unit 3 Flashcards

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1
Q

all imp formulas

A

μ=Vd/E, μ=eZ/m (mobility)
μ=1/pne, P=RA/L

Vd=eEz/m
z=λ/vth
E= Vd/μ

λ=vtz
λf=VfZ
Vf=√2Ef/m

J=σE, v=jpl, Vd=j/ne

time btw collision = 1/z
NOTE: take area as 1mm or 2mm, they expect to take dimension
J=neVd

V=Vde^-t/z => imp
t=z, V=Vde^-1

1/2mvth^2= 3/2kT
vth = √3KbT/m

Average velocity of e- due to thermal energy
random motion doesnt contribute to a net drift of electrons

CURRENT DENSITY
J = I/A=(ne^2τ/m)Ε = σΕ
σ=conductivity
σ= (ne^2t)/m = neμ
RESISTIVITY
J/E = σ= (ne^2t)/m = neμ
TOTAL
p= 1/σ= m/(ne^2t)/= 1/neμ
σ = ne^2λ/√3mKbT

electon drift m*dv/dt =
eE − kmv

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2
Q

Merits and demarks of CFET(Classical free electron theory)

A

CFET (Drude, Lorentz)
they treated electrons as if they were like molecules of an ideal gas

MERITS OF CFET
- explained conduction in terms of free e-
- showed that current is due to drift motion of e-
- showed that resistance is due to obstruction faced by e- due to vibrating atoms
-showed that as T increases J
- showed ohms law is valid
-> free electron gas

DEMARKS
->CFET could not give correct dependence of resistivity(σ) on T(Temp).
but experiment σ ∝ 1/√T
σ ∝ 1/T

2)experimentally values of specific heat of electrons in a metal is 1% of theoretical values
Ce ≈ 10^-4RT

3) Conductivity variations with electron concentrations.
σ= ne^2τ/m
nCu<nAl

8.4x1028/m^3 and Al is 18.1x1028/m^3

σcu>σAl
Copper with lesser electronic concentration has a higher electrical conductivity as compared to Aluminum.

CANNOT MAKE A STATEMENT LIKE THAT

expected linear dependency on the free
electron concentration

But, experimental results show no linear dependence

CFET COULDNT PROVE K/σ = LT

As per CFET resistivity arises due to scattering mechanism with
stationary ionic centers

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3
Q

how does current flow in a metal, explain in terms of drift velocity

A

the average velocity experienced by the electrons is called drift velocity

In a metal, free electrons move randomly in all directions due to thermal energy.
This random motion does not result in a net flow of charge

When an electric field is applied across the metal the electrons experience a force that causes them to drift in a direction opposite to the electric field (since electrons are negatively charged).

The electrons still move randomly due to thermal motion, but now there is a slight overall bias in their movement, thus net flow of charge. aka drift velocity

FORCE ON e-
Fe=-eE
a = Fe/m a=acceleration

drift velocity
Vd=etE/m= μE

μ=et/m = vd/E
drift velocity per unit electric field

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4
Q

particles (distinguishable, indistinguishable) using stats and Somerfield

A

exclusive : alone not together
non-exclusive: together like crows
distinguishable:
P(E) ∝ e^-E/Kbt (maxwell-Boltzmann)

indistinguishable:
(particles of light)=> non exclusive (BOSE-EIENSTIEN) BOSSON
exclusive: e- photons neutrons (FERMI-DIRAC) Fermious

IN THE FORMULA TAKE EF AS 0, and if E > EF its ‘+’, else E is ‘-‘ for ex -0.05, ot 0.05
ISA: if the energy levels are symmetrically located to the fermi level, probability will be one

sommerfield: the arrangement of particles together/not together in energy levels

Boltzman: no spin property
Bossons: spin integral values(0,1,2)
Fermions spin half integral(1/2,3/2,5/2…)

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5
Q

Fermi-Dirad Statistics/fermi energy

A

Occupation probabilities of valence electrons estimated using
Fermi Dirac statistics for Fermions

