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Chapter One
Atomic Structure
Slide 1 of 58
The abundances of the elements in the
universe
Slide 2 of 58
Subatomic particle of relevance to chemistry
Particle
Symbol
Mass/u*
Mass
number
Charge/
e#
Spin
Electron
e-
5.486 x 10-4
0
-1
½
Proton
p
1.0073
1
+1
½
Neutron
n
1.0087
1
0
½
Neutrino
ν
c. 0
0
0
½
Positron
e+
5.486 x 10-4
0
+1
½
α particle
α
4
2+
2 He
nucleus 4
+2
0
β particle
β
e- ejected from 0
nucleus
-1
½
γ photo
γ
Electromagnetic 0
radiation from
nucleus
0
1
*Masses are expressed in atomic mass units, u, with 1 u = 1.6605 x 10-27 kg
# Elementary charge e = 1.602 x 10-19 C.
Slide 3 of 58
Nuclear Binding Energy
for Helium
Ebind = (∆m)C2
Slide 4 of 58
Average Binding Energies
Slide 5 of 58
Rutherford’s Alpha Scattering
Experiment
Slide 6 of 58
Rutherford’s Model
• Ernest Rutherford discovered the positive charge of
an atom is concentrated in the center of an atom,
the nucleus
– An atom, can be visualized as a giant indoor
football stadium
– The nucleus can be represented by a pea in the
center of the stadium,
– The electrons are a few bees buzzing throughout.
The roof of the stadium prevents the bees from
leaving.
Slide 7 of 58
Continuous
Spectra
Line
Spectra
Slide 8 of 58
Emission Spectrum of Hydrogen
in Visible Light Region
Slide 9 of 58
Line Spectra of Some
Elements
Slide 10 of 58
Planck’s Constant
• Planck’s quantum hypothesis states that energy can be
absorbed or emitted only as a quantum or as whole
multiples of a quantum, thereby making variations
discontinuous, changes can only occur in discrete
amounts.
• The smallest amount of energy, a quantum, is given by:
E = hv
as Planck’s constant, h = 6.626 X 10-34 J s.
Slide 11 of 58
The Photoelectric Effect
• Albert Einstein considered electromagnetic energy
to be bundled in to little packets called photons.
Energy of photon = E = hv
– Photons of light hit surface electrons and transfer
their energy
hv = B.E. + K.E.
hv
e- (K.E.)
– The energized electrons overcome their attraction
and escape from the surface
Slide 12 of 58
Bohr’s Hydrogen Atom
Postulations
• Rutherford’s nuclei model
• The energy of an electron in a H atom is quantized
• Planck & Einstein’s photon theory
E = hv
• Electron travels in a circle
• Classical electromagnetic theory is not applied
Z
v
r
• e-
Orbit
Slide 13 of 58
(1) Classical physics
centripetal force = Coulombic attraction
mv2/r = Ze2/r2
(2) Total energy
E = 1/2 mv2 - Ze2/r
(3) Quantizing the angular momentum
mvr = n (h/2π )
Quantum number
E = - ( 2π2 mZ2e4)/(n2 h2)
when n =1, E (1) = - (2π2 mZ2e4)/( h2)
E = E (1) /n2
r = (n2 h2)/ (4π2 mZe2)
Slide 14 of 58
Bohr’s Hydrogen Atom
• Niels Bohr found that the electron energy (En) was
quantized, that is, that it can have only certain specified
values.
