固体物理Chapter 9 problems

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高等固体物理(基泰尔)Su9

（x a) i cos
3 ( x a) 3x ika e i cos a a

a
eika 1
电子在此状态中的简约波矢为
Reduced Zone Scheme
(1) (2)
(3)
(4)
All the energy bands in the first Brillouin zone; An energy band is a single branch of energy vs k surface; Two wavefunctions at the same k but of different energies will be independent of each other; The wavefunctions will be made up of different combinations of the plane wave components
Example
一维周期场中电子波函数k(x)应当满足布洛赫定理，若晶格常数为a，电子波函数为， 3 x k x i cos
a
试求电子在此状态中的简约波矢。 ik Rn （r Rn ) e (r )
（x a) eika ( x)
n,k=GCn(k+G)exp[i(k+G)· r],
Periodic Zone Scheme
repeat a given Brillouin zone periodically through all of wavevector space The engery of a band is a periodic function in the reciprocal lattice

《固体物理学》概念和习题答案

《固体物理学》概念和习题固体物理基本概念和思考题：1.给出原胞的定义。

答：最小平行单元。

2.给出维格纳-赛茨原胞的定义。

答：以一个格点为原点，作原点与其它格点连接的中垂面(或中垂线)，由这些中垂面(或中垂线)所围成的最小体积(或面积)即是维格纳-赛茨原胞。

3.二维布喇菲点阵类型和三维布喇菲点阵类型。

4. 请描述七大晶系的基本对称性。

5. 请给出密勒指数的定义。

6. 典型的晶体结构（简单或复式格子，原胞，基矢，基元坐标）。

7. 给出三维、二维晶格倒易点阵的定义。

8. 请给出晶体衍射的布喇格定律。

9. 给出布里渊区的定义。

10. 晶体的解理面是面指数低的晶面还是指数高的晶面？为什么？11. 写出晶体衍射的结构因子。

12. 请描述离子晶体、共价晶体、金属晶体、分子晶体的结合力形式。

13. 写出分子晶体的雷纳德-琼斯势表达式，并简述各项的来源。

14. 请写出晶格振动的波恩-卡曼边界条件。

15. 请给出晶体弹性波中光学支、声学支的数目与晶体原胞中基元原子数目之间的关系以及光学支、声学支各自的振动特点。

（晶体含N个原胞，每个原胞含p个原子，问该晶体晶格振动谱中有多少个光学支、多少个声学支振动模式？）16. 给出声子的定义。

17. 请描述金属、绝缘体热容随温度的变化特点。

18. 在晶体热容的计算中，爱因斯坦和德拜分别做了哪些基本假设。

19. 简述晶体热膨胀的原因。

20. 请描述晶体中声子碰撞的正规过程和倒逆过程。

21. 分别写出晶体中声子和电子分别服从哪种统计分布（给出具体表达式）？22. 请给出费米面、费米能量、费米波矢、费米温度、费米速度的定义。

23. 写出金属的电导率公式。

24. 给出魏德曼-夫兰兹定律。

25. 简述能隙的起因。

26. 请简述晶体周期势场中描述电子运动的布洛赫定律。

27. 请给出在一级近似下，布里渊区边界能隙的大小与相应周期势场的傅立叶分量之间的关系。

28. 给出空穴概念。

29. 请写出描述晶体中电子和空穴运动的朗之万（Langevin）方程。

固体物理学部分习题解答

《固体物理学》部分习题解答1.3 证明：体心立方晶格的倒格子是面心立方；面心立方晶格的倒格子是体心立方。

解由倒格子定义2311232a a b a a a π⨯=⋅⨯ 3121232a a b a a a π⨯=⋅⨯ 1231232a a b a a a π⨯=⋅⨯体心立方格子原胞基矢123(),(),()222a a aa i j k a i j k a i j k =-++=-+=-+ 倒格子基矢231123022()()22a a a ab i j k i j k a a a v ππ⨯==⋅-+⨯+-⋅⨯202()()4a i j k i j k v π=⋅-+⨯+-2()j k a π=+ 同理31212322()a ab i k a a a aππ⨯==+⋅⨯ 32()b i j a π=+ 可见由123,,b b b 为基矢构成的格子为面心立方格子面心立方格子原胞基矢123()/2()/2()/2a a j k a a k i a a i j =+=+=+ 倒格子基矢2311232a a b a a a π⨯=⋅⨯ 12()b i j k a π=-++ 同理22()b i j k a π=-+ 32()b i j k aπ=-+ 可见由123,,b b b 为基矢构成的格子为体心立方格子1.4 证明倒格子原胞的体积为03(2)v π，其中0v 为正格子原胞体积证倒格子基矢2311232a a b a a a π⨯=⋅⨯3121232a a b a a a π⨯=⋅⨯1231232a a b a a a π⨯=⋅⨯倒格子体积*0123()v b b b =⋅⨯3*23311230(2)()()()v a a a a a a v π=⨯⋅⨯⨯⨯ 3*00(2)v v π=1.5 证明：倒格子矢量112233G hb h b h b =++垂直于密勒指数为123()hh h 的晶面系。

固体物理习题解答Chapter 4 problems

Supposing
Cp
Asin pk0a pa
with A and k0 are constants and p runs over all integers,
we have
2
2 M
Cp(1cospKa)
p1
2A sin pk0a (1cospKa) M p1 pa
Then
2
K
2M Ap 1sinpk0asinpKa
1 Cu2
1
2
dx[(1 cosKa)2 cos2(x sKa) sin2 Kasin2(x sKa)
2
s 2 0
sin Ka(1 cosKa)sin 2(x sKa)]
1 NM2u2 1 NCu2[1 (1 cosKa)2 1 sin2 Ka]
4
2
2
2
N [M2u2 Cu2(2 2cosKa)]
rest while the other lattice moves
4. Kohn anomaly
Considering the interactions between p nearest planes, we have
the dispersion relation
2 M 2 pj10Cj(1cosjK)a
固体物理习题解答Chapter 4 problems
(b) The time-average total energy per atom
Substitute us = u cos(t sKa) into the expression
E 1 M [ u sin( t sKa )] 2
2s
24C(K/a2)2v2K2
M
where

