crystal physics lectures- 2

8/18/2019 Crystal physics lectures- 2

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(1) Derivation of equilibrium numberof Schottky defects (vacancydefects) in elemental solids:-

1


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(1) Derivation of equilibrium number of

Schottky defects (vacancy defects) inelemental solids:-

• At any fnite temperature, some o thelattice sites which are normallyoccupied by metal ions may be vacant.

• i.e. vacancy deects are present atfnite temperatures.

• E.g. in copper metal, some o thecopper atoms/ ions may be missingrom some o their regular lattice

positions. 2


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• At some constant temperaturesome o the vacancies are

created and some o thevacancies are destroyed.

•So euilibrium is reached at thattemperature.

• !.e. euilibrium number o

vacancy deects are present attemperature ".

#


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$


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So i,

n % number o vacancies at

temperature "

Ev % average energy reuired to create

a vacancy

& % "otal number o atoms

" %Absolute temperature in 'elvin.

)


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"hen, n * + &, Ev, ".

-e will derive this relation now.

• "he creation o a deect reuires energy,

i.e. the process is endothermic, i.e. heatis absorbed by the system.

• As a conseuence o a deect ormation,

the energy gain is more thancompensated by the confgurationdisorder introduced due to deect

ormation.


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• or low concentration o deects, auantitative relation between n, &, Ev 0 "

may be obtained through the use oawell 3 4olt5mann statistics.

• "he use o awell 3 4olt5mann statisticsis 6ustifed since the lattice sites aredistinguishable although the metal ionsoccupying them are not.

• "he interaction between deects isneglected.

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• "he total number o di8erent ways +- inwhich we can pic' up n3 atoms rom thecrystal consisting o &3 atoms is given by

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( 1)( 2)...........( 1)

!

N N N N nW

n

− − − +

=

!

( )! !

N

W N n n= −


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• &ow the entropy o the system changeswith this process.

-here,'4 % 4olt5mann constant * :/&A.

'4 *1.#9 × 1;32# 3= 31.

?

ln B

s k W ∆ =


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• A% @elmholt5 ree energy

1;

!ln

( )! ! B

N s k

N n n

∆ =

−


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Energy

11

% !nternal Energy E

n/N0,0

3"S

A%@elmholt5reeenergy


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* A B "S * E

A * 3 "S * E C "S

&ow the @elmholt5 ree energy change

+ ∆A is given by

12

A = E -

S


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1#

0T

An∂∆ ∴ = ÷∂


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∆A * ∆E 3 "∆S

• "a'e partial derivative o ∆A withrespect to n at constant temperature0 euate to 5ero. Dsing Stirlingsapproimation ln F * ln 3 G

1$

!ln( )! !

v B N A nE Tk

N n n ∆ = − −


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1)

!ln

( )! !ln ! ln( )! ln !

ln ( ) ln( ) ln

ln ( ) ln( ) ln

N

N n n

N N n n

N N N N n N n N n n n n

N N N n N n n n

− = − − −

= − − − − + − − +

= − − − −

[ ]ln ( ) ln( ) lnv B A nE Tk N N N n N n n n∆ = − − − − −


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1

[ ]{ }v(nE )

ln ( ) ln( ) ln

1 1{0 ( ) ( 1) ln( )( 1) ln }( )

{1 ln( ) 1 ln }

B T T T

v B

v B

ATk N N N n N n n n

n n n

E k T N n N n n n N n n

E k T N n n

∂∂∆ ∂ ∴ = − − − − − ÷ ÷∂ ∂ ∂

= − − − × − − − − − × −−= − + − − −

( ){ln } 0

( )ln

ln

exp

exp

v B

T

v B

v

B

v

B

v

B

A N n E k T

n n N n

E k T n

E N n

k T n E N n

n k T

E n

N n k T

∂∆ − = − = ÷∂

−∴ =

− ∴ = ÷

−∴ = ÷

∴ = − ÷−


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exp

exp

v

B

v

B

E n

N k T

E n N k T

= −

÷

= − ÷


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• sually the number o vacancies n is avery small raction o the total number ometal ions / atoms, so that above

euation can be used to calculate thenumber o raction o vacancies so longas the temperature is ar less rom

melting point o crystal.e.g.

