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Solutions Supplement to Accompany | QUANTUM PHYSICS OF ATOMS, MOLECULES, SOLIDS, NUCLEI, AND PARTICLES Second Edition [Walxoyalra) = 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 towarê 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 becsuse 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 Shurtle£f Institute cf Tecinology) for preparing the solutions to the problems in Chapter 18. Preparation of the supplement, including choles of problema, was left to the undersigned, ho vas also his om typíist 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 41 Montocmery Drive Plymouth, MA 02360 CHAPTER ONE 1-2 The radiart energy contaired in volume dy that is moving toward A at any time, in the frequency interval v, vidy is CE (uldy = pplvidy É ay, where fi is the solid angle subtended at dy by A. With & = Rcosg/r? aná du = rêsinddrdsas, the enerqy becomes dEptvdy = | npfv)év Asinecossáresas, The energy in this frequency interval that crocses À in time t fror the entire upper hemisphere is ferir Epfuldo = apluldv É ( N sirtcostdtagdr = Jg=0)x=0 Eptu)dy = i Pp tuldy act. Hence the energy that passes through a unit area in unit time from the upper hemisphere is tuldy = Enlvóviat = E polvidu. rp Er q Pr nn 2 2 P «afro = Ppfuldo = Ac lo) Bi 1 8 vw = cf = EOBEX IO . 5.4500 x 1014 mz, 5.50 x 107 BZ ip = Am, =c = pr “a 2 s.51x107 Therefore, Yap mM) + v9) = 5.46 x 1014 may do = vo = 9.9 x 10 ne, Eince ent) = Ci ta 2 yr, mumerically; entmd,/o? = 1.006 x 19/33, hu = 4.37, EraE 3 = 78.04, ôriv = (1.006 x 10722) (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) (9.8 x 2015, Pe 7,51 W. 15 0 Lx anrêgrÊ = qui? x 208)205.67 x 108) (570099, L = 3.685 x 102, Le Gm? 2 26 dn. L 3 de = 5 - dao = 8.094 x 10º g/s. (3 x 10) (bl The mass lost in one year is mem DE ta (4.094 x 10) (3.156 x 107) = 1.252 x 1077 g. The desired fraction is, then, 3 £ = 8 2 1.292 x 101 4. =6.5 x 10 2.0 x Fr lo la) The solar constant S Ls defined by L s=EE, am? Fr = 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 mé, CÍ arrival Of energy at the earth's Surface is od tm = - MB 1s, ER amR 238 Wmê É 401352 W/m) = 338 wjn?. Since, for À ur x = 4.565, these solutions give Me 1.815 ur da = D6LMA 1 Let à" = 200 nm, A” = 400 mm; then, E FA = (3.82) É AR he/rkT " Sep = a.s26ão?. 1 e - Mumerically, o] 3, +E = testa posso x 2) À aço, (4 x 107") (1.38 x 10 É) so that 35587/T e = 5 SA = (3.82) (4)” = 0.1194. Let x = 035887, enon, a q = 0.n194 = 1, x = 7.375; SBT 2 9.398, T = os = 18,000 E. 15 CRAPTER THO The photoclectric equation is With V, Hence, o ho = eNgA + wpd. = 1.85 UV for à = 300 rm, and v5 = 0.82 V for à = 400 mm, he = 8.891 x 108 + 3x 10upa 26 he =5.255x 10 +4 x 107%. 