固体物理课后答案

第一章、晶体的结构习题1．以刚性原子球堆积模型，计算以下各结构的致密度分别为：（1）简立方，6π; （2）体心立方, ;83π（3）面心立方，;62π（4）六角密积，;62π（5）金刚石结构，;163π[解答]设想晶体是由刚性原子球堆积而成，一个晶胞中刚性原子球占据的体积与晶胞体积的比值称为结构的致密度，设n为一个晶胞中的刚性原子球数,r表示刚性原子球半径，V表示晶胞体积，则致密度ρ=Vrn334π（1）对简立方晶体，任一个原子有6个最近邻，若原子以刚性球堆积，如图1.2所示，中心在1，2，3，4处的原子球将依次相切，因为,,433aVra==面1.2 简立方晶胞晶胞内包含1个原子，所以ρ=6)(33234ππ=aa（2）对体心立方晶体，任一个原子有8个最近邻，若原子刚性球堆积，如图1.3所示，体心位置O的原子8个角顶位置的原子球相切，因为晶胞空间对角线的长度为,,433aVra==晶胞内包含2个原子，所以ρ=ππ83)(*2334334=aa图1.3 体心立方晶胞（3）对面心立方晶体，任一个原子有12个最近邻，若原子以刚性球堆积，如图 1.4所示，中心位于角顶的原子与相邻的3个面心原子球相切，因为3,42a V r a ==，1个晶胞内包含4个原子，所以ρ=62)(*4334234ππ=a a .（4）对六角密积结构，任一个原子有12个最近邻，若原子以刚性球堆积，如图1。

5所示，中心在1的原子与中心在2，3，4的原子相切，中心在5的原子与中心在6，7，8的原子相切，图 1.5 六角晶胞图 1.6 正四面体晶胞内的原子O 与中心在1，3，4，5，7，8处的原子相切，即O 点与中心在5，7，8处的原子分布在正四面体的四个顶上，因为四面体的高h =223232c r a == 晶胞体积 V = 222360sin ca ca =, 一个晶胞内包含两个原子，所以ρ=ππ62)(*22233234=ca a .（5）对金刚石结构，任一个原子有4个最近邻，若原子以刚性球堆积，如图1.7所示，中心在空间对角线四分之一处的O原子与中心在1，2，3，4处的原子相切，因为,8 3r a=晶胞体积3aV=,一个晶胞内包含8个原子，所以ρ=163)83(*83334ππ=aa.2．在立方晶胞中，画出（102），（021），（122-），和（201-）晶面。

固体物理课后习题与答案

第一章金属自由电子气体模型习题及答案1. 你是如何理解绝对零度时和常温下电子的平均动能十分相近这一点的？[解答] 自由电子论只考虑电子的动能。

在绝对零度时，金属中的自由（价）电子，分布在费米能级及其以下的能级上，即分布在一个费米球内。

在常温下，费米球内部离费米面远的状态全被电子占据，这些电子从格波获取的能量不足以使其跃迁到费米面附近或以外的空状态上，能够发生能态跃迁的仅是费米面附近的少数电子,而绝大多数电子的能态不会改变。

也就是说，常温下电子的平均动能与绝对零度时的平均动能十分相近。

2. 晶体膨胀时，费米能级如何变化？[解答] 费米能级3/222)3(2πn mE o F= ，其中n 单位体积内的价电子数目。

晶体膨胀时，体积变大，电子数目不变，n 变小，费密能级降低。

3．为什么温度升高，费米能反而降低？[解答] 当K T 0≠时，有一半量子态被电子所占据的能级即是费米能级。

除了晶体膨胀引起费米能级降低外，温度升高，费米面附近的电子从格波获取的能量就越大，跃迁到费米面以外的电子就越多，原来有一半量子态被电子所占据的能级上的电子就少于一半，有一半量子态被电子所占据的能级必定降低，也就是说，温度生高，费米能反而降低。

4．为什么价电子的浓度越大，价电子的平均动能就越大？[解答] 由于绝对零度时和常温下电子的平均动能十分相近，我们讨论绝对零度时电子的平均动能与电子的浓度的关系。

价电子的浓度越大，价电子的平均动能就越大，这是金属中的价电子遵从费米—狄拉克统计分布的必然结果。

在绝对零度时，电子不可能都处于最低能级上，而是在费米球中均匀分布。

由式3/120)3(πn k F =可知，价电子的浓度越大费米球的半径就越大，高能量的电子就越多，价电子的平均动能就越大。

这一点从3/2220)3(2πn m E F=和3/222)3(10353πn mE E oF ==式看得更清楚。

电子的平均动能E 正比于费米能o F E ，而费米能又正比于电子浓度32l n。

固体物理学解答_黄昆原著_韩汝琦改编

d 2 a 2 (h 2 k 2 l 2 ) ， 1.6、对于简单立方晶格，证明密勒指数为 (h, k , l ) 的晶面系，面间距 d 满足：
其中 a 为立方边长；并说明面指数简单的晶面，其面密度较大，容易解理。解：简单立方晶格： a1 a2 a3 ， a1 ai , a2 aj , a3 ak 由倒格子基矢的定义： b1 2 倒格子基矢： b1
2U N r m n 1 [( m 1 n 1 ) ] 2 V 2 V r r r 3NAr 2
2U V 2 N 1 m 2 n 2 m n [ m n m n ] 2 9V02 r0 r0 r0 r0
V V0
由平衡条件
2U V 2 2U V 2
4 3 x 3
（3）对于面心立方：晶胞面对角线BC= 2a 4r, a 2 2 r n=4，Vc=a3
x 4 4 3 4 r 4 r 3 2 3 3 0.74 3 3 6 a (2 2r )
（4）对于六角密排：a=2r晶胞面积：S=6 S ABO 6 晶胞的体积：V= S C
4 3 4 3 r r 3 3 ∴x 3 0.52 3 6 a 8r
nV Vc
4 3
（2）对于体心立方：晶胞的体对角线BG= 3a 4r a n=2, Vc=a3 ∴x
2 4 3 4 r 2 r 3 3 3 3 0.68 3 8 a 4 3 3 ( r) 3
c a
8 3
a a ( j k) 1 2 a 证明：（1）面心立方的正格子基矢（固体物理学原胞基矢）： a2 ( i k ) 2 a a3 2 (i j )

