Solutions Of Selected Problems And Answers PDF

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Solutions of Selected Problems and AnswersChapter 1Problem 1.5sThe sphere and the probability distribution have both inversion and rotationsymmetry; the first implies x y z 0 and the second in combinationwith the first implies 1 2 Δx2 x2 Δy 2 y 2 Δz 2 z 2 r .3Hence,13 2 19 23Δp2x . εk 2m2m 4 Δx28m r2 For a uniform probability, within the sphere of radius r0 and volume V (4π/3) r03 , r2 (3/5) r02 (3/5) (3/4π)2/3 V 2/3 0.2309 V 2/3 . Thus εk 4.87 2/mV 2/3 .Problem 1.9sThe quantity λm must depend on:(a) , since black body’s radiation is of quantum nature(b) c, since it is an electromagnetic phenomenon(c) kB T , since T is the only parameter in the spectral distribution of thisradiation; furthermore, absolute temperature is naturally associated withBoltzmann’s constant, kB , as a product kB T with dimensions of energyOut of , c, kB T , there is only one combination with dimensions of length c/kB T (remember that c has dimensions of energy times length). Hence,λm c1 c,kB Twhere the numerical constant c1 1.2655

780Solutions of Selected Problems and AnswersProblem 1.10sThe scattering cross-section has dimensions of length square. The photonscattering by electron must depend on:(a) The electron charge, e, since, if this charge was zero, there would be nointeraction and no scattering.(b) The velocity of light, since we are dealing with an electromagnetic phenomenon.(c) The mass of the electron, me , since if the mass was infinite, the electronwould not oscillate under the action of the photon electromagnetic fieldand would not emit radiation.(d) The energy of the photon, ω.The quantity e2 /4πε0 me c2 has dimensions of length (since e2 /4πε0 r hasdimensions of energy). Hence, the cross-section, σ, is of the form σ e24πε0 me c2 2 ω,fm e c2where the function f of the dimensionless quantity ω/mec2 cannot be determined from dimensional analysis; it turns out that f (x) is a monotonicallydecreasing function of x with f (0) 8π/3, and f ( ) 0. (The formula forσ is known as the Klein–Nishina formula).Problem 1.11sThe natural linewidth, ΔE, has dimensions of energy. It must depend on:(a) The dipole moment p e · r (as suggested by the question).(b) The velocity of light, c since the decay is due to the emission of a photon.(c) The frequency of the emitted photon, since only an oscillating dipole emitsradiation.Hence, by finding the only combination of p, c, ω with dimensions ofenergy, we obtainΔE c14πε0 c3 e2 r 2 ω 3 and the lifetime tl ,34πε0 cΔEc1 e 2 r 2 ω 3where the “constant” c1 depends on the details of the initial and the finalatomic level, since actually r2 is the square of the matrix element of an appropriate projection of r between the initial and the final states. We expect that,for the transition 2p to 1s in hydrogen, r2 to be of the order of a2B . Choosing,arbitrarily, c1 1 and r2 a2B , while ω 13.6 1 14 eV 10.2 eV we findfor this transitiontl 1.17 10 9 s,while a detailed advanced calculation gives tl 1.59 10 9 s.

