpCOSO + Mv, c0s6; myvgsina = mypeinê 0 = (Dum96) = micr2) jo, Using classical expressions for K also, this last gives O = [2(25.98260) - SL.94051Juc?, (2x7) "sine O = (0.02069) (031.5 Mey) = 23.0 Mey. = (May) 15-12 Terefore, (a Du imo + Mg Mom no (x n)" = (xa) Posso = (Mr cos, O = (2.0141022 + 3.0160500 - 4.0026033 - 1,0086654) (uc2), 2 O = 00.0188835) (ucê) = (0,0188835) (931.5 Me) = 17.59 Mey, (pg ooele = 287, + By m coe2s — 2166 Rm m) cos. : (b) In Bg.15-16, let a =H), b=n,A=H', B= Bel; then, But, , Ba 2 sin?o 2 ME = q Mag = 4 tmgm/mg)! = 2/4; K, = 0.5 May. sinCa = =1- cos 4, With these, Eg.15-16 becomes 2 2 - sino K cos q = . 35.68 = 5K /2 - Ki; K, = 15.9 Mv. Equating the tro expressions for cos gives 15-13 = = The b Ema + Er, = Rpg — 20,68,8,) "0086 = 0. indirg energy dE is ae: 8E = (Pm, + Ra, - Matom) (02), 2 0-K, + ar = (6(1.0078252) + 6/1.0086654) - 12.0900000Jucê, AF = (0.098943) (931,5) = 92.165404 MeV. The average binding eneray per nucleon becomes SE - SEIS 68 mov. 15-14 fa) The energy release is E= (Mg - mae? = (2(2.0141022) - (4.0026033) Juc?, E = (0,0256011) (931.5) = 23.8 Mev. (b) With the nuclei just touching, their centers are 3 F apart: 2 9, =19,2 a E: (3x 107Pp (1,6 x 1072) 15-15 (a) vith rt = LD go É m, 22 Z 3.7 3 e 3/3 a920M/3 vm e 5 due" "E (âneçd (1.1 x 10) f this has the same fcrm as the Coulomb term of the rass formila. (É) The energy coefficient in V above is -15,2 a = Bt8.988 x 10º)— —Mutg2 x 10 Do 2 0.7854 Meu, tai x 1072) (1,602 x 102) In mass units, a = 97854 331.5 compared to a, = 0.0007€63 u cf the mass formula, = 0.000843 u, 165 15-16 (a) The bindirs enersy of Sgll às, with masses in u, 1Eg = (my + 6m = npc? = (11.091118 - 11.009305) (931.5 Mey), E, = 76.209 Mey. Similarly, the binding eneray cÉ E ss ER = tm + , -moe = (11.090278 — 11.011432) (931.5 Mev), E- = 73.445 MeV- The difference is 2.764 MeV. (b) From Problem 15-15, V= (D.TAM Mev) E ror fell 2 =6 and for “Bl ga 5. Assming the ame r', it is required that DES s% = 2.764; 1º = 3.13P, [c] This is someshat larger than the mean radius (= 2.5 F) of the charge distribution of €cl? aiven in Pig.15-6. 1517 (a) mor pe? 2 = 26, A =56, N = 30. The mese formia is Bg. 15-30 and the terms are the following: Mass of separate parts: 1.007825(26) + 1.008665(30) =56.4634 u. Volume term: -ajA = -0.01691(56) = -0.94696 u. Surface tem: pa (0.01911) (6928 — O.Z1927 ue Ooulonb term: a,5 = So. 00976) (26) (58) = 0.134816 u. Asymetry tom: a, (2 = SA/A = (0.10175) (237/56 = 0.007268 u. Paíring tem: 2 even, N even so this tem 15 aà = —(0.012)//56 = -0.001604 0. (b) To convert to eneray, multipiy by 931.5 Mev/u. To fom the average binding eneray per nuclecr, crit the first term, add 166 and divido by As = = [=15.7517 + 4.6528 + 2.24252 + 0.1205 - 0.02668]. volre surface coulob asymetry pairing There is qc0d agreement with Fig.15-12, except thar the paíring tem is too small to discriránate from the craph, (e) The atomic mass is the sum cf the terms in lab: to wit, 55,93664 us (8) From (bl, AE/A = 8,762 Mev, and agrees well with Fig. 15-10. IS-1E (a) The binding energy of 2 45 (Gm; + 60, = mp) (931.5 Mev/u) = 92.1660 Mev, and thot E “Be! is (mg + 2 = Mo) (931.5 Moju) = 28.2970 May. The difference, BE(C) - 3BE(He) = 7.275 MeV, and equals the binding energy of C cn the atpha-partíicie model. tb) The binding energy of *0!8, calculated as above, is just 127.62463 Mev. Then, ten, BE(0) - 4BE(Be) = 14,44 MeV, amd is the binding energy of O on the alpha-particle mdel. (c) Teg muter of bonde de NIN = 19/21 = 3 for clê (1=3) and 6 for *0lÉ q = 4). 