固体物理学1~6章习题解答

3.9按德拜近似，试证明高温时晶格热容
证明：由书可知
在高温时，，则在整个积分范围内为小量，因此可将上式中被积函数化简为
将上式代入的表达式，得
将
代入上式得
3.10设晶格中每个振子的零点振动能为，试用德拜模型求三维晶格的零点振动能
解：由讨论由一个N个原子组成的二维晶格的比热，证明在低温下其比热正比于
（1）这种晶格属于哪种布拉维格子？
（2）原胞的体积和晶胞的体积各等于多少？
答：（1）因为a=3i，b=3j，而c=1.5（i+j+k）=1/2（3i+3j+3k）=1/2（a+b+c′）式中c′=3k。显然，a、b、c′构成一个边长为3*10-10m的立方晶胞，基矢c正处于此晶胞的体心上。因此，所述晶体属于体心立方布喇菲格子。
《固体物理学》习题解答
第一章
1.1有许多金属即可形成体心立方结构，也可以形成面心立方结构。从一种结构转变为另一种结构时体积变化很小.设体积的变化可以忽略，并以Rf和Rb代表面心立方和体心立方结构中最近邻原子间的距离，试问Rf/Rb等于多少？
答：由题意已知，面心、体心立方结构同一棱边相邻原子的距离相等，都设为a：
（2）晶胞的体积= = =27*10-30(m3)
原胞的体积= = =13.5*10-30(m3)
1.7六方晶胞的基失为：，，
求其倒格子基失，并画出此晶格的第一布里渊区.
答：根据正格矢与倒格矢之间的关系，可得：
正格子的体积Ω=a·（b*c）=
那么，倒格子的基矢为，，
其第一布里渊区如图所示：(略)
答：根据题意，由于OA、OB和OC分别与基失a1，a2和a3重合，那么
1.3二维布拉维点阵只有5种，试列举并画图表示之。

固体物理课后习题答案

(
) )
)
1 3 a 4
a 2
(
(
)
2π ⎧ b a 2 × a3 1 = ⎪ Ω ⎪ 2π ⎪ a 3 × a1 ⎨b 2 = Ω ⎪ 2π ⎪ ⎪b3 = Ω a1 × a 2 ⎩
(
) ) )
(
(
Ω = a1 ⋅ a 2 × a 3 =
i a a 2 × a3 = 2 a 2 j 0 a 2
(
k 0 a =i a 2 2 0
(
)
⎞ 2π k⎟= −i + j + k 同理 ⎠ a
(
)
(
)
(
)
2π ⎧ ⎪b1 = a −i + j + k ⎪ 2π ⎪ i− j+k ⎨b 2 = a ⎪ 2π ⎪ ⎪b3 = a i + j − k ⎩
(
)
(
)
(
)
由此可得出面心立方格子的倒格子为一体心立方格子；所以体心立方格子和面心立方格子互为正倒格子。 2.2 在六角晶系中，晶面常用四个指数（hkil）来表示，如图所示，前三个指数表示晶面族中最靠近原点的晶面在互成 1200的共面轴 a1 , a2 , a3 上的截距为
设两法线之间的夹角满足
K 1 i K 2 = K1 i K 2 cos γ
K 1iK 2 cos γ = = K1 i K 2 2π 2π (h1 i + k1 j + l1 k )i (h2 i + k2 j + l2 k ) a a 2π 2π 2π 2π (h1 i + k1 j + l1 k )i (h1 i + k1 j + l1 k ) i (h2 i + k2 j + l2 k )i (h2 i + k2 j + l2 k ) a a a a

固体物理(胡安)第二版课后习题答案 Word版完整版校核版精品

Word版完整版校核版第一章晶体的结构及其对称性1.1石墨层中的碳原子排列成如图所示的六角网状结构，试问它是简单还是复式格子。

为什么？作出这一结构所对应的两维点阵和初基元胞。

解：石墨层中原子排成的六角网状结构是复式格子。

因为如图点A 和点B 的格点在晶格结构中所处的地位不同，并不完全等价，平移A →B,平移后晶格结构不能完全复原所以是复式格子。

1.2在正交直角坐标系中，若矢量k l j l i l R l321++=，错误！未找到引用源。

i ，j ，k为单位向量。

错误！未找到引用源。

为整数。

问下列情况属于什么点阵？（a ）当i l为全奇或全偶时；（b ）当i l之和为偶数时。

解： 112233123l R l a l a l a l i l j l k=++=++ 错误！未找到引用源。

()...2,1,0,,321±±=l l l当l 为全奇或全偶时为面心立方结构点阵，当321l l l ++错误！未找到引用源。

之和为偶数时是面心立方结构1.3 在上题中若=++321l l l 错误！未找到引用源。

奇数位上有负离子，=++321l l l 错误！未找到引用源。

偶数位上有正离子，问这一离子晶体属于什么结构？解：是离子晶体，属于氯化钠结构。

1.4 （a ）分别证明，面心立方（fcc ）和体心立方（bcc ）点阵的惯用初基元胞三基矢间夹角相等，对fcc 为错误！未找到引用源。

，对bcc 为错误！未找到引用源。

（b ）在金刚石结构中，作任意原子与其四个最近邻原子的连线。

证明任意两条线之间夹角θ均为'1c o s109273a r c ⎛⎫-= ⎪⎝⎭'1c o s109273a r c ⎛⎫-= ⎪⎝⎭解:（1）对于面心立方()12a a j k =+ 错误！未找到引用源。

()22a a i k =+ ()32a a i j =+13222a a a a ===()1212121602a a COS a a a a ⋅⋅===()2323231602a a COS a a a a ⋅⋅===()1360COS a a ⋅=（2）对于体心立方()12a a i j k =-++ ()22a a i j k =-+ ()32a a i j k =+-12332a a a a ===()12'12121129273a a COS a a a a ⋅⋅==-=()'1313131129273a a COS a a a a ⋅⋅==-=()'2312927COS a a ⋅=（3）对于金刚石晶胞()134a i j k η=++()234a i j k η=--()2212122122314934a COS a ηηηηηη-⋅⋅===-错误！未找到引用源。

固体物理8-9讲习题参考答案

习题8.1 单原子线型晶格：考虑一个纵波，在原子质量为M 、晶格常数为a 和最近邻力常数为C 的单原子线型晶格中传播。

(a) 试证该波的总能量为22111()22s s s s s du E M C u u dt + =+∑∑−其中求和指标s 遍历所有的原子。

(b) 将u s 代入这个表达式，证明每个原子的时间平均总能量为2222111(1cos )422M u C ka M ωω+−=u 其中最后一步采用了一维布拉伐晶格的色散关系式。

解：（a ）第s 个原子的位移为cos()s u u t ska ω=−（1）动能为212s du M dt ，晶体的总动能为212s s du dt =∑T M 。

晶体的总势能为''()ss ss RφΦ=∑，其中'ss R 为s ，s’原子间的距离，'''ss ss s s R r u u =+−，'ss r 为s ，s’原子间的平衡距离。

