材料物理性能部分课后习题8页

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材料物理性能课后习题答案_北航出版社_田莳主编(仅供借鉴)

材料物理习题集第一章固体中电子能量结构和状态（量子力学基础）1. 一电子通过5400V 电位差的电场，（1）计算它的德布罗意波长；（2）计算它的波数；（3）计算它对Ni 晶体（111）面（面间距d =2.04×10-10m ）的布拉格衍射角。

（P5）12341311921111o '(2)6.610 =(29.1105400 1.610)=1.67102K 3.7610sin sin 2182hh pmE m d dλπλθλλθθ----=⨯⨯⨯⨯⨯⨯⨯=⨯==⇒=解：（1）=（2）波数=（3）22. 有两种原子，基态电子壳层是这样填充的;;s s s s s s s 2262322626102610（1）1、22p 、33p （2）1、22p 、33p 3d 、44p 4d ，请分别写出n=3的所有电子的四个量子数的可能组态。

（非书上内容）3. 如电子占据某一能级的几率是1/4，另一能级被占据的几率为3/4，分别计算两个能级的能量比费米能级高出多少k T ？（P15）1()exp[]11ln[1]()()1/4ln 3()3/4ln 3FF F F f E E E kT E E kT f E f E E E kT f E E E kT=-+⇒-=-=-=⋅=-=-⋅解：由将代入得将代入得4. 已知Cu 的密度为8.5×103kg/m 3，计算其E 0F 。

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

（Na 的摩尔质量M=22.99，.0ρ⨯33=11310kg/m ）（P16）220323426233311900(3/8)2(6.6310) 1.01310 =(3 6.0210/8)291022.99=5.2110 3.253 1.085FF h E n mJ eVE E eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯===解：由由 6. 若自由电子矢量K 满足以为晶格周期性边界条件x x L ψψ+()=()和定态薛定谔方程。

