This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 2, due January 29, 1999 Problem 1. Ashcroft—Mermin 5.2 dl=a§3,a’2:%§c+ afg753202 #2 X 43 = —%3}+ “cg/git *3 x [£1 = acg} 51 X #2 = [ﬁg/32A“ (7:1 X 62 #3 = (Raﬁ 31 = 2%(9‘0 ﬁr?) — —f”3:e)+ “2—7; 35:
52 2 “47:32) 332279 The angle between the ﬁrst two vectors is again 60 degrees, but they are rotated by 30 degrees compared to the original lattice. _27r _ 47r
Cree — 7 aarec— a—ﬁ c _ \/§ 0 —1
(3)rec — The c over a ratios are the same when (5) = 9
If the direct lattice is ideal = % and ($7190 = 43% Trigonal basis 61 = (10(22, 1, 1) , 552 = a0(1,x,1) ,(‘1’3 = a0(1,1,;1:)
The length of these vectors is a = a0 \/ 2 + $2 The angle between the vectors is cos(6) = 31561 '52 = ——:c 61 x52 =a3(1—$,1—w,m2—1)
(1'1 X (2’2 133 =a3(x($2 — 1) +2(1—a:)) =ag(m— 1)(:z:2 +m—2) = (13(3: — 1)2(m + 2) Hence 53 = WU —m,1 — m,;r2 — 1) or
53 = War—(1 +33) It is easy to see that (cyclic permutations) 5’1 = Wen +w),1,1) 52 = mu, —(1 +m),1) This is again a trigonal lattice. The length of each vector is b = 2m/2+(1+w)2 _ 2m/2—1—7‘/2+(1+m)2 a0(1—x)(x+2) _ a(1—ac)(ac+2) _ 2 1+ +1 _ 2m~1 cos(0') — — —2+(1__z)2
Compare 7 _ 277:1:2
cos 1((9) _ In”
and
cos—109’) = _—$2;;2fl+3
This shows
COS—1(6) + cos—1(6") = 13:02:31 = _1
from which the required relation follows. Also _ 1 2 2 1 _ 1+2m 2 ROM) COS”) — ‘2ﬂg2ml—w)? — ‘2—r(2+x()(ﬁ—2w )+2;c+3 2 i 2
1 + 2cos(0) cos(0’) = which gives the required relation between a and b. Problem 2. Ashcroft—Mermin 6.1
The formula to use is K = 2k sin(%q§) where k = (a) For the bcc structure the reciprocal lattice is fcc and the ﬁrst four distances
K are (see problem 4.7): , , 2T7H/5, For the fcc structure the reciprocal lattice is bee and the ﬁrst four distances
Kareafr 3 2—” 4,2T"\/§,% 11 7 a
For the diamond structure we have a bcc reciprocal lattice, but (see book)
peaks at (2,0,0), (2,2,2), etc are extinguished , , {TM/11 ,
This gives
bcc: sin(%¢2) = x/2sin(%¢1) ,
sin(%¢3) = x/gsin(%¢1) , sin(%¢4) = x/ZTsin(%¢1)
fcc: Sln(%¢2) = ﬁsingm) , [email protected]) = ﬁSin(%¢1) ,sin(%¢4) = %sin(%¢1) diamond: Sln(%¢2) = gsin(%¢>1) ,
Sin<§¢3) I \/13—15in(%¢1) : Sin(%¢4) = V 13—6 Sin(%¢1)
Therefore and Azfm7 B=bcc, C=diarnond. We have A = 1.5Ang. Using K1 : 2ksin(%¢1) and k = 2T” we get
bcc.K1=r‘:T7r 20razi—tﬂzmﬂ
fcc K1 = 277% = diamond K1 2 277H/3 = which leads to structure A (fcc) a = 3.6Ang structure B (bcc) a = 4.3Ang structure C (diamond) a = 3.6Ang Now we go back to an fcc structure and the second peak shows, therefore
the angles would be 42.8 , 49.8 , 73.2 , 89.0 Problem 3. Ashcroft—Mermin 6.3 We have cl: §Eil + %(:i2 + éd’g (use ﬁgure 4.20)
and I? = 771151 + 711252 + 717,333 With ‘ 37' = 27113“ In standard notation: SI; = 1 — Gigi: 1 —— €27Ti(mT+T2+T)
01‘
SE = 1 —— and 2ml + 21712 + 3mg can take all integer values. 6%i(2m1+2m2+37n3) (b) This is the plane given by 7713 = 0, in which case
SR 2 1 + asi<m1+m2> and this is never zero, since that would require m1 + m2 = %(1 + 219) with
k integer. This is not possible. SI? 2 0 requires 27711 + 27712 + 37713 = 3 + 6k , k integer. This has solutions
only when 7713 is odd. These are the points I? = m353 with m3 odd. We have for these points
SI? 2 1 + 6%13m3 = 1 + 6”"‘3 which is zero. Take, for example, the case m3 = 1. The points left in this plane form a
two dimensional structure given by P 2 m1 b1 +m2 b2 with m1+m2 = 3k+1
0r m1 + m2 = 3k + 2, k integer. The collection of all points P = mlgl + 7712132 forms a triangular net, since
b1 and b2 have equal lengths and have a 120 degree angle. The collection of points 13 = mlgl + mggg with m1 + m2 = 33k, kainteger
also form a triangular net, with basis vectors b1 + 2132 and 2b1 + b2. From Figure 4.17 we see that the collection of all heavy dots and centers
of the hexagons form a triangular net and that the collection of center
points forms a triangular net with one third of the points, identical to our
situation. Therefore, the set of remaining points is a honeycomb net. ...
View
Full Document
 Spring '10
 dont
 Prime number, Reciprocal lattice, Dual space

Click to edit the document details