Valence electrons above the Fermi level contribute to conductivity

No contribution from valence electrons below the Fermi level
* Thus, all valence electrons are not conduction electrons
* The effective number of electrons above the Fermi level:
neff=n*KT/Ef
Thus a small fraction of valence electrons excited above the Fermi
level only contribute to conductivity

g(E) = π/2(8m/h^2)^3/2E^1/2

P(E) ∝ e^-E/Kbt

P(E) = 1/(1+e^E-Ef/KbT )
f(E) = Fermi factor
Ef => fermi level

Fermi energy- energy corresponding
to the highest occupied level at 0K

At 0K all the states below the Fermi energy are filled and all the states above are empty

highest energy level electrons can fill in 0K
No electrons can occupy energy levels beyond Ef

P(E)*g(E)=> number of electrons with a given E
=> total num of e-

Particles with spin ±
𝟏/𝟐 are classified as Fermions

for E<Ef T-0K
f(E) = 1/e-∞+1 =1 E-EF=’-‘
at 0k all e- states below fermi level are filled
for E>Ef
f(E) = 1/e∞+1 =0; T is 0 on denominator thus its ∞
g(E)*f(E) = N(E) (energy states density)
for T>0K and E=Ef
1/e^0+1=1/2; E-EF=0
fermi level: Prob of a level getting occupied is 50% at Ef,

n=π/3(8m/h^2)^3/2Ef^3/2 (electon concentration)

Ef=(3/π)^2/3(h^2/8m)n^2/3
n is the number of electrons per unit volume

Ef=1/8(3/π)(h^2/8m)^3/2n^2/3
0.121h^2/mn^2/3
Ep(Cu)=7ev

vp=√2Fp/m

vf=10^6
vp=10^-4

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6
Q

concept of fermi temperature
fermi mean free path

A

fermi temperature L
KbTf = Ef
this temp = fermi energy

QFET: λf = vf2
CFET:λ = vth
vf>vth

when a wave moves through orderly obstacles, it will move without collisions, when impurity is present the electron waves collide, thus resistance is formed

As per quantum free electron model, valence electrons near
Fermi energy only are excited into conduction band
* At temperature T, thermal energy available is kBT
* Thermal energy required to excite the last electron at the bottom
of the energy band at temperature Tf
is kBTf
Then,
𝑬𝒇 = 𝒌𝑩𝑻𝒇
* This temperature is termed Fermi temperature

Fermi velocity is greater than the thermal velocity and drift
velocity (from CFET) of electrons

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7
Q

does vf depend on T’
does Ef depend on T’

fermi energy temp difference

A

only small change thus it doesnt depened
Ef(T) = EF(0)[1-π^2/12*(KbT/Ef(0)^2]

0 K has the highest energy of electrons
At T=0, the system is in its ground state, and all electrons occupy the lowest possible energy states. The energy levels are filled up to Ef

which becomes the maximum energy for electrons.

There is no thermal excitation at 0 K, so electrons cannot jump to higher energy states.
This creates a sharp cutoff at EF​, meaning no electron has energy greater than EF

QFET valence e- near
Fermi energy only are excited into conduction band
At temperature T, thermal energy available is kBT
* Thermal energy required to excite the last electron at the bottom of the energy band at temp Tf
is kBTf =
Tf= ISA!!(fermi energy)
SIGNIFICANCE OF THIS TEMP= BOTTOM e- to conduction band
Then, Ef = kBTf

  • This temp is termed Fermi temp
  • An example - For Copper (Ef = 7 eV)

Fermi temperature ≈ 81000K

all valence electrons cannot be conduction electrons

.

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8
Q

what does an e- travelling through a metal detect when it approaches an obstacle

A

KBT ∝ R^2; KBT ∝ πR^2
No of collisions ∝ πR^2
No of collisions of ∝ KbT Mean Free Path
1/No of collisions
projectile is a circle

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9
Q

which QFET assumption was wrong,
and rectangular an array, of an 1D finite potential

Bloch function

A

Free e- are not free from electrostatic potentials
* Move in a periodic P.E approximated by rectangular P.E due to the regular arrangement of the ionic centers
If V(x) is the P.E at x then, then 𝑽 𝒙 + 𝒂 = 𝑽 𝒙 , the periodic P.E is invariant under translation through lattice parameter (a)

The P.E is lower close to ‘+’ ions in lattice
* The valence e- of diff atoms experience
similar P.E

ktan(ka/2) =a
kcot(ka/2)=-a
(Q^2K^2/2QK)sinhθbsinka + coshθbcoska+
coshθb
coska=cosk(a+b)

QFET could not explain the behavior of semi-conductors and insulators which have energy gap btw conduction band and valance band,