材料物理性能第八章课后作业课后习题英文版

Third Edition (© 2001 McGraw-Hill) Chapter 88.1 Inductance of a long solenoid Consider the very long (ideally infinitely long) solenoid shown in Figure 8.69. If r is the radius of the core and is the length of the solenoid, then >> r. The total number of turns is N and the number of turns per unit length is n = N/ . The current through the coil wires is I. Apply Ampere's law around C, which is the rectangular circuit PQRS,and show thatB≈μoμr nIFurther, show that the inductance isL≈μoμr n2V core Inductance of long solenoid where V core is the volume of the core. How would you increase the inductance of a long solenoid?Figure 8.69What is the approximate inductance of an air-cored solenoid with a diameter of 1 cm, length of 20 cm, and 500 turns? What is the magnetic field inside the solenoid and the energy stored in the whole solenoid when the current is 1 A? What happens to these values if the core medium has a relative permeability μr of 600? SolutionWe use Ampere's law in Equation 8.15. Consider Figure 8.9. If H is the field along a small length d along a closed path C, then around C, ⎰Hd= total threaded current = I total = NI.Figure 8.9:Ampere’s circuital lawAssume that the solenoid is infinitely long. The rectangular loop PQRS has n (PQ ) number of turns where n is the number of turns per unit length or n = N / (See Figure 8.69). The field is only inside the solenoid and only along the PQ direction (long solenoid assumption) and therefore the field along QR , RS and SP is zero. Assume that the field H is uniform across the solenoid core cross section. Then the path integral of the magnetic field intensity H around PQRS is simply is H = H (PQ ). Ampere's law ⎰ Hd = I total is then H (PQ ) = I (nPQ ) i.e.H = nIThe dimensions of the solenoid are such that length >> diameter. We can assume that H field isrelatively uniform at all points inside the solenoid. Note: The approximate equality sign in the text (equation for B ) is due to the fact that we assumed H is uniform across the core and, further, along the whole length of the solenoid from one end to the other. The ends of the solenoid will have different fields (lower). Let A be the cross-sectional area of the solenoid. The magnetic field B , the flux Φ and hence the inductance L are B = μo μr H ≈ μo μr nI∴ Φ = BA ≈ μo μr nAI = μo μr (N / )AIand L = (N Φ)/I = N [μo μr (N / )AI ]/I = μo μr (N 2/ )A = μo μr n 2( A ) ∴L = μo μr n 2V corewhere V core is the volume of the core. Inductance depends on n 2, where n is the number of turns per unit length, on the relative permeability μr and on the volume of the core containing the magnetic flux. For a given volume inductor, L can be increased by using a higher μr material or increasing n , e.g. thinner wire to get more turns per unit length (not so thin that the skin effect diminishes the Q -factor, quality factor; see §2.8). The theoretical inductance of the coil is L = (4π ⨯ 10-7 H/m)(1)[(500)/(0.2 m)]2(0.2 m)(π)[(0.01 m)/(2)]2 ∴ L = 1.23 ⨯ 10-4 H or 0.123 mHB ≈ (4π ⨯ 10-7 Wb A -1 m -1)(1)[(500)/(0.2 m)](1 A) = 3.14 ⨯ 10-3 T The energy per unit volume is,E vol = B 2/(2μo ) = (3.14 ⨯ 10-3 T)2/[2(4π ⨯ 10-7 Wb A -1 m -1)] ∴E vol = 3.92 J / m 3The total energy stored is then,()()()J 61.6tot μm 2.02m 01.0J/m 92.3Area Length 23vol =⎪⎭⎫ ⎝⎛=⨯=πE E Suppose that μr = 600 and suppose that the core does not saturate (an ideal ferromagnetic material) then, ()()H 0.0738=⨯≈-600H 1023.14L()()T 1.88=⨯≈-600T 1014.33Band()()()372vol J/m 2344600m Wb/A 1042T 88.1=⋅⨯≈-πEso that E tot = (E vol )(Volume) = 36.8 mJThis is a dramatic increase and shows the virtue of using a magnetic core material for increasing theinductance and the stored magnetic energy.8.2 Magnetization Consider a long solenoid with a core that is an iron alloy (see Problem 8.1 for therelevant formulas). Suppose that the diameter of the solenoid is 2 cm and the length of the solenoid is 20 cm. The number of turns on the solenoid is 200. The current is increased until the core is magnetized to saturation at about I = 2 A and the saturated magnetic field is 1.5 T.a . What is the magnetic field intensity at the center of the solenoid and the applied magnetic field, μo H , forsaturation? b . What is the saturation magnetization M sat of this iron alloy?c . What is the total magnetization current on the surface of the magnetized iron alloy specimen?d . If we were to remove the iron-alloy core and attempt to obtain the same magnetic field of 1.5 T inside thesolenoid, how much current would we need? Is there a practical way of doing this?Solutiona. Applying Ampere’s law or H = NI we have,()()m2.0A 2200==NI H Since I = 2 A gives saturation, corresponding magnetizing field is H sat ≈ 2000 A/mSuppose the applied magnetic field is the magnetic field in the toroid core in the absence of material. ThenB ap p =μo H sat =4π⨯10-7 Wb A -1 m -1()2000 A/m () ∴ B app = 2.51 ⨯ 10-3 Tb. Apply()sat sat sat H M B o +=μ∴ A/m 2000mA Wb 104T5.11-1-7sat satsat -⨯=-=-πμH B M o∴ M sat ≈ 1.19 ⨯ 106 A/m c. Since M is the magnetization current per unit length,I m = M sat ≈ 1.19 ⨯ 106 A/mThen I surf ace = Total circulating surface current:∴I surface =I m =1.19⨯106 A/m ()0.2 m ()=2.38⨯105 ANote that the actual current in the wires, 2 A is negligible compared with I surf ace . d. Apply, B ≈μo nI (for air)()⎪⎭⎫⎝⎛⨯≈-m 2.0200m A Wb 104T5.11-1-7πI = 1194 ANot very practical in every day life! Perhaps this current (thus field B = 1.5 T) could be achieved byusing a superconducting solenoid.8.3 Paramagnetic and diamagnetic materials Consider bismuth with χm = -16.6×10-5 andaluminum with χm = 2.3×10-5. Suppose that we subject each sample to an applied magnetic field B o of 1 T applied in the +x direction. What is the magnetization M and the equivalent magnetic field μo M in each sample? Which is paramagnetic and which is diamagnetic?SolutionBismuth: χm = -16.6×10-5 χm is negative and small. Bismuth is a diamagnetic material.M = χm H = χm B o /μo∴M = (-16.6×10-5) (1 Wb m -2)/(4π×10-7 Wb m -1 A -1) = -132.1 A m -1Negative sign indicate – x direction.Magnetic field = B o + μo M = B o + χm B o = B o (1 + χm ) = (1 - 16.6×10-5)(1 T) = 0.999834 TAluminum: χm = 2.3×10-5 χm is positive and small. Aluminum is paramagnetic material.M = χm H = χm B o /μo∴M = (2.3×10-5) (1 Wb m -2)/(4π×10-7 Wb m -1 A -1) = 18.3 A m -1Positive sign indicates + x direction.Magnetic field = B o + μo M = B o + χm B o = B o (1 + χm ) = (1 +2.3×10-5)(1 T) = 1.000023 T Author's Note: Both effects are quite small.8.4 Mass and molar susceptibilities Sometimes magnetic susceptibilities are reported as molar ormass susceptibilities. Mass susceptibility (in m 3 kg -1) is χm /ρ where ρ is the density. Molar susceptibility (inm 3 mol -1) is χm (M at /ρ) where M at is the atomic mass. Terbium (Tb) has a magnetic molar susceptibility of 2 cm 3 mol -1. Tb has a density of 8.2 g cm -3 and an atomic mass of 158.93 g mol -1. What is its susceptibility, masssusceptibility and relative permeability? What is the magnetization in the sample in an applied magnetic field of 2 T?Solutionχm = Molar susceptibility (ρ/M at ) = (2 cm 3 mol -1) (8.2 g cm -3)/(158.93 g mol -1) = 0.1032 μr = 1 + χm = 1 + 0.1032 = 1. 