Soution of HW2
description
Transcript of Soution of HW2
1
Solution of HW2 Ex 3.
a) The parameter a λ=1.54 Å, the plane is (111), θ=19.2° First we need to obtain d from Bragg’s law
λ = 2dsinθ d= λ/(2sinθ) d=1.54/(2 sin19.2°) d= 2.34 Å
We obtain a from d by using:
we use n=1 for the first order a=b=c in the cubic structure Thus
d = 11+1+1a2
=a3
a = d 3 = 4.05 Å
B) We obtain the Avogadro’s number from The density (𝜌) = !!! !"## !"# !"#$ !"## (!)
!"#$%& !"# !"#$ !"## (!)
V= a3= (2.91 ×10-‐8)3= 66.430125 ×10-‐24 cm3 M= the number of atoms per unit cell × mass per atom The mass per atom= atomic weight (MA)/ Avogadro’s number NA The number of atoms in FCC= 4 atoms Thus
M=4✕(MA/NA)
d = nh2
a2+k2
b2+l2
c2!
"#
$
%&
1/2
2
Then,
ρ =MV=1V×4MA
NA
NA =MV=1V×4MA
ρ
NA =4×27
66.43012×10−24 ×2.7= 6.02×1023 per mole
3
Ex 12 In order to obtain the structure factor of BCC of 1/8 atoms on the corners and one in the middle we need to find their coordinates shown in figure 1. Eight 1/8 atoms on the corners with (000), (111), (100),(001),(010),(110),(101),(011) and One in the middle (½ ½ ½ )
Figure 1 The structure factor
Will be written as
Fhkl = fa e(i2π (h/2+k/2+l/2)+
18ei2π (0)+ ei2π (h)+ ei2π (k )+ ei2π (l )+ ei2π (h+l )+ ei2π (h+k )+ ei2π (l+k )+ ei2π (h+k+l )!" #$
%&'
()*
Fhkl = fajj∑ exp2πi(ujh+ vjk +wjl)
(½½½)
(000)
(100)
(010)
(110)
(001)
(110)
(011)
(111)
4
Fhkl = fa e(iπ (h+k+l )+
181+ ei2π (h)+ ei2π (k )+ ei2π (l )+ ei2π (h+l )+ ei2π (h+k )+ ei2π (l+k )+ ei2π (h+k+l )!" #$
%&'
()*
1/8 factor in the second term because we consider only 1/8 of corner atom inside the unit cell as shown in figure 1 (the small figure). The second term always gives = 1/8 (+8) = +1 because it has “2π” in the power of the exponential. Therefore the structure factor depends on the first term “Exp(iπ(h+k+l)) “ which has “π” in the power of the exponentials because it can be either +1 when k+h+l=even Fhkl = fa (1+1), the diffraction of the plane will be shown -‐1 when h+k+l=odd Fhkl = fa (1-‐1)=0, the diffraction of the plane will not shown This agrees with what we found in the text.
Second term First term
5
Ex16 CsCl
fCs = 3fCl If Cs in (000) position Cl in (½ ½ ½ ) position In this case the structure factor will be Fhkl = fCs expi2π (0)+ fCl expi2π (h+ k + l)
Fhkl = fCs + fCl expi2π (h+ k + l)
In this case the structure factor it has two solutions If h+k+l = even Fhkl = fCs + fCl = positive number (strong diffraction ) If h+k+l = odd Fhkl = fCs − fCl = Positive but smaller number because fCs = 3fCl (weak diffraction). In this case, if h+k+l = odd, diffraction does not vanish because the atomic factor of Cs and Cl are not equal (FCs ≠ FCl) . In this case the diffraction will look like this There is a famous reference related to this example that used a lot as an example of this deviation. “CsCl diffraction from “The Crystal Structure of two Rare Halogen Salts. W.P Davey,. F.G. Wick, (1921) Physical Review 17, 403-‐404”.
Fhkl = fajj∑ exp2πi(ujh+ vjk +wjl)
6
CsCl: If one atom in 000 position and the other one in ½ ½ ½ position You can see that the diffraction k+h+l = odd such as (111), (001), (113) (023) do not vanish. It is shown in the figure below
But for CsCs BCC as shown below the diffraction when h+k+l=odd is not shown For example (111), (001), (113) (023) are not shown
0 0
1
0 1
1
1 1
1
0 0
2
0 1
2 1 1
2
0 2
2
0 0
31 2
2
0 1
3
1 1
3
2 2
2
0 2
31
2 3
0 0
4
0 10 20 30 40 50 60 70 80 90 100
CsCl (Davey, W.P.;Wick, F.G. (1921) Physical Review 17, 403-404)�Lambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: P m -3 m Direct cell: 4.1200 4.1200 4.1200 90.00 90.00 90.00
0 0
1
0 1
1
1 1
1 0 0
2
0 1
2
1 1
2
0 2
2
0 0
31
2 2
0 1
3
1 1
3
2 2
2
0 2
3 1 2
3
0 0
4
0 10 20 30 40 50 60 70 80 90 100
"CsCs" - mythical bcc material�Lambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: P m -3 m Direct cell: 4.1200 4.1200 4.1200 90.00 90.00 90.00
7
Ex20 a) First we need to drive the volume of real unit cell (Ωreal):
= a.(bxc)
(see the proof in the last page) a=4Å, b=6 Å, c=8 Å and α=β=90° and γ=120° The volume of monoclinic using the above equation:
Therefore it is monoclinic structure.
