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```Infinite Potential Well … bottom line
d ψ 2m
 2 (E  V)ψ  0
2
dx

Schrodinger Equation
2
V(x)

V= 
0
Energies are quantized, defined by
one single quantum number,
n = 1, 2, 3, 4 …
n 2h 2
En 
8ma
Electron
V=
V=0
0
a
x
Tunneling
“ … If an electron comes up a potential barrier greater than its energy …
there is a finite probability that it will “pass” through the barrier…”
D
C
A
B
E
… for an electron !
We place an electron in region I… with energy E less than VO (E<VO)
… what is the probability the electron will be in I … II … III ??
V(x)
Vo
I
III
II
x=0
x=a
x
How do we calculate the probability ??
… we need to solve Schrodinger’s equation … apply boundary conditions etc.
Tunneling
d 2 ψ 2m
 2 (E  V)ψ  0
2
dx

V(x)
Vo
ψ I (x)  A1e jkx  A 2 e -jkx
ψ II (x)  B1ex  B2 e -x
ψ III (x)  C1e
jkx
 C2e
-jkx
I
III
II
x=0
x=a
2m(VO - E)
2mE
2
with k  2 and α 

2
2
We now need to apply BC’s at x=0 and x=a …
The properties of ψ require that it be continuous and single valued
x
Tunneling
ψI (x) ... ψII (x) ... ψIII (x)
V(x)
Vo
I
A1
A2
yII
Incident
yIII
Reflected
I
III
II
x=0
x=a
x
Therefore … the solution suggests that the electron can be found beyond
the barrier VO … EVEN THOUGH its energy E is less than VO!
Tunneling
T
ψIII (x)
ψI (x)
2
2
2
VO
1
T
where D 
2
1  Dsinh (α a)
4E(VO - E)
2m(VO - E)
α 
2
2
What are the important factors that influence the tunneling probability ??
… the energy of the electron… the width and height of the barrier
For a wide or high barrier …
α a  1 and T  TO e
V(x)
Vo
I
( 2 α a)
16E(VO  E)
with TO 
2
VO
A1
A2
yII
Incident
yIII
Reflected
I
III
II
x =0
x=a
x
Application of Tunneling
Metal
(a)
y(x)
Vacuum
Second Metal
Itunne
Probe
Vo
V(x)
Scan
x
Itunnel
Material
surface
l
y (Å)
x (Å)
(b)
Tunneling
current gray
scale value
(nA)
x
The Potential “Box”
If you confine an electron in a box … what would you expect the
wave-function to be?
Think of it as a combination of 3 one-dimensional infinite
potential wells… and therefore
the general solution will have the form of:
ψ(x, y,z)  Asin(kx x)sin(k y y)sin(kz z)
n3 π
n1π
n2 π
where kx 
, ky 
, kz 
a
b
c
V= 
c
V =0
V= 
x
z
0
a
V= 
V= 
y
b
The Potential “Box”
2
E n1 , n 2 , n 3
h
2
2
2