1032 M = χm H = χm B o /μo∴M = (0.1032)(2 Wb m -2)/(4π×10-7 Wb m -1 A -1) = 1.642×105 A m -1Note: The magnetic field in the sample iso r o r B H B μμμ===1.1032( 2 T) = 2.206 T.8.5 Pauli spin paramagnetism Paramagnetism in metals depends on the number of conductionelectrons that can flip their spins and align with the applied magnetic field. These electrons are near the Fermilevel E F , and their number is determined by the density of states g (E F ) at E F . Since each electron has a spin magnetic moment of β, paramagnetic susceptibility can be shown to be given byχpara ≈ μo β 2 g (E F )Pauli spin paramagnetismwhere the density of states is given by Equation 4.10. The Fermi energy of calcium, E F , is 4.68 eV. Evaluate the paramagnetic susceptibility of calcium and compare with the experimental value of 1.9 ⨯ 10-5.SolutionApply,()()E h m E e 23228⎪⎭⎫⎝⎛=πg(Equation 4.10)so that ()()()()()J/eV 10602.1eV 68.4s J 10626.6kg 10109.928192323431---⨯⎪⎪⎭⎫⎝⎛⋅⨯⨯=πF E g∴g E F ()=9.197⨯1046 J -1 m -3Then, χpara ≈ μo β 2 g (E F )()()()314622241-1-7para m J 10199.9m A 10273.9m A Wb 104----⨯⨯⨯≈πχ∴χpara ≈ 0.994 ⨯ 10-5This is in reasonable agreement within an order of magnitude with the experimental value of 1.9 ⨯ 10-5.8.6 Ferromagnetism and the exchange interaction Consider dysprosium (Dy), which is arare earth metal with a density of 8.54 g cm -3 and atomic mass of 162.50 g mol -1. The isolated atom has theelectron structure [Xe] 4f 106s 2. What is the spin magnetic moment in the isolated atom in terms of number of Bohr magnetons? If the saturation magnetization of Dy near absolute zero of temperature is 2.4 MA m -1, whatis the effective number of spins per atom in the ferromagnetic state? How does this compare with the number of spins in the isolated atom? What is the order of magnitude for the exchange interaction in eV per atom in Dy if the Curie temperature is 85 K?SolutionIn an isolated Dy atom, the valence shells will fill in accordance with the exchange interaction:4f 106s 2Obviously, there are 4 unpaired electrons. Therefore for an isolated Dy atom, the spin magnetic moment = 4β.Atomic concentration in dysprosium (Dy) solid is (where ρ is the density, N A is Avogadro’s number and M at is the atomic mass):()()3283-12333m 10165.3kg/mol1050.162mol 10022.6kg/m 1054.8--⨯=⨯⨯⨯==atAat M N n ρSuppose that each atom contributes x Bohr magnetons, then βx n M at =sat()()2243286sat mA 10273.9m 10165.3A/m104.2--⨯⨯⨯==βat n M x = 8.18 This is almost twice the net magnetic moment in the isolated atom. Suppose that the Dy atom in the solid loses all the 4 electrons that are paired into the "electron gas" in the solid. This would make Dy +4 have 8 unpaired electrons and a net spin magnetic moment of 8β (this is an oversimplified view).Exchange interaction ~ kT C =8.617⨯10-5 eV/K ()85 K ()=0.00732 eV The order of magnitude of exchange interaction ~ 10-2 eV/atom for Dy (small).8.7 Magnetic domain wall energy and thickness The energy of a Bloch wall depends on twomain factors: the exchange energy E ex (J /atom) and magnetocrystalline energy K (J m -3). If a is the interatomicdistance, δ is the wall thickness, then it can be shown that the potential energy per unit area of the wall isδδπK a E U +=2ex2wall Potential energy of a Bloch wallShow that the minimum energy occurs when the wall has the thickness2/1ex 22⎪⎪⎭⎫ ⎝⎛='aK E πδBloch wall thicknessand show that when δ = δ', the exchange and anisotropy energy contributions are equal . Using reasonable values for various parameters, estimate the Bloch energy and wall thickness for Ni. (See Example 8.4)SolutionδδπK a E U +=2ex2wall∴K a E d dU +-=2ex2wall 2δπδ Minimum energy occurs, when 0wall=δd dU ∴ 022ex 2=+'-K a E δπ ∴2/1ex 22⎪⎪⎭⎫ ⎝⎛='aK E πδAt δδ'=δδδδπ'=''='=K K a E U 2ex 2exchange 2Andexchange anisotropy U K U ='=δFor Ni, T C = 631 K and K = 5 mJ cm -3 = 5×103 J m -3 E ex = kT C = (1.38×10-23 J K -1) (631 K) = 8.71×10-21 J ∴⎥⎦⎤⎢⎣⎡⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛='---)m J 10(5 m)103.0(2)J 1071.8(23392122/1ex 2ππδaK E = 1.69×10-7m or 169 nm And m)10(1.69)m J 105(m)10m)(1.69103.0(2 J) 108.71(27-337-9-212ex 2wall ⨯⨯+⨯⨯⨯='+'=--πδδπK a E U= 1.69×10-3 J m -2 or 1.69 mJ m -2.*8.8 Toroidal inductor and radio engineers toroidal inductance equationa . Consider a toroidal coil (Figure 8.10) whose mean circumference is and has N tightly wound turnsaround it. Suppose that the diameter of the core is 2a and >> a . By applying Ampere's law, show that if the current through the coil is I , then the magnetic field in the core isNIB r o μμ=[8.30]where μr is the relative permeability of the medium. Why do you need >> a for this to be valid? Doesthis equation remain valid if the core cross section is not circular but rectangular, a ⨯ b , and >> a and b ? b . Show that the inductance of the toroidal coil isAN L r o 2μμ=Toroidal coil inductance [8.31]where A is the cross-sectional area of the core.c . Consider a toroidal inductor used in electronics that has a ferrite core size FT-37, that is, round but with arectangular cross section. The outer diameter is 0.375 in (9.52 mm), the inner diameter is 0.187 in (4.75 mm), and the height of the core is 0.125 in (3.175 mm). The initial relative permeability of the ferrite core is 2000, which corresponds to a ferrite called the 77 Mix. If the inductor has 50 turns, then using Equation 8.31, calculate the approximate inductance of the coil. d . Radio engineers use the following equation to calculate the inductances of toroidal coils,6210)mH (N A L L =Radio engineers inductance equation [8.32]where L is the inductance in millihenries (mH) and A L is an inductance parameter, called an inductance index , that characterizes the core of the inductor. A L is supplied by the manufacturers of ferrite cores and is typically quoted as millihenries (mH) per 1000 turns. In using Equation 8.32, one simply substitutes the numerical value of A L to find L in millihenries. For the FT-37 ferrite toroid with the 77 Mix as the ferrite core, A L is specified as 884 mH/1000 turns. What is the inductance