将φ展开，只取简谐近似，得：''2'''1()()()2ss ss ss s s 'R r u φφφ=+−u ，其中''2''''2'()ss ss ss ss ss R R φφ=∂= ∂R r 为力常数。

若只考虑最近邻原子的作用，则总位能为8.2 连续介质弹性波的波动方程：证明对于长波长，一维布拉伐格子晶体的运动方程22()s p s p s pd u M C u u dt +=−∑ 约化为连续介质弹性波的波动方程：2222u u t x υ2∂∂=∂∂，其中υ为声速。

（见讲义）8.3 孔氏异常(Kohn anomaly) 在立方晶体中，沿[100]、[110]、[111]方向传播的格波，整个原子平面作同位相的运动，其位移方向平行或垂直于波矢方向。

可用一单一坐标u s 来描述平面s 离开平衡位置的位移。

固体物理总结 Charles Kittel第七版

● A solid is said to be a crystal if the atoms are arranged in such a way that their positions areexactly periodic●Bravais lattice : all lattice points are equivalent,and hence by necessity all atoms in the crystal areof the same kind.●Non-Bravais lattice:some of the lattice points are nonequivalent●Choose the origin of coordinates at a certain lattice point, say A●Position R=n1a+n2b (n1,n2)is a pair of integers values depend on the lattice point.●Lattice vetors: the set of all vectors expressed by this equation.●Primitive Unit cell: the area of the parallelogram whose sides are the basis vectors a and b is aunit cell of the lattice.●TIPS:1) All unit cells have the same area2)each unit cell has one lattice point●Nonprimitive cell:1,the area of the non primitive cell is an integral multiple of the primitivecell.2,no connection should be drawn between nonprimitive cells and no-bravais lattices.●14 bravais lattice ,7crystal systems.●In all non simple lattices the unit cells are nonprimitive●Inversion center. Acell has an inversion center if there is a point at which the cell remainsinvariant when the mathematical transformation r-->-r is performed on it.All bravais lattices are inversion symmetric.●Reflection plane : A plane in a cell such that, when a mirror reflection in this plane is performed,the cell remains invariant.●Rotation axis: this is an axis such that, if the cell is rotated around it through some angle,the cellremains invariant. The axis is called n-fold if the angle of rotation is 2pi/n.● A nonbravais lattice is one in which, with each lattice site, there is associated a cluster of atomscalled the basis.●The symmetry of the basis ,called point-group symmetry,refers to all possible rotations(includinginversion and reflection)which leave the basis invariant. 32 different groups●SPACE-GROUP SYMMETRY: combine the rotation symmetries of the point groups with thetranslation symmetries. 72.●CRYSTAL DIRECTION: choose an origin on the line. Choose the latttice vector joining A to anypoint on the line. R=n1a+n2b+n3c the direction is now specified by the integral triplet[n1,n2,n3].the triplet [n1,n2,n3]is the smallest integer of the same relative ratios.●Collective all directions to the[n1,n2,n3] by <n1,n2,n3>●MILLER INDICES: 1,find intercepts with the axes along the basis vectors a,b,c, Let theseintercepts be x,y,and z. Fraction triplete (x/a,y/b,z/c),invert (a/x,b/y,c/z).●Curly bracket {hkl}represents to all the plane labeled by the same Miller indices(hkl)●● A Bravais lattice is a lattice in which every lattice point has exactly the same environment.●Five Bravais lattices in two dimensions 5、Hexagonal lattice1、Square lattice2、Rectangular lattice3、Oblique lattice4、Centered Rectangular lattice●Unit cell :1、primitive:Single lattice point per cell .Smallest area in 2D, or Smallest volume in3D ; 2、Conventional&Non-primitive: More than one lattice point per cell .Integral multiples of the area of primitive cell SC both 1 and 2●Seven crystal systems divided into fourteen Bravais lattices:Triclinic三斜1：Simple Angles and edge unequalMonoclinic单斜2:Simple,Base-Centered One2-foldrotation axis Orthorhombic正交Tetragonal正方Cubic立方Trigonal三角Hexgonal六角●3Common Unit Cell:SC BCC FCCSimple BC FCCV olume,conventional cellLattice points per cellV olume,primitive cellLattice points per unit volumeNumber of nearest neighborsNearest-neighbor distanceNumber of second neighborsSecond neighbor distancePacking fraction●To distinguish a lattice direction from a lattice point, the triple is enclosed in square brackets [ ...]is used.[n1n2n3]which is the smallest reduced integer of the same relative ratios.As to negative directions we should add a bar on n●Cubic has highest symmetric directions●Miller Indices are a symbolic vector representation for the orientation of an atomic plane in acrystal lattice.●Coordination Number (CN)(配位数) : The Bravais lattice points closest to a given point are thenearest neighbours：A simple cubic has coordination number 6; a body-centered cubic lattice, 8;and a face-centered cubic lattice,12.●Atomic Packing Factor (APF) is defined as the volume of atoms within the unit cell divided bythe volume of the unit cell.●BCC lattice + single atom basis=SC lattice + basis of 2 atoms at (0,0,0) and (1/2,1/2,1/2)Li NaK Rb Cs Fe Nb W●FCC lattice + single atom basis=SC lattice + basis of 4 atoms at (0,0,0), (1/2,1/2,0)(1/2,0,1/2),and (0,1/2,1/2) Cu Ni Pb●Sodium chloride structure consists of equal numbers of sodium and chlorine ions placed atalternate points of a simple cubic lattice. Each ion has six of the other kind of ions as its nearest neighbours. e.g.LiF,NaBr,KCl,LiI,●Cesium chloride consists of equal numbers of cesium and chlorine ions, placed at the points of abody-centered cubic lattice so that each ion has eight of the other kind as its nearest neighbors.●Diamond structure:The diamond lattice is consist of two interpenetrating face centered bravaislattices.There are eight atom in the structure of diamond.Each atom bonds covalently to 4 others equally spread about atom in 3d.●Zinc Sulfide structure● perofskite structure(钙钛矿结构) CHAPTER 2● The condition of constructive interference(相长干涉) ● Bragg Law● diffraction condition● (1)Show that the reciprocal lattice vector G=hb1+kb2+lb3 is orthogonal to he plane representedby Miller indices (h,k,l) (2) prove that the distance between two adjacent parallel planes of the direct lattice is d hkl =2π/|Ghkl|. (3)Show that the diffraction condition is equivalent to the Bragg law .● Reciprocal lattice vector correspond to a set of crystal planes.● Brillouim zone surface describes all k vectors that are constructively diffracted by the crystal. ● Properties of reciprocal lattice:(1) Reciprocal lattice is also a Bravais lattice in the samesymmetry class.(2)Every crystal structure has two lattices associated with it, the crystal lattices and reciprocal lattice. (3)Reciprocal lattice vector correspond to a set of crystal planes.(4) Reciprocal lattice of the reciprocal lattice is the original direct lattice. CHAPTER 3 ● Cohesive (binding) energy U:the energy that must be addedto the crystal to separate its components into neutral free atoms at rest ； Energy of free atoms - Crystal energy • Magnitude ~ 1-10eV except for the inert gas crystals (0.02-0.2eV) • U ≤Eion(Ionization energy = Binding energy of valence electrons) • U controls the melting temperature and bulk modulus ●V(r): the net potential energy of interaction as function of r. r : he distance between atoms, ions, or molecules.a,b: proportionality constant of attraction and repulsion, respectively. m, n: constant characteristics of each type of bond and type of structure. ● Types of bonds (a) Van der Waals (Molecular) Electrons localized among atoms (Inert gas crystals : He, Ne, Ar, Kr, Xe, Rn ) Fluctuating dipole –dipole interactionOn average spherically symmetric distribution of electron charge with the positive nucleus in the center.But thermal fluctuations (finite T) cause instantaneous electric dipole momentTransparent Insulators –completely filled outer electron shells \\FCC structures except for He3and He4\\Weakly bonding –van der Waals bondingMODEthe Lennard-Jones potential(b) Covalent Electrons shared by the neighboring atoms HCl, NH3Covalent bonding takes place between atoms with small differences in electronegativity which are close to each other in periodic tableThe covalent bonding is formed by sharing of outer shell electrons (i.e., s and p electrons) between atoms rather than by electron transfer.Gd π2=()m na b V r r r-=+(c) Metallic Electrons free to move through sample Metallic bonding is the typeof bonding found in metal elementsThis is the electrostatic force of attraction between positively charged ions and delocalized outer electrons(d) Ionic Electrons transferred to adjacent atoms （Alkali halides 金属卤化物Li, Na, K, Rb, Cs ，F, Cl, Br, I ）Electron configuration(电子组态) closed electronic shellsionic bonding is produced whenever an element w/. a relatively low ionization energy is combined with an element w/. a high electron affinity.Madelung constant α: geometric sum(depends on relative distance, number, and sign of neighboring atoms )CalculationNotice:Higher coordination number gives larger Madelung constant\\It depends on the structure of the crystal but not unit cell dimensions. CHAPTER 41.Born-Oppenheimer approximation adiabatic approximation(绝热近似)2.Monoatomic Chain● elastic waves.:A solid is composed of discrete atoms, however when the wavelength is very long,one may disregard the atomic nature and treat the solid as a continous medium. ●● The force on the sth atom;:The force to the right: The force to the left: F=● Our wavelike solutions on the other hand are uncoupled oscillations called normal modes; each khas a definite w given by above eqn. and oscillates independently of the other modes.So the number of modes is expected to be the same as the number of equations N. ;● Periodic boundary conditions (Born-Karman) : (i)N atom form a cirque, which make all atoms tobe equivalent.(ii)The motion of atom can be seen as linearity, due to numbers of atoms are large. ● In three dimensionsp atoms per primitive cell \3p vibration branchesAcoustic (3) : LA (longitudinal TA1(transverse) TA2(transverse) Optic (3p-3) : LO (longitudinal) TO(transverse)N primitive cell: 3pN vibration modes ；Acoustic (3N) : LA (longitudinal)TA1(transverse)TA2(transverse) Optic ((3p-3)N): LO (longitudinal) TO(transverse)● Phonon ：Specifying the vibrational states of the crystal by specifying number of phonon in each()()222()()...........2r a r a r a d V d V V r V a r a dr dr ==-⎛⎫⎛⎫=+-++ ⎪ ⎪⎝⎭⎝⎭∑∑∑∑⎪⎭⎫ ⎝⎛+==j k j j j kj k n k E E )(21)(ωstate ki 。