Solutions of Selected Problems and Answers781Problem 1.12sThe effective absolute temperature, T , would appear as the product kB T ofdimensions of energy. It must depend on:(a) The product GM , where G is the gravitational constant. The reason isthat the strong gravitational field responsible for this radiation dependson the product GM .(b) Planck’s constant , since the phenomenon is of quantum origin.(c) The velocity of light, c, since we are dealing with electromagnetic radiation.The reader may convince himself that out of GM , and c there is onlyone combination to give dimensions of energy, namely, c3 /GM . HencekB T c1 c3 /GM,where the numerical constant c1 turns out to be equal to 1/8π.Problem 1.13sIt is more convenient for dimensional analysis to employ the G-CGS system(to get rid of ε0 and μ0 ). The skin depth will depend on:(a)(b)(c)The frequency ω (see the statement of the problem).The velocity of light c (EM phenomenon).Theσ (we are dealing with a good conductor). But [σ] 2 conductivitye / aB [t] 1 . Hence,δ c1ωωf; it turns out that fωσσσ1 .2πChapter 2Problem 2.3sAccording to the book by Karplus and Porter [C65], the minimum of the curveappears at d 0.74 A 1.4 a.u. and it is equal to 4.75 eV. Assuming thatat d 5 a.u., the energy is still given by the van der Waals, we haveεd̄ 5 6.48/d 6 a.u. 6.48/56 a.u. 11.3 meV.At d 0.5 a.u., the total energy (excluding the proton–proton repulsion)is expected to be slightly higher than the total electronic energy of the Heatom. The latter energy is equal to minus the sum of the first and the secondionization potential of He, i.e., 24.587 54.418 eV 79 eV. To this energywe must add 2 13.6 eV to be consistent with our choice of the zero of energy.

782Solutions of Selected Problems and Answers2Thus the total energy at d 0.5 a.u. is higher than 79 27.2 4πεe 0 d 51.8 27.20.5 eV 2.6 eV.In Fig. 2.5, we plot the experimental curve and the two points weestimated above.Fig. 2.5. Interaction energy of two hydrogen atoms vs their separationProblem 2.5s 11 ψ(r) 2 Ae 2r/a ,4πr2 dr ψ 2rr0 2 2r/adrre1/ (2/a)10 3 a,2 2r/adrr e2/ (2/a) 0 4 2r/a 2 drr e4!/ (2/a)5ar 0 3 4 3 22 2r/a2!/ (2/a)0 drr e2 3a2 ,

Solutions of Selected Problems and Answersp22m 783 d4π 2dψdψr24πr2 dr ψr2drdr2mdrdr 2 d r/a dr dr edr 2 0 e r/a dr 2 2 a/4 ,32m2m 2a /82ma2drr2 e 2r/a0 22m ψ1 dr2 dr ε 4πε0 2 0 a 2 aB , ae m1 2 1 2p Δp2i ,33 a21 2 3 21r a a2 , Δp2i Δx2i 2 ,Δx2i 333 2e2, 2ma24πε0 a Δr r2 3aB , ε vs. Δp2i Δx2i 24 .Problem 2.6sSolving the systemr r 1 r 2,R (m1 r 1 m2 r 2 ) / (m1 m2 ) ,with respect to r 1 , r 2 , we findm2m1r, r2 R r.MMHence, taking into account that p 1 m1 ṙ1 , p 2 m2 ṙ2 , we haver1 R 2μ2 ṙ2m1 Ṙp 21 μṘṙ,2m122m1Summing the two equations we find2μ2 ṙ2p 22m2 Ṙ μṘṙ.2m222m2p2p211P2p 212 , QED. 2 M Ṙ μṙ 2 2m12m2222M2μProblem 2.8sβFrom virial theorem E 1 β2 V̄ , where V (x) x . According to thecorrespondence principle (valid for large n), to go from the level n to leveln 1, V̄ must go from V̄ (x) to V̄ (x δx), where δx λ; but λ 2 EK E.Thus V̄ V̄ 1 .δE E(n 1) E(n) λ x x EButδE (dE/dn) αE/n E/E 1/a E 1 (1/a) , V̄βλ β x / x λ V̄ 1 (1/β) E 1/2 . xsince E na ,