of the toroidal inductor in part (c ) from the radio engineers equation in Equation 8.32? What is the percentage difference in values calculated by Equations 8.32 and 8.31? What is your conclusion? (Comment : The agreement is not always this close).SolutionFigure 8.10: A toroidal coil with N turns.a. As in Figure 8.10, if we choose a closed path C that runs along the geometric center of the core and if I is the current, N is the number of turns and = mean circumference, then: NIH d H Ct ==⎰ or H = (NI )/∴NIH B r o r o μμμμ==This particular derivation only applies in the special case of >> radius (a ); that is, for a very long, narrow core. If the core is very long and narrow, it may be safely assumed that the magnetic flux density B is uniform across the entire width (2a ) of the core. If B was not uniform, then applying Ampere’s Law to different (concentric) closed paths would yield different results.This above derivation for B is valid for a rectangular cross-sectioned core of area a ⨯ b , provided that >> a and >> b . The magnetic field is then,B =μo μrNIb. The inductance by definition is given by,A N I NAI N I BA N I N L r o r o 2)(μμμμ=⎪⎭⎫ ⎝⎛==Φ=c.Figure 8Q8-1: Toroidal core with rectangular cross section.Given, outer radius = r outer = 0.00476 m, inner radius = r inner = 0.002375 m and height = H = 0.003175 m.We take the mean circumference through the geometric center of the core so that the mean radius is: mm 5675.32inner innerouter =+-=r r r r ∴()()m 1042.22m 105675.32233--⨯=⨯==ππ rWidth = W = r outer - r inner = 0.002385 mA = Cross sectional area = W ⨯ H = (0.002385 m)(0.003175 m) = 7.572 ⨯ 10-6 m 2. Since >> a and >> b (at least approximately) we can calculate L as follows:AN L r o 2μμ=∴ ()()()()m1042.22m 10572.7502000H/m 10432627---⨯⨯⨯=πL ∴L = 0.00212 H or 2.12 mH (2)d. Using the radio engineer’s equation,()()()mH 2.21mH ===62621050mH 88410N A L L (3)so that 4.07%=⨯-=%100mH21.2mH12.2mH 21.2difference %Equations 2 and 3 differ only by 4.1% in this case. This is a good agreement.*8.9 A toroidal inductora . Equations 8.31 and 8.32 allow the inductance of a toroidal coil in electronics to be calculated. Equation8.32 is the equation that is used in practice. Consider a toroidal inductor used in electronics that has a ferrite core of size FT-23 that is round but with a rectangular cross section. The outer diameter is 0.230 in (5.842 mm), the inner diameter is 0.120 in (3.05 mm), and the height of the core is 0.06 in (1.5 mm). The ferrite core is a 43-Mix that has an initial relative permeability of 850 and a maximum relative permeability of 3000. The inductance index for this 43-Mix ferrite core of size FT-23 is A L =188 (mH/1000 turns). If the inductor has 25 turns, then using Equations 8.31 and 8.32, calculate the inductance of the coil under small-signal conditions and comment on the two values. b . The saturation field, B sat , of the 43-mix ferrite is 0.2750 T. What will be typical dc currents that willsaturate the ferrite core (an estimate calculation is required)? It is not unusual to find such an inductor in an electronic circuit also carrying a dc current? Will your calculation of the inductance remain valid in these circumstances? c . Suppose that the above toroidal inductor discussed in parts (a) and (b) is in the vicinity of a very strongmagnet that saturates the magnetic field inside the ferrite core. What will be the inductance of the coil?Solutiona. Provided that the mean circumference is much greater than any long cross sectional dimension (e.g. >> diameter or >> a and >> b ), then we can use,AN L r o 2μμ≈Note that μr is the initial permeability. We need the mean circumference which can be calculated from the mean radius r ,mm 97.132 that so mm 223.22inner innerouter ===+-=r r r r r π Width = W = r outer - r inner = 0.001396 m and Height = H = 0.0015 m.A = Cross sectional area = W ⨯ HSince >> a and >> b (at least approximately) we can calculate L as follows:)(2WH N L r o μμ≈()()()()()m1097.13m105.1m 10396.125850H/m 10433327----⨯⨯⨯⨯≈πLL ≈ 0.10 mHAnd, the radio engineer’s equation, Equation 8.32, (radio engineers toroidal inductance equation, pg. 6.25 and data pg. 24.7 in The ARRL Handbook 1995)()()mH 0.118===62621025mH 18810N A L L There is a 15% difference between the two inductances; limitations of the approximation are apparent. b. To estimate H sat , we’ll take the maximum relative permeability μr max = 3000 as an estimate in order to findH sat (see Figure 8Q9-1). We know that sat max sat and H B NIH o r μμ==∴satmax sat NI B r o μμ=∴ ()()()m1097.13253000m A Wb 104T 2750.03sat-1-17--⨯⨯=I π∴I sat = 40.8 mAwill saturate the core.Figure 8Q9-1: B versus H curve for 43 - Mix ferrite .If the core is saturated, the calculation of inductance is of course no longer valid as it used the initialpermeability. The inductance now will be much reduced. Since under saturation ∆B = μ0∆H , effectively μr = 1. The use of an initial permeability such as μr = 850 implies that we have small changes (and also reversible changes) near around H = 0 or I = 0.When an inductor carries a dc current in addition to an ac current, the core will operate “centered” at a different part of the material’s B -H curve, depending on the magnitude of the dc current inasmuch as H ∝ I . Thus, a dc current I 1 imposes a constant H 1 and shifts the operation of the inductor to around H 1. μr will depend on the magnitude of the dc current.c. The same effect as passing a large dc current and saturating the core. When the core is saturated, theincrease in the magnetic field B in the core is that due to free space, i.e. ∆B = μo ∆H . This means we can use μ = 1 in Equation 8.31 to find the inductance under saturation. It will be about 0.10 mH / 850 or 0.12 μH , very small.Author’s Note: Static inductance can be defined as L = Flux linked per unit dc current or L = N Φ/I . It applies under dc conditions and I = dc current. For ac signals, the current i will be changing harmonically (or following some other time dependence) and we use the definition⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛Φ=Φ=dt di dt d N i N L δδ Further, by Faraday’s law, since v is the induced voltage across the inductor by the changing total flux Nd Φ/dt , we can also define L by,⎪⎭⎫ ⎝⎛=dt di L vMoreover, since L = N δΦ/δi we see that the ac L is proportional to the slope of the B versus H behavior.