E(k)–k diagram for the system show discontinuity in the energy
at zone boundary of 𝒌 = ±𝒏(𝝅/a)

the assumption which is wrong is

e- in a metal are treated as if they are in an infinity 3D potential well

look at potential well using the V formula diagram: e^ikx(from diagram)

the graph made results in prob density of electrons graph, and will prob be same across adjacent points >

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10
Q

Energy of electron and KP model and simplified boundary condition in terms of k and q

KP FAILURE

A

an array of 1D finite potential adjacent boundaries can have similar properties ex ψII=ψII’ but not the same

a periodic pattern but the b(barrier gap, is stretched very thin, this its 0) and Vo goes to infinite

approximated the periodic potential as a long chain of coupled finite rectangular wells,

calculated by solving time-independent 1D
Schrödinger’s wave equations for the two regions I and II

ψx = C1e^ikx+C2e^-ikx
ψII=C3e^Qx+C4e^-Qx

Using Bloch theorem and the boundary conditions for continuity of the
wave function, can be obtained.
* A transcendental equation gives variation of E with propagation k, has discontinuities (forbidden gap)
* Allowed regions, e- are free to move with 𝑬 =ℏ^2K^2/2m

E of electron will appear in Band, graphs proves here E(Free e-) and E-k diagram, break in energy and k=+-nπ/a

BOUNDARY Condition
x=0
ψI = ψII
dψI/dx = dψII/dx

x=a
ψI= ψII’
dψI/dx = dψII’/dx

PsinKa/Ka + cosKa=cosKa

some K values allowed, some k values not allowed
Bands of k=> sinK and E are same Bands of E

A transcendental equation with solutions gives variation of E with
propagation k, has discontinuities (forbidden gap)
* Allowed regions, electrons are free to move with energy 𝑬 =ℏ^𝟐𝑲𝟐

Using Bloch theorem and the boundary conditions for continuity of the
wave function, the solution can be obtained.

Completely filled lowest inner band followed by a band of
forbidden energy states
* The highest band of allowed states represents valence band
* The upper most occupied states form conduction band

Thus, there are allowed and forbidden energy states for the
electrons in solids
Materials are then classified as metals, semiconductors or
insulators on the basis of energy band structure

Conduction and valence bands in reduced k space (Brillouin zone)

look at the graph made here, the graph says, some K values are allowed, some arent allowed

Bands of K -> sinK and E are same Bands of E

KP failure
cant give right value of band gap

P vs T resistivity vs temp
p=p(1+aΔT)
ISA: explain the the curve
P=Pi+Pph
Pi= intrinsic => temp independent
Pph => Matthiessen’s rule
ph=> phonon(temp based) Quantized elastic energy, on collision based opposite of photon

Resistance is because of scattering of matter ways, very small temp scatter is minimal

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11
Q

consequence of electron interacting with a periodic potential
names of zones ISA!!!!

CONCEPT OF EFFECTIVE MASS

A

-π/a<k<+π/a 1st (Brillouin Zone)
-2π/a<k<-π/a => 2nd Zone

Completely filled lowest inner band followed by a band of
forbidden energy states
* The highest band of allowed states represents valence band
* The upper most occupied states form conduction band

are allowed and forbidden energy states for the
e- in solids
explains why insulators/semi conductors see valence band and conduction band

CONCEPT OF EFFECTIVE MASS

Motion of electrons in the crystal is governed by the E=ħ^2k^2/2m* m*=>

shows E is nonlinearly dependent on the propagation constant k

m* is not a constant and depends on the non
linearity of E

effective mass
charge carriers have an effective mass which depends on the
curvature of E-k
“Free” e- => apply electric field
F=ma=eE, a=eE/m how to measure a?, change in velocity/motion of a body

“Bound Electron e-“ => Bound to metal, forces holding it

eE+?=ma ?=Fc (forces due to surroundings on the electron)

eE=m(variable mass=> mass is not ∞)*a(push it mass wont budge)

you push it in one direction, it goes into an other diagram, so is mass (‘-‘), no!!
ISA: effective mass concept, just to capture effects of the electron, and mass aint negative and not ∞ heavy

Effective mass of electrons can be higher or lower than the rest mass
of the electrons and depends on the position of electron in the
particular band