2 q iusis Ji Mo!B, E bonis <> (4) The energy per bond for (12 4a 7.275/3 = 2.43 mew; Pol6 gives 14.44/6 = 2.41 Mev,virtually identical. 15-21 cl, Protons: an even number; 3 = 0, P even. Neutrons: 5, 3 in lp; 1 = 3/2, P= (tubo odd; hence, (3/2, cad). = Pven E, even Z: (0, even). ta) 157 B461 3 even, 3=0, Peven. N= 33, last single neustron in 1Es,9' 3 = 5/2, P= (IJ = 1, civing (5/2, od). Ra73 Z eve, 3=0, Peven. F=ad4l, last síngie nestron in 165,9) 1 = 9/2, Po (1)! = 4], predicting (9/2, even) . 28,61 (b) Discrepancy is : observed is (3/2, odd), and predicted is (5/2, cd). The 1fs,ar 223,2 levels are very close together, So cne neutron in 2p,,, couples with the single nestron in 1f + leaving a mucleon in 2p, giving l + 3/9 cultas: tias 5/2 far 5/2. Parity is unaffected since (-1)* = (-nl = (p?=a. 15-23 ta) By Example 15-10, spin £ = 5/2. (b) By Berple 15-11, parity (-Dº = (-1)? às ever. (e) Urpaired neutron 1s in a 1d, state: Le 2, j=5/20 j= +. By Fig.15-19, the lower Scheidt line gives -1.6y;, 1.e., negative. td) 3 = 8, which is a magic munber; hence, q = d. 15-24 (ap PSL nas odd Z, even 5; VD res close to the upper Q=2 +) line, so that on this basis g=)-h=7/2-1/2m 3 is expected. (b) By the shell model, an even rumber of neutrons couple in pairs; the last proton is in 1f,,,; the f-level has £ = 3, which agrees with (a). a (8) Prom Pig. 15-20, q/2r'? = 0.09 > 0, The desired ratio is 14 q/rr'2 = 1.09. 172 18 For À atores, Rem o ma, The number of E atoms increases due to adéiticnal À atoms that to B ators, but decrenses as B atoms decay to C atoms; SE = OS RSA A + Ma, FÊ Meltiply by e'Eº and integrate: [os + besta = [peruca o fEngna - (enctennta, o o (Essyt na = Mo = oa ER Rg Bet No = O by assumption. Therefore, se em, 1 173 lé-10 tal B=12,N=15 E= =14 E=18,NEI3 1.007825 Z 12.0939 13.101725 14.10955 1.008665 [A=E) 15.129975 14,12131 13. 112645 o =0.45657 =D. 45657 =), 45657 apê 0.17199 o.17159 0.17199 aguada 0366219 0,0425823 0.0498493 aptz-a/22/A O.O0847916 0.00054213 0.00094213 ElOtaçias O o o Summing gives: 42,27 = 26,98097 u; Es, = 75135.965 Mev; M,3,27 = 26.982379 u; Es, = 25134,086 Mev; Ma 7) * 26-988406 us = 25139,700 Mev, 40 Bum (b) The smallest M is tha most stable: this belongs to Z = 13. (c) Electron rest mass m = 0,0005486 u; rest energy = mc” = 0.511 Mey. Masses in (a) are atoric messes, and so the various decay posstnilitios ares (1) Electron enission by 2 = 12 E =, »-My = 25125.965 = 25134.086 = 1.88 Mov. (ii) Electron capture by 2 = 14; E = Mn" Ma, = 25139.7 - 25134.086 = 5.61 May. (dt) Positron enission by 2 = 14; E = 04, » My po = 5.614 - 2(0.511) = 4.59 Mev. 16-11, ta) By the conservation cf momentum, with the initial mementum equal to zero, E Wee 174 with E = neutrino energy = (0.00093) (931.5 Mev) = 0.8663 MeV, ignoring tha kinetic energy WrW2 cf the rucleus. Evidently then, -13 v = É - — (OS6ED 0d A 1) 2 3,98 x 10º ja. Ne cm 0.661 x 107) (2.898 x 10º) (bj ho process may be monitored by detecting the x-ray eniesicn as another electron êrcps into the hole created by absorption cf the E-shell electron, and other transitions. 