*8.10 The transformera . Consider the transformer shown in Figure 8.70a whose primary is excited by an ac (sinusoidal) voltage offrequency ƒ. The current flowing into the primary coil sets up a magnetic flux in the transformer core. By virtue of Faraday's law of induction and Lenz's law, the flux generated in the core is the flux necessary to induce a voltage nearly equal and opposite to the applied voltage. Thus,dtNAdBdt d ==)linked f lux Total (υwhere A is the cross-sectional area, assumed constant, and N is the number of turns in the primary.Show that if V rms is the rms voltage at the primary (V max = V rms √2) and B m is the maximum magnetic field in the core, thenV rms = 4.44 NAƒB mTransformer equation[8.33]Transformers are typically operated with B m at the "knee" of the B -H curve, which corresponds roughly to maximum permeability. For transformer irons, B m ≈ 1.2 T. Taking V rms = 120 V and a transformer core with A = 10 cm ⨯ 10 cm, what should N be for the primary winding? If the secondary winding is to generate 240 V, what should be the number of turns for the secondary coil?b . The transformer core will exhibit hysteresis and eddy current losses. The hysteresis loss per unitsecond, as power loss in watts, is given byP h = KƒB m n V coreHysteresis loss[8.34]where K = 150.7, ƒ is the ac frequency (Hz), B m is the maximum magnetic field (T) in the core (assumed to be in the range 0.2 - 1.5 T), n = 1.6 and, V core is the volume of the core. The eddy current losses are reduced by laminating the transformer core as shown in Figure 8.70b. The eddy current loss is given bycore 22265.1V d B f P me ⎪⎪⎭⎫⎝⎛=ρEddy current loss [8.35]where d is the thickness of the laminated iron sheet in meters (8.70b) and ρ is its resistivity (Ω m). Suppose that the transformer core has a volume of 0.0108 m 3 (corresponds to a mean circumference of 1.08 m). If the core is laminated into sheets of thickness 1 mm and the resistivity of the transformer iron is 6 ⨯ 10-7 Ω m, calculate both the hysteresis and eddy current losses at f = 60 Hz, and comment on theirrelative magnitudes. How would you achieve this?Figure 8.70(a) A transformer with N turns in the primary. (b) Laminated core reduces eddy current losses.Solutiona. The induced voltage either at the primary or the secondary is given by Faraday’s law of induction (the negative sign indicates an induced voltage opposite to the applied voltage), that is,dtdB NA-=υ Suppose we apply this to the primary winding. Then υ = V m sin(2πft ) where f = 60 Hz and V m is the maximum voltage, that is V m = V rms (√2). ThendtdB NAft V m -=π)2sin( It is clear that B is also a sinusoidal waveform. Integrating this equation we find,)2cos()2(ft f NA V B mππ=which can be written asB = B m cos(2πft )so that the maximum B is B m given byfNAV fNA V B m m ππ222rms==where we used V m = V rms (√2). Thus,V rms =4.44NAfB mLet N P = Primary winding turns and assume f = 60 Hz, then()()()()T 2.1Hz 60m 1.044.4V12044.42rms ==m AfB V P N = 38 turns on primary Secondary winding turns are()()()()T 2.1Hz 60m 1.044.4V2402=S N = 75 turns on secondary b. Part a gives B m = 1.2 T. Then the hysteresis loss isP h =KfB m nV co reB m = 1.2 T()()()()36.1m 0108.0T 2.1Hz 607.150=h P = 131 WEddy current loss iscore 22265.1V d B f P me ⎪⎪⎭⎫ ⎝⎛=ρB m = 1.2 T()()()()W 154=⎪⎪⎭⎫ ⎝⎛Ω⨯=-37222m 0108.0m 106m 001.0T 2.1Hz 6065.1e P With 1 mm lamination and at this low frequency (60 Hz) hysteresis loss seems to dominate the eddycurrent loss. Eddy current loss can be reduced with thinner laminations or higher resistivity core materials (e.g. ferrites at the expense of B max ). Hysteresis loss can be reduced by using different core material but that changes B . Eqn. 8.34 is valid for only silicon steel cores only which have the required typical B m for power applications.8.11 Losses in a magnetic recording head Consider eddy current losses in a permalloymagnetic head for audio recording up to 10 kHz. We will use Equation 8.35 for the eddy current losses.Consider a magnetic head weighing 30 g and made from a permalloy with density 8.8 g cm -3 and resistivity 6 ⨯ 10-7 Ω m. The head is to operate at B m of 0.5 T. If the eddy current losses are not to exceed 1 mW, estimate the thickness of laminations needed. How would you achieve this?SolutionWe will apply the eddy current loss equation in this problem though this equation has a number of assumptions so that answer is only an estimate . The eddy current loss iscore 22265.1V d B f P me ⎪⎪⎭⎫⎝⎛=ρwhere the core volume is3633core m 10409.3 or cm 409.3g/cm8.8g30Density Mass -⨯===V Then,core22265.1V B f P d m e ρ=∴ ()()()()()()362273m10409.3T 5.0Hz 000,1065.1mW 106W 101---⨯⨯⨯=d ∴d = 2.07 μmVery thin (a page of a textbook is typically about 50 - 100 μm).*8.12 Design of a ferrite antenna for an AM receiver We consider an AM radio receiverthat is to operate over the frequency range 530 - 1600 kHz. Suppose that the receiving antenna is to be a coil with a ferrite rod as core, as depicted in Figure 8.71. The coil has N turns, its length is , and the cross-sectional area is A . The inductance, L , of this coil is tuned with a variable capacitor C . The maximum value of C is 265 pF, which with L should correspond to tuning in the lowest frequency at 530 kHz. The coil with the ferrite core receives the EM waves, and the magnetic field of the EM wave permeates the ferrite core and induces a voltage across the coil. This voltage is detected by a sensitive amplifier, and in subsequent electronics it is suitably demodulated. The coil with the ferrite core therefore acts as the antenna of the receiver (ferrite antenna). We will try to find a suitable design for the ferrite coil by carrying out approximate calculations - in practice some trial and error experimentation would also be necessary. We will assume that the inductance of a finite solenoid is2AN L o ri μγμ=Inductance of a solenoid [8.36]where A is the cross-sectional area of the core, is the coil length, N is the number of turns, and γ is a geometric factor that accounts for the solenoid coil being of finite length. Assume γ ≈ 0.75. The resonant frequency f of an LC circuit is given by()2/121LC f π=[8.37]a . If d is the diameter of the enameled wire to be used as the coil winding, then the length ≈ Nd . If we usean enameled wire of diameter 1 mm, what is the number of coil turns, N , we need for a ferrite rod given its diameter is 1 cm and its initial relative permeability is 100? b . Suppose that the magnetic field intensity H of the signal in free space is varying sinusoidally, that isH = H m sin(2πƒt )[8.38]where H m is the maximum magnetic field intensity. H is related to the electric field E at a point by H = E /Z space where Z space is the impedance of free space given by 377 Ω.. Show that the induced voltage at the antenna coil is2ππ377CfdE V m m =Induced voltage across a ferrite antenna [8.39]。