8.891 x 1026 4 3 x 107u, = 5.255 x 1028 4 4 x 1074, wo = 3.636 x 1022 3 = 2.27 ev. tb o Therefore, he = 8.891 x 1028 + (3 x 1077) (3.636 x 10), he = 19.799 x 10 É Jem, ta) : 26 nã n=EBxID 6,604 x 10º gos. 2.998 x (ia) “a” her 3.636 x 102? = 19.799 x 1072871, dp = 5.445 x 107 m = 544.5 1 x In a magmetio field E = mu/0B. p=v=erB=(1.602x 10 1)(.88 x 104, p= 2.012 x 1023 km/s, -23, p= x ad0A x, = 105637 Mev e(1.602 x 10 * Also, Pets, E = 10.05637)? + (0.511)2, E = 0.5141 Mev, Hence, ta) E =E- E, = 0.514] - O.5I10 = 0.003] Me. (bl The Photon eneray às EO = JE = JE = 0.0175 My; “Gr RpcE= 5 31 = 4 ey operate, apely conservation of mass-energy and of mementum: gra [2] RR hu + E, =E+E + hum Es & o] m.o= q t0=P These equations taken together imply that p= Ke. [9] But, for an electron, Pest, + E2= po + Ei, 9 p= (tê + 2eRi/o. o) (*] and (**) can be satisfied tosether only if E, £ O, which 15 not true for an electron. tb) Tn the Compton effect, a photen is present after the cottision; this allows the conservation laws to hold without contradiction. 2-4 Let n = maber of photons per unit volume. In time t, ali protons initially a distance < ct will cross area À normal to the beam direction. Thus, n Alot, 1 BE o DEAR e re = BÉ, For tvo besms cf vavelengtha À, and à, with T = 1. he, nã, ny dy and therefore md da The energy density is p = nhv = nhe/d. Since this differs from 1 e the same for both beams), then if Set E, = 20 kV, Ep = 0; K = electron kínetic energy aftor the first decejeration; then Er - desu -m= oa a ido a Ag mM + Ms with dA = 0.13 mm. Since he = 1.2400 kev-m, 2-33 The number of particles stopped/scattered betwomn distances x and x+dx is dl(x) = clitpáx. 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 PRE op i the Limits cn all x integrais being x= 0 tox== CHAPTER THREE st É = pis = Ta Do do ele Pax =31 RR 1 a 1.813 x 107! = a b, a = 1.675 x 102 rg; evidentiy, the particle is a nevtron. mm ta) E = plo + Ei (6 + E)? =p 4 ER, er poda? + capta + ot, Bit E = evandE, = mc, so that 2 2(eu) (mc) 1 ES 1 Cj UE, j= 2 Jd = ( pe y = “Uma, and KB, = evfome?, Thezefore, à = = great e da (5) Monrezsesviseio Limit: ev cemc?; set 2 + aU/2mpe? = 1 to get à = h/(mgen 5 = /tomgo! = ar B M 38 jo ML au, te pe = vie", eaêo, impe ) (vie) Numerical ly, -34 po = (6.626 x 107º pos) (2,996 x 20º m/0), (1.602 x 10 JfMev) (O * mir) ho = 1,240 x 10 mem; bence, a E . A trm) = E topo, 219 ta) =34 p =D. Six ds (2990 Jg no). 0,2400100, uclme « E e 1,602 x 10 3/mev) Pepe E, E = (0.120)? + (0.51)? + E = 0.5258 mv; E=E-E,=0.5258- 6.510 = 0.014 MV = 14.8 key, [15] p = Selzsgo mem . Sh * E o 124 e, 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- 1 dis om o p = S228 x 197 pams 2.998 x 10Ú m/s D.S868 key e 1.602 x 10 pay 8 E= (p%2 + eb)! = (10.566)? + (5112)% = 511.00095 key; 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 =; hence, p = 9.868 Mev/c, from (a). qi + ed) = (9.868? + 0.512) = 9.B812 Mey; X=E- E, = 9.812 - 0.510 = 9.37 Mev. 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, = (pie emp" (5.868? + 9367) = 0.052 Mev. = 338.052 Mev; K=E- E = 938.052 - 338 This last result 15 much less than the average binding energy per nucleon; thus the uncertainty principle is consistent with finding these particles confined inside nuclei. d=az la) Since p, > sp, and x 2 dx, for the smallest E use p= fp, and x= &x to cbtain « EiiçãE 2 E FE ltP,) + AOlax) With Mp dx = SH = E the minimum energy becomes 41 Consider am electron oscillating along a dismeter. When at à Gentanico 1, Eru tos CONDE: DE tis (abas AT Soros (ii ate Fm (qa mer? where o = e/(4180/3) > O