固体物理导论基泰尔第九版答案

固体物理导论基泰尔第九版答案1、5．交警用电子检测设备检测汽车是否超速时测得的速度是平均速度．[判断题] *对错(正确答案)2、51．下列关于物质结构和运动的说法中正确的是（）[单选题] *A．原子核带正电，也是运动的(正确答案)B．组成固体的分子或原子是静止不动的C．组成液体的分子可以是运动的也可以是静止的D．原子核是单一粒子不可再分3、孔明灯俗称许愿灯，放孔明灯是我国的一种民俗文化。

如图58所示，孔明灯在点燃后加速上升的过程中，忽略其质量的变化，则孔明灯的（）A．重力势能减少，动能减少B．重力势能减少，动能增加C．重力势能增加，动能减少D．重力势能增加，动能增加(正确答案)4、4．子弹以速度v从枪口射出，v指瞬时速度．[判断题] *对(正确答案)错5、人耳听不到次声波，是因为响度太小[判断题] *对错(正确答案)答案解析：次声波和超声波的频率超过了人耳的听觉范围6、小刚是一名初中生，他从一楼跑到三楼的过程中，克服自己重力所做的功最接近下面哪个值（）[单选题]A．3JB．30JC．300JD．3000J(正确答案)7、5．推着自行车前行时前轮和后轮所受摩擦力的方向相同．[判断题] *对(正确答案)错8、4．骑着自行车前行时前轮和后轮所受摩擦力的方向相同．[判断题] *对错(正确答案)9、闭合电路的导体在磁场中做切割磁感线运动时，导体中就会产生感应电流[判断题] *对错(正确答案)答案解析：闭合电路的一部分导体在磁场中做切割磁感线运动时，导体中就会产生感应电流10、2．物体的加速度a＝0，则物体一定处于静止状态．[判断题] *对错(正确答案)11、下列有关力做功的说法中正确的是（）[单选题]A.用水平力推着购物车前进，推车的力做了功(正确答案)B.把水桶从地面上提起来，提水桶的力没有做功C.书静止在水平桌面上，书受到的支持力做了功D.挂钩上的书包静止时，书包受到的拉力做了功12、57．彩色电视机荧光屏上呈现各种颜色，都是由三种基本色光混合组成的，这三种基本色光是（）[单选题] *A．红、橙、绿B．红、绿、蓝(正确答案)C．蓝、靛、紫D．红、黄、蓝13、若跳高运动员竖直向下蹬地的力大于他的重力，运动员就能从地上跳起来[判断题]*对(正确答案)错答案解析：运动员竖直向下蹬地的力与地面给他的支持力是一对相互作用力，大小相等。