《材料物理性能》课后习题答案.doc

1-1 一圆杆的直径为2.5 mm、长度为25cm并受到4500N的轴向拉力，若直径拉细至2.4mm,且拉伸变形后圆杆的体积不变，求在此拉力下的真应力、真应变、名义应力和名义应变，并比较讨论这些计算结果。

解：真应力OY = — = ―"°。

—=995(MP Q)A 4.524 xlO-6真应变勺=In — = In — = In^v = 0.0816/0 A 2.42名义应力a = — = ―4°°°_ 一= 917(MPa)A) 4.909x1()2名义应变£ =翌=& —1 = 0.0851I。

A由计算结果可知：真应力大于名义应力，真应变小于名义应变。

1- 5 一陶瓷含体积百分比为95%的/\12O3(E = 380 GPa)和5%的玻璃相(E = 84 GPa), 试计算其上限和下限弹性模量。

若该陶瓷含有5%的气孔，再估算其上限和下限弹性模量。

解：令Ei=380GPa, E2=84GPa, V^O. 95, V2=0. 05o则有上限弹性模量=E]% +E2V2 = 380 X 0.95 +84 X 0.05 =365.2(GP Q)下限弹性模量战=(¥ +3)T =(?料+誓尸=323.1(GP Q)E]380 84当该陶瓷含有5%的气孔时，将P二0. 05代入经验计算公式E=E O(1-1. 9P+0. 9P2) 可得，其上、下限弹性模量分别变为331.3 GPa和293. 1 GPa。

1-6试分别画出应力松弛和应变蠕变与时间的关系示意图，并算出t = 0, t = oo和t二£时的纵坐标表达式。

解：Maxwell模型可以较好地模拟应力松弛过程：其应力松弛曲线方程为：b⑴=贝0光必则有:<7(0) = b(0);cr(oo) = 0;<7(r)= a(0)/e.Voigt模型可以较好地模拟应变蠕变过程：其蠕变曲线方程为：的)=火(1 -广")=£(00)(1 _g")E则有：£(0)=0； £(OO)= 21;冶)=%1-(尸).以上两种模型所描述的是最简单的情况，事实上山于材料力学性能的复杂性，我们会用到用多个弹簧和多个黏壶通过串并联组合而成的复杂模型。

材料物理性能部分课后习题..

材料物理性能部分课后习题..课后习题第⼀章1.德拜热容的成功之处是什么？答：德拜热容的成功之处是在低温下，德拜热容理论很好的描述了晶体热容，CV.M∝T的三次⽅2.何为德拜温度？有什么物理意义？答：HD=hνMAX/k 德拜温度是反映晶体点阵内原⼦间结合⼒的⼀个物理量德拜温度反映了原⼦间结合⼒，德拜温度越⾼，原⼦间结合⼒越强3.试⽤双原⼦模型说明固体热膨胀的物理本质答：如图，U1（T1）、U2（T2）、U3(T3）为不同温度时的能量，当原⼦热振动通过平衡位置r0时，全部能量转化为动能，偏离平衡位置时，动能⼜逐渐转化为势能；到达振幅最⼤值时动能降为零，势能打到最⼤。

由势能曲线的不对称可以看到，随温度升⾼，势能由U1(T1）、U2（T2）向U3（T3）变化，振幅增加，振动中⼼就由r0'，r0''向r0'''右移，导致双原⼦间距增⼤，产⽣热膨胀第⼆章1.300K1×10-6Ω·m4000K时电阻率增加5%由于晶格缺陷和杂质引起的电阻率。

解：按题意:p(300k) = 10∧-6 则: p(400k) = (10∧-6)* (1+0.05) ----(1)在400K温度下马西森法则成⽴，则: p(400k) = p(镍400k) + p(杂400k) ----(2) ⼜: p(镍400k) = p(镍300k) * [1+ α* 100] ----(3) 其中参数: α为镍的温度系数约= 0.007 ; p(镍300k)(室温) = 7*10∧-6 Ω.cm) 将(1)和(3)代⼊(2)可算出杂质引起的电阻率p(杂400k)。

2.为什么⾦属的电阻因温度升⾼⽽增⼤，⽽半导体的电阻却因温度的升⾼⽽减⼩？对⾦属材料，尽管温度对有效电⼦数和电⼦平均速率⼏乎没有影响，然⽽温度升⾼会使离⼦振动加剧，热振动振幅加⼤，原⼦的⽆序度增加，周期势场的涨落也加⼤。

材料物理性能课后习题答案_北航出版社_田莳主编

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

(最新整理)材料物理性能答案

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)（E k →第一章：材料电学性能1 如何评价材料的导电能力?如何界定超导、导体、半导体和绝缘体材料？用电阻率ρ或电阻率σ评价材料的导电能力。

按材料的导电能力（电阻率），人们通常将材料划分为:）（）超导体（）（）导体（）（）半导体（）（）绝缘体（m .104m .10103m .10102m .1012728-828Ω〈Ω〈〈Ω〈〈Ω〈---ρρρρ2、经典导电理论的主要内容是什么？它如何解释欧姆定律？它有哪些局限性？金属导体中，其原子的所有价电子均脱离原子核的束缚成为自由电子,而原子核及内层束缚电子作为一个整体形成离子实。

所有离子实的库仑场构成一个平均值的等势电场,自由电子就像理想气体一样在这个等势电场中运动。

如果没有外部电场或磁场的影响，一定温度下其中的离子实只能在定域作热振动，形成格波，自由电子则可以在较大范围内作随机运动，并不时与离子实发生碰撞或散射，此时定域的离子实不能定向运动,方向随机的自由电子也不能形成电流。

施加外电场后，自由电子的运动就会在随机热运动基础上叠加一个与电场反方向的平均分量，形成定向漂移，形成电流.自由电子在定向漂移的过程中不断与离子实或其它缺陷碰撞或散射，从而产生电阻.E J →→=σ，电导率σ= (其中μ= ，为电子的漂移迁移率,表示单位场强下电子的漂移速度），它将外加电场强度和导体内的电流密度联系起来,表示了欧姆定律的微观形式。

材料物理性能课后习题答案_北航出版社_田莳主编

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

材料物理性能课后习题答案_北航出版社_田莳主编

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

材料物理性能课后习题答案北航出版社主编

材料物理性能课后习题答案_北航出版社_主编材料物理习题集第一章固体中电子能量结构和状态（量子力学基础）1.一电子通过5400V电位差的电场，（1）计算它的xxxx波长；（2）计算它的波数；（3）计算它对Ni晶体（111）面（面间距d=2.04×10-10m）的布拉格衍射角。

（P5）hh?=1?）解：（1p)mE(22?3410?6.6 =1?3119?)?10?(2?9.1?10?54001.6211?m?10=1.67?211103.76?2（）波数K=?????sin）（32d?'o??18??sin2?d22.有两种原子，基态电子壳层是这样填充的，请分别写出n=3的所有电子的四个量子数的可能组态。

（非书上内容）3.如电子占据某一能级的几率是1/4，另一能级被占据的几率为3/4，分别计算两个能级的能量比费米能级高出多少kT？（P15）4.已知Cu的密度为8.5×103kg/m3，计算其（P16）5.计算Na在0K时自由电子的平均动能。