Curvature of the E-k is ‘+’ in the conduction band the
effective mass is ‘+’
* Curvature in the valence band is ‘-‘ with a ‘-‘ m, and indicates the concept of hole conduction in the valence band
* Effective mass of e- can be higher or lower than the rest mass of the e-and depends on the position of
e- in the particular band
* Concept of effective mass helps to explains mobility of charge carriers
* Expression for electrical conductivity, 𝛔 =𝐧𝒆𝒇𝒇𝒆^𝟐𝝉/𝒎∗

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12
Q

why a break in graph effective mass

Matthien’s rule, super conductivity and Meissiens effect

A

why a break in the εk graph at k= +-=nπ/a

ISA: shape and of graph breaks because of standing waves are results of superposition as it obeys Bragg’s law

similar anology
2dsinθ = nλ
θ=90, 2d=nλ

Matthien’s rule, super conductivity and Meissiens effect

An unusual property of certain metals,
alloys, and ceramics in which electrical
resistance drops to zero when the
temperature is reduced below a critical
value (Tc - known as the transition
temperature)

tot temp: intrinsic temp+phonon temp based
p=m/ne^2z => p=m/ne^2z
=m/ne^2zi+m/ne^2zPh

1/z=1/zi+1/zph

super conductor= mercury Tc=4.2K, kammerlingh Ounes

T<Tc>Tc normal conductor</Tc>

Type 1 SC
TC=1-10K
Type 2
TC=10-30K
Type 3
Tc>30K
258K record

HOLY GRAIL Tc=300K

Superconductors exhibit unique features other
than their ability to perfectly conduct current
Super conductors expel magnetic fields during
the transition to the superconducting state- like
a perfect diamagnetic material
This property is called Meissner effect

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13
Q

Sommerfield’s quantum Free e- QFET

A

Valence e- occupy discrete energy states following Pauli’s exclusion principle. Only e- close to Fermi level participate in conduction process; Valence e- = n≅ 𝟏𝟎^𝟐𝟖𝒎^-3

energy states split into discrete and closely spaced to accommodate all the valence e-; At ‘0K’ arrangement leads to a sea
with small separation
* Occupation probability = Fermi Dirac distribution
function applicable to fermions

  • The electrostatic interactions, the electron – electron and the
    electron – ion are negligible

Sommerfield’s quantum free e- QFET
- moving electrons are waves, when impurity is present the e- waves collide, thus resistance is there, but normally it through orderly obstacles, it moves without collision
- e- in a metal are treated as if they are in an infinity 3D potential well

En = (nx^2+ny^2+nz^2)h^2/8ma^2

  • Merit of QFET
    1) Correct evaluation of electronic specific heat – from valence e- close to the Fermi level
  • Heat absorption happens due to this small fraction of e-
    kBT/Ef is a fraction less than 1% and temperature dependent
    CuTF approx = 8.110^-4K
    it could explain the result CV = 10^-4
    dU/dT=(π^2/2)
    neff*Kb^2T/Ef

2)scattering mechanism with thermally vibrating ionic array 𝝀 ∞ 𝟏/T
σ not only depends on
the number of e- per unit volume but also depends
on the λ ∝ 1/T
σ =ne^2τ/m=ne^2λ/mvf

*Conduction e- move in array of ‘+’ ions, colliding with ionic centers and other e- resulting in R

Amplitude of vibrations increase with T followed by
increase in e- scattering
*cross sectional area of scattering=πr^2
𝝆 ∞ T (mean free path)

3) It doesnt open with the statement σ ∝ 1/T
σ not only depends on
the number of e- per unit volume but also depends
on the λ/vF

σ =neff^2τ/m=neff^2λ/mvf

4) wiedman - Frans law
K(Thermal conductivity)/σ(electrical conductivity)
∝ T
K/σ = LT L=> Lorenz const
L=π^2/3e^2Kb^2T
(2.4510^-8) WΩK-2
Thermal Conductivity
K=1/3
Cel/VvL

5) for a mole of electron gas, the fraction of e- gain energy KbT = neff=Na/Ef*KbT
6) As per QFET, the actual number of valence electrons depends on
concept of density of states; (g(E).dE) gives number of available electron states per unit volume per unit energy range at a certain energy
level, E

(g(E).dE) gives number of available electron
states per unit volume per unit energy range at a certain energy
level, E

DEMARKS QFET
metals are supposed to have HALL coefficients which are negatice
Rh = 1/n^e <0
Zn, Be etc = its positive R