(n

n

n
1
2
3)
2
8ma
The solution to the electron in a “box” problem results in 3
quantum numbers
A specific solution or eigenfunction i.e.
ψ1,1,2 or ψ2,1,2 or ψ1,4,2 or ψ2,1,3 etc.
is called a state …
Note that the electron energy is quantized and depends on 3
quantum numbers
The H-Atom … An Overview
Describe the H-atom …
i.e. what does the nucleus look like?
how much charge is there at the nucleus?
i.e. Z=1
how many electrons?
The H-atom represents the simplest system we can use to have
a look at a real quantum physics example
The H-Atom … Force & PE
Obviously the electron is being attracted to the nucleus because
of the …
… Coulombic attraction between two opposite charges!
The force between two charges is:
Q1  Q2
F
4π  εo r 2
and the potential energy is given by:
- e2
V(r) 
4π  εor
The H-Atom … Spherical Coordinates
Due to the spherical symmetry of the H-atom … it makes sense
to work in the spherical coordinate system instead of the
cartesian one … i.e. x, y, z  r,θ,φ
z
P(r,q,f)
e
q
r
Nucleus
+Ze
x
y
f
V(r)
r
- e2
V(r) 
4π  εor
Ze2
V(r) =
4peor
+Ze
The H-Atom … Wavefunction
No need to go through the solution in detail …
… we do however need to understand the origin of certain
parameters and functions!
Obtaining the wavefunction for the H-atom electron can be done
by solving …
… in 3-dimensions
i.e. one would expect to get
… 3 quantum numbers!
And the general wave function looks like:
ψ n,l,m l (r, θ, φ)  R n,l (r)  Yl,m l (θ , φ)
The H-Atom … Quantum Numbers
ψ n,l,m l (r, θ, φ)  R n,l (r)  Yl,m l (θ , φ)
Two functions … R a function of r and Y a function of θ and φ
Three quantum numbers! … n, l, ml
The spherical part i.e. R depends on n and l
… while the angular or spherical one, Y depends on l and ml
n=1,2,3,4,…… is the Principal Quantum Number
l=0,1,2,3 ……(n-1) is the Orbital Angular Momentum Quantum
Number
ml=-l, -(l-1), -(l-2), ……-2, -1, 0, 1, 2, …+l is the Magnetic
Quantum Number
… for now MEMORIZE these!
The H-Atom … Quantum Numbers
l
n
0-> s 1-> p 2-> d 3-> f 4-> g
1
1s
2
2s
2p
3
3s
3p
3d
4
4s
4p
4d
4f
5
5s
5p
5d
5f
5g
Let’s check the validity of these Energy States
3,2,2
2,3,1
2,3,0
1,2,4
2,3,-1
2,0,1
1,1,0
1,2,3
4,1,2
1,0,0
Energy !
Electron energies depend on n only … given by:
4
2
me Z
(13.6eV)Z
En   2 2 2  
2
8ε o h n
n
2
What does this energy represent ?
… the energy required to remove the electron from the n=1
state (i.e. to free the electron)
also known as the ionization energy …
E1  13.6 eV
Electrons prefer to minimize their energy … therefore most likely
to be found in n=1 state known as the ground state!
An electron with velocity 2.1E6 m/s strikes a H atom. Find the n th
energy level the electron will excite to. Calculate the wavelength of
the light as the electron returns to ground state.
K.E. = 12.5eV;
n = 3.51 -> 3; ΔE = 12.09 eV;
λ = 102.6 nm
Energy !
Electron energy, En.
0
0.54
0.85
1.51
5
4
3
3.40
2
n=
Excited states
E = KE Continuum of energy. Electron is free
5
Ionization
energy, EI
10
13.6 eV
15
1 Ground state
n
n=1
Orbital Angular Momentum
Just like energy (En) … angular momentum L is also quantized… by ‘l’
Ll  [l(l  1)]
1/2
… what happens when l=0 ?
Bexternal z
Lz
q
L
y
x
Orbiting electron
Orbital Angular Momentum
For l=2 … ml would be …
…-2, -1, 0, 1, 2
LZ  ml
z
Bexternal
Bexternal z
ml
2
1
l =2
Lz
L=  2(2+1)
q
L
0
y
1
2
x
Orbiting electron
Selection Rules …
Electron has momentum … also photons have intrinsic momentum
When photons are absorbed … in addition to the energy
conservation momentum must also be conserved …
Selection rules: Δl=±1 … Δml=0, ±1
i.e. if electron is in ground state 1,0,0 … (n,l,ml)
If enough energy is gained to move up to n=2 then what are l
and ml?
l …0, 1 … and ml … -1, 0, 1
Therefore … n=2, l=1, and ml=-1,0,1
Selection Rules …
Energy
0
l
l
=1
l
=2
l
=3
n
5 5s
5p
5d
5f
4
4s
4p
4d
4f
3s
3p
3d
2 2s
2p
3
13.6eV
=0
1
1s
Photon
l
Selection Rule Example
An electron in State (3,2,-2). What are the energy states in
Spin … (intrinsic angular momentum)
Spin: last quantum number required to fully describe an electron!
S  [s(s  1)]
1/2
with s  1/2
SZ  m s  with m s  1/2
The component of the spin along a magnetic field is also
quantized (i.e. if B-field is in Z-direction)
The Quantum Numbers
ψ n,l,m l (r, θ, φ)  R n,l (r)  Yl,m l (θ , φ)
n=1
n=2
R 1,0
n=1
R 20
n=2
r 2 |R 2,0 |2
r 2 |R 1,0 |2
2s
2s
0
1s
0
R 21
1s
0 .2
r (nm)
0 .4
0
0
2p
2p
0
0
r 2 |R 2,1 |2
0 .2
0 .4
0 .6
0 .8
0
0 .2
r (nm)
0
0 .4
0
0 .2
0 .4
0 .6
r (nm)
r (nm)
maximum
0 .8
“Angular” Probability
z
z
y
y
x
x
2
|Y|2 for a 1 s orbital |Y| for a 2 px orbital
z
z
y
y
x
|Y|2 for a 2 py orbital
s-states symmetrical
p-states directional
x
|Y|2 for a 2 pz orbital
(m l = 0)
Multi electron atom : He
-e
Electron 1
r1
r12
r2
Nucleus
+Ze
-e
Electron 2
A helium-like atom. The nucleus has a charge of +Ze, where for He Z = 2.
If one electron is removed, we have the He+ ion which is equivalent to the
hydrogenic atom with Z = 2.
Energy states for multi electron atom
O
Energy
5g
5f
N
6p
4f
4d
M
3d
5d
6s
5p
5s
4p
4s
3p
3s
L
Energy depends on both
n and l
2p
2s
K
1s
1
2
3
4
5
6
n
Pauli Exclusion Principle & Hund’s Rule
NO two electrons can have the same set of quantum numbers …
i.e. if one electron in ψ1,0,0,1/2 then a second electron in the
same system will have … ψ1,0,0,-1/2
Electrons in the same n, l orbital “like” to have “parallel" spins …
-1
L
(n=2)
K
(n=1)
1 = ml
p
s
K
(n=1)
L
(n=2)
0
H
He
Be
B
Li
s
p
Electronic configurations for the first five elements. Each box represents
an orbital y (n, l, ml ).
C
O
N
p
L
s
K
s
F
Ne
p
L
s
K
s
Electronic configurations for C, N, O, F and Ne atoms. Notice that Hund's
rule forces electrons to align their spins in C, N and O. The Ne atom has
all the K and L orbitals full.
```
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