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Atomic Structure
 Planck's SWE
Quantum
Rutherford’s Theory
Model

Heisenberg’s
de Broglie’s Uncertainty
Bohr’s hypothesis principle
Model
J.J. Thomson’s
Model
 Discovery of electron

Cathode rays

In 1891, George Johnstone Stoney named the

fundamental unit of electricity as 'electron'

J. J. Thomson and his team identified electron as a

particle in 1897
 Discovery of electron
Discharge tube: Cylindrical hard glass tube fitted with two
metallic electrodes connected to a battery.
Air at very low pressure

Cathode Anode

High voltage To Vacuum

generator pump
 Discovery of electron
Observations
Cathode Anode
Readings of electric current
was observed

Anode end of the tube showed a

greenish glow on the ZnS screen
 Discovery of electron

Why are Gases used under LOW pressure?

At high pressure, more number of gas
molecules are present and so there is
more obstructions in the paths of
electrons which prevents electrons from
reaching the anode.
 Discovery of electron

Observation and Characteristics 3. Cathode rays are more

efficiently observed with the help
1. Cathode rays move from of a fluorescent or
cathode to anode. phosphorescent material like ZnS.

4. Cathode rays rotate the light

2. Cathode rays travel in a straight paddle wheel placed in their
line with high velocity path. It shows that the particles
in the absence of electric & of cathode ray particles are
magnetic fields. material particles which have
mass and velocity.
 Discovery of electron

Light paddle wheel

Cathode Anode

High voltage
generator To Vacuum
pump
 Discovery of electron
Observation and Characteristics
5. Cathode rays are deflected in 6. Cathode rays are deflected in
the presence of an electric field. the presence of a magnetic field.

Anode + Anode

Cathode
Cathode

-
 Discovery of electron
Conclusions

Cathode rays consist of

negatively charged particles and
identified as electrons.
 Charge to mass ratio

In 1897, J.J. Thomson

Anode +
● Measured the charge (e) to mass
Cathode
(m) ratio of an electron.
● Electric & magnetic fields
were applied perpendicular to
each other & to the path of
electrons
-

e
m = 1.758820 × 1011 C/kg
Charge to mass ratio

Charge to mass ratio is

the same irrespective of

Nature of Material of
the gas Cathode

Electrons are fundamental

particles
 Discovery of Anode Rays
Anode
Perforated
Discovered by Goldstein Cathode

He repeated experiment with a

discharge tube by using a perforated
cathode.
Red glow is due to anode
Existence of positively charged particles particles which passes
was shown using anode rays. through perforated cathode
and strikes the wall of the
tube at the cathode side.
 Discovery of Anode Rays
Green colour
fluorescence Perforated
Anode Perforated
observed due Cathode
to Cathode Red colour
rays fluorescence
observed due
to Anode rays
Observations and Characteristics

1. Anode rays possess positive charge

Concluded by their directions of

deflections in the presence of
electric & magnetic fields

2. Anode rays travel in straight lines in

the absence of both electric and
magnetic fields.
 Observations and Characteristics

3. e/m ratio of the canal rays is

In 1919, Rutherford discovered
different for different gases
that the smallest and the lightest
positive ions are obtained from
Properties of anode rays hydrogen and called them
depends on nature of the gas protons
taken in the discharge tube
 Discovery of Neutrons

James Chadwick

Named the electrically neutral

Discovered neutrons in 1932
particles emitted as neutrons

Mass of neutrons is slightly greater Bombarded a thin sheet of beryllium

9
than that of protons ( 4 Be) with alpha particles ( 24He2+)
 Discovery of Neutrons
Beryllium Paraffin Wax

𝞪 Source Detector and counter

for protons

4 9 12 1
2
He++ + 4
Be 6
C + 0
n

Z=2 Z=4 Z=6 q=0

A=4 A=9 A = 12 A=1
 Thomson’s Model

Raisin
pudding
model

Plum
pudding Watermelon
model model
 Thomson’s Model

1. An atom has a spherical shape 4. Mass of the atom is assumed to

(radius～10–10 m) be uniformly distributed all over it

2. Positive charge is uniformly

distributed throughout the
sphere

3. Negatively charged electrons

are embedded in it like raisins in
a pudding Plum Pudding Watermelon
 Thomson’s Model

Electrons are embedded in an

atom in such a way that Explains the overall
the most stable electrostatic neutrality of an atom
arrangement is achieved.

Drawback:
Negatively Positively
Charged Not consistent with the
Charged Matter
Electrons results of later experiments
 Rutherford’s Experiment

A stream of high energy 𝞪–particles was

directed at a thin gold foil (thickness ∼ 100 nm)

Radioactive ZnS
Source Screen

Gold Foil
Observations of Rutherford’s Experiment

When an 𝞪–particle strikes the screen, a glow

was produced at that point on the screen
Observations of Rutherford’s Experiment

Gold Atoms

Nucleus
Observations of Rutherford’s Experiment

1. Most of 4. Small
them 𝞪–particles fraction was
passed deflected by
undeflected small angles

3. Very few
2.Very small
were
fraction was
deflected by
deflected by
180° (∼1 in
large angles
20,000)
 Observation Conclusion

Most 𝞪-particles
Presence of large
passed through the
empty space
foil without
in the atom.
deflection.
Positive charge is
Few 𝞪-particles
concentrated in a
were deflected by
very small region.
small angles.
Small positively
Very few 𝞪-particles
charged core at
(∼1 of 20,000)
the centre.
deflected at 180°.
 Nucleus

Atom consists of
a small positively It has negligible
charged core at volume
the center which compared to the
carries almost the volume
entire mass of the atom.
of the atom
 Nucleus

Both, protons and neutrons present

1
in the nucleus are collectively called R = R0 A 3

nucleons.

