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Answered Same Day Dec 22, 2021

Solution

David answered on Dec 22 2021
133 Votes
Problem 1
In this review nanocrystal band structure is explained from the starting point ie Molecular bonding
theory and gradually we end up with nanocrystal band structure , also a detailed understanding of Brus
equation which is used to calculate the band gap of a nanocrystal .
Molecular Bonding Theory
Bonding
Bonding between two atoms can be described as overlapping of two hy
id o
itals, which represent the co
ect
valence in geometry.
The hy
id o
ital is a linear combination of AOs (atomic o
itals) of a single atom and different linear
combination will result in different geometries
Example
1) Beryllium hydride  5 2 2 1Be 1s ,2s ,2p
Linear combination of one 2s and one 2p o
itals results to sp hy
id o
itals
z z
1 1
sp(1) (2s 2p ) and sp(2) (2s 2p )
2 2
   
2) Methane
The 2s and three 2p o
itals of ca
on result four sp3 hy
id o
itals
Frontier o
itals
Many properties of the molecules can be interpreted to the use of the MO model and in particularly by looking at
the frontier o
itals
 the HOMO (highest occupied MO o
ital) --- Donor level
 the LUMO (lowest unoccupied MO o
ital) ---- acceptor level


 
N
E f (N) E decreasesas Nincreases
0 Conductor eg.graphite
lim E
0 Semiconductor / Insulator eg.hexagonal boron nitride
  
 
  
 

Energy levels
With increasing number of o
itals, the energy levels gets closer and closer and the energy difference between
HOMO and LUMO is decreases .
Cyclic structure (Benzene and Graphite )
In Benzene each ca
on atom undergoes SP2 hy
idisation, and the remaining 6 P-o
itals combined to give 6 de-
localised MO ∏ o
itals
Benzene graphite
Graphite is like Benzene stretching to infinity having a hexagonal structure. Graphite consist of layers of fused six-
membered ca
on ring with an interlayer spacing of 335pm .The remaining unhy
idized p - o
itals participate
in extensive ∏ bonding, with electron density delocalised over the layers.
Band Structure for bulk materials
At very high pressure graphite can be converted to diamonds. In diamond all ca
on atoms undergo sp3
hy
idisation. All the four sp3 o
itals overlap to the maximum , resulting extensive valency electron delocalised in
3-D. Diamond has tetrahedral crystal structure
Solids are extreme case of de-localisation.
Even the silicon crystal is having the same tetrahedral crystal structure and valency electrons are delocalised as
that of diamond but the sp3 o
itals overlap is are not as large and bonds are weak
(Source : NSF Nanocenter on Electrical Conductivity of Single Molecules and Ca
on Nanotubes)
Classification of materials according to their size
Materials having diamond and Zinc-blend lattices (like Si ,Ge ,GeAs ,CdS ,InSb , ect...) can be classified in term of
their size , as shown below
L E Brus Equation
Brus equation describes the lowest electronic transaction in quantum dots and a pertu
ation that results from
the columbic interaction between the electrons and the holes .
The Brus equation is given as
2
NC g 2
e h o
h 1 1 1.8e
E E (bulk) polarization term
8R m m 4 R
 
     
 

2k2 2
NC g k2
k 1e h o
h 1 1 1.8e e S
Or E E (bulk)
8R m m 4 R R R


   
        
   

Bruce equation is derived based on the following assumptions
 effective mass approximation
 the potential energy outside the radius R is infinity -- The radius defines the confining boundary of the box
 The interior of the nanocrystal consist of a uniform medium
 effective mass of charge ca
ier (excitons) and dielectric constant of the solid are constant with reference to
size
Variables and constants in Brus equation
First term gE (bulk) --- Band gap of the Bulk crystal
Second term
2
e h
h 1 1
8R m m
 
 
 
---- This term is the kinetic energy of the electron in an infinite potential well of
adius R.
Where me = effective mass of electron
mh = effective mass of hole
R = radius of the nanocrystal
h = plank’s constant
Third term
2
o
1.8e
4 R
---- This term is the coulomb interaction energy between electron and hole .
Where e = charge of an electron
 = permittivity of free space
o
 = dielectric constant of the nanocrystal
R= radius of the nanocrystal

Fourth term polarization energy
Further localisation
Localisation used to imply electron confinement within a cluster, in a state that has population density
throughout the cluster .
If the cluster has point effects and/or surface states, then further localisation is possible
Bulk material defect states are of two types
Shallow traps --- lies within few meV of the co
esponding band edge and can be described by a hydrogenic
model where the mobile charge o
its the trap in a larger 1S o
it of radius 
o o
o
a where a Bohr radius
m*
m

  
Deep traps -- essentially localised in space at a lattices site defect and lie in the middle of the band gap
The below figure shows the co
elation between the cluster states to bulk crystal states
(Source : The Journal of Physical Chemistry,Vol 90 , No 12, 1986)
Problem 2
The Brus equation is given as
2
g g 2 * *
e h o
h 1 1 1.8e
E E (NC) E (bulk) polarization term -------(1)
8R m m 4 R
 
       
 
Where E =shift in energy , gE (NC) =band gap of nanocrystal
me = effective mass of electron
mh = effective mass of hole
R = radius of the nanocrystal
h = plank’s constant
e = charge of an electron
 = permittivity of free space
o
= dielectric constant of the nanocrystal
For CdS , we have
g e e h eE (CdS Bulk) 2.58eV, m * 0.19m , m * 0.80m , 5.7    
Given that diameter =60 A R 3nm  and polarization term = +0.05
Substituting the values in the eq(1) , we have
2
2 * *
e h o
34 19 2
9 2 31 12 9
34
9 2
h 1 1 1.8e
E 0.05
8R m m 4 R
6.6x10 1 1 1 1 1.8x(1.6 x10 ) 1
= 0.05
8x(3x10 ) 0.19 0.8 9.1x10 eV 4 x5.7x(8.854 x10 )x3x10 eV
6.6x10 1 1 1
=
8x(3x10 ) 0.19 0.8...
SOLUTION.PDF

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