784Solutions of Selected Problems and AnswersSubstituting in δE ( V̄ / x)λ the last two relations and taking into accountthat E V̄ , we have11111E 1 a V 1 β E 2 E 1 β 2 ,or111 a a2 β 11 2 β 1.(1)In spite of the hand-waving character of “deriving” (1), the latter is valid(for every n) for β 2 (harmonic potential) and for β 1 (Coulombpotential), since the exact results are a 1 and a 2 respectively. a The WKB approach (see Q34, p.447) gives that E n 12 .Problem 2.11sWe introduce the quantities q and k as follows: 2 q 2 /2m εb and 2 k 2 /2m ε εb .Then the ground state has the form:ψ AJ0 (kr) , r a; ψ BK0 (qr) , r a;(1)(1)(K0 (z) iπ2 H0 (iz) is the modified Bessel function of zero-order; seeTable H.18). The continuity of the logarithmic derivative ψ /ψ at r a leadsto the following relationqK0 (qa)kJ0 (ka) ,J0 (ka)K0 (qa)(2)where the prime denote differentiation with respect to the corresponding argument ka or qa respectively. For small values of εb and ε (in comparison withE0 2 /ma2 ), ka and qa are much smaller than one. Expanding J0 (ka) andK0 (qa) we have 1(3)J0 (ka) 1 (ka)2 O k 4 a4 ,4 22 (qa)(qa)qa 4 4 qa γ 1 O lnq a .K0 (qa) ln2442Substituting in (2) we have k 2 a21 1 . 2 ln 2 eγ qa(4)Taking into account that q 2 a2 2εb / E0 and k 2 a2 2( ε εb )/ E0 2 ε / E0 , we obtain the relation 22E0.εb 2γ E0 exp e ε

Solutions of Selected Problems and Answers785Chapter 3Problem 3.1sAccording to (3.1) the viscosity η is equal to μs t, where μs is the shear modulus and t is a characteristic time of motion of each water molecule; t isexpected to be of the order of the period of molecularvibration T in ice: 2t c1 T 2πc1 / ω, where ω c2 / me a2B r̄wme / mw and c1 , c2 arenumerical constants of the order of one. Substituting mw 18 1823 meand r̄w (2.68 18)1/3 3.64 we have ω c2 1.72 1013 rad/s. The shearmodulus μs (in ice) is expected to be around 0.3 B where B is the bulk mod5 9.2 109 N/m2 (the numericalulus of water, where B 0.2 2 / me a5B r̄wcoefficient was taken 0.2 and not 0.6 as usually, because the hydrogen bond ismuch weaker than the strong bonds for which the 0.6 is a reasonable choice).Hence, the result for η isη (2πc1 / c2 )(0.3 9.2 109 /1.72 1013 ) (c1 / c2 )10 3 kg/ms,which coincides with the experimental result if c1 c2 .Problem 3.2sH2 O has larger cohesive energy, because H2 O possesses a dipole moment(being non-linear), while CO2 is non-polar (being a symmetric linear molecule).Problem 3.12sConsider a rotation of a Bravais Lattice, by an angle θ, around the axis z, ofan orthogonal Cartesian system, passing through a lattice point. The matrix(θ) implementing this rotation has the form cos θsin θ 0 (1)(θ) sin θ cos θ 0 . 001 denote the same rotation in the system a 1 , a 2 , a 3 by the matrix Now we (θ). If(θ) is compatiblewith the translational symmetry of the lat nawill be mapped to another lattice pointtice,eachlatticepointiii (θ) [ni ]; where [n i ] and [ni ] are column matrices. Fori ni a i : [ni ] {n i } (i 1, 2, 3) to be integers, for any set of three integers n 1 , n2 , n3 the (θ), must be integers. Since(θ) and(θ) describematrix elements ofthe same rotation in different coordinate systems, they must be related by atransformation of the form (θ) S 1(θ)S,(2)