～10–10 R = Radius of nucleus of an element

Radius of the atom m
A = Mass number of element

～10–15 m Radius of the nucleus R0 = 1.11 x 10-15 m to 1.44 x 10-15 m

 Extranuclear part

Electrons and nucleus

are held together by
electrostatic forces of
attraction. FCentripetal

Nucleus is surrounded by
revolving electrons. FElectrostatic = FCentripetal
 Drawbacks of Rutherford’s Model

It could not explain stability of the atom.

It could not explain line spectrum of the H atom.

It could not explain the electronic structure of the atom.

 R.A. Millikan’s Oil drop experiment
Charge on oil droplets This experiment was
measured and found to conducted to get the
be an integral multiple charge on electron.
of magnitude of charge
on an electron (e). Atomizer

Ionizing Charged metal plate ( + )

Radiation
Charged oil droplets
Light
source Viewing microscope
Charged metal plate ( - )

Charge on electron – 1.602176 × 10–19 C

Mass of the electron
Charge from
Thomson’s e/m ratio
Millikan’s experiment

1.75882 × 1011 C kg−1 – 1.602176 × 10–19 C

From Thomson’s experiment, e/m ratio

calculated and from Oil drop experiment,
charge of electron calculated. Using the
data from these two experiments, mass
of the electron was determined.
 Subatomic Particles

Subatomic Absolute Subatomic Relative Absolute

Mass (u)
Particles Mass (kg) Particles Charge Charge (C)

Electron 0.0005 9.1 × 10-31 Electron -1 -1.602 x 10-19

Proton 1.007 1.6722× 10-27 Proton +1 1.602 x 10-19

Neutron 1.008 1.6749 ×10-27 Neutron 0 0

 Quantization of Charge

q = n (1.6 x 10-19 C)

q = n (4.8 x 10-10 esu)

n = 1 , 2 , 3 ...

The charge can’t have continious range of

values but only take values in multiple of
charge on one electron. The magnitude of
q = n (e) charge on an electron is the smallest unit
and denoted as “e”. Thus charge on an
electron is -e and on a proton, it is +e.
 Electrostatic Force

q1 q2
1 Nm2
F12 = F21 = K
r2 K = 4πƐ0 = 9 x 109
C2

F12
+q1 +q2 F21
C
Ɛ0 = 8.854 x 10-12
Vm
F12 F21
+q1 -q2

r
Ɛ0 = Permittivity of vacuum
 Potential Energy

P.E. = qxV

q = Charge of the particle

V = Potential of surface

1 q1 q2 q1 q2
P.E. = 4πƐ0 r
P.E. = K r
 Closest Distance Of Approach

1 3 2
When two charged particles of
similar nature approach each other,
V𝛂 the repulsion between them
m𝛂 increases but due to initial kinetic
R
energy, the particles come closer to
each other (Recall the bombarding of
R1 positively charged alpha particle on
Closest distance of gold foil where the alpha particle
approaching positively charged
approach( in alpha particle
nucleus of gold atom).
scattering by gold nucleus),
R = √4KZe2/m𝝰 V𝝰
 Closest Distance Of Approach

1 3 2
But at a certain distance between
them, the relative velocity becomes
V𝛂 zero and after that due to repulsion,
m𝛂 the particles starts going away from
R
each other. This distance between
the particles where velocity once
R1 becomes zero is called Closest
distance of approach and can easily
be calculated using conservation of
Closest distance of approach, energy concept.
R = √4KZe2/m𝝰 V𝝰
 Electromagnetic Waves

Electric Field

Oscillating
Electric &
Magnetic field

Magnetic Field
 Properties of Electromagnetic Waves

Electric & magnetic field

oscillate perpendicular to each other

Both oscillate perpendicular to the

direction of propagation of wave

Do not require any medium for

propagation

Can travel in vacuum c = 3 x 108 ms-1 (in vaccum)

Propagate at a constant speed i.e. with
the speed of light (c)
Characteristics of Electromagnetic waves

Wavelength Time Period

Frequency Wavenumber

Velocity Amplitude
 Characteristics of Electromagnetic waves
Crest
Wavelengt Amplitude
h Mean
position

Trough
Velocity of the wave

Frequency: Number of times a

wave oscillate from crest to
trough per second
 Characteristics of Electromagnetic waves

Frequency (𝛎)
Wavelength (λ)
Number of waves passing
a given point in one second
Distance between two SI unit : Hertz (Hz), s-1
consecutive crests or troughs

Related to time period as:

SI unit : m
1
𝛎 = T
Characteristics of Electromagnetic waves

Velocity (c or v)

Distance travelled by a wave in

one second SI unit : ms-1

Related to frequency (𝛎) &

wavelength (λ) as:

c = 𝛎λ
 Characteristics of Electromagnetic waves

Time Period (T) Wavenumber (ν)

Time taken to Number of waves per unit

complete one oscillation length
SI unit : m-1

SI unit : s Amplitude (A)

Height of the crest or the

depth of the trough from the
mean position
SI unit : m
 Characteristics of Electromagnetic waves

1
𝛎 = λ

Consists of radiations
having 𝛎 = c𝛎
different wavelength or
frequency

1
𝛎 = T
Electromagnetic Spectrum

Electromagnetic radiations are

arranged in the order of

Decreasing Increasing
or
frequency wavelength
 Electromagnetic Spectrum

Gamma rays

X-rays

Ultraviolet
Frequency Wavelength
decreases Visible decreases
Infrared

Microwave
Radiowave
 Electromagnetic Spectrum
Short Long
wavelength wavelength

Gamma rays X-rays UV Infrared Microwave Radio wave

High Low
frequency frequency

Visible light
EM Radiation: Wave or Particle?