• Each specified energy value is called an energy level of
the atom
En = - B/n2
– n is an integer, and B is a constant which equals
2.179 x 10-18 J
– The energy is zero when the electron is located
infinitely far from nucleus
– The negative sign represents the forces of attraction
Slide 15 of 58
The Bohr Model
Slide 16 of 58
Bohr Explains Line Spectra
• Bohr’s equation is most useful in determining the
energy change (∆Elevel) that accompanies the leap of
an electron from one energy level to another
• For the final and initial levels:
Ef = -B / nf2
Ei = -B / ni2
The energy difference between nf and ni is:
∆Elevel = Ef - Ei
= ( -B / nf2 ) – (-B / ni2 )
= B(1/ni2 – 1/nf2)
Slide 17 of 58
Energy Levels and Spectral Lines for
Hydrogen
IR
Visible
UV
Slide 18 of 58
Ground States and Excited States
• When an atom has its electrons in their lowest possible
energy levels, it is in its ground state
• When an electron has been promoted to a higher level, it
is in an excited state
– Electrons are promoted through an electric discharge,
heat, or some other source of energy
– An atom in an excited state eventually emits photons
as the electron drops back down to the ground state
Slide 19 of 58
Problems of
Bohr’s Model of Atom
• The energy levels of Bohr’s H atom cannot
be applied to other atoms.
• The orbit of electrons cannot be defined.
Slide 20 of 58
The Uncertainty Principle
• Werner Heisenberg’s uncertainty principle states
that we can’t simultaneously know exactly where a
tiny particle like an electron is and exactly how it is
moving
h
h
h=
(∆Px) (∆x) = h/4π =
Px = mvx
2
2π
• The act of measuring the particle actually interferes
with the particle
• In light of the uncertainty principle, Bohr’s model of
the hydrogen atom fails, in part, because it tells more
than we can know with certainty.
Slide 21 of 58
Uncertainty Principle Illustrated
Slide 22 of 58
De Broglie’s Equation
• Louis de Broglie speculated that matter can behave
as both particles and waves, just like light
• He proposed that a particle move with a mass m
moving at a speed c will have a wave nature
consistent with a wavelength given by the equation:
p = mc
E = mc2 = pc = hν
p = hν /c = h/ λ
λ = h/p = h/mc
• De Broglie’s prediction of matter waves led to the
development of the electron microscope
Slide 23 of 58
What is the speed of an electron to have a wavelength of
X-ray?
Wavelength of X-ray ~ 0.1 nm = 1 x 10-10 m
Mass of electron = 9.11 x 10-31 kg
Planck constant h = 6.626 x 10-34 Js (kg m2/s)
<Answer>
λ= h/p = h/mv
v = h/m λ = (6.626 x10-34)/[(9.11 x10-31)(1 x 10-10)]
= 7 x106 m/s
Slide 24 of 58
How to achieve the speed of an electron of 7 x106 m/s?
V
<Answer>
E= eV= ½ mv2
V = ½ mv2 /e = ½ (9.11 x10-31)(7 x106)2 /(1.6022 x 10-19)
V = 140 V
(1 eV = 1.6022 x 10-19 J)
Slide 25 of 58
Electron diffraction
The experimental confirmation of de Broglie’s wave
hypothesis was first made in 1927 by Davisson and
Germer of the Bell Laboratories who investigated the
scattering of electrons from various surfaces.
Slide 26 of 58
Wave Functions
• Quantum mechanics, or wave mechanics, is
the treatment of atomic structure through the
wavelike properties of the electron
• Wave mechanics provides a probability of
where an electron will be in certain regions of
an atom
Slide 27 of 58
Erwin Schrödinger developed a wave equation to
describe the hydrogen atom
• An acceptable solution to Schrödinger’s wave
equation is called a wave function
• A wave function (ψ) is characterized by an energy
state of the atom
ψ ( x, y , z )
∫
+∞
−∞
2
: the probability of finding an electron
at (x, y, z) position in an atom
ψ ( x, y, z) dτ = 1
2
The probability of finding
an electron in the universe
is equal to 1.
Slide 28 of 58
y
x
Traveling wave
y(x, t) = y0 sin(kx−ωt)
y: amplitude of the wave
If y is a function of x only,
then, y(x) = y0 sinkx
n=1
n=2
n=3
L x
0
L
Standing wave
y(x) = y0 sinkx
Boundary condition:
y = 0, when x = 0
y = 0, when x = L
kL = nπ, k = nπ/L
n = integers (quantum number)
y(x) = y0 sin[(nπ/L) x]
Slide 29 of 58
Standing Waves
&
Quantum Number
Slide 30 of 58
Wave Mechanics
Schrödinger equation
one-dimensional
P.E.
d 2ψ 8π 2 m
+
[E − U (x )]ψ = 0
2
2
dx
h
Η̂ψ = Eψ
Hamiltonian operator
• ψ : amplitude of the wave
• The probability that a particle will be detected is
2
proportional to ψ .
h 2 d 2ψ
− 2
+ U (x )ψ = Eψ
Operator for K.E.