These demerits reflects that real potentials of ionic centers in metal was ignored in the development of quantum free electron gas model

could not explain the behavior of se

Failed to explain
- differences in conduction in metal, semiconductor and insulator

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14
Q

comparison of fermi velocity and drift velocity

A

Fermi velocity is greater than the thermal velocity and drift velocity (from CFET) of electrons
Thermal velocity, typical order is 10^5 m/s
Typical order of Vd
for an electric field of 1 V/m is 10-4 m/sFg

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15
Q

classifications of solid

A

Case of conductors
Thus conductors (metals) characterized by a partially filled conduction
band (no band gap)
* Case of semi-conductors
Completely filled valence band and completely empty conduction band
Energy gap of 3-5 eV
At normal temperatures, possible for electrons in the valence band to
move into the conduction in the case of metals and semiconductors
* Case of insulators
Materials with energy band gap greater than 5eV
Electron conduction is impossible and attempts to excite lead to a
dielectric breakdown

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16
Q

Magnetism

A

Fields: source-moving charge(current)

magnetic field is measured as magnetic flux denisty (B) Wb/m^2 (T)

H(magnetic field intensity) or A/m (ampere/m)
-> M(Intensity of Magnestism) A/m

source: of Magnetic Field:
Ideal Solenoid: lenght of solenoid is far greater than diameter L»d
Bo=μnI; μ=free space, I
current is added for Flux linkage

H=nI => ISA, P and H which is depended and independent of medium??

=> B=μoH=> empty linkage in vacuum
=> B=μrμonI=> medium/material present
μr= relative permeability

(total permiability)μnI=> μrBo

17
Q

what are the three vectors of magnetism and chi , and materials with chi values

A

Bo = μo(H+M)
μrμoH=μoH+μoM

μrH=H+M
(μr-1)H=M

Xb=magnetic susceptibility
M/H=Xb

1) paramagnetic(most metals, except Cu)
Xb>0 and small ≈ 10^-5

2) Ferromagnetic
=> Fe,Ni,Co,Gd,Dy
Xb»»>0 ≈ 10^5

3) Ferrimagnetic Xb»»0
Ferrites and ≈ 10^3

4) Anti Ferromagnetic Xb>0 and 10^-5
Fe3O4

5) Dia magnetic Xb<0 and ≈ 10^-5
NaCl, Cu, H20, inert gases

ISA!!!!, they will ask these values
These 2 aspect of all materials
= M-H curve
= Xb Vs T

18
Q

is an atom a bar magnet

A

atom-> nucleon => pand electron => n

μ=> dipole moment coil equivalent form
μ=I*area = Am^2

the electron a dipole; e- inside an atom has l (orbital motion) and 2 spin motion

Here charge is going round and round, thus current is produced, and e- is a diople

μL = e/2mL; L=(angular momentum)
L= psedo vector as it never points in the direction in which the body moves,
=> always perpindicular, only mass matters not charge

and also direction of e- and I are opposite

μL/L=> gyro magnetic ratio
orbital motion gives e- a dipole motion

vector μL=-e/2m*L (e- is a dipole)

according to Quantum Mechanics
L=√l(l+1)ℏ=> Scrodingers Model
μL=e/2m√l(l+1)ℏ
=eℏ/2m√l
(l+1)

eℏ/2m=> very common

Bohrs atom
L=nℏ
can move in stationary models only 1st orbit = Lℏ

μL=e/2mL=e/2m
μb=eℏ/2m => Bohrs magnetism
μb=9.27*10^-24 Am^2

ISA

μl=μb*√6 CBT
Significance of this:
Kind of Unit for any Dipole Moment

19
Q

vector model and B to dipole

A

l=z, Ml=2,1,0,-1,2
Lz=meℏ
+l to -l
possible orientations of L where μL same whether orientations

Apply B to dipole
Torque T = μB
polarised energy = -μ
B
T= μL*B
T is outwards of the page

motion of L = > motion of plate whens its moving sidways on the ground, L is doing dandanakka

dp/dt=F, dL/dt (rate of change of angular momentum)
Lsindθ/dt = Lsinθ = Wp

dT -> very small value

=> motion => precision magnetic field produced torque

20
Q

prescession frequency IMPP, 1m ISA question

Spin motion,
Orbital L

CARD IS NOT DONE IGNORE!