Wave nature of the EM radiation explains

Diffraction

Interference
 EM Radiation: Wave or Particle?

Electromagnetic wave theory could not explain

Black-body radiation

Photoelectric effect

Variation of heat capacity of

solids with temperature

Line spectrum of Hydrogen

Continuous vs Discrete

Mass = N (mass of 1 water molecule)

Where N ∈ + I

It seems that mass of water (or any other matter)

can take any values (suppose we can go till 30
decimal points) and so we can say that mass has
continuous range of values. But on microscopic
level, we can observe that mass of water is
always an integral multiple of 1 molecule of water.
i.e., mass is quantized or we can say that
quantization is a property of matter.
Back to EM waves

Low temperature
Low frequency
Longer wavelength

High temperature
High frequency
Shorter wavelength
 What is a black body?

Idealized system

Absorbs & emits all frequencies

Absorbs regardless of the angle

of incidence
 Why the name, Black Body?

Radiatively Black
Visually Black body vs
Body

A true black body appears black

because it is not reflecting any
electromagnetic radiation.

However, everything you see to

be black can not ber called as
blackbody because there could
be radiation coming out which is
not in the visible range.
 Wavelength-Intensity relationship
UV Visible
14
5000 K
This graph shows intensity as a
12 Classical Theory function of wavelengths emitted from
(5000 K)
10 a black body. It shows how bright it is
at what wavelength. It shows
8
Intensity

quantization nature of energy and

6 hence favours particle nature of light.
4000 K
4

2
3000 K
0
0 0.5 1 1.5 2 2.5 3
Wavelength (μm)
 Particle Nature of Radiation

Planck’s quantum theory explains

Variation of
Quantisation
intensity with
of Energy
wavelength
 Planck’s Quantum theory

The smallest packet or bundle of energy

(quantum of radiation) is called a photon.

This is the smallest quantity of energy that

can be emitted or absorbed in the
form of EM radiation.
 Quantum theory of Radiation
Energy (E) of a photon is proportional to its
frequency (𝝂)

E ∝ 𝝂Radiation E = nh𝝂

c
E = h𝝂 = h
λ n = number of photons
= 0, 1, 2, 3, ….
h = Planck’s constant
= 6.626 × 10-34 Js
 One electron volt (eV)

Energy gained by an electron when

it is accelerated from rest through
a potential difference of 1 V
 Important Conversions

1 eV = 1.6 × 10-19 J

12,400
E (eV) = λ (Å)

(
E
kJ
mole ) = E ( eV
particle ) × 96.48
 Photoelectric Effect

When a radiation of sufficient

Phenomenon of
frequency falls on the metal
electrons ejection
surface

Electrons are ejected with the

Photoelectrons
aid of light
 Photoelectric Effect

When radiation of
Metal plate Photoelectrons sufficient energy falls on
the metal plate, there
starts emission of
electrons called
photoelectrons.
 Observations

Electrons are ejected as soon as the beam of light

of sufficient frequency strikes the metal surface

Instant transfer
of energy to the
electron when a
photon of sufficient
frequency strikes the
metal atom
 Threshold frequency (𝝂O)

Minimum frequency
Each metal has a
required to eject a
characteristic
photoelectron from a
threshold frequency
metal surface
 Observations

Intensity of light

Number of electrons ejected ∝ or

Brightness of light
 Observations

No electron is ejected,
regardless of the intensity of light
𝝂incident < 𝝂0

𝝂incident > 𝝂0 Even at low light intensities,

electrons are ejected immediately
 Particle nature of light

One photon is absorbed by only one electron in a

single interaction. Not more than one photon can
be absorbed by an electron.

If intense beam of light is used, large number of

photons are available and large number of
electrons are ejected. This observation shows
particle nature of light.
 Observations

When 𝝂incident > 𝝂0

K.E.Ejected electron ∝ 𝝂Incident

Energy Transfer of K.E. of the

possessed by energy to the ejected
the photon electron electron
 Photoelectric Effect

0 ≤ K.E.Ejected electrons ≤ K.E.Max

K.E. is independent of
the intensity of radiation

Striking photon’s
energy = h𝝂

Work function = 𝝓
 Work Function (𝝓)

𝝓 = h𝝂0 Minimum energy

required to eject
an electron from the
metal surface
𝝓 = W0 = Work function
 Photoelectric Effect

1 2
EPhoton - 𝝓 = K.E.Max h𝝂 = h𝝂0 + 2
mevmax

From the Law of Energy

Conservation
me = mass of the electron
EIncident = 𝝓 + K.E.Max
vmax = maximum velocity of the electron
 Photoelectric Effect

K.E. of the ejected electron is given as

K.E.Max = h𝝂 - h𝝂0

c c
K.E.Max = hλ - hλ
o
 Plotting K.E. vs Frequency

From the plot of kinetic energy vs

frequency, it shows linear variation
according to the equation:

K.E.Max = h𝝂 - h𝝂0
K.E.