2
P.E.
8π m dx
 h2 d 2

− 8π 2 m dx 2 + U ( x )ψ = Eψ


2
2
h
d
Ηˆ = − 2
+ U (x )
2
8π m dx
Slide 31 of 58
three-dimensional Schrödinger equation
∂ 2ψ ∂ 2ψ ∂ 2ψ 8π 2 m
+
+ 2 +
[E − U ( x, y , z )]ψ = 0
2
2
2
∂x
∂y
∂z
h
Η̂ ψ = Eψ
2
2
2
2


h
∂
∂
∂
ˆ

Η = − 2  2 + 2 + 2  + U (x, y , z )
8π m  ∂x
∂y
∂z 
Slide 32 of 58
Particle in a Box
One-dimensional Schrödinger equation
d ψ 8π m
+ 2 [E − U (x )]ψ = 0
dx2
h
2
2
Η̂ ψ = Eψ
2
2
h
d
ˆ=−
Η
+ U (x )
2
2
8π m dx
In the box, 0 < x < L and U = 0
2
2
2
2
h
d
h
d
ˆ=−
Η
=−
2
2
8π m dx
2m dx2
d 2ψ 8π 2m
+ 2 [E − U (x )]ψ = 0
dx2
h
The probability that the
particle will be detected is
proportional to ψ 2
h=
h
2π
h 2 d 2ψ
−
= Eψ
2
2m dx
Let ψ = A sin( kx) and A, k are constants
h2 d 2
−
[A sin( kx)] = E [A sin( kx)]
2
2m dx
Slide 33 of 58
h2 d 2
−
[A sin( kx)] = E [A sin( kx)]
2
2m dx
Since
d2
2
[
A
sin(
kx
)
]
=
−
k
[A sin( kx)]
2
dx
h2 2
k [ A sin( kx) ] = E[ A sin( kx) ]
2m
h2 k 2
E=
2m
Boundary conditions for the particle in the box:
1. The particle cannot be outside the box. ψ ( 0) = 0 and ψ ( L) = 0
2. In a given state, the total probability of finding
L
the particle in the box must be 1.
3. The wave function must be continuous.
∫
0
ψ ( x ) 2 dx = 1
Slide 34 of 58
ψ ( x ) = A sin( kx)
Boundary condition ψ (0) = 0 and ψ ( L) = 0
ψ ( L) = A sin kL = 0
nπ
kL = nπ , or k =
, where n is intergers
L
nπ
ψ ( x) = A sin
x
L
Boundary condition
∫
L
0
0
ψ ( x ) 2 dx = 1
nπ 2
x ) dx = 1
∫0
L
L
nπ 2
1
∫0 sin( L x ) dx = A2
ψ ( x ) dx =
2
∫
L
L
A 2 sin(
Slide 35 of 58
nπ 2
L 1
∫0 sin( L x) dx = 2 = A2
2
A=
L
L
ψ ( x) =
2
nπ
sin
x
L
L
h 2 k 2 h 2  nπ 
h 2  nπ 
E=
=

 = 2 

2m 2m  L 
8π m  L 
2
E=
2
2
nh
8mL2
2
ψ
ψ2
Slide 36 of 58
Boundary condition for solving ψ in
Schrödinger eq. for an electron in an atom:
•
∫
+∞
−∞
ψ ( x, y , z ) d τ = 1
2
(normalization)
• ψ(x,y,z) is a single valued function w.r.t.
the coordinates
• ψ(x,y,z) is a continuous function
• ψ(x,y,z) is a finite function
Slide 37 of 58
For H atom
z
θ
Z= 1
x
φ
r
ey
− Ze 2 − e 2
U =
=
r
r
To solve the equation more easily,
Cartesian coordinates x, y, z are
transformed to polar coordinates r, θ, φ.