A

Prescession
wp=gle/2mB=(μL/L)B
wp=gs
e/2mB=(μS/S)B
wp=gje/2mB=(μJ/J)*B

B=2T
(e/2m)B = 2e/2mB
2
μL/L=μS/S

J=√j(j+1)ℏ
j=1+s, 1-s,

A magnetic field of 1 T is applied to an orbiting electron. estimate the presectional frequency

B= IT; wp=e/2m 1.610^-19/29.1110^-31
= rads^-1= 8.78
10^10 rads^-1

Spin motion
S= spin angular momentum
S= √S*(S+1)ℏ s=> spin quantum number
S=1/2

μs=e/mS =>
μs=-2e/zm
S
μs=2√S(S+1)μb =>

μsB/s=2eB/zm

spinning electron precision is 2X faster in S
μL =e/2mL

Sz=msℏ
ms=+s to -s
(1/2,-1/2) => ms
wp(orbital) => μl/L*B

wp spin=μsB/s=2eB/zm

ORBITAL L
thus total =>
J=> μj = gJe/2mJ

-e/2m*L(J+S)
μj=projection of J on the other side sum of L and S

spin S => μs = (2e/zm)*S

An e- has both orbital and spin motion

total angular momentum J=L+S
√J(J+1)ℏ
J=-mjℏ, mg=+j to -j ,
j=> l+s, l-s

mj=5/2, 3/2, 1/2, -1/2, -1/2, -3/2, -5/2

21
Q

gJ formula and all magnetism and M defintion

A

gJ=(j(j+1)+S(S+1)-l(l+1))/
ZJ*J(J+1)

μj=gle/2mL
μs=gle/2mS
μl=gle/2mJ

No spin: J=l => gJ=l=gL
no orbital; j=S =>gJ=z=gs

for atom we need to consider L for all e-, s for all e- and hence J for all e-

L’ =sumL
S’=SumS
J’=Sum J= sum(L+S)

μl=> dipole moment of an atom
if μj =0 atom is not a dipole
if μJ =!0 atom is a dipole

fara,ferro,feri,antiferro μj =! 0
dia
μj=0 => inert gases,no extra e- thus μj=0
dia magnet: suppress the magnetic field of other electrons for magnetic

TEMP HAS NOT INFLUENCE ON ELECTRON MOTION ON ORBITAL

Para magnetism
atom->dipole
=> condition atom should be dipole, paramagnetic sample => dipoles

M= summation of projected dipole moment

22
Q

μJz

A

takes projected value of N on z axis in terms of μJz

take projected value of N energy are degenerate same energy level

P.E=-μJ*Bcos()

the pe of each dipole value is diff as each has diff cos value

Labelling of energy level:

using mj=, -j,+j
limits of mj atoms have these possible level

N atoms would have distributed fully in all levels

sum Ni=N (energy level)
AVERAGE ENERGY
if equal height up and down its 0
μjz=gjμbbmJ
PE=-gJμb
mj*B

23
Q

classical dipole

A

quantum: says its discrite, classical says no!
can have all orientations, can take all possible values
mj=+j,+0,-j
j=7/2
7/2,5/2,3/2,1/2,-1/2,-3/2,-5/2,-7/2
M=MsB(j,y)

B(j,y)= 2j+1/2jcoth((2j+1)y/2j))-1/2jcoth*y/2j

La(y) =cothy=1/y
paul Langeus

24
Q

classifications of magnetic materials

A

Diamagnetic materials

  • No unpaired e- in the orbitals
  • Atoms/molecules that have zero magnetic moment hence no
    permanent magnetism
    In presence of external field, interaction between field and
    electrons
  • This effect is similar to that due to Lenz’s law induced magnetic fields tend to oppose the change which created
    them.
  • Diamagnetic materials have ‘-‘ susceptibility
  • Magnetic Susceptibility is:
    Xdia=Ne^2μ0/(6m)<r^2>

Paramagnetic
Atoms have non-zero magnetic moment due to
unpaired electrons
* Aluminium
have ‘+’ susceptibility
H>0, μ>0
Magnetic moments of unpaired electrons are in random directions
* Applied field induces magnetization
* State of magnetization is disturbed due to thermal agitation
* 𝞆 depends on temperature