ν0
Frequency
Acceleration and Deceleration of Charged Particles

Acceleration:
If a positive charge moves from higher to
lower potential (like an electron moves
from cathode to anode) or a negative
charge moves from lower to higher
potential.
Deceleration is just opposite to the
acceleration.
 Stopping potential(Vs)

Minimum opposing potential required to stop the

photoelectron having the maximum K.E.

eVS = K.E.max
Accelerating potential voltage (V)

Voltage applied to increase the K.E. of

an emitted electron

Minimum K.E. eV

K.E.Max + eV Maximum K.E.

Photocurrent v/s Frequency of the Radiation

Photocurrent (I)

Frequency
Photocurrent v/s Frequency of the Radiation

For ν > ν0

K.E.

Intensity
 Photocurrent vs Collector Plate Potential
at different intensity of radiation
I : Intensity
Photocurrent at

Photocurrent
I3 > I2 > I1 different intensity of
radiation. Here
I3 frequency of radiation
I2 is same.
I1

Saturation Current
Stopping potential

-V0 0
Retarding Potential Collector Plate Potential
 Photocurrent vs Collector Plate Potential
at different intensity of radiation

Photocurrent
Photocurrent at different
frequencies of radiation.

Saturation
Here intensity of radiation

Current
𝛎3 > 𝛎2 > 𝛎1 is same.
𝛎3
𝛎2
𝛎1
-V03 -V02 -V01 0
Retarding Potential Collector Plate Potential
Bohr Atomic Model
 Postulates
n=3
Stationary
n=2
orbits
n=1
Concentric circular orbits around the K L M
nucleus

These orbits have fixed value of energy

Electrons revolve without radiating Stationary Energy states /

or
energy orbits levels
 Postulates

Quantization of Angular momentum nh

mvr = 2π

n = 1, 2, 3...
Angular momentum of the
electron in these orbits is
h Planck’s constant
always an integral multiple of
h
2𝜋 m Mass of electron

v Velocity of electron

r Radius of orbit
 Postulates
Energy (h𝜈)
absorbed
Electron can jump from lower to higher orbit by
absorbing energy in the form of photon

Energy
= E3 - E2
Absorbed

Electrons can jump from higher to lower orbit Energy (h𝜈)

released
by releasing energy in the form of photon

Energy
Released = E2 - E1
 Postulates

Energy change does takes place in a discrete manner

ΔE = En2 - En1

n1 Initial energy state

n2 Final energy state

 Bohr’s Frequency Rule
Frequency (𝝂) of a radiation absorbed
or
emitted when a transition occurs

ΔE E2 - E1
𝝂 = h
= h

E1 = Energy of lower energy state

E2 = Energy of higher energy state

Bohr’s Atomic Theory

Applicable only
for single
electron species
like
H, He+, Li2+, Be3+
Mathematical Analysis

Radius of Bohr orbit

Velocity of an electron
in Bohr orbit
Time period of an
Calculating electron in Bohr orbit
Frequency of an
electron in Bohr orbit
Energy of an electron
in Bohr orbit
 Postulate

FElectrostatic

Electron revolves in a circular orbit

Required centripetal force is provided

by electrostatic force of attraction
Calculating the radius of Bohr orbit

Equating both the forces, mv2 KZe2

r = r2
F Centripetal = F Electrostatic

On rearranging,

mv2 KZe2
r
= KZe2
r2 v2
= i
mr
 Calculating the radius of Bohr orbit
According to Bohr's Postulates,
On comparing equation (i) and (ii),
nh
mvr = 2π
KZe2 n2h2
mr = 4π2r2m2
nh
v = 2πrm
n2h2
r =
4π2mKZe2
2 n2h2
v = 2 2 2
ii
4π r m
Calculating the radius of Bohr orbit

Putting the value of constants,

n2
rn = 0.529
Z
Å
rn ∝ n2

rn = Radius of nth Bohr orbit

1
n = Energy level rn ∝ Z

Z = Atomic number
 Calculation of velocity of an electron
in Bohr orbit

Angular momentum of the electron

n2h2
revolving in the nth orbit r = ii
4π2mKZe2
nh
mvr = 2π

nh
v = 2π r m i
Calculation of velocity of an electron
in Bohr orbit
Putting equation (ii) in equation (i),

nh 4π2mKZe2
v = 2πm
×
n2 h2

2πKZe2
v = nh

Z ms-1
vn = 2.18 x 106
n

vn = Velocity of the electron in nth Bohr orbit

Relation between vn, n and Z

vn ∝ Z

1
vn ∝ n
 Time period of Revolution (T)

Time period of revolution of an electron in its orbit

Circumference 2πrn
T = = vn
Velocity

n3 s n3
T = 1.5 x 10 -16 T ∝
Z2 Z2
 Frequency of Revolution (f )

Frequency of revolution of an electron in its orbit

1 vn
f = T
= 2πrn

Putting the value of constants, rn & vn

Z2 Z2
f = 6.6 x 10 15
3
n
Hz f ∝ n3
 Calculation of Energy of an electron
Total energy (T.E.) of an electron revolving in
a particular orbit
1
K.E. = mv2
T.E. = K.E. + P.E. 2

2
KZe
K.E. Kinetic energy P.E. = _
r

Potential
P.E. 2
energy 2 KZe
T.E. = mv + _
r
2
Calculation of Energy of an electron

Centripetal force = Electrostatic force

mv2 KZe2
r = r2

mv2 KZe2
K.E. = 2 = 2r
Calculation of Energy of an electron

T.E. = K.E. + P.E.