ψ ( x , y , z ) = ψ (r , θ , φ )
= R (r )Θ(θ )Φ (φ )
= R (r )Υ (θ , φ )
Slide 38 of 58
Wavefunctions of Hydrogen Atom
Slide 39 of 58
Slide 40 of 58
Quantum Numbers and
Atomic Orbitals
• The wave functions for the hydrogen atom contain three
parameters that must have specific integral values called
quantum numbers.
• A wave function with a given set of these three quantum
numbers is called an atomic orbital.
• These orbitals allow us to visualize the region in which
there is a probability of find an electron.
Slide 41 of 58
Quantum Numbers
When values are given to quantum numbers, a specific
atomic orbital is defined
The principal quantum number (n)
– Can only be a positive integer
n = 1, 2, 3 ····
– The size of an orbital and its electron energy
depend on the n number
– Orbitals with the same value of n are said to be in
the same principle shell
Value of n
1
2
3
4
5
Shell
K
L
M
N
O
Slide 42 of 58
Quantum Numbers (continued)
• The orbital angular momentum quantum number (l)
– Can have positive integral values
0≤ l ≤ n-1
– Determines the shape of the orbital
– All orbitals having the same value of n and the same
value of l are said to be in the same subshell
– Orbitals and subshells are also designated by a letter:
Value of l
Orbital or subshell
0
s
1
2
3
p
d
f
Slide 43 of 58
Quantum Numbers (continued)
• The magnetic quantum number (ml):
– Can be any integer from -l to +l
-l ≤ ml ≤ l
– Determines the orientation in space of the orbitals of
any given type in a subshell
– The number of possible value for ml = 2l + 1, and
this determines the number of orbitals in a subshell
Slide 44 of 58
The relationship between quantum
numbers
0≤ l ≤ n-1
1s- orbital
For example: n= 1, l= 0
2p, 2s- orbitals
n= 2, l= 1, 0
n= 3, l= 2, 1, 0 3d, 3p, 3s- orbitals
-l ≤ ml ≤ l
For example: l= 1, ml = -1,0,1 px, py, pz- orbitals
l= 2, ml = -2,-1,0,1,2
dxy, dyz, dzx, dz2, dx2-y2
orbitals
Slide 45 of 58
Slide 46 of 58
Quantum Numbers Summary
Slide 47 of 58
The 1s Orbital
Υ0, 0(θ,φ) = 1/2π1/2
• The 1s orbital has spherical symmetry.
• The electrons are more concentrated near the center
Slide 48 of 58
The 2s Orbital
Υ0,0(θ,φ) = 1/2π1/2
(+)
r
0
(-)
0
node
0
Slide 49 of 58
The 2s Orbital
• The 2s orbital has two regions of high electron
probability, both being spherical
• The region near the nucleus is separated from the outer
region by a spherical node- a spherical shell in which
the electron probability is zero
node
Slide 50 of 58
Slide 51 of 58
The Three p Orbitals
Slide 52 of 58
The Five d Orbital Shapes
Slide 53 of 58
The Seven f Orbital Shapes
Slide 54 of 58
Electron Spin – the 4th Quantum
Number
• The electron spin quantum number (ms) explains
some of the finer features of atomic emission spectra
– The number can have two values: +1/2 and –1/2
( ms= ½ , -½
)
– The spin refers to a magnetic field induced by the
moving electric charge of the electron as it spins
– The magnetic fields of two electrons with opposite
spins cancel one another; there is no net magnetic
field for the pair.
Slide 55 of 58
The Stern-Gerlach Experiment
Slide 56 of 58
Hydrogen atom and Schrö dinger
equation
• energy is quantized (n)
• magnitude of angular momentum is
quantized (l)
• the orientation of angular momentum
is quantized (ml)
Electron has spin (ms)
Slide 57 of 58
Z 2 me 4
E=− 2 2
2n h
hydrogen atom
Slide 58 of 58