《固体物理(英文)》教学大纲

“Solid State Physics I” Course Syllabus1、General information2、Course Aim(1)General Aim:Solid State Physics I is a basic course of Applied Physics. The subject focuses on the relationship between the solid microstructure and particles and the law of their motion. The subject illustrates the solid properties and application, especially solid state theory and band theory. The topics include: Crystal structure, Binding of solid, Lattice vibration, Electronic energy band, Electron dynamics in solid.(2)Specifications:Course Aim 1: Be aware of the history and evolution of Solid State Physics; Important developments and applications in other science and technology communities, such as materials, nanotechnologies, and electronics; and the scope of application.Course Aim 2: Grasp the fundamental principles and methods of Solid State Physics. Know well the Categories of crystals according to their geometry, atomic bindings and electronic structures. Be capable of estimating crystal properties according to their categories.Course Aim 3: Master the analytical methods used in Solid State Physics, Master the method to calculate the binding energies, the phonon spectra, the electronic energy structures of typical crystals. Be capable of analyzing the transport properties of typical crystals and figuring out the electronic motion in electric field.(3) Correspondence of the Course Aims to its Contents and the Graduation requirementsTable 1: Correspondence of Aims to Contents and graduation requirements3、Teaching Targets and ApproachesChapter 1 Crystal Structure1.Teaching AimIn this chapter students study the basic concepts and methods to describe crystals. Students are required to be familiar with the concepts including Bravais Lattice and Primitive Vectors; Simple, Body-Centered, and Face-Centered Cubic Lattices; Primitive Unit Cell, Wigner-Seitz Cell, and Conventional Cell; Crystal Structures and Lattices with Bases; Symmetry Operations; Lattice Planes, and Miller Indices. Students should know how to Classify Bravais Lattices and understand the Seven Crystal Systems and Fourteen Bravais Lattices. Using the basic concepts, they can identify typical crystal structures such s Hexagonal Close-Packed and Diamond Structures; Sodium Chloride, Cesium Chloride, and Zincblende Structures.2.Difficult PointsBravais Lattice and Primitive Vectors; Symmetry Operations; Miller Indices3. Teaching ContentsSection 1. Periodic Array of Atoms(1) Lattice Translation VectorsKey points: Understand translation symmetry and know how to find the lattice translation vectors of a crystal. Grasp the concepts basis and the crystal structure. Know how to find the primitive lattice cell of a crystal.(2) Fundamental Type of LatticesKey points: Know the two-dimensional lattice types and the three-dimensional lattice types.(3) Index Systems for Crystal PlanesKey points: Know how to find out the Miller Indices of a given crystal lattice.Section 2. Fundamental Types of Lattices(1) Two-Dimensional Lattice typesKey points: Understand the classification of lattice types of two-dimensional crystals.(2) Three-Dimensional Lattice typesKey points: Understand the classification of lattice types of three-dimensional crystals.Section 3. Index Systems for Crystal PlanesKey points: Understand the index systems for crystal planes using primitive lattice vector coordination and conventional lattice vector coordination systems. Be familiar with the Miller Index system.Section 4. Simple Crystal Structures(1) Sodium Chloride StructureKey points: Be familiar with the Sodium Chloride Structure. Know how to figure out the corresponding lattice structure, primitive cells, basis, and lattice type.(2) Cesium Chloride StructureKey points: Be familiar with the Cesium Chloride Structure. Know how to figure out the corresponding lattice structure, primitive cells, basis, and lattice type.(3) Hexagonal Close-Packed Structure (hcp)Key points: Understand the meaning of ‘Close-Packed Structures’. B e familiar with the difference between hcp and fcc structures.(4) Diamond StructureKey points: Be familiar with the Diamond Structure. Know how to figure out the corresponding lattice structure, primitive cells, basis, and lattice type.(5) Cubic Zinc Sulfide StructureKey points: Be familiar with the Cubic Zinc Sulfide Structure. Know how to figure out the corresponding lattice structure, primitive cells, basis, and lattice type.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 2 Wave Diffraction and the Reciprocal Lattice1. Teaching AimIn this chapter students study the wave diffraction of crystal lattice and the concept of reciprocal lattice. Students are required to be familiar with the experiments of X-ray diffraction by crystals. Master the Bragg Law and Equation as well as Laue Equation or Formulation. Understand the equivalence of Bragg’s and Laue’s views of the X-ray diffraction in crystals. Understand the reason and the background for the introduction of reciprocal lattice. Master the relations between the direct and the reciprocal lattices.2. Difficult PointsEstablish a clear and full picture of crystal structure, crystal (direct) lattice, and reciprocal lattice.3. Teaching ContentsSection 1. Diffraction of Waves by Crystals(1) Bragg LawKey points: Understand why the crystal planes can be treated as mirrors with specular reflection. Be able to derive the Bragg equation considering two neighbor crystal planes.Section 2. Scattered Wave Amplitude(1) Fourier AnalysisKey points: Understand the Fourier components of physical properties in crystals.(2) Reciprocal Lattice VectorsKey points: Master the method to obtain reciprocal lattice vectors.(3) Diffraction ConditionsKey points: Understand the diffraction conditions of an incident wave in crystals.(4) Laue EquationsKey points: Be able to derive the Laue Equations and understand their equivalence to the Bragg Law.Section 3. Brillouin ZonesKey points: Be familiar with the reciprocal lattices to the sc, bcc, and fcc lattices. Grasp the method to find the Brillouin Zones of reciprocal lattice.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 3 Crystal Binding and Elastic Constants1. Teaching AimIn this chapter students study the interatomic coupling for different elements. The energetic, mechanical, and geometric properties are discussed for typical binding types of atoms to construct crystal structures. Students are required to be familiar with the concept of binding energy and be able to estimate the binding energy, equilibrium lattice constant, and elastic constants for some simple cases.2. Difficult PointsEstimate the energy versus the lattice constant of crystal from the interatomic coupling for an atom pair.3. Teaching ContentsSection 1. Crystals of Inert Gases(1) Van der Waals-London InteractionKey points: Understand the physical mechanism of Van der Waals-London Interaction and be familiar with the simple model for the interaction.(2) Repulsive InteractionKey points: Understand the origins of the interatomic repulsive interaction.(3) Equilibrium Lattice ConstantsKey points: Estimate the equilibrium lattice constants for a known crystal structure from the interatomic coupling.(4) Cohesive (Binding) energyKey points: Be familiar with the definition of Binding energy and be able to apply it for analysis of crystal stability.Section 2. Ionic Crystals(1) Electrostatic or Madelung EnergyKey points: Understand the importance of direct Coulomb interaction in ionic crystals.