KZe2 KZe2 KZe2

T.E. = 2r + r = 2r

1
T.E. = K.E. = 2
P.E.
 Calculation of Energy of an electron

KZe2
T.E. = 2r

Substituting the value of ‘r’ in the equation of T.E.

KZe2 4𝝿2Ze2mK 2𝝿2Z2e4mK2

T.E. = = 2 - x n2h2 = n2h2

2𝝿2me4K2 Z2
T.E. = = h2 n2
 Calculation of Energy of an electron
Putting the value of constants we get:

Z2
T.E. = En = 13.6
n2
eV/atom

Negative sign of T.E. shows attraction

between electrons & nucleus.

Electron in an atom is more stable than a

free electron
Energy of an electron

Z2
En ∝ n2

Z↑ En ↓

n↑ En ↑
 Energy of an electron

EK < EL < EM < EN E∞ = 0

Distance of electron
↑
from the nucleus

Energy ↑
 Energy of an electron
Z2
En -13.6 2 eV/atom
n
2 2
-11-18Z Z
E
Enn -2.18
-2.18x x1010 2 2 erg/atom
J/atom
nn
22
-313.6 Z
Z kcal/mol
En -1312 22 kJ/mol
n
n

Z2
En -2.18 x 10 -11
erg/atom
n2
2
-313.6 Z kcal/mol
En
n2
 Energy of an electron

At n = ∞

T.E. = 0

P.E. = 0

K.E. = 0
Energy Difference

2
ΔE = En
2
En1

2 2
eV _ 13.6 Z _ 13.6 Z
ΔE atom = n2 2 n1 2

1 _ 1 eV
ΔE = 13.6 Z 2
n1 2 n2 2 atom

1 _ 1 J
ΔE = 2.18 x 10-18 Z2
n1 2 n2 2 atom
Energy level diagram

E=0
n=∞

n=5
n=4 Higher
energy
n=3
states

n=2

Ground state
n=1
Energy level diagram
n=∞ E=0

n=5 E5 = -0.54 eV
n=4 ΔE = E4- E3 = 0.66 eV E4 = -0.85 eV
n=3 E3 = -1.51 eV
ΔE = E3- E2 = 1.89 eV
n=2 E2 = -3.4 eV

ΔE = E2- E1 = 10.2 eV

Ground state
n=1 E1 = -13.6 eV
Definition For Single Electron
Systems
 Ground state (G.S.)

Lowest energy state of any atom or ion n=1

G.S. energy of H-atom -13.6 eV

G.S. energy of He+ ion -54.4 eV

 Excited state

States of atom or ion other than the ground state n≠1

n=2 First excited state

n=3 Second excited state

n=m+1 mth excited state

 Ionisation energy (I.E.)

Minimum energy required from

to remove an electron n = 1 to n = ∞

1 _ 1 eV
2
ΔE = 13.6 Z
n1 2 n2 2 atom

Putting n2 = ∞ & n1 = 1

eV
ΔE = 13.6 Z2 atom
Ionisation energy (I.E.)

I.E.H = 13.6 eV

I.E.He+ = 54.4 eV

I.E.Li 2+ = 122.4 eV
 Ionisation potential (I.P.)

Potential difference through

which a free electron must such that its K.E. = I.E.
be accelerated from rest

I.P. = 13.6 Z2 V I.P.H = 13.6 V

I.P.He + = 54.4 V
 Excitation Energy

Energy required to
move an electron from
Excitation
n = 1 to energy of
any other state For H atom 2nd state
n = 1 to n = 2

10.2 eV

Excitation 1st
energy of excitation
1st E.S. energy
 Excitation potential

Potential difference through Excitation

which an electron must be potential
accelerated from rest such For H atom of 3rd
that its n = 1 to n = 3 state
K.E. = Excitation energy

12.09 V
Excitation 2nd
potential excitation
of 2nd E.S. potential
Binding or Separation energy

Energy required to
move an electron from any state to
n=∞

B.E.Ground state = I.E.Atom or Ion

 Summary

Ground State n=1

Excited State n>1

From
Ionization Energy
n = 1 to n = ∞
From
Binding Energy
any state to n = ∞

From n = 1 to
Excitation Energy
any other state
 Spectrograph
Spectroscopy

Spectrum
Spectrogram
 Spectroscopy

Spectroscopy Spectrograph/
Branch of SpectroscopeIns Spectrogram
trument
science that Spectrum of
used to separate
deals with the radiation of the given
study of different radiation
spectra wavelengths
Classification: Based on Origin

