Order 1262647: CMP2-2

Q1. For plane A: Intercepts= , Reciprocals=3 ,2, -2, Indices: (3 2 -2 )

For plane B: Intercepts = - , Reciprocals = -2, 0, 2, Reduction= -1, 0, 1, Indices: (-1 0 1)

Q2. For copper the crystal structure will be FCC and those peaks with the h,k and l that are integers will be seen. So, the planes with miller indices (111) will produce the first peak by diffraction

a) Interplanar spacing is found by using

d =

n= 1, λ= 0.1542 nm and θ is approximately 42.5⁰/2 = 21.3

So, for plane (111)

d111 = 1x 0.1542/2sin(21.3°) = 0.2123 nm

b) Lattice constant from each peak is given by:

R = =

For the first peak h=1, k=1, l= 1

And d111 = 0.2123

So

R= 0. 2600

For the second peak

Peak index is 200, so

d 111 = 2 x 0.1542/ 2sin(25°)

= 0.1824 nm

Therefore R= 0.1824 x / 1.14142

= 0. 1290 nm

For the third peak

d = 3 x 0.1542/ 2sin(37.5°)

= 0.1267 nm

R = 0.1267x 1.7321/ 1.4142

= 0.1552 nm

For the fourth peak

d = 0.1542/ 1.4142

= 0.1090 nm

R= 0.1090 x 1. 7321/ 1.4142

= 0.1336 nm

Q3.

Given that Nd = 1x107 cm-3 and Na = 3 x10 cm -3

Then it is assumed that all donor and acceptor impurity atoms are ionized. Therefore from the charge neutrality condition,

N+NA = p + ND

Where n is electron and p is hole concentration

For the n type semiconductor n is greater than p

Therefore

N ≈ ND – NA 1

np = ni2

Implying that

P= ni2/ (ND –NA ) 2

ni is intrinsic carrier concentration

For the p type p is greater than n, so;

P ≈ NA 3

n = ni2/p

Electron concentration = ni2/(NA –ND ) 4

From equation 3, p= 2x1017 cm-3

Therefore, for ni for Si at 450 K will be;

ni (450) = ni (300)Eg /2KB (1/300 -1/450)

Eg = 1.12x 1.6 x10-19 and 2KB = 2x 1. 38 x10-23 and ni(300) = 1.5 x 1010 e

Substituting gives

ni ( 450) as 2.04 x 1013 cm-3

Using equation 2

ni = 2.04 x 1013 cm-3 and p = 2x1017 cm-3

n = 2.1 x 109 cm-3

Q4.

a) X= µ0M/Bex = µ0µB2g(Ef)

Y0 = 4 x 107(units)

YB = -9.27 x 10-24

g(Ef) = 3n/2Ef ; Ef = 3.24 eV

Therefore X = 4pi x 107 x -9.27 x 10-24 x 3 x 2.5 x 1028 / 2 x 3.24 x 1.6 x 10-19

= 7.905 x 108

b) Bin = µ0M

M = nµB2H/kbT

Magnetic Flux B = µ0(Hχ)H

M = Bex/ µ0, where Bex = 1

M = 1/ 4pi x 107

= 7.8539 x 10-8

Bin = 7.8539 x 10-8 x 4pi x 10-7

= 9.4 x 10-14T

c) Bin = µ0M

M = Nµ/v

but N/v = n, so

Bin = µ0µni

For Bin > Bex

ni >> 2.53 x 1028

Bni = Bex

Thus: 1 - 9.4 x 101.4n = 2.53 x 1028

n = 2.53 x 1028

9.99922n x 10-1 = 2.53 x 1028

n = 3.66 x 1013m-3

Therefore, n should be higher by

3.66 x 1013m-3

Q5.

Superconductivity

Superconductivity is phenomena in which materials are able to conduct current at zero resistance. Superconductivity is associated with extremely low temperatures.

Materials that conduct current at almost zero resistivity are referred to as superconductors.

The most notable types of super conductors depend on the transition of the material’s resistivity to zero. This phenomenon groups the superconductors into type I and Type II superconductors. .Type I are exhibited in pure metals such as mercury and lead . Type II are found in alloys like Niobium-Tin and in those with high critical Temperature TC . Such as BaCuO.

At critical temperature the material is also associated with Critical Magnetic Field. the behavior of Superconductors are explained by the BCS theory which is based on the Bandgap of electron pairs together with the Isotope effect and heat capacity. The interaction of superconductors with magnetic fields is explained by Meisser Effect which is associated with characteristics lengths which produce Magnetic levitation.

Applications of superconductivity include ; superconducting magnets, SQUID magnetometer, NMR imaging and in Josephson devices.