E> *1 e> " * 1;;; =, '4 * 9.2)×1;3) e>3= 31

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1?

5

6

6 5

exp

1exp

8.625 10 / 1000

exp( 11.6) 9.2 10

10 10 10

1

100000

100000 1

v

b

E n

N k T

eV

eV K K

for N atoms n vacancy

−

−

− −

= − ÷

= − ÷× ×

= − = ×≈ × =

=

∴ = ⇒ =


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or 1;) atoms → 1 vacancyor 1 mole atoms → +H >acancy

1 mole * .;2 × 1;2# atoms.

2;

23

235

1 6.02 10 6.02 1010

× × = ×!n 1 mole o atoms. !.e. or & *.;2

1;2# atoms, n * .;2 1;19 vacancies.rom the above relation we can say thatthe euilibrium number o vacancies

increases with increase in temperature.


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Euilibrium number o vacancies inionic crystals or diatomic solids

(!) Derivation of equilibriumnumber of Schottky defects(vacancy defects) in ionic solids

or diatomic solids:-

• !n ionic crystals ormation o pairedvacancies is most avored.

• i.e. an eual number o Bve and Cveion vacancies are produced.

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• "he ormation o pairs ma'es itpossible to 'eep the surace o the

crystal electro3statically neutral.• "he number o vacancy pairs can be

related to the total number o atoms

present in the crystal on ollowingthe same procedure as adoptedabove.

• "he number o di8erent ways inwhich n separated vacancy pairs canbe ormed are given by

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2#

for -ve ion vacanciesfor ve ion vacancies

2

! !

( )! ! ( )! !

!

( )! !

N N W

N n n N n n

N W

N n n

= × − −

=

−


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• So that increase in entropy will begiven by

2$

2

ln

!ln( )! !

B

B

s k W

N s k N n n

∆ =

∆ = − Now the Helmholtz free energy change ( ∆ A)

is given by


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2)

∆A * ∆E 3 "∆S

• -here Ep is the energy o ormationo vacancy pair

• "a'e partial derivative o ∆A withrespect to n at constant temperature

0 euate to 5ero. Dsing Stirlingsapproimation ln F * ln 3 G

2

!ln( )! !

p B N A nE Tk

N n n ∆ = − −


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• Substituting the above epression or

ree enerergy, we get

2

[ ]

[ ]

[ ]

2

!ln

( )! !2 ln ! ln( )! ln !

2 ln ( ) ln( ) ln

2 ln ( ) ln( ) ln

N

N n n N N n n

N N N N n N n N n n n n

N N N n N n n n

− = − − −

= − − − − + − − +

= − − − −

[ ]2 ln ( ) ln( ) ln p B A nE Tk N N N n N n n n∆ = − − − − −


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• &ow di8erentiating the aboveepression with respect to n, we get

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[ ]2 ln ( ) ln( ) ln p B A nE Tk N N N n N n n n∆ = − − − − −

[ ]{ } p(nE )

2 ln ( ) ln( ) ln

1 12 {0 ( ) ( 1) ln( )( 1) ln }

( )

2 {1 ln( ) 1 ln }

B T T T

p B

v B

ATk N N N n N n n n

n n n

E k T N n N n n n N n n

E k T N n n

∂ ∂∆ ∂ ∴ = − − − − − ÷ ÷∂ ∂ ∂

= − − − × − − − − − × −−

= − + − − −


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!ree ener"# in $%er&al e'iliri& a$$aine*

a$ $e&pera$re + is cons$an$, so $%a$

( )2 {ln } 0

( )2 ln

ln2

exp 2

exp2

p B

T

p B

p

B

p

B

p

B

A N n E k T n n

N n E k T

n E N n

k T n

E N n

n k T

E n

N n k T

∂∆ − = − = ÷∂ −

∴ =

− ∴ = ÷

−∴ = ÷

∴ = − ÷−


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• &ow n JJ N. So we can write

2?

( ) exp2

exp 2

p

B

p

B

E n N nk T

E

n N k T

= − − ÷

∴ = − ÷


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• or &aKl crystal Ep*2.;2 e> and at

room temperature

• &ow calculations o ren'el deectsor elemental solids and ionic crystalsare remaining.

#;

6 310 c&n −≈

crystal physics lectures- 2

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