（Na的摩尔质量M=22.99，）（P16）材料物理性能课后习题答案_北航出版社_主编6.若自由电子矢量K满足以为晶格周期性边界条件和定态xx方程。

试证明下式成立：eiKL=17.已知晶面间距为d，晶面指数为（h k l）的平行晶面*?角入射，试证明，一电子波与该晶面系成的倒易矢量为r hkl*??r/cos的轨迹满足方程K2。

产生布拉格反射的临界波矢量K hkl8.试用布拉格反射定律说明晶体电子能谱中禁带产生的原因。

（P20）9.试用晶体能带理论说明元素的导体、半导体、绝缘体的导电性质。

答：（画出典型的能带结构图，然后分别说明）10.过渡族金属物理性质的特殊性与电子能带结构有何联系？（P28）答：过渡族金属的d带不满，且能级低而密，可xx较多的电子，夺取较高的s带中的电子，降低费米能级。

补充习题为什么镜子颠倒了左右而没有颠倒上下？ 1.只考虑xx力学，试计算在不损害人体安全的情况下，加速到2.光速需要多少时间？已知下列条件，试计算空间两个电子的电斥力和万有引力的 3.比值画出原子间引力、斥力、能量随原子间距变化的关系图。

材料物理性能课后习题答案_北航出版社_田莳主编

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

《材料物理性能》王振廷版课后标准答案页

《材料物理性能》王振廷版课后标准答案页《材料物理性能》王振廷版课后答案页————————————————————————————————作者：————————————————————————————————日期：1、试说明下列磁学参量的定义和概念：磁化强度、矫顽力、饱和磁化强度、磁导率、磁化率、剩余磁感应强度、磁各向异性常数、饱和磁致伸缩系数。

a、磁化强度：一个物体在外磁场中被磁化的程度，用单位体积内磁矩的多少来衡量，成为磁化强度Mb、矫顽力Hc：一个试样磁化至饱和，如果要μ=0或B=0，则必须加上一个反向磁场Hc，成为矫顽力。

c、饱和磁化强度：磁化曲线中随着磁化场的增加，磁化强度M或磁感强度B开始增加较缓慢，然后迅速增加，再转而缓慢地增加，最后磁化至饱和。

Ms成为饱和磁化强度，Bs成为饱和磁感应强度。

d、磁导率：μ=B/H，表征磁性介质的物理量，μ称为磁导率。

e、磁化率：从宏观上来看，物体在磁场中被磁化的程度与磁化场的磁场强度有关。

M=χ·H，χ称为单位体积磁化率。

f、剩余磁感应强度：将一个试样磁化至饱和，然后慢慢地减少H，则M也将减少，但M并不按照磁化曲线反方向进行，而是按另一条曲线改变，当H减少到零时，M=Mr或Br=4πMr。

（Mr、Br分别为剩余磁化强度和剩余磁感应强度）g、磁滞消耗：磁滞回线所包围的面积表征磁化一周时所消耗的功，称为磁滞损耗Q（J/m3）h、磁晶各向异性常数：磁化强度矢量沿不同晶轴方向的能量差代表磁晶各向异性能，用Ek表示。

磁晶各向异性能是磁化矢量方向的函数。

i、饱和磁致伸缩系数：随着外磁场的增强，致磁体的磁化强度增强，这时|λ|也随之增大。

当H=Hs时，磁化强度M达到饱和值，此时λ=λs，称为饱和磁致伸缩所致。

2、计算Gd3+和Cr3+的自由离子磁矩？Gd3+的离子磁矩比Cr3+离子磁矩高的原因是什么？Gd3+有7个未成对电子,Cr3+ 3个未成对电子.所以, Gd3+的离子磁矩为7μB, Cr3+的离子磁矩为3μB.3、过渡族金属晶体中的原子（或离子）磁矩比它们各自的自由离子磁矩低的原因是什么？4、试绘图说明抗磁性、顺磁性、铁磁性物质在外场B=0的磁行为。

《材料物理性能》课后习题答案

.《材料物理性能》第一章材料的力学性能1-1一圆杆的直径为2.5 mm 、长度为25cm 并受到4500N 的轴向拉力，若直径拉细至 2.4mm ，且拉伸变形后圆杆的体积不变,求在此拉力下的真应力、真应变、名义应力和名义应变，并比较讨论这些计算结果。

解：由计算结果可知：真应力大于名义应力，真应变小于名义应变。

1-5一陶瓷含体积百分比为95%的Al 2O 3 (E = 380 GPa)和5%的玻璃相(E = 84 GPa)，试计算其上限和下限弹性模量。

若该陶瓷含有5 %的气孔，再估算其上限和下限弹性模量。

解：令E 1=380GPa,E 2=84GPa,V 1=0.95,V 2=0.05。

则有当该陶瓷含有5%的气孔时，将P=0.05代入经验计算公式E=E 0(1-1.9P+0.9P 2)可得，其上、下限弹性模量分别变为331.3 GPa 和293.1 GPa 。

0816.04.25.2ln ln ln 22001====A A l l T ε真应变)(91710909.4450060MPa A F =⨯==-σ名义应力0851.0100=-=∆=A A l l ε名义应变)(99510524.445006MPa A F T =⨯==-σ真应力)(2.36505.08495.03802211GPa V E V E E H =⨯+⨯=+=上限弹性模量)(1.323)8405.038095.0()(112211GPa E V E V E L =+=+=--下限弹性模量1-6试分别画出应力松弛和应变蠕变与时间的关系示意图，并算出t = 0,t = ∞ 和t = τ时的纵坐标表达式。

解：Maxwell 模型可以较好地模拟应力松弛过程：Voigt 模型可以较好地模拟应变蠕变过程：以上两种模型所描述的是最简单的情况，事实上由于材料力学性能的复杂性，我们会用到用多个弹簧和多个黏壶通过串并联组合而成的复杂模型。