786Solutions of Selected Problems and Answerswhere S is a 3 3 matrix, connecting the two coordinate systems. By takingthe trace of (2), we have (θ)S TrSS 1(θ) Tr(θ).(3)Tr(θ) TrS 1 The Tr (θ)2 cos θ 1 and the Tr (θ) is an integer, since all matrix (θ) are integers. Henceelements of2 cos θ integer,from which it follows that θ 2π/n, n 1, 2, 3, 4, 6.Chapter 4Problem 4.7tsFor Cu, ζ 2.57, r̄c 1.113, η 0.6025, a 4.429 6.2687 10.70,γγ 1.97 5.944 7.916, r̄a 2.70, B 1.16 Mbar, PP 294 4πr̄a4185.2 3.48 Mbar.r̄a4Problem 4.1sBinding energies in eV(Th.: Theory: Exp.: Experiment) Na (Th. 4.92;Exp. 6.25), K (Th. 4.07; Exp. 5.27), Mg (Th. 19.98; Exp. 24.19), Ca(Th. 15.73; Exp. 19.82), Fe (Th. 56.29; Exp. 59.02), Al (Th. 52.49; Exp. 56.65),Ti (Th. 69.36, Exp. 96.01).Problem 4.2sDebye temperature in degrees K according to the RJM for some solids:Al (419) , Cu (300) , Au (142) , Fe (497) , Pb (80) , Mg (344) , Be (1322) .Problem 4.3sHint: Combine (C.25) with (4.100). At T ΘD use (4.51) for B and (4.66)for ΘD .Problem 4.7s The constant electronic charge density is ρ 3ζe / 4π ra3 rc3 , rc r ra ;from Gauss theorem the electric field ε(r) is 4πε0 ε(r)r2 (4π / 3)ρ r3 r c3 .The potential φ(r) ε(r)dr is 4πε0 φ(r) (2π / 3)ρr2 4πρrc3 /3r

Solutions of Selected Problems and Answers7873const. const. is determined from 4πε0 φ(ra ) ζe / ra : const 3ζera / 2 3 The3ra rc . The classical electronic Coulomb self-energy isEe e1 2 raφ(r)ρ(r)d3 r rc 36 5ζ 2 e243 xx, 2 4πε0 (1 x2 )2 ra 1010x rc.raThe electrostatic Coulomb electron-ion interaction is ra 3 ζ 2 e2 ra2 rc2ζe .d r4πε0 r2 4πε0 (ra3 rc3 )3Ee i ρrcAdding Ee e and Ee i , we obtain the classical electrostatic energy per atomin agreement with the given formula.Chapter 5Problem 5.4sValues of ωpf (in eV) for some solids according to (5.27).Li (8.39) , Na (6.62) , K (4.87) , Rb (4.45) , Mg (11.21) , Al (14.98) , Ag (9.33) .Problem 5.9sFor Si and from Table 4.4 (p. 98), we have cl 8945 m/s and ct 5341 m/s.We shall choose c 7500 m/s. The Debye temperature ΘD 645 K, sothat T / ΘD 300 / 645 0.465 and CV 0.82 3Na kB ; V / Na 20 A3 ; ph 300 A. The result, according to (5.134), is Kph 127 Wm 1 K 1 1.27 Wcm 1 K 1 vs. 1.48 Wcm 1 K 1 experimentally.Problem 5.11sThe Fourier transform of f (r ) exp ( ks r) /r is f (k ) d3 r exp ( ik · r ) f (r ) 1d (cos θ) exp ( ikr cos θ) drr2 f (r) 2π 1 dr sin kr exp ( ks r) 4π/ k 2 ks2 . (4π/k)0