16-12 (a) For the electron, E = pi? + mão!, The kinetic energy E = E - mc? so that Rs? - 2-0 Ke (ne! + pio! me, zf p= me, then E = m/0 +07) - 2 putting in the indicated runbers gives the following: pino 2.8 4.5 6.9 R 35 500 250 moiR/p)* 6.916 4.563 2.292 jm? L93 4 s.972 (b) From the gragh, p.175, RT tmcÊL4s) = (4 Mev) (E) = 4 Mev, when K, the kinetic energy cf the electron, = RJ j the enercy of the antineutrino Ls zero, so that the decay energy E = 4 MV. 15 metro ME oo. tomo z 2 q 5 g Tm? 16-14 (a) From Fig.16-12, K = 0.8 MeyV. Fig.16-13 has log” = 0,3; Fenes, Soma " P=2 T= 10008; FT=Z2000s. In sctuality, FT = 2340 8. (b) The decay is a little siover tha that cf ly), for which PT = 1200 a. 18-15 (a) Use FT = 2000 8 and assume Mº = 1. Example 16-5 with FT = 1200 s gets É = 3.7 x 10-62 Jem), Since 2 = LET, the value of f for the present case will be 2.9 x 192 genê, tt] The result for this process is a little less then for Example 16-5 dus to the larger FT. 176 tc) The shell structure cf the initial mucleus, a neutron, is identical with that of the final rucleus, a proton energy 1s absent with only a single proton as the final nucleus) and thus tha eigenfunctions are identical. J6-16 o e cent , EE Let a = characteristic distance between charges, j = cusrent density and p = charge per mit volime, u=sa=sa= (gala? = ja, poemas (oa = pa!, and thus “ BeB/JÍE. E. E iz Sa where E = cB for a plane electromaqnetic wave in a vacam. 1€17 Cosa asp * art? Err. 1£ E = Epsin2rff - vt) « Ego E x E = (ar coszr tê - vt) = Ena, implying that Vuad/Uap * ar egfAgrtEy = 1'/3. lé-18 Foltow the reasoning en p.580. (2,even) to (O,even): |iy - 4g] = 2 = Ly since parity does not change and L is even, radiation is electric quadrupole (L = 2). 17 (Lida) to (Oeven)s li, - Lg] = 1 = L; L is Odd and the parity ; indicating electric dipole. (Lodd) to (2,even)s ba=sA == L; Lis cdd aná the parity changes; as above, this Indicates electric gdipole radiation. 16-20 The integral in Eg.16-26 is The parity of x is even; Sf U$ , is Of 064 parity, the partty of the integrand is odd and the integral vanishes. For tha integral to be different frem zero, Wj must have tha eme parity as py. The other integral being considered is [uv ra Bu or. These parities are: Pl! = —y, odã ptBl) = BEE) 2 dE, siso ods Thus the parity of the cperator, irvolving products of those above, is ever, Às a consequence, Yg must have the some parity as 4, or the integral vanishes. 163 The net reaction is e” 4 pl, Mg, all, 2a? That is, ue? + mel, - mude!) - meme) 3 Orotay * 20.91 May, welágs3L, 302 = 20.91 182 put 0=0 m, = 12,m, =m =1, 0 = 90º, K = 7 MV, to get o=n0+-na-Ta: MK, = SK tad E, mL MV; E = 0.154 Me. (b) E, = 0.001 Mevs Hm 154 GU. (e) E = 3X7/2 = 318.617 x 1072) (500) /2 = 0.06463 ev. : (8) Since ' "Fa a Kao" os a ss it follows that £p = 0.06863 eU = (591 MEN), Ê -16.5546 = nlnHh = -0.167ny mm 59. 