Note: α between b and c β between a and c γ between a and b For reciprocal cell:
Ωreal
Ωreal = abc 1− cos2α − cos2 β − cos2 γ + 2cosα cosβ cosγ
Ωreal = abc 1− cos2 90− cos2 90− cos2 γ + 2cos90cos90cosγ
Ωreal = abc 1− cos2 γ = abcsinγ
a* = 2πΩreal
(b× c) = 2πabcsinγ
bcsinα = 2πasin120°
sin90° =1.8Å−1
b* = 2πΩreal
(c× a) = 2πabcsinγ
acsinβ = 2πbsin120°
sin90° =1.209Å−1
c* = 2πΩreal
(b× a) = 2πabcsinγ
absinγ = 2πc= 0.785Å−1
8
b) The volume of real cell:
The Volume of the reciprocal cell;
Ω* =(2π )3
Ωreal see the proof in the last page.
Ω* =(2π )3
Ωreal
=(2π )3
abcsinγ=(2π )3
166.27=1.49Å−3
c ) We need to determine the reciprocal victor G to calculate d (the space between two planes of (210) by using the relation:
We use the square of the two sides of the equation to solve the problem :
|Ghkl |2=(2π )2
(dhkl )2
|Ghkl |2=(2π )2
(dhkl )2
Where |Ghkl |
2= (ha* + kb* + lc*).((ha* + kb* + lc*) l=0 |Ghkl |
2= (ha* + kb*).(ha* + kb*) = a2 + b2 + 2abcosγ * (Note: a.a=a2, b.b=b2, a.b=ab cosγ) where γ* is the reciprocal angle between a* and b* From the figure below in one dimension, we can determine the angle γ* The red vectors are a and b vectors and the black vectors are a* and b* are the reciprocal vectors. The total angle of any triangle = 180o Thus γ*= 180 o -‐90 o -‐30 o = 60 o
Ωreal = abcsinγ = 4×6×8sin120° =166.27Å3
€
|Ghkl |=2πdhkl
9
Hence |Ghkl |= a2 + b2 + 2abcosγ * = (1.8)2 + (1.209)2 + 2×1.8×1.209cos60° = 18.78 = 4.334Å−1
Thus, the distance between (210) dhkl
|Ghkl |2=(2π )2
(dhkl )2
(dhkl )2 =
(2π )2
|Ghkl |2
dhkl =2π|Ghkl |
=2π4.334
=1.449Å
1-‐ Driving the volume of the unit cell:
The volume of the unit cell:
Ωreal = a.(bxc)= a (bxc) cosϕ = abc cosϕ sinα (1*) Where ϕ is the angle between the plane (bc) and a We need to solve the volume as a function of α,β and γ only. We use : a x(bxc)=c(a.b)-‐b(a.c) abc sinα sinϕ=(acb cos γ –acb cos β) we square the both sides (abc sinα sinϕ)2=(acb cos γ –acb cos β)2 (acb cos γ –acb cos β)2= (acb cos γ –acb cos β). (acb cos γ –acb cos β)2
Ωreal = abc 1− cos2α − cos2 β − cos2 γ + 2cosα cosβ cosγ
γ=120o
90o
γ*
90o
30
o
b
a
b* a*
10
Note: a.a=a2, a.b=ab sin γ, a.c=ac sin β, the same for b and c. (abc sinα sinϕ)2=(acb cos γ –acb cos β)2= (abc)2[(cosβ)2 -‐ 2cos βcos γ cosα + (cos γ)2] (sinα)2 (1-‐cos2ϕ) =(cosβ)2 -‐ 2cos βcos γ cosα + (cos γ)2 sin2α cos2ϕ=sin2α – (cos2β-‐ 2cos βcos γ cosα + cos2γ) =1-‐ cos2α -‐ cos2β + 2cos βcos γ cosα -‐ cos2γ Thus,
sinα cosϕ=[ 1-‐ cos2α -‐ cos2β + 2cos βcos γ cosα -‐ cos2γ]1/2 (2*) by using eq.(2*) into e.q.(1*) Ωreal = abc cosϕ sinα = abc [ 1-‐ cos2α -‐ cos2β + 2cosβ cosγ cosα -‐ cos2γ]1/2
2-‐ driving
Ω* =(2π )3
Ωreal
We use again: a x(bxc)=c(a.b)-‐b(a.c) a.(bxc)=b.(axc)=c(axb) we will use also: a*, b* and c* equations.
Ω* = a*.(b* × c*) = 2π (c×b).(b* × c*)
a.(b× c)=2πb.[c× (b* × c*)]
a.(b× c)=2πb.[(c.c*)b* − (c.b*)c*]
a.(b× c)
we use : b.b*= 2π and b.c*=0 and definition of a*, b*
Ω* =2πb.[(c.c*)b* − (c.b*)c*]
a.(b× c)=(2π )2b..b*
a.(b× c)=
b.(c× a)[a.(b× c)]2
=(2π )3
a.(b× c)
Ω* =(2π )3
Ωreal
Ωreal = abc 1− cos2α − cos2 β − cos2 γ + 2cosα cosβ cosγ
11