如采用四元件模型来表示线性高聚物的蠕变过程等。

材料物理性能课后习题答案解析北航出版社田莳主编

（P5）12341311921111o '(2)6.610 =(29.1105400 1.610)=1.67102K 3.7610sin sin 2182hh pmE md dλπλθλλθθ----=⨯⨯⨯⨯⨯⨯⨯=⨯==⇒=解：（1）=（2）波数=（3）22. 有两种原子，基态电子壳层是这样填充的;;s s s s s s s 2262322626102610（1）1、22p 、33p （2）1、22p 、33p 3d 、44p 4d ，请分别写出n=3的所有电子的四个量子数的可能组态。

（P16） 2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

（Na 的摩尔质量M=22.99，.0ρ⨯33=11310kg/m ）（P16）22323426233311900(3/8)2(6.6310) 1.01310 =(3 6.0210/8)291022.99=5.2110 3.253 1.085F F h E n mJ eVE E eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯===解：由由 6. 若自由电子矢量K 满足以为晶格周期性边界条件x x L ψψ+()=()和定态薛定谔方程。

材料物理性能课后习题答案解析_北航出版社_田莳主编

（P16）2203234262333118(3/8)2(6.6310)8.510 =(3 6.0210/8)291063.5=1.0910 6.83Fh E n m J eVππ---=⨯⨯⨯⨯⨯⨯⨯⨯=解：由5. 计算Na 在0K 时自由电子的平均动能。

新版材料物理性能课后习题答案_北航出版社_田莳主编-新版.pdf

=5.21 10 19 J 3.25eV
2
6.02 1023 / 8 ) 3
由E0
3 5
E
0 F
1.08eV
6. 若自由电子矢量 K 满足以为晶格周期性边界条件试证明下式成立： eiKL =1
( x)= ( x L ) 和定态薛定谔方程。
解：由于满足薛定谔定态方程
(x ) AeiKx
又满足周期性边界条件
A1e( W /kT ) B 3000
13.
6
本征半导体中，从价带激发至导带的电子和价带产生的空穴共同电导，
激发的电子数 n可以近似表示为：
n N exp( E g / 2kT )
式中： N 为状态密度， k为波尔兹曼常数， T为热力学温度（ K ），试回答（ 1）设 N =1023cm-3， k=8.6 10-5 eV k 1时， Si（ Eg 1.1eV） ,
=1.09 10 18 J 6.83eV
2
6.02 1023 / 8 ) 3
5. 计算 Na 在 0K 时自由电子的平均动能。（Na 的摩尔质量 M=22.99 ， =1.013 103kg/m3 ）
（ P16）
2
解：由
E
0 F
h2 (3n / 8
2
)3
2m
(6.63 10 34 )2 1.013 106 = 2 9 10 31 (3 22.99
12.
实验测出离子型电导体的电导率与温度的相关数据，经数学回归分析得出关系为 lg A B 1
T (1)试求在测量温度范围内的电导激活能表达式； (2) 若给出 T1=500K 时， 1=10-（9 m）-1，
T2 1000K时， 2=10-（6 m）-1 计算电导激活能的值。

(完整word版)《材料物理性能》课后习题答案

《材料物理性能》第一章材料的力学性能1-1一圆杆的直径为2.5 mm 、长度为25cm 并受到4500N 的轴向拉力，若直径拉细至2.4mm ，且拉伸变形后圆杆的体积不变,求在此拉力下的真应力、真应变、名义应力和名义应变，并比较讨论这些计算结果。

解：由计算结果可知：真应力大于名义应力，真应变小于名义应变。

1-5一陶瓷含体积百分比为95%的Al 2O 3 (E = 380 GPa)和5%的玻璃相(E = 84 GPa)，试计算其上限和下限弹性模量。

若该陶瓷含有5 %的气孔，再估算其上限和下限弹性模量。

解：令E 1=380GPa,E 2=84GPa,V 1=0.95,V 2=0.05。

则有当该陶瓷含有5%的气孔时，将P=0.05代入经验计算公式E=E 0(1-1.9P+0.9P 2)可得，其上、下限弹性模量分别变为331.3 GPa 和293.1 GPa 。

解：Maxwell 模型可以较好地模拟应力松弛过程：V oigt 模型可以较好地模拟应变蠕变过程：以上两种模型所描述的是最简单的情况，事实上由于材料力学性能的复杂性，我们会用到用多个弹簧和多个黏壶通过串并联组合而成的复杂模型。

如采用四元件模型来表示线性高聚物的蠕变过程等。

材料物理性能课后习题答案北航出版社田莳主编.docx

材料物理习题集第一章固体中电子能量结构和状态（量子力学基础）1.一电子通过 5400V 电位差的电场，（1）计算它的德布罗意波长；（2）计算它的波数；（ 3）计算它对 Ni 晶体（ 111）面（面间距 d =× 10-10 m ）的布拉格衍射角。

（ P5）解：（ 1） =hh1p(2 mE) 2= 6.6 10 341(29.1 10 31 5400 1.6 10 19 ) 2=1.67 10 11 m（2）波数 K = 23.76 1011（ 3） 2d sinsin2o 18'2 d2.有两种原子，基态电子壳层是这样填充的（1）1s 2、2s 2 2p 6、3s 2 3p 3;，请分别写出 n=3 的所有电子的四个量（2）1s 2、2s 2 2p 6、3s 2 3p 63d 10、 4s 2 4p 6 4d 10;子数的可能组态。

（非书上内容）3.如电子占据某一能级的几率是的能量比费米能级高出多少1/4 ，另一能级被占据的几率为k T ？（ P15）3/4 ，分别计算两个能级1解：由 f ( E)EF ]exp[E1kT E E F11] kT ln[f ( E )将 f (E) 1/ 4代入得 E E F ln 3 kT将 f (E)3/ 4代入得 EE Fln 3 kT4.已知 Cu 的密度为× 10 3kg/m 3，计算其 E 0F 。

（ P16）解：h22(3n / 8) 3由 E F2m= (6.6334262 1031)(38.5 10 6.02 1023 / 8 ) 3291063.5=1.0910 18J 6.83eV5.计算 Na 在 0K 时自由电子的平均动能。

（ Na 的摩尔质量 M=，=1.013103 kg/m3）（P16）解：由 E F0h22 (3 n / 8) 32m= (6.6334262 1031)(3 1.013 10 6.021023 /8 )3291022.99 =5.2110 19J 3.25eV由E03E F0 1.08eV 56.若自由电子矢量K 满足以为晶格周期性边界条件( x)= ( x L)和定态薛定谔方程。

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