16-34 (a) o = tada? - une) - mood, O = (0.00351) (932.5) = 3,77 Mev. (b) One Meçaton yLelds 2.6 x 10? Mey; hence 48 Megatons gives 1.248 x 100 me. The number of required fusions ís 30 4 = EMBALA. 0,38265 x 2030, Since to hydzosen atoms-are required for each fusion, the minimum mass cf hydrogen necded is m = (0.28165 x 1030) (2)((2) (1.66 x 10273) m= 254 ks. CEAPTER SEVENTEEN 124 (a) Eq.7=17, with & = O às É iE Be -uneo. (b) mith R(z) = ulr)/r, E uv,iê dem a r Sê S-acsetá Substituting these into tha equation for R(r) given in (a) EB eve fc) E3.5-43, the time independent one-dimensional equation is Schrodinger BBemes, identical in form to the equatimm 0 (b). (d) Since utu = (r*Rº) (8) = r2 (Ren), E, E É A [ a a 1 o = qi ferotability of the neutror-eoton separaticn being between F md co. 163 184 te) The reduced mass is ve Use of the reduced mass reduces the problem to cone of relative motion; 1.€., one nucleon is at the center or origin cf the coerdinates 17=2 (a) Assume a bond state, EV, . Forr<r'V=0ané I I 1 I E a | Em e RÉ —E = — md In the vegion r > E, Vim v n 1 so that 1 = 4 -=—3 E Sem E s (b) Replace E with V, - 4E; the equations above and their solutions will ber E 2 rar's: E + ivo = ama = 03 u= Asinkr + Bcosk,r; r>r's É - Bju = 0» u = Cear + pelir, te) From Cb) rer's * = Bio, - 88 r>r's K = Bum. 1-3 (a), (b) In the solutions given in Problem 17-2, (ij) D=0, othemviseu+=asr+e, (ii) B= O, otherwise R(r) = u(r)/fr + Ímas E + 0, For the remaining, the conditions at r = r' are x (1414) u continucus at r's Asink,r' = ar, Civ) E contimscus at r': KyAcosk r* - = gpatar”, These last two conditions thus imply that host (k rt) =. Substituting the expression for k, and k, from Problem 17-2(c) gives the desired relation. 19=4 Making the substitutions indicated in Problems 17-1 (8) and 17-3 results im cottfitm vg) et a - 0%) = EA, with x = SEN. Mmericaliy, with r' = 2F=2x 105 m and E Vo = 36 May, dtmgvoes = 1.86, Fence, the equation becomes cot(1.86(1 = md) - ds from the text values, cr grester precision calca In any event, 2.2 Mey will be used in the following problem to maintain conformity with the text. 186 as (a) With 4E = 2.2 Me, Vo = 36 Mv, 2º = 2 F, the constanta k, el 1 k, = 0.90 F 1 k = 0.23 P . u = Asíni0.90r), r<r'=2F; =0.23r 4 u=Ca r>r' = 27. These expressions mist be equal at r'; that is, =0.46 asin(1.60) = Ce 0-46, c/a = 1.54. Also, it is necessary that [emana E [ema -1. o o Putting ir the expressiora for u, and € in terms cf A yields eotifunt ooo + fo sotetet =1. Evaluating the integrais, 2 = [5220.0000 = 1,123; (ptetca = 0.866. o 2 - Therefore, auto + (1.54)2(0.866)) = 13 A = 0.16. Thus, the firal results are, with os dio F, utz) = (0.16)sin(0,90r), = < 2, utr) = (age UBE sa. (&l For v(r), see Problem 17-2. mM posou " Tre radial probability density P(r) = utu = uí(r). Vl) 0.12 D.u2 1 1 0.05 : Em 4 D.0ms ; E 1 É > E 1 E 1 4 =(2)
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