Soution of HW2

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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)

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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

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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)

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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

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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

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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

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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


γ=120o

90o

γ*

90o

30

o

b

a

b* a*

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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


Soution of HW2

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