8/8/2019 Sol Prob on Semi- & Super-cond

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Solution of Problems on Semiconductors & Superconductors

1. 12.7)09.010505.0108(106.1)( 192019 =+=+= np npe ohm1 m1

( ) ( )005.0

)09.010505.0108(106.1

09.010505.0108

)(

12192019

219220

2

22

=+

=

+

=

np

np

Hnp

np

eR

2.p

n

n

p

np

np

np

Hn

p

npnp

np

eR

===+

=22

2

22

0)(

1

Current density due to electrons alone, EneJ nn =

Total current density, EnpeJ np )( +=

Fraction of current carried by electrons,np

p

p

n

n

pnp

ne

n

pnp

n

J

Jf

+=

+

=

+

=+

==

1

1

1

1

3.kTEgeA

2/=

7.0

2

)0(

)20(

)2730(2/

)27320(2/

==

+

+

kE

kE

g

g

e

e

C

C

(given)

=

7.0

2ln

293

1

273

1

2 k

Egwhich gives

=

7.0

2ln

20

293273

106.1

1038.1219

23

gE eV = 0.724 eV

4. Intrinsic concentration of charge carriers,

( )

( )

( ) ( )3001038.12/106.17.04/331312/3

234

23

2/4/3**

2/3

2

2319

101.94.0101.907.0

10625.6

3001038.1142.322

22

=

=

e

emmh

Tkn

kTE

hei

g

= 2.3 x 1018 /m3

+=

+=

07.0

4.0ln

106.1

3001038.1

4

3

2

7.0ln

4

3

2 19

23

*

*

e

hg

fm

mTk

EE = 0.384 eV above the Valence Band

5. Density of Ge atoms = 281041.4 /m3

Density of As atoms = 0.1 % of density of Ge atoms = 251041.4 /m3Intrinsic carrier density, ni = 2.37 x 10

19 /m3

Electron density, ne = density of donor (As) atoms as all of these are ionized

= 251041.4 /m3

( ) 1325

21922 1027.1

1041.4

1037.2=

===

e

i

hihen

nnnnn /m3

Resistivity of intrinsic Ge, 47.0)18.038.0(106.11037.2

1

)(

111919

=+

=+

==

heii

ien

ohm-m

Resistivity of doped Ge,

7

1325191073.3

)18.01027.138.01041.4(106.1

1

)(

11

=+

=+

==hheeex

exnne

ohm-m

8/8/2019 Sol Prob on Semi- & Super-cond

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6. Critical field varies parabolically with temperature

( )

( ) 11237

2

2

2

2

52.34811025104

7)(

21.237

61,6

5.87252

7

2

221

0

0

000

=

=

=

==

=

=

===

===

=

KkmolJTd

BdTTTatC

mTBBKatfieldCritical

mTTd

BdTB

TTforBT

BT

T

Td

Bd

T

TBB

c

o

c

cel

cc

TT

cc

c

cc

c

c

c

c

c

cc

c

7. Critical field at any temperature T,

( )

( ) mAKH

mAHfromThen

KTT

T

TbyDividing

THand

TH

TTHH

c

c

cc

c

c

cc

cc

ccc

/1005.2047.14

2.41109.21)2.4(

/109.21

47.14141

104.1),1(

47.145.209

141

131

3:)1()2(

)2(131102.4

)1(141104.1

1

52

5

52

5

2

2

2

2

5

25

2

0

0

0

0

=

=

=

=

==

=

=

=

=

LL

LL

8. Each lead atom contributes 1 free electron,

No. of free electrons per unit volume = No. of atoms per unit volume= Avogadros no. x density /Atomic weight

OR 28326

1028.3

19.207

103.1110023.6=

=n /m3

London penetration depth at 0 K,

7

219287

312/1

210293.0

)106.1(1028.3104

101.9

=

=

=

en

m

o

o m = 29.3 nm

London penetration depth at any other temperature,

4

1

=

c

o

T

TT

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Penetration depth at 3.61 K,

( )3.30

22.761.31

3.29

461.3 =

= nm

Percentage increase in %4.31003.29

3.293.30=

=

9. Energy gap, 2 4104.3 = eV; Fermi velocity,61002.2 =Fv m/s

Intrinsic coherence length, 6194

634

109.3106.1104.3

1002.21005.1

2

=

=

= F

vh m = 3.9 m

10. The solenoid is at the temperature of liquid He, i.e., 4.2 K

Critical field at this temperature, ( ) 5252

10255.15.9

2.411056.11 =

=

=

ccc T

THHo

A/m

Each turn of wire in the solenoid is under the effect of the field produced by the solenoid,

Transverse applied filed,H= 6 x 104 A/mFrom Silsbees rule for such a situation, )2(2 HHrI cc =

Minimum diameter of wire = 345

1015.1)106210255.1(142.3

20

)2(2

=

=

=HH

Ir

c

c

m = 1.15 mm