sínca the net charge on atem-electron is te. Therefore, 2 Pad F. Bed This force is attractive: i.e., directed toward the center oÉ the atm. Hence, F=ma, a & “Taças em ut modo If the electron revolves ín a circular orbit of radius R, orbits of radius less than R, since the charge exterior to the amplitude or radiuvs exerts zero spherically symetric charge distributicns. j ; R Kinetic energy conservation: O The momentum equaticrs gives EMCOS4 = Mfv = UCcosg) msinp = Musind. Bence, ne? = mê? - zuvcoso + u?). The energy equation yields nã = Meto? - vu), Equating the tuo expressions for mÃ?: 1242 = Juvcosê + u?) = Mat? - vd, cos TO-D+Slem. (Since m O, a minisum for cosé, a caximm for €. Substitute this value of u into the equatior for cosó to get 2 O" Ep. Since m << M, this implies that 142, =1- wênê, Sox = 8=7ê5* (5) tee ye = = The mesimm force felt by the alpha-particle in passage through the atom occurs at the atom's surface: p -—1 tes m “Teo For meximm deflection, suppose this force imparts memertum dp perpendicular to the original direction of motion: A ad, = Ap = fdt = Flat) = Rc 2 - q SE k tp Ames Then, anticipating a small deflection 6, po. A o? Uount mt ra) Fa For gold, E = 75; suppose E = 5 MeV; then =19,2 6 = (8.988 x 10)-(2L 0210.6022 1010. 4,55 x 107 rod, ) (1079) (5) (1.602 x 10 Hence, the deflection Gue both to the positive and negative charge in a Thomson atom cach are about 0.000) rad, so that tha overall deflection is about 0.000] rad also. Only if all the deflections dus to the electrons are in the same direction could a larger deflection, about 0,01 zad, be obtained. t-1 ty Ega 48,9, á Tc mê O qmito/m Vho solid angle of the detector is 2 af = da/rê = 1.0/(10)? = 1072 strad. ALEC n = (4 muclet per cn”) (thickness) , nm bZ e ng8) = 5.898 x 102) qm? (197) (1.661 x 109) tience, by direct numerical substitution, am = 6.7920 x 10) —l— a, sin (8/2) The number of counts per hour is E = (36008 = 0.2045 ——— sinêg/2 This gives: 6 =10º &= 423; G= 45" E = LA. eu L=nfs m, 7.382 x 108 = Lt6.626 x 10724) n=7. a 24 2 With recoil: (lh Now take into acouunt the reduced mass pt -=19 2.4 24 MA ME ( (02089) (0602 x 10) e ug e 0? 1,2 Mute 1,2 Mello mo HF F) =27. F] ” E = J E] = ) = t a o Mo (201.673 x 1027) (2.998 x 108) o e ao Ce Go quo MR A = 661 mm. mm = mid) Lo! LE «m-&; - ema-. ” da mta, "Pe Ro me" TE" Pe a sn Therefore, The kinetic energy of the electron is Hge > Hyr K = (0.511 Mov) ds - D. dci Te = Rael = do > mta - do. With 3 yo ne mM ngm 8 = LEXIO . o qu0g, 2.998 x 10 Hence, compared to the hydrogen Ho line, the helium 6+4 Line wavelength is a litete shorter.. this cives K = 409,3 ev, For helium, the second ionization E 2 ” Cb) Since à « q” (the factor Sé is combined with Lin = 1/ni potential from the around state is to aive equal values for E and He), “ i 2 z 3.6 E 13.6) Zion À DGE UBS 2 04.4 7, o = = a dg ge Ha Hence, = 54.4 + 409.3 = 463,7 ev, 1- En 1 B-a-toger = dr = dB ao nao, à = 258 = 2,674 mm. dy m “mM : SA = (4.084 x 1079) [656.3 ne) = 0.268 mm. á=38 (a) Hydrogen Has aq = Ratd =D. Atz The momentum associated with the angle 6 is L = Iy. The total -1 energy E is Betim, 5 =2: A == RT SO pstuneicE a TE Ag = Agr them L 45 independent of 6 for a freely rotating object. Henoe, by 2=ngdi + np=4, 3engdo o n-6. 