Emission

Spectra

Absorption
 Emission Spectrum

Spectrum of radiation emitted by a substance

Continuous

Emission
Spectrum
Discrete
 Continuous Spectrum

Continuous distribution of colours (VIBGYOR)

such that each colour merges into the next one
 Continuous Spectrum

Screen

Prism

Light Source
 Classification: Based on Nature

Line

Discrete

Emission Band
Spectrum
Continuous
 Line Spectrum

Ordered arrangement of lines of a particular

wavelength separated by dark space

Line spectrum
 Emission Spectra
Screen

Light emitted by
gas

Prism

Gas in Excited
State in
Discharge
Tube at low
Pressure
Application of Line Spectrum

Each element has a unique

line spectrum

Identification of unknown atoms

 Band Spectrum

Continuous bands separated by

some dark space

Molecular spectrum
 Classification : Based on Origin

Emission

Spectrum

Absorption
 Absorption Spectrum

The gas in ground state

Scree
n absorbs radiation of particular
wavelengths and rediations of
remaing wavelength passes
through the gas which scatter
through a prism and appears
Prism as bright lines on the ecreen.
The dark lines shows the
Gas in missing wavelengths which
Ground State are already absorbed by the
gas and no more available to
pass through the prism.
 Absorption Spectrum

Some dark lines in the

continuous spectrum

Represent absorbed radiations

Line Spectrum of Hydrogen

Emission Spectral Lines/

De-Excitation Series

Due to de-excitation of electron

from higher to lower orbit
 Energy Level Diagram for H atom
n=∞ 0 eV

-0.54 eV

3.4 eV
n=5
n=4 -0.85 eV
n=3 -1.51 eV

n=2 -3.4 eV

12.09 eV

12.75 eV

13.06 eV
10.2 eV
13.6 eV

n=1 -13.6 eV
Rydberg’s Formula

Electron makes transition from n2 to n1

hc 1 _ 1 J
ΔE = λ = 2.18 x 10-18 Z2
n1 2 n2 2 atom

1 2.18 x 10-18 Z2 1 _ 1
= m-1
λ hc n1 2
n2 2
Rydberg’s Formula

1 1 _ 1 -1
= 1.09678 x 10 x Z7 2 m
λ n1 2 n2 2

1 1 _ 1
cm-1
λ = 109678 x Z2
n1 2 n2 2

n1 1, 2, 3, ... n2 n1 + 1, n1 + 2, ... RH Rydberg constant

RH = 1.09678 x 107 m-1 ≈ 1.1 x 107 m-1

 Rydberg’s Formula

For any atom

1 1 _ 1
= RH Z2 x
λ n1 2 n2 2

n1 1, 2, 3, ... 912
Å
λ = 1 _ 1
Z2
n2 n1 + 1, n1 + 2, ... n1 2 n2 2
 Rydberg’s Formula

For H atom

1 1 _ 1
λ
= RH x
n1 2 n2 2

n1 1, 2, 3, ...

n2 n1 + 1, n1 + 2, ...
Spectral series of H atom

Lyman Balmer Paschen

Brackett Pfund Humphrey

 Lyman Series

For an electron present in H atom

1 1 _ 1
λ = RH x
12 n2 2
1st spectral series.
Found in UV region
by Lyman
n1 = 1 (Final state)

n2 = 2, 3, 4 …
(Initial states, n2 > 1)
Lyman Series

st
n2 = ∞
1 line of 2→1
Lyman series (𝛂 line)
Wavelength n12
=
of last line RH
2nd line of
3→1
Lyman series (𝛃 line)
1
λLyman = RH
Last line of Lyman
∞→1
series (Series limit)
Lyman Series

10.2 eV ≤ (ΔE)Lyman ≤ 13.6 eV

12400 12400
Å ≤ λLyman ≤ Å
13.6 10.2
Lyman Series

Longest 12400
wavelength
line = λlongest or λmax = Å
ΔEmin

Shortest 12400
line
wavelength = λshortest or λmin = Å
ΔEmax

Where, E in eV
 Lyman Series

1st spectral line λmax

Last spectral line λmin

Series limit
Limiting/last line of any spectral series

Since n2 = ∞ 𝜈last line = RH

 Lyman Series
n=∞

n=6
n=5
n=4
n=3

n=2

λmax λmin

n=1
Lyman series
 Balmer Series

2nd spectral series

Found in visible region by Balmer

For H atom Only first 4 lines belongs to visible region

Rest belongs to UV region

 Balmer Series

1 1 _ 1
λ
= RH x
22 n22

n1 = 2 (Final state)

n2 = 3, 4, 5 …
(Initial states, n2 > 2)
 Balmer Series
n=∞

n=6
n=5
n=4
n=3
λmax λmin
n=2
Balmer series

n=1
 Paschen Series

3rd spectral series; Found in IR region by

Paschen

1 1 _ 1
λ
= RH x
32 n22

n1 = 3 (Final state)

n2 = 4, 5, 6 …
(Initial states, n2 > 3)
 Paschen Series
n=∞

n=6 λmin
n=5
n=4 λmax
n=3

n=2
Paschen series

n=1
 Brackett Series

4th spectral series; Found in IR region by

Brackett

1 1 _ 1
λ
= RH x
42 n22

n1 = 4 (Final state)

n2 = 5, 6, 7 …
(Initial states, n2 > 4)
 Brackett Series
n=∞

n=6 λmin
n=5 λmax
n=4
n=3

n=2
Brackett series

n=1
 Pfund Series

5th spectral series; Found in IR region by

Pfund

1 1 _ 1
λ
= RH x
52 n22

n1 = 5 (Final state)

