Chapter 1:

Atomic structure
 Sub-atomic Particles
 Thomson discovered electrons, Goldstein the protons
 Rutherford discovered the nucleus containing protons
 Chadwick discovered neutrons: neutral particles to
minimize the repulsion between protons
 Light as Electromagnetic Radiation
Light is a form of energy. It is an electromagnetic
radiation characterized by a frequency (wavelength)

c


Wavelength ()
Distance between two peaks
Frequency ()
How often waves goes through a complete cycle (1 sec-1 = Hertz)
Speed (c)
How fast wave propagates c = 3 x 108 m/sec
Electromagnetic Radiation Domain

• Wavelengths cover a
wide range of values
• Visible domain:
400 nm <  < 800nm
Example 1.1
Calculate the wavelength of red light with a
frequency of 4.62 x 1014 s−1.
Given:  = 4.62 x 1014 s-1
Find:  (nm)
Conceptual  (s-1)  (m)  (nm)
Plan:

Relationships: ∙ = c, 1 nm = 10-9 m

Solve:

Check: The unit is correct; the wavelength is

appropriate for red light.
Example 1.2
Calculate the wavelength of a radio signal
with a frequency of 100.7 MHz.
Given:  = 100.7 MHz
Find:  (m)
Conceptual  (MHz)  (s-1)  (m)
Plan:

Relationships: ∙ = c, 1 MHz = 106 s-1

Solve:

Check: The unit is correct; the wavelength is

appropriate for radio waves.
 The Nature of Matter: History
 At the end of the nineteenth century: matter
and energy were distinct.
 Matter was thought to consist of particles, whereas
energy in the form of light (electromagnetic
radiation) was described as a wave.
 Particles were things that had mass and whose
position in space could be specified.
 Waves were described as massless and delocalized;
that is, their position in space could not be specified.
 It also was assumed that there was no intermingling
of matter and light.
Max Planck Experiments (1858-1947)
 Heated a piece of iron to 1000 K (727 C)
 It begins to emit light as you can see in the soft red
glow of smoldering coal.
 At about 1500 K (1227 C), the light is brighter and
more orange, like that from an electric heating coil.
 At temperatures greater than 2000 K (2227 C), the
light is still brighter and whiter, as from the filament
of a light bulb.
 Planck Conclusions
 The energy absorbed by the iron piece is directly
proportional to the frequency of the light emitted
 Matter could absorb or emit any quantity of energy.
The energy is quantized
 Energy can be gained or lost only in whole-number
multiples of the quantity h called quanta

𝐄 = 𝐧. 𝐡. 
n: an integer (1, 2, 3, . . .),
h: Planck’s constant Planck’s constant, determined by experiment
6.6261034 J.s.
: the frequency of the electromagnetic radiation absorbed or emitted
 Albert Einstein vs Max Planck
 Einstein proposed that electromagnetic
radiation is itself quantized
 Electromagnetic radiation can be viewed as a
stream of “particles” called photons
 The energy of each photon is given by:
𝒉. 𝑪
𝐄𝐩𝐡𝐨𝐭𝐨𝐧 = 𝐡. =

h: Planck’s constant
: the frequency of the radiation
: the wavelength of the radiation.
Example 1.3
Calculate the number of photons in a laser pulse with
wavelength 337 nm and total energy 3.83 mJ.
Given:  = 337 nm, Epulse = 3.83 mJ
Find: number of photons
Conceptual (nm)  (m) Ephoton number
Plan: photons

Relationships: E = hc/, 1 nm = 10-9 m, 1 mJ = 10-3 J, Etotal= Ephoton  # photons

Solve:
Example 1.4
What is the frequency of radiation required to
supply 1.0 x 102 J of energy from 8.5 x 1027 photons?
Given: Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027
Find: 
Conceptual number
photons
Ephoton  (s-1)
Plan:

Relationships: E = h, Etotal = Ephoton∙# photons

Solve:
Photoelectric Effect
Electrons are emitted from the
surface of a metal when light
strikes it. The following
observations characterize the
photoelectric effect.
1. When the frequency of the light is varied: no electrons emitted
by a given metal below a specific threshold frequency, 0.
2. For frequency lower than the threshold frequency, no electrons
are emitted regardless of the intensity of the light
3. For frequency greater than the threshold frequency, the number of
electrons emitted increases with the intensity of the light
4. For frequency greater than the threshold frequency, the kinetic
energy of the emitted electrons increases linearly with the
frequency of the light.
Photoelectric Effect
 Photoelectric effect = Ejection of electrons
 Condition: E > E0
 Conservation of energy:
E = E0 + Ek
E: Total energy of light hitting the metal
E0: Minimum energy required to remove electron from the metal (Threshold energy)
Ek: Kinetic energy of electron ejected
 Louis De Broglie Theory
 Light is a wave: =c/
 Light is flow of photons E=h
 Wave model of the atom: =c/
 Particle Model : E = h = hc/ 
 Planck: E = h = hc/
 By Einstein E = mc2
 E = E  h (c / ) = m c2 for radiation.
Consider a particle in movement of mass m and speed
v. By De Broglie, its movement could be imagined as a
wave of wavelength :
 Motion of Electron vs Human

me = 9.111031 Kg me = 80 Kg
v = 106 m.s1 v = 10 m.s1
6.621034

𝐡 6.621034 𝐡 6.621034

 8010 =8.2710 m  0
= 9 = 36
𝐦.𝐯 9.111031.106 =72.610 m = 72.6 nm 𝐦.𝐯

𝐓𝐡𝐞 𝐦𝐨𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧 𝐜𝐚𝐧 𝐛𝐞 𝐚𝐬𝐬𝐨𝐜𝐢𝐚𝐭𝐞𝐝 𝐭𝐨 𝐰𝐚𝐯𝐞

Example 1.5
Calculate the De Broglie Wavelength of an
electron moving at 1.00 % the speed of light.
c = 3108m/s, m = 9.111031kg

h
mv 

h 6.636 x1034 J  s
   24nm
mv 9.11x10 kg  0.01  3.00 x10 s 
31 8 m
 Light Principle
 Thomson theory: When an atom at
the gaseous state receives determined
energy, a light is emitted, example:
Neon lamps
Hg He H
How can this phenomena be explained?
 Gas receives a high-energy spark, the molecules absorb energy,
and some of the bonds are broken.
 The resulting gas atoms are excited (electrons are excited by gaining
energy)
 Gas atoms contain excess energy, which they release by emitting light of
various wavelengths (electrons returned at their normal state)
 Each wavelength corresponds to radiation emitted when
electrons return from excited to a normal state (ground state)
Atomic Line Spectra of hydrogen gas
 light is passed through
a prism: continuous
spectrum
 Hydrogen spectrum:
only few lines

What is the significance of the line spectrum of hydrogen?

 Certain energies are allowed for electron in hydrogen atom (quantized)
 The energy of the electron in the hydrogen atom is quantized (the
postulates of Max Planck)
 Changes in energy between discrete energy levels in hydrogen will
produce only certain wavelengths of emitted light
 A given change in energy from a high to a lower level would give a
wavelength of light that can be calculated from Planck’s equation
 Hydrogen Line Spectrum
• Balmer Series
– Discovered in 1885
– Explained presence of visible lines
– Calculated the wavelength 1 11
Balmer – Rydberg formula:  RH( 2  2 )
 n n'
Spectral ni nf Spectral
n < n’
series region
n: initial state of the electron
Lyman 1 2,3,4,5,6,7,… UV n’: final state (excited) of electron
Balmer 2 3,4,5,6,7,… Visible n and n’ = 1, 2, 3, 4, 5, 6 …
Pashen 3 4,5,6,7,… IR 
1/ = = : wave number
𝐶
Bracket 4 5,6,7,… IR
RH = 1.09737 x 107 m-1
Pfund 5 6,7,… IR
Previous Atomic Model: Inconvenients
Electrons Inconvenient:
around
the nucleus

Electrons are attracted

to the nucleus, atom
Nucleus (protons, neutrons) will collapse
With time electron
loose energy (-e2/r), so
r , electron falls down
to the nucleus, atom
will collapse
 The Bohr Theory
 The energy of a hydrogen atom is
quantized and that the amount of
energy in the atom is related to the
electron’s position in the atom.
• Quantized means that the atom can only have
very specific amounts of energy.
n = 1,2,3,...
 The electrons travel in orbits that are at a fixed
distance from the nucleus.
 Stationary states
 Therefore, the energy of the electron is proportional
to the distance of the orbit from the nucleus.
 The Bohr Theory
Electrons revolves around the nucleus in
definite energy levels called orbits or
shells in an atom without radiating energy
As long as an electron remain in a shell it
never gains or losses energy.
The gain or loss of energy occurs within orbits only due
to jumping of electrons from one energy level to
another energy level.(E = Ef  Ei = h)
The angular momentum of an
electron is multiple to nh/2π.
𝒏𝒉
m.v.r =
𝟐
n = number of the orbit, h: Planck’s constasnt,
r: radius of the orbit
Bohr Calculations: Radius of Orbit
 Consider an atom 𝐀𝐙𝐗 having an electron e moving around
the nucleus having charge +Ze
 Two forces applied to e: Coulomb vs centrifuge forces
 The forces are equal (e on orbit)
𝐊𝐪.𝐪′ 𝒌.𝒁𝒆.𝐞 𝒗𝟐
𝐅𝐜 = = 𝐅𝐜𝐞𝐧𝐭 = 𝐦
𝐫𝟐 𝒓𝟐 𝐫
𝒌. 𝒁𝒆𝟐 𝒗𝟐 𝒌. 𝒁𝒆 𝟐

𝟐
= 𝐦 𝒎. 𝒗𝟐
=
𝒓 𝐫 𝒓
1
𝟐
𝒌. 𝒁𝒆 k = 4
0
𝒓 =  (vacuum permittivity
0 constant) =
𝒎. 𝒗𝟐 12
8.8410 C /J.m 2
Bohr Calculations: Radius of Orbit
𝒏𝒉 𝒏𝒉 𝒏 𝟐 𝒉𝟐
𝒎. 𝒗. 𝒓 = 𝒗 =  𝒗𝟐 =
𝟐 𝟐𝒎𝒓 𝟒  𝒎𝟐 𝒓𝟐 𝟐

𝒌.𝒁𝒆𝟐 𝒌.𝒁𝒆𝟐 𝟒𝟐𝒎𝟐 𝒓𝟐 𝒌.𝒁𝒆𝟐 𝟒𝟐𝒓𝒎

𝒓 = =  𝟏 =  𝟐 𝟐
𝒎.𝒗𝟐 𝒎 𝒏𝟐 𝒉𝟐 𝟏 𝒏 𝒉

𝒏𝟐 𝒉𝟐 𝒂. 𝒏𝟐 a = 0.529 Å
𝒓 = 𝟐 𝒓(Å) = = 0.53 Å
𝟒 . 𝒎. 𝒌. 𝒁𝒆𝟐 𝒁
𝟓𝟑. 𝒏𝟐
𝒓(pm) =
𝒁
For hydrogen: Z= 1 and a = 0.529 Å
𝒓(Å) = 𝟎. 𝟓𝟐𝟗. 𝒏𝟐
 Bohr Calculations: Energy of e
 The energy of an electron in an orbit is the sum of its
potential and kinetic energy
𝟏 𝒌.𝒁𝒆𝟐
𝐄𝐭 = 𝐊𝐄 + 𝐏𝐄 = 𝒎. 𝒗𝟐 
𝟐 𝒓
𝟏 𝒌.𝒁𝒆𝟐 𝒌.𝒁𝒆𝟐
𝐄𝐭 = . 
𝟐 𝒓 𝒓
𝟏 𝒌. 𝒁𝒆𝟐 𝐧𝟐 𝐡𝟐
𝐄𝐭 =  . 𝐚𝐧𝐝 𝐫 = 𝟐.
𝟒 𝐦. 𝐤. 𝐙𝐞𝟐
𝟐 𝒓
𝟏 𝒌. 𝒁𝒆𝟐 . 𝟒𝟐. 𝐦. 𝐤. 𝐙𝐞𝟐 𝟐𝒌𝟐 . 𝒁𝟐 𝒆𝟒 . 𝟐. 𝒎
𝐄𝐭 =  . 𝐄𝐭 = 
𝟐 𝐧𝟐 𝐡𝟐 𝐧𝟐 𝐡𝟐

𝟐.𝟏𝟖𝟏𝟎−𝟏𝟖 2 𝟏𝟑.𝟔 2
𝐄𝐭 =  .Z in Joules per e 𝐄𝐭 =  𝟐 .Z in e.V
𝐧𝟐 𝐧
Bohr Calculations: Excitation and Emission
The first energy level when n=1 is
called ground state of H atom. All
others are called excited states.
E = Ef  Ei = h
Ef  Ei 𝟐𝒌𝟐 . 𝒁𝟐 𝒆𝟒 . 𝟐. 𝒎
= 𝐄𝐭 = 
ℎ 𝐧𝟐 𝐡𝟐
𝟐𝐤 𝟐 .𝐙 𝟐 𝐞𝟒 .𝟐 𝐦
..
𝟏 𝟏
= 𝟑 . ( 𝟐 − 𝟐)
𝐡 𝐧𝐟 𝐧𝐢
𝟐𝐤 𝟐 .𝐙 𝟐 𝐞𝟒 .𝟐𝐦 𝟏 𝟏
= . ( − )
𝐡𝟑 𝐧𝟐𝐟 𝐧𝟐𝐢
𝐂 𝟐𝐤 𝟐 . 𝐙𝟐 𝐞𝟒 . 𝟐. 𝐦 𝟏 𝟏
= . ( 𝟐 − 𝟐)
𝛌 𝐡𝟑 𝐧𝐟 𝐧𝐢
𝟐𝐤 𝟐 .𝐙𝟐 𝐞𝟒 .𝟐.𝐦
𝟏 𝟐𝐤 𝟐 . 𝐙𝟐 𝐞𝟒 . 𝟐. 𝐦 𝟏 𝟏 o agrees with Rydberg constant 109740 cm1
𝐡𝟑 .𝐂
= . ( 𝟐 − 𝟐) o Deals with Balmer Formula
𝛌 𝐡𝟑 . 𝐂 𝐧𝐟 𝐧𝐢 o Explain the line spectra model of hydrogen
Example 1.6
Calculate the wavelength of light emitted when the
hydrogen electron transitions from n = 6 to n = 5.
Given: ni = 6, nf = 5, RH = 2.18 x10−18 J
Find:  m
Conceptual ni , nf Eatom Ephoton 
Plan: Eatom = −Ephoton

Relationships: E = hc/, En = −2.18 x 10-18 J (1/n2)

Solve:

Ephoton = −(−2.6644 x 10-20 J) = 2.6644 x 10-20 J

Another method:
Balmer-Rydberg
Check: Units are correct; the wavelength is in the infrared, which is
formula
appropriate because it's less energy than 4→2 (in the visible).
 The Bohr Theory: Importance
The model correctly fits the quantized energy levels of
the hydrogen atom and postulates only certain allowed
circular orbits for the electron
As the electron becomes more tightly bound, its energy
becomes more negative relative to the zero-energy
reference state (electron being at infinite distance from
the nucleus)
Bohr assumed electrons could ‘jump’ orbits if it
absorbed the corresponding E. Light was energy
associated with ‘jump
When atom absorbs energy, electron returns
from the excited state to the ground state, it
emits the energy absorbed as light
 Defects 0f Bohr Model
 According to Bohr, the radiation results when an electron
returns from one energy orbit to another energy orbit, but he
did not explained how this radiation occurs.
 Bohr explained the existence of various lines in H-spectrum
(few lines exist). Later on it was realized that the spectral lines
that had been thought to be a single line was actually a
collection of several lines very close to each other.
 Bohr successfully explained the observed spectra for systems of
one electron: hydrogen (H) or hydrogen like (hydrogenoide)
(He+, Li2+, Be3+..)
 Defects 0f Bohr Model
 According to Bohr, the radiation results when an electron
returns from one energy orbit to another energy orbit, but he
did not explained how this radiation occurs.
 Bohr explained the existence of various lines in H-spectrum
(few lines exist). Later on it was realized that the spectral lines
that had been thought to be a single line was actually a
collection of several lines very close to each other.
 Bohr successfully explained the observed spectra for systems of
one electron: hydrogen (H) or hydrogen like (hydrogenoide)
(He+, Li2+, Be3+..)
 Defects 0f Bohr Model
 If a substance which gives line emission spectrum is placed in a
magnetic field, it get split up into a number of closely spaced
lines (Zeeman effect). Bohr has no explanation

 Same observation in presence of electrical field

(Stark effect)
 Bohr suggested circular orbits of electron
around the nucleus of H – atom but later it was
proved that the motion of electron is not in a
single plane, but in three dimensional space.
 Bohr determine the position and the speed of electron
simultaneously (later proved not true)
 The Helium Atom: Polyelectronic System
 Consider a two-electron atom with nuclear charge Z.
 Neglecting electron repulsion, each electron is in a
hydrogen-like orbital with energy:

−2.18 × 10−18 2
𝐸= × 𝑍
𝑛2
Heisenberg’s Uncertainty Principle
 The characteristic wavelength of an electron in a
hydrogen atom is 240 pm (1 pm = 10-12 m). The size of
an isolated H atom is about 240 pm. This leads to an
uncertainty in the location of the electron
You can not simultaneously know both the
position and momentum of an electron (that
does agree with Bohr concepts)
x : the uncertainty in a particle’s position
(mv) : the uncertainty in a particle’s momentum
h: Planck’s constant.

The more accurate we know where something is, the less accurate
we know where it is going
Modern Quantum Mechanical Model
 Heizenberg: Position and
speed are not known at the Z M (x,y,z)
same time x

- At M(x, y, z)  speed VM  Y
De Broglie wave M X
- At N(x’,y’,z’)  speed VN 
De Broglie wave N

At any position, the movement of the electron is a

wave  wave function (x,y,z)
 Schrodinger Equation
 Bohr model uses Orbits (circular position)
 Quantum mechanics uses Orbitals (2, probability
of presence)
Schrödinger (1926) set up a differential equation that define the
location of an electron in space for atoms having one or more
electrons
Ĥ   E  2      8 m
2 2 2 2
 2  ( E  V )  0
x2
y2
z h
m: mass of the electron, E: its total energy, V: its potential energy
 : Wave function is a solution of the differential
equation of Schrodinger called an Orbital
 For a position P(x,y,z)  (x,y,z)  [(x,y,z)]2 is
probability of presence
 Modern Atomic Model
• By this model: electron does not
have a definite position
• Atom does not have a definite shape
• At a given position M(x,y,z):
calculation of the wave function
(x,y,z)
[(x,y,z)]2 - Probability distribution
function, describes the probability of
finding an electron in a specific location
for a given energy state
The probability distribution for the hydrogen electron (n=1) wave
function(orbital). The more times the electron visits a particular point,
the darker the negative becomes. Thus the darkness of a point indicates
the probability of finding an electron at that position.
This diagram is also known as an electron density map
Physical Meaning of Wave Function
Suppose we have two positions in space, P1(x1,y1,z1)
and P2(x2,y2,z2)
The relative probability of finding the electron at
positions 1 and 2 is given by substituting the values of x,
y, and z for the two positions into the wave function
Squaring the function value, and computing the
following ratio:

The quotient N1/N2 is the ratio of the probabilities of

finding the electron at positions 1 and 2.
For example, if the ratio N1/N2 = 100, the electron is 100
times more likely to be found at position 1 than position 2.
 Quantum Numbers
: is a solution of Schrödinger equation
It depends on M(x,y,z)  (x,y,z)  (n,l,ml)
2 ways to describe the
position of this person:
1- Coordinates x, y and z

the center is on the ground
2- Coordinates:
Floor, apartment, room

n l ml
n - Principle Quantum Number
 n characterizes the energy of the electron in a
particular orbital.
corresponds to Bohr’s energy level
 n can be any integer >1.
 The larger the value of n, the more energy the
orbital has.
 Energies are defined as being negative.
An electron has E = 0 when it just escapes the
atom.
 The larger the value of n, the larger the orbital.
 As n gets larger, the amount of energy between
orbitals gets smaller.
ℓ – Angular Momentum Quantum Number
 Describes the shape of the orbital
 ℓ can have integer values from 0 to (n – 1)
 Each value of ℓ is called by a particular letter that
designates the shape of the orbital.
 s orbitals are spherical.
 p orbitals are like two balloons tied at the knots.
 d orbitals are mainly like four balloons tied at the knots.
 f orbitals are mainly like eight balloons tied at the knots.
-The 1st principle level has ℓ = 0 1s
-the 2nd has ℓ= 0,1
-the 3rd has ℓ = 0,1,2 2s, 2p
-the 4th has ℓ= 0,1,2,3 3s, 3p, 3d
-all the rest have 4 or more 4s, 4p, 4d, 4f
 mℓ - Magnetic Quantum Number
 Describes orientation of orbital in space relative to
the other orbitals in the atom
  ℓ  mℓ + ℓ. There are 2ℓ+1 values of mℓ for each
type of ℓ

1 type of s (l = 0) orbital
3 types of p (l = 1) orbitals
5 types of d (l = 2) orbitals
7 types of f (l=3) orbitals
 Electron Spin
• In 1928, it was discovered that an electron has an
intrinsic angular momentum, or spin.
• In a magnetic field, the rotation axis has only two
possible orientations.
 Orbital Shapes and Energies
 An orbital is represented most clearly by a probability
distribution
 Each orbital has a unique probability distribution
 Representing an orbital is by the surface that surrounds 90%
of the total electron probability These three types of
representations

“Slices” of the three-dimensional electron density The surfaces that contain 90% of the total electron probability

2s and 3s orbitals contain areas of high probability separated by areas of

zero probability. These latter areas are called nodal surfaces, or simply
nodes. The number of nodes increases as n increases.
 l = 0, s orbital
 Each principal energy state
has 1 s orbital.
 lowest energy orbital in a
principal energy state
 spherical
 number of nodes = (n – 1)
 l = 1, p orbitals
 Each principal energy state above n
= 1 has 3 p orbitals. ml = −1, 0, +1
 Each of the three orbitals points
along a different axis px, py, pz
 Second lowest energy orbitals in a
principal energy state
 Two-lobed. Node at the nucleus; A cross section of the
electron probability
total of n nodes distribution for a 3p orbital
 l = 2, d orbitals
 Each principal energy state above n = 2 has 5 d
orbitals.
 ml = −2, −1, 0, +1, +2
 Four of the five orbitals are aligned in a different
plane.
 The fifth is aligned with the z axis, dz squared.
 dxy, dyz, dxz, dx squared – y squared
 third lowest energy orbitals in a principal energy state
 mainly four-lobed
 One is two-lobed with a toroid.
 planar nodes
 Higher principal levels also have spherical nodes.
 l = 2, d orbitals

(a) Electron density plots of selected 3d orbitals

(b) The boundary surfaces of all five 3d orbitals, with the signs (phases) indicated
 l = 3, f orbitals
 Each principal energy state above n = 3 has 7 d
orbitals.
ml = −3, −2, −1, 0, +1, +2, +3
 Fourth lowest energy orbitals in a principal energy
state
 Mainly eight-lobed
some two-lobed with a
toroid
 Planar nodes
Higher principal levels
also have spherical
nodes.