26 the Eilson-Somerfelá rule, É Lão = nh, CHAPTER FIVE LS 00 = Li2m) = CZE) (Zn) = mb, 53 UE) = nd, In) The time-dependent part of the wavefunction is sê eit/tem 1 ABM O qritnt Es. Timrofore, 16% = 2m * vo! hj Since E = ho = Zeho, E= mg". (1) The Límiting x can be fomd frm ta According to Example 5-6, the normalizing integral is 4 c) ceyoh 20% dx =P Ma É 1=2B sin | e URE/O = EE e eEjo" o 1a sêoS + pê= tcmid, Er] Problem 5-3tc) provides the limits on x; the wavefunction is 1/8 2 Eae 7 = tom A ot tri) 27 30 states equals this also, Since all of the space vave functions are simple sines or cosines, this equality is understandable. ES The wavefunction is. v= Bsintonça e TEM, and therefore +a/2 E= de sin? (arx/ajéx = 0. =a/2 As for *: pc +n/2 E 2 - 22 cin? (owxyajár = 2 loeintuas = Ltd - aa, a 2 43 -a/2 x = 0070672. 5-12 The linear mementum operator is -1y É and therefore +a/2 = 2 2 a p= SjuineE cp « Si com 0 -a/à Similariy, = ta/2 2 p= Sum Es totnÊB) ax, ax -a/2 iaimaii - pê = Bai? a) Apt = ab? = 8º. o Eni Let = 08", PRA a (a) Problems 5-11 and 5-12 yield Jd dS. do x = na, n =5 Em a Fence, etp = tra) E) = ami) = (62 - 5 É 3.34 É, (bh Tn the ground state, tp = (0.180) (By = 1.13 É. In the first excited state the uncertaínties in position and mementum both increase over the ground state values, due to the higher energy of the particle, 5-14 The normalized vavefunction is 1/8 2 e (Onix ima eiEtA, += (up) 14 with E = &M/(C/m). (a) Since the Kinetic energy is E2/2m the correspondirg eperator is Ea Ts- DÊ. “O Therefore, Ee 1 a Es 2 Ze mea eta HG! te [o] EO a AR PR 2 a an 2 er, - Similarly for the potential energy U = 10: 32 E t Cm), | 2? 7 = tm dio e ax = HE) Ee du, ie É Te MO! Tot (b) This ssme relation, U = T= bE, is obeyed by the classical oscillator also. 5-15 Vse the notation be), = (pede [E cane E am é Ci da 1z ao Clearly te bp), = meo e + Tdx= (mel; — dk, implying that (xp) and (xp), cannot both be real. Also, by integrating by . me +. bel; - ate [o Eae) = app SE ax. Thus, tel, = teplj. T£ (xp), is real, this last relation says that Gp), 15 real also, which contradicts the first Finding above. Bence xl, às complex and therefore so is Ge). Moe try == = 2,2 , =] sobre ds + Bor eu — = = 4lbe), + teplsh = Sith, + Gplgd, *» = Reixp); so that this new Xp 1s real, as desired. del With V = 0, the energy cf the photon is E= po. Replacing the energy E and momentum p by their operators gives Mm ET dio Nou set Fix,t) = plbdTIt) and divide the equation by 7 to get niZ- e lã-r where K is independent of x and t. Write E = kjc and the tuo equations directly above become EE = suor + q «qt, É = my + peelx Hence, for the photon, Ta lt te-et) E EmA (a), (b) The curvature of 4 15 proporticnal to |V - E]: where tu = 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, Leading to an odd function since V 1s symmetrical about the origin. The wavefunction decays exponential ly wherever WE, the classicalIy forbidden region. For further discussion, see Exemple 5-12, which treats the símilar simple harmento cacillator potential. 