n2 = 6, 7, 8 …
(Initial states, n2 > 5)
 Pfund Series
n=∞
λmin
n=6 λmax
n=5
n=4
n=3

n=2
Pfund series

n=1
 Humphrey Series

6th spectral series; Found in IR region by

Humphrey

1 1 _ 1
λ
= RH x
62 n22

n1 = 6 (Final state)

n2 = 7, 8, 9 …
(Initial states, n2 > 6)
 Humphrey Series
n=∞
n=7 λmax λmin
n=6
n=5
n=4
n=3

n=2
Humphrey series

n=1
Line Spectrum of Hydrogen
n=∞
n=7
n=6
n=5
n=4
n=3

n=2
Balmer Paschen Brackett Pfund Humphrey
series series series series series

n=1
Lyman series
Maximum Number of Spectral Lines

Maximum number of lines

=
Maximum number of different
types of photons
Maximum Lines for the transition from 4 to 1
4 to 3
4 to 2 3 to 2
4 to 1 3 to 1 2 to 1
n=4
n=3

n=2

n=1
Maximum Lines for the transition from 5 to 1

5 to 4
5 to 3 4 to 3
5 to 2 4 to 2 3 to 2
5 to 1 4 to 1 3 to 1 2 to 1
n=5
n=4
n=3

n=2

n=1
 Maximum Number of Lines

(nH - nL+ 1) (nH - nL) (Δn + 1) (Δn)

Spectral Lines = 2
= 2

Δn nH - nL

nH Higher energy level

nL Lower energy level

 Maximum Number of Lines

For transition upto n = 1 or n = nSeries

In Lyman series
(nHigher > 1) = nHigher - 1 For a particular
series

In Balmer series Spectral

(nHigher > 2) = nHigher - 2
lines
= nHigher - nSeries

In Paschen series
(nHigher > 3) = nHigher - 3
Maximum Number of Lines

In Brackett series
(nHigher > 4) = nHigher - 4

In Pfund series
(nHigher > 5) = nHigher - 5

In Humphrey series
(nHigher > 6) = nHigher - 6
 Example
Number of spectral lines in Lyman series
from 4th shell = nH - 1 = 4 - 1 = 3

n=4
n=3

n=2

n=1
 Example

4 to 3
3 to 2
2 to 1 4 to 2 3 to 1 4 to 1

n=4
n=3

n=2

n=1

1st Atom 2nd Atom 3rd Atom 4th Atom

Pathway to Quantum Mechanical Model

Dual Nature
of Matter
Important
Developments
Heisenberg’s
Uncertainty Principle

These developments shows that electron has sufficient

wave character and it can not be assumed as a particle
only. So, its position or trajectory can not be determined
as shown by Bohr model.
 Dual Nature of Matter

de Broglie proposed that particle has dual nature

Particle Wave Louis de Broglie

nature nature

Einstein suggested that light has dual nature i.e.,

particle nature as well as wave nature.
de Broglie Hypothesis
de Broglie Hypothesis

h
λ=
p
Wave
associated with
moving
particles
 h
How λ = ?
p
 de Broglie Hypothesis

hc
Planck’s equation E = λ

Einstein’s Mass
Energy relationship
E = mc2
 de Broglie Hypothesis

Equating both
hc
λ
= mc2

For photon
h
λ = mc

By same analogy, de Broglie proposed

h
For matter λ = mv
de Broglie Wavelength (λ)

h h
λ = mv = p

Momentum Velocity
p v
of particle of particle

Mass Planck’s
m h
of particle constant
de Broglie Wavelength (λ)

h
λ = p

0
p
Relativistic Mass

mO

m =
√1-( ) v 2
c

Velocity
Velocity
v w.r.t. c
of light
the observer

Dynamic
m mO Rest mass
mass
 Relativistic Mass

√1-( )
v 2
m0 = m Χ
c

If v = c

mO = 0

Rest mass of photon is zero

Davisson and Germer’s Experiment

Experimental verification of
de Broglie’s prediction

It was observed that an electron

beam undergoes diffraction
Wavelength of a ball & an electron!
h
de Broglie wavelength: λ =
mv
Cricket ball Electron

m = 150 g m = 9.1 × 10-31 kg

v = 25 m s-1 v = 2 × 103 m s-1
6.626 × 10-34 6.626 × 10-34
λ= λ=
(150 × 10-3) × 25 (9.1 × 10-31) × 2 × 103

λ = 1.767 × 10-34 m λ = 0.364 × 10-6 m

= 364 nm
λ is insignificant. λ is significant.
 Matter Waves

Wave nature because its wavelength is

can’t be detected too short due to larger
for macroscopic object mass.

1
λ ∝ m
de Broglie’s Equation & Kinetic Energy

1
K.E. = mv2
2

Multiplying both sides by m & rearranging

m2v2 = 2 K.E. × m

mv = √ 2 K.E. x m
 de Broglie’s Equation & Kinetic Energy

de Broglie h h
equation
λ = mv = p

Since mv = √ 2 K.E. x m

h
λ =
√2 K.E. × m
 de Broglie’s Equation & Kinetic Energy

A charged particle
accelerated from rest
across a potential
| K.E. | = |qV|
difference of V

∵ mv = √ 2m × K.E.