When field is removed the magnetic moments go
back to random orientations

When the field is removed the magnetic moments go
back to random orientations

FERROMAGNETC materials
large’+’ susceptibility values
* Internal dipole moments
* Magnetic dipole

Magnetic domains regions where the dipoles are aligned

When B is applied - alignment of all the domains along B
* long range alignment of spins due to strong
exchange interaction

Even after B is removed -remnant magnetization
* Magnetization of these materials decreases with
increase in temperature
* 𝞆 depends on temperature

moments of atoms arising from the spin of e-
1) Spin-> spin interaction leading to spontaneous magnetization even large regions
2) characteristic temp, Tc or theta c (Curies temp), IMP!! curies units = K
beyond which ferro turns parra
3) M-H curve dispalys hysteresses

Energy of interaction = W=-2JklSkSl

soft and hard ferro:
soft: low loss, high permeability
ex: carbon steels, iron-silicon alloys, iron-Al alloys, nickel iron alloys

Permanent: hard magnetic materials strongly resist demagnetization once magnetised its hard to demagnetize

FERRI magnets
magnetic moments of the sublattices are not equal.

Anti Ferro: net effect cancels overal, but overall up is greater than down

thus feablemagnetisation

25
Q

SUPER CONDUCTOR

A

An unusual property of certain metals,
alloys, and ceramics in which electrical R drops to 0 when the temp is reduced below a critical
value (Tc - known as the transition temperature)

expel magnetic fields during the transition to the superconducting state - like a perfect diamagnetic material = Meissners effect

Normal B =! to 0, superconducting state B=0

CRITICAL STATE:
strong external fields cause destruction of
superconducting property
Magnetic field at which the material loses it superconducting state - Critical Magnetic Field (Hc)

The critical field strength is temperature dependent
and is given by => Hc=Ho[1-(T/Tc)^2]
Ho is the magnetic field required to destroy the superconducting property at 0K

Super conductors are perfect diamagnets

Xb<0, Xb=10^-5, M/H = 10^-5, Xe=-1, M=-H
(Perfectly Opposed)

TYPE 1 superconductors:
exhibit complete Meissner
effect
In the presence of external magnetic field H < Hc
material in superconducting state

As soon as H exceeds Hc material become normal

Type I material has very low values of Hc
Ex: Al Pb, Indium etc.

TYPE 2:
two critical fields
Hc1 and Hc2 and practically important
Behave as perfect superconductor up to Hc1
Above Hc1 magnetic flux starts to penetrate
(mixed state –Vortex state) up to Hc2; Above Hc2 material behave as normal

phi= n*h/2e => unit of flux=>

Hc1 and Hc2 – lower critical field and upper
critical filed

Vortex state – partial flux penetration through filaments

Ex: niobium, Si and Va.

Current in the superconductor persists for a long time.
➢ Not observed in Mono valent metals.
➢ Exhibited by metals for which the valence e-
num are bet 2 & 8.

➢ in metals having a higher P at normal temp .
➢ Destroyed by applying high magnetic fields or excessive currents.
➢ Ferro and anti ferromagnetic materials are not superconductors.

26
Q

BCS theory

A

➢Based on the formation of Cooper pairs
➢During electron flow, because of opposite polarity between
electron and ion core, results in lattice distortion (phonons)
➢Lattice distortion results in the interaction of another e- as interaction between two e- via lattice

➢If the e- have equal and opposite spins and opposite momentum, e-lattice (phonon field)-e- interaction exceed Coulomb repulsive force forming Cooper pairs (quantum
pairs – bosons)

➢The energy required to separate the pairs is far too large compared to the thermal energy available
- COOPER PAIR=> 2 e-:
describing a bound pair of e- (or fermions) that move through a lattice without resistance, 2 e- attracted ‘+’ ion s thus in that region its more ‘+’

ISA: oppsite spin and momentum, cooper pairs carry current in super conductor, cooper pair is a Boisson, doesnt follow exclusion principle

its a giant wave packet, so no scattering, 10^6 pairs giving Boissions, 0 reisitance

➢ Cooper pairs collectively move through the lattice with small velocity
➢Low speed reduces collisions and decreases resistivity, which
explains superconductivity
➢The attraction between the e- in the Cooper pair can be separated by a small increase in temp, which results in a
transition back to the normal state
➢MAGLEV VEHICLES – Magnetic levitation based on Meissner
effect, MRI