固体物理习题解答Chapter7problems.ppt

1、1.Square lattice,free electron energies(a)Two-dimensional simple square latticeAt the corner of the first Brillouin zone:Point C kC=(/a,/a)At the midpoint of a side face of the first Brillouin zone:Point A kA=(/a,0)OThe reciprocal lattice of a two-dimensional simple square lattice is a two-dimension

2、al simple square lattice with the lattice constant 2/a.CBAThe kinetic energy of a free electron in two dimensions isThen and Thus C=2A(b)Three-dimensional simple cubic latticeThe reciprocal lattice of a three-dimensional simple cubic lattice is a three-dimensional simple cubic lattice with the latti

3、ce constant 2/a.At the corner of the first Brillouin zone:Point C kC=(/a,/a,/a)At the midpoint of a side face of the first Brillouin zone:Point A kA=(/a,0,0)The kinetic energy of a free electron in three dimensions isThen Thus C=3A(c)Discussion for the conductivity of divalent metalsThere are N allo

4、wed electron wavevectors in the first Brilouin zone,which allows 2N independent orbitals.For a divalent metals,each atom contributes two valence electrons.N primitive cells have 2N two valence electrons.Therefore the volume of the Fermi sphere is equal to the volume of the first Brilouin zone.i.e.If

5、 the edge of the second band B is lower than C,the electrons will go into the second band instead of the corner of the first Brilouin zone.For the divalent metals,the energy of the second band edge is smaller than the energy at the corner of the first Brilouin zone so that an overlap in energy of th

6、e first and second bands.There are valence electrons in the second band,which contribute to the electric conductivity.2.Free electron energies in reduced zoneThe free electron energy of a wavevector iswhere is a wavevector inside the first Brillouin zone and is a reciprocal lattice vector.The recipr

7、ocal lattice of an fcc crystal lattice is a bcc lattice with lattice constant 4/a.In the 111 direction,In the first Brillouin zone,1/2 l 1/2.The reciprocal lattice vectorwhere u,v,and w are integers.with Therefore(1)For the first energy band,u=v=w=0,At the first Brillouin zone bourdary(point L),l=1/

8、2.The bottom of the energy band(kx,ky,kz,)=(0,0,0),(0,0,0)=0.In the 111 direction(2)For the 2nd and 3rd energy bands,u=v=w=1,At(kx,ky,kz,)=(0,0,0),In the 111 direction(3)For the 4th,5th,and 6th energy bands and u=v=0,w=1,and v=w=0,u=1,At(kx,ky,kz,)=(0,0,0),u=w=0,v=1,In the 111 direction(4)For the 7t

9、h,8th,and 9th energy bands and u=v=0,w=1,and v=w=0,u=1,At(kx,ky,kz,)=(0,0,0),u=w=0,v=1,In the 111 directionv=0,u=w=1,and w=0,u=v=1,At(kx,ky,kz,)=(0,0,0),and u=0,v=w=1,(5)For the 10th,11th,and 12th energy bandsIn the 111 directionv=0,u=w=1,and w=0,u=v=1,At(kx,ky,kz,)=(0,0,0),and u=0,v=w=1,(6)For the

10、13th,14th,and 15th energy bandsIn the 111 directionwhereand03.Kronig-Penney modelFor the delta-function potential,the Kronig-Penney model gives(a)At k=0,Set then ThenFor the lowest energy band,n=0.In the limit P 1,Then we obtainThe energy of the lowest energy band is(b)At k=/a,With the same process,

11、we haveThen For the lowest energy band,n=0.In the limit P 1,From Eq(2),we haveThe band gap at k=/a isThe solutionis not correct because it gives the energy close to the bottom of the energy band.4.Potential energy in the diamond structure(a)There are two atoms in the basis of the diamond structure:(

12、0,0,0)and(a/4,a/4,a/4).Then the crystal potential may be written aswith whereIf is a basis vector in the reciprocal lattice referred to the conventional cubic cell,thenThe diamond structure is a fcc crystal lattice.Its reciprocal lattice is a bcc lattice with the lattice constant 4/a.Therefore(b)At

13、the Brillouin zone boundary,k=G/2.In the first-order approximation,only CG/2 are considered.The center equation becomes(page 190)As we calculated above,U=UG=0.Then =,and the energy gap vanishes at the zone boundary.In a higher order approximation we would go back to the center equations where any no

14、n-vanishing UG enters.5.Complex wavevectors in the energy gapAt the Brillouin zone boundary,we have the wavevector at the center of the energy gap isThen The secular equation(page 190 Eq.(46)becomes:where k G.At the center of the energy gapThus6.Square latticeThe crystal potential of a square lattice in 2d where ,and the other UG=0.The center equations givesAt the corner point of the Brillouin zone,In a simple approximation,assume the wavefunction isIgnoring the contributions from ,we obtainwhereThe determinantal equation