∴ mv = √ 2m × q V
de Broglie’s Equation & Kinetic Energy

h
λ = 2m × q V
√

Charge on a
λ Wavelength (m) q
particle (C)

Mass of charged Planck’s

m h
particle (kg) constant (Js)
 Electron as a Wave
In phase Out of phase
n = Number of waves made
in Bohr’s orbit

r
An integral number of
complete wavelengths must
fit around the circumference
of the orbit.
Electron exist Electron don’t exist

Circumference, Circumference,
2πr = nλ 2πr ≠ nλ
 Electron as a Wave

When electrons nh
are in phase,
2𝜋r = mv

Bohr’s Postulate nh
verified mvr = 2𝜋

n = Energy level
 Heisenberg’s Uncertainty Principle (H.U.P.)

Exact position and

momentum of a
microscopic particle
cannot be determined
simultaneously

Werner Heisenberg
Heisenberg’s Uncertainty Principle

h
Δx . Δp ≥ 4𝜋

h
Δx . m . Δv ≥ 4𝜋

Uncertainty in Uncertainty in
Δx Δp
position momentum

Mass of Uncertainty in
m Δv
particle velocity
 Principle of Optics

If a light (wavelength ‘λ’) is used to locate the

position of a particle, then

Minimum error in the

position measurement (Δx) = -
+λ
Heisenberg’s Uncertainty Principle

Since Δx = λ

For accurate
Δx 0 λ 0
position

For a photon

E = hc
λ

λ 0 E ∞
Heisenberg’s Uncertainty Principle

High energy photon

Δp
strikes particle

Similarly

For accurate momentum Δx

Heisenberg’s Uncertainty Principle

For an electron

h
Δx . m . Δv ≥ 4𝜋

6.626 × 10-34 Js
Δx . Δv = 4 × 3.14 × 9.1 × 10-31 kg

Δx . Δv ≅ 10-4 m2s-1
Heisenberg’s Uncertainty Principle

If Δx = 10-8 m then Δv = 104 ms-1

High Conclusion:
Position Δx is small
accuracy
Heisenberg’s
Velocity Uncertain Δv is large uncertainty
principle is
meaningless for
If Δv = 10-8 ms-1 then Δx = 104 m
bigger particles.
High
Velocity Δv is small
accuracy

Position Uncertain Δx is large

Energy - Time Variant of H.U.P.

h
Δx . Δp ≥ 4𝜋

Multiplied and divided by Δt

Δp h
Δt
. Δx . Δt ≥ 4𝜋

Δp
= Rate of change in momentum = F
Δt
Energy - Time Variant of H.U.P.

h
F . Δx . Δt ≥ 4𝜋

h
ΔE . Δt ≥ 4𝜋

ΔE Uncertainty in Energy

Δt Uncertainty in Time
 Significance of the Uncertainty Principle

Not an instrumental error, Precise statements of

1 rather conceptual error
position& momentum
of an electron replaced with
Rules out the existence of probability
2 definite paths of electrons
Forms the basis of
Quantum Mechanical Model
Introduced concept of probability of atom
3
of finding the electrons
 Limitations of Bohr Model
Could not explain the line spectra of atoms
1 containing more than one electron

Could not explain the presence of

2
doublet i.e. two closely spaced lines
 Limitations of Bohr Model

Unable to explain the splitting of spectral lines

3 in the presence of magnetic field (Zeeman effect)
and electric field (Stark effect).

No conclusion was given for the principle

4
of quantisation of angular momentum

Unable to explain de Broglie’s concept

5
& Heisenberg’s Uncertainty Principle
Quantum Mechanical Model
Schrodinger Wave Equation (SWE)

∂2𝚿 ∂2𝚿 ∂2𝚿 8𝝅2m ( E - V ) 𝚿

∂x2
+
∂y2
+
∂z2
+
h2 = 0

● x, y, z = Cartesian coordinates

● 𝚿 = Amplitude of the electron wave or Wave function

● h = Planck’s constant

● V = Potential energy of the electron

● E = Total energy of the electron

 Wavefunction

SWE is solved to get values of 𝚿 and

their corresponding energies

A function that contains all the

𝚿 dynamical information about a
system
SWE can be solved for H like species
more easily in
Spherical polar coordinates
(r,𝛳,ɸ)
 Spherical Coordinate System

z
x = r sin𝜃 cosɸ

P
r y = r sin𝜃 sinɸ

z = r cos𝜃 𝜃

O y
z = r cos𝜃

ɸ
rs
in
𝜃

x Pl
 Wavefunction
When SWE is solved for H like species, the obtained values
of 𝚿 could be factorized into one containing only ‘r’ and the
other containing (𝜃, ɸ).

Wave Radial part of Angular part of

function wave function wave function

𝚿 (r, 𝜃, ɸ) = 𝚿 (r) × 𝚿 (𝜃, ɸ)

n, l l, ml
 Schrodinger Wave Equation

𝚿 corresponds to atomic orbital

First three quantum

numbers (n, l, m) were
Characterized by a set of quantum numbers derived from Schrodinger
equation. The spin
quantum number added
later.

n l ml s

✔ ✔ ✔ х
 What is an Orbital?

3D region Probability of
Orbital around the finding an electron
nucleus is maximum
 What is an Orbital?

Does not define a

Defines the probability
definite path of
of finding an electron
electrons
 Quantum Numbers

1 Principal