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# cha5n6 - Homework 2 due Problem 1 Ashcroft—Mermin 5.2...

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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. ...
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cha5n6 - Homework 2 due Problem 1 Ashcroft—Mermin 5.2...

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