788Solutions of Selected Problems and AnswersProblem 5.13s dσ2The resistivity in SI is ρ υF / ε0 ωpf , where 1 ns dΩ(1 cos θ)dΩ.We have k 2 4kF2 sin2 (θ / 2) 2kF2 (1 cos θ), 2kdk 2kF2 d(cos θ), dΩ 2πd(cos θ) 2πkdk / kF2 . Moreover, dσ / dΩ (m2 / 4π 2 4 )(e2 ne / ε0 )2 2 2 4 22e ne / 4π 2 4 ε20 (V ωk nk / B)/(k 2 ε(k))2 k · 4u /k ε(k) 2 2 mλ0 ks /ρFV (m /4π 4 ) (V ωk nk )/(k22 ε(k))2 . We took into accountin the expressionthat λ0 / ρFV Es2 / B and Es e2 ne / ε0 ks2 . Substituting 1 1422 4 1for (mwehave λk/ρ/4π )VV ω nk k 2 /2kF20FVks 2πkdk/kF2 /(k 2 ε(k))2 λ0 ks4 m2 / 4πkF4 4 ρFV dkk 3 ωk nk /22(kor, by defining y β c̄k, 1 (λ0 m2 ks4 /4π 4 kF4 ρFV )(24 kF4 /y04 ) ε(k))3dyy ωk k4 ε12 (k) ey1 1 , where y0 2β c̄kF . We have taken into accountthat EF 2 kF2 /2m, EF ρFV (3/4)ne (3/4)(kF3 /3π 2 ) kF3 /4π 2 , andk 2 ε(k) k 2 ks2 ks2 [1 (k 2 /ks2 )] and we have 2 1/ks2where k given by (D.28).λo m 4π 2 16kB T 8π pF y04 y0dyy 4[1 0(2by 2 /y02 )]2 (ey2 1)22(k /kF2 )(kF2 /kTFf ) (2y 2 /y02 )(2kF2 /kTFf)22We write b 2kF /kTF f . ThusmυF 8π 2 λ0 4kB TυF ρ 2 2ε0 ωρfpF 4πε0 ωpfy04 y00,and f (k/kf ) isdy y 4,[1 (2by 2 /y02 )]2 (ey 1)which coincides with (5.59) since mυF pF .Problem 5.15smυ̇ eE eυ B,cemB υ̇ eB E B (υ B) ,cB (υ B) υB 2 B (υ · B) ,e 2υB B (υ · B) ,mB υ̇ eB E cee 2B υ mB υ̇ eB E (υ · B) B,ccmc 1c1υ B υ̇ 2 B E 2 (υ · B) B,e B2BBmc 1cB 0 υ̇ B 0 E (υ · B 0 ) B 0 ,υ e BBmcB 0 υ̇ υ 0 (υ · B 0 ) B 0 ,υ eBmcB 0 υ̇ υ 0 , υ (υ · B 0 )B 0 Const.υ eB

Solutions of Selected Problems and Answers789Chapter 6Problem 6.5sWe implement the successive transformations shown in Fig. 6.16a and at eachstage we calculate the matrix elements of the Hamiltonian11χ1n 1 (sn 1 px,n 1 ) , χ2n (sn px,n ) ,221112χn (sn px,n ) , χn 1 (sn 1 px,n 1 ) .22For the matrix elements of Ĥ we have ε ε ps, i n 1, n, n 1,χ1i Ĥ χ1i χ2i Ĥ χ2i 2 εp εs χ1i Ĥ χ2i χ2i Ĥ χ1i V1 , i n 1, n, n 1,2 1 sn 1 Ĥ sn sn 1 Ĥ px,n px,n 1 Ĥ snχ1n 1 Ĥ χ1n 2 px,n 1 Ĥ px,n 2 2[ 1.32 1.42 1.42 2.22] 0.45.22mdmd2 Similarly, χ2n Ĥ χ2n 1 χ1n Ĥ χ2n 1 ψbn 1,n ψa,n 1,n bbHn,n 1 aaHn,n 1 abHn,n 1 22md22 [ 1.32 1.42 1.42 2.22] 0.45 md2, 2 2[ 1.32 1.42 1.42 2.22] 3.19,2md2md2 1 1 χ1n 1 χ2n , ψbn,n 1 χ1n χ2n 1 ,22 1 11 12 χn 1 χn , ψan,n

Solutions of Selected Problems and Answers 785 Chapter 3 Problem 3.1s According to (3.1) the viscosity η is equal to μst,whereμs is the shear mod- ulus and t is a characteristic time of motion of each water molecule; t is expected to be of the order of the period of molecular vibration T in ice: t = c1T =2πc1 /ω,whereω = c2 /mea2 B