(2) Evaluation of the Madelung ConstantKey points: Be familiar with the calculation of Madelung constant for simple crystals.Section 3. Covalent CrystalsKey points: Be familiar with the features of covalent crystals.Section 4. MetalsKey points: Be familiar with the features of metallic crystals.Section 5. Hydrogen BondsKey points: Be able to identify hydrogen bonds and typical materials with hydrogen bonds.4. Teaching methodPPT lectures, handwritten derivation, discussion, tutorial after class, self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 4 Phonon I. Crystal Vibrations1. Teaching AimIn this chapter students study the dynamics of atoms in crystals and derive vibration modes of atoms in simple cases. Students are required to understand the physical origin of the adiabatic approximation and the harmonic approximation, be able to extract the Hook constant from the interatomic coupling, and be familiar with the derivation of vibration modes in one-dimensional crystals. Students should also establish the connection between the degree of freedom and the mode number and understand the reason why only the modes are distributed in the first Brillouin zone.2. Difficult PointsSwitch between real space and k-space when discussing atomic dynamics in crystals. Distinguish the group velocity from the phase velocity.3. Teaching ContentsSection 1. Vibrations of Crystals with Monatomic Basis(1) First Brillouin ZoneKey points: Establish the dynamic equation of atoms connected via spring to neighbors in the one-dimensional monatomic crystal chain, and derive the vibration modes in the first Brillouin zone.(2) Group VelocityKey points: Figure out the dispersion relation of vibration modes and derive the group velocity of each mode.(3) Long Wavelength LimitKey points: In the long wavelength limit estimate the sound velocity and compare it with the result of elastic wave.Section 2. Two Atoms per Primitive BasisKey points: Derive the phonon dispersion for one-dimensional diatomic chain.Section 3. Quantization of Elastic WavesKey points: Phonon is quasi-particle due to the quantization of vibration wave.Section 4. Phonon MomentumKey points: Understand the phonon momentum and know how to use it.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 5 Phonon II. Thermal Properties1. Teaching AimIn this chapter students study the density of states and the thermal properties of phonons. Students are required to derive the heat capacities of phonons in the Einstein and the Debye models. Comparing the results with those obtained in classical gases, students should understand the quantum effects on thermal properties and the result of zero point energy.2. Difficult PointsUnderstand the density of states and use it in the calculation of thermal properties. Obtain the total energy by integrating the energies of phonons and derive the limit at low and high temperatures.3. Teaching ContentsSection 1. Phonon Heat capacity(1) Bose-Einstein (Plank) DistributionKey points: Be familiar with the Bose-Einstein (Plank) Distribution for phonons.(2) Normal Mode EnumerationKey points: Transfer the summation over states into integration with the density of states.(3) Density of States in One DimensionKey points: Density of states in energy space can be derived from the constant density of states in k-space. They are connected by the group velocity which can be obtained from the dispersion.(4) Density of States in Three DimensionKey points: Similarly, density of states in three dimension is obtained from the constant density of states in k-space and dispersion.(5) Debye Model for Density of StatesKey points: Density of states in obtained by assuming linear dispersion or constant sound velocity.(6) Debye T3 LawKey points: Heat capacity is proportional to cubic power of temperature in the Debye model.(7) Einstein Model of the Density of StatesKey points: Density of states in obtained by assuming constant energy/frequency of phonon. The Heat capacity decreases exponentially at low temperature in the Einstein model.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 6 Free Electron Fermi Gas1. Teaching AimIn this chapter students study the quantum description of free electrons with Coulomb interaction between electrons neglected. Students are required to derive the Fermi energy and understand the temperature dependence of the Fermi energy in systems of different dimensions. Thermal and electric properties of free electron Fermi gas are discussed.2. Difficult PointsConnect the thermal & electric properties with the system parameters such as electron mass, density of states, density of electrons, dimensionality, and temperature, etc.3. Teaching ContentsSection 1. Energy Levels in One dimensionKey points: Solve the Schrodinger equation to obtain the energy levels of standing electrons in one-dimensional square well. Energy levels of propagating electrons can be obtained using the periodic boundary condition.Section 2. Effect of Temperature on the Fermi-Dirac DistributionKey points: Be familiar with the variation of Fermi-Dirac distribution at different temperature.Section 3. Free Electron Gas in Three DimensionsKey points: Extend the one-dimensional results to three-dimensional cases. Notice thek-space difference between the standing and the propagating states and the difference in density of states between three- and one-dimensional systems.Section 4. Heat Capacity of the Electron GasKey points: Calculate the energy of electron gas versus temperature with the temperature dependence of Fermi-Dirac distribution.Section 5. Electrical Conductivity and Ohm’s LawKey points: With the collision time known, electrical conductivity can be derived by estimating the momentum deviation under an electric field.Section 6. Thermal Conductivity of MetalsKey points: Thermal conductivity can be estimated by assuming a temperature gradient.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 7 Energy Bands1. Teaching AimIn this chapter students study the energy bands – the electronic structure of electron gas in periodic potential. Students should be familiar with the Bloch theorem and the Bloch functions. Students are required to understand well the physical mechanism behind the formation of energy bands.2. Difficult PointsProve and understand the Bloch theorem. Be familiar with the features of Bloch functions.3. Teaching ContentsSection 1. Nearly Free Electron Model(1) Origin of the Energy GapKey points: Understand the origin of the energy gap by considering redistribution of electrons in periodic potential.(2) Magnitude of the Energy GapKey points: Estimate the energy difference of two standing waves in a periodic potential.Section 2. Bloch FunctionsKey points: Be familiar with the form of Bloch functions.Section 3. Kronig-Penney ModelKey points: In the Kronig-Penney Model the energy bands can be solved exactly from the Schrodinger equation. This convinces us the existence of energy bands..Section 4. Wave Equation of Electron in a Periodic Potential(1) Restatement of the Bloch TheoremKey points: Exact proof of the Bloch Theorem is given.(2) Crystal Momentum of an ElectronKey points: Understand the crystal momentum of an electron in a periodic potential.(1) Solution of the Central EquationKey points: Understand the form of the central equation for energy bands and its solutions.(2) Kronig-Penney Model in Reciprocal SpaceKey points: Describe the KP model in k-space with the help of Fourier expansion.