tugas fisika zat padat chapter 2

7/24/2019 Tugas Fisika Zat Padat Chapter 2

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Nama Kelompok:

Dewi Hardianti (130322615541)

Fahyrinda !r"iari#mana (130322615545)

$mam %antowi (130322615540)

&i#a 'malia ari (13032261554)*hammad Firman#yah (13032261550+)

,ini $#mi -anti (13032261553)

.hapter 2

1. The crystal plane with Miller hkl is a plan defined by the points a1/h, a2/k,

and a3/l.

a. Two vectors that lie in the plane may be taken as a1/h a2/k and a1/h -

a3/l. !t each of these vectors "ives #ero as its scalar prod!ct with

/ha1 ka2 la3, so that /m!s be perpendic!lar to the plane hkl.

b. $f n is the !nit normal to the plane, the interplanar spacin" is n

a1h. !tn

// whence d%hkl& ' (.a1/h)() ' 2*/)()

c. +or a simple c!bic lattice / %2*/a&%h ^x

k y

l ^z

&, whence

222

2

2

222

2

2

2 -

1

lkh

ad

a

lkhG

d

++=

++==

2.

( ) ( ) zcayaxaayaxaa ..2.23.2.23 321

22

1

1 =+

=+

=

a. 0ell vol!me a1.a2a3

ca

c

aa

aa

axaa

2

321

32

1

2

13

2

1

2

13

2

1

=

=


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b. rimitive translation of the reciprocal lattice

zc

aa

aa

zyx

caxaaa

xaab

yxa

c

aa

zyx

caxaaa

xaab

yxa

c

aa

zyx

caxaaa

xaab

.2

2

13

2

1

2

13

2

1

...

3

-2

.2

1.

3

2

2

13

2

1

...

3

-2

..3

12

2

13

2

1

...

3

-2

2321

213

2321

312

2321

321

=

==

=

+=

==

=

+=

==

=

c. 4i vectors in the reciprocal lattice are shown as solid lines. The

broken lines are the pendic!lar bisectors at the midpoints. The

inscribed hea"on forms the first rillo!in 5one.

3. y definition of the primitive reciprocal lattice vectors6

( )( ) ( ) ( )( )

( ) ( )

( ) c

BZ

V

xaaaxaaa

xaaxaaxaaV

3

321

3

3

321

2113323

2

/22

=

=

=

+or the vector identity, see (. 7. 8orn and T. M. 8orn, Mathematical

handbook for scientists and en"ineers, Mc(raw-9ill, 1:;1, p. 1


3/4

.

( )[ ]( )[ ]kai

kaiMF

=ep1

ep1

x

xx

MM

M

M

=

1

11

a. This follows by formin"

( )[ ]( )[ ]

( )[ ]( )[ ]

( )( )

( )

( )ka

kaM

ka

kaM

kai

kaiM

kai

kaiMF

=

=

=

2

1sin

2

1sin

cos1

cos1

ep1

ep1

ep1

ep1

2

2

2

b. The first #ero in2

1sin

M=occ!rs for

M 2=. That is the correct

consideration follows from

MMhMMhhM2

1sincos

2

1cossin

2

1sin +=

+

>.

( ) ( ) ++=j

vzvyvxi jjjefvvvS 3212

321

?effered to an fcc lattice, thr@e basis of diamond is -

1

-

1

-

1

. Th!s in the

prod!ct

( ) ( ) ( )basisSxlaticefccSvvvS =321,

Ae take the lattice str!ct!re factor from %=&, and for the basis

( ) ( )321

2

1

1vvvi

ebasisS++

+=

now 4%fcc& ' only if all indices are eeven or all indices are odd. $f all

indices are even str!ct!re factor of the basis vanishes !nless v1 v2

v3'n, where n is an inte"er. +or eample, for the reflection %222& we have

( ) 31 iebasisS +=, and this reflection is forbidden

5ero as

Mh isan

B


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;.

( ) ( )

( ) ( )( )( ) ( )

( )222

22

2

3

3

3

3

13

2

-1;

1--

2epsin-

2epsin-

aG

aGrGaaG

GaxxxdxaG

drarGrGrarfG

+=

+==

=

The inte"ral is not diffic!lt it is "iven as Cwi"ht =;,=1. Dbserve that f'1

for (' and f E 1/(for (aFF1.

tugas fisika zat padat chapter 2

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