(3) Empty Lattice ApproximationKey points: Reduce the energy band of free electron into the first Brillouin zone.(4) Approximate Solution near a Zone BoundaryKey points: Consider only the coupling between states k and k+G near k= G.(5) Number of Orbitals in a BandKey points: Each primitive cell contributes one independent value of k to each band.(6) Metals and InsulatorsKey points: Check the existence of overall energy gap near the Fermi energy.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 8 Semiconductor Crystals1. Teaching AimIn this chapter students study the elementary properties of semiconductors – the gapped crystals. Students are required to understand the equation of motion for electrons in energy bands and be familiar with the widely used concepts in semiconductors such as hole and effective mass.2. Difficult PointsDescribe electronic dynamics in both the real space and the k-space.3. Teaching ContentsSection 1. Band GapKey points: Understand the projection of energy bands in real space and be familiar with the conduction band, forbidden band and valence band.Section 2. Equations of Motion(1) Physical Derivation of the equations of motionKey points: Establish the picture that the force on an object results in the change of momentum.(2) HolesKey points: Vacancies in valence band can conduct current as well electrons in conduction band.(3) Effective MassKey points: Parabolic fitting of the bottom and the top of bands.(4) Physical Interpretation of the Effective MassKey points: Understand the mass ‘change’ of electrons in a crystal.(5) Effective Masses in SemiconductorKey points: Be familiar with the usage of effective mass for electrons and holes.(6) Silicon and GermaniumKey points: Be familiar with the typical element semiconductors.(7) Intrinsic Carrier ConcentrationKey points: Understand the difference between Intrinsic and extrinsic (doped) semiconductors.(8) Intrinsic MobilityKey points: Be able to evaluate the mobility of electrons.(9) Impurity ConductivityKey points: Understand the effects of impurities in semiconductors.(10) Thermal Ionization of Donors and AcceptorsKey points: Be able to distinguish between donor and acceptor impurities.Section 3. Thermoelectric EffectsKey points: Understand the thermoelectric phenomena in semiconductors..4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 9 Fermi Surfaces and Metals1. Teaching AimIn this chapter students study mainly the isoenergetic surface of energy bands at the Fermi energy and the tight-binding method for energy bands. Students are required to be familiar with the reduced, the extended, and the periodic schemes of Brillouin zones and be able to figure out briefly the shape of Fermi Surfaces in energy bands of metals. Students should be able to calculate the energy bands from a tight-binding Hamiltonian and derive the motions of electrons under a constant electric field.2. Difficult PointsEstablish the three-dimensional picture of energy bands in crystals. Be familiar with the solutions of eigen equations in matrix form.3. Teaching ContentsSection 1. Schemes of Brillouin zonesKey points: Be familiar with the transform among the reduced, extended, and periodic zone schemes.Section 2. Construction of Fermi SurfacesKey points: Have a clear picture of the boundaries of Brillouin zones, the energy bands, and the Fermi Surfaces.Section 3. Calculation of Energy BandsKey points: Understand the physical origin of tight-binding method and be familiar with the method.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 10 Superconductivity1. Teaching AimIn this chapter students study the phenomena of superconductivity and the corresponding theoretical description and understandings. Students are required to be familiar with the London Equation and the BCS description for superconductivity.2. Difficult PointsEstablish the picture of macroscopic quantum transport phenomena.3. Teaching ContentsSection 1. Experimental SurveyKey points: Be familiar with the phenomena and features parameters for superconductivity.Section 2. Theory SurveyKey points: Understand the London equation and the BCS ground state for superconductivity.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 11 Diamagnetism and Paramagnetism1. Teaching AimIn this chapter students study briefly the diamagnetism and paramagnetism and the corresponding theoretical understandings.2. Difficult PointsUnderstand the physical mechanisms of diamagnetism and paramagnetism.3. Teaching ContentsSection 1. Langevin Diamagnetism EquationKey points: Understand the Langevin equation for diamagnetism.Section 2. Quantum Theory of Diamagnetism of Mononuclear SystemsKey points: Understand the quantum explanation of diamagnetism.Section 3. Quantum Theory of ParamagnetismKey points: Understand the quantum explanation of paramagnetism.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class,self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.Chapter 12 Ferromagnetism and Antiferromagnetism1. Teaching AimIn this chapter students study briefly the ferrognetism, the ferrimagnetism, and the antiferromagnetism plus the corresponding theoretical understandings.2. Difficult PointsUnderstand the physical mechanisms of ferromagnetism and antiferromagnetism.3. Teaching ContentsSection 1. Ferromagnetic OrderKey points: Understand the phenomena and physical mechanism of ferromagnetism.Section 2. Ferrimagnetic OrderKey points: Understand the phenomena and physical mechanism of ferromagnetism.Section 3. Antiferromagnetic OrderKey points: Understand the phenomena and physical mechanism of antiferromagnetism.4. Teaching methodPPT lectures, Handwritten derivation in class, discussion in class, tutorial after class, self-directed learning.5. EvaluationVia homework assignments, thinking questions in class, quizzes.4、Class Hour AllocationTable 2: Contents and class hours per chapter5、Course ScheduleTable 3: Course schedule6、Textbook and References1．Charles Kittel, Introduction to Solid State Physics (Wiley: New York, 2004).2. 黄昆. 固体物理学. 北京：人民教育出版社，1988.3. 中译本: 基泰尔“固体物理导论”，项金钟、吴兴惠译，化学工业出版社4. 胡安，章维益，“固体物理学”，高等教育出版社，北京，2010.7、Teaching Philosophy and Methods1. All students are regarded as potential discoverers of new facts rather than as receptacles for memorizing previously developed knowledge. A teacher is a facilitator for this discovering process. It is the teacher’s responsibility to design the framework in which learning can take place, and then stimulate and nurture the students' development, giving help in terms of knowledge, techniques, and encouragement.2. The students learn in different ways and all students in a class should be included in the learning process. Different teaching styles will be introduced to the class addressing different learning styles and students will be encouraged to participate in a mixture of lecture, discussion, and group activities.3. PPT presentation, blackboard hand written derivation, and other teaching aid are used in the teaching practice.7、Course Examination7.1 The Correspondence between the Course Examination and the Course AimTable 4: Correspondence between Examination and Aim7.2 Grading Method(1) GradingMulti-process evaluation: in class and off class assessment, processive and final exam.Calculation: Overall Grade = (Attendance and Homeworks)*20% +(Quiz1)*10%+(Midterm Exam)*20%+(Quiz 2)*10%+(Cumulative Final Exam)*40%(2) Weight of Course Aims in GradingTable 5: Weight of Course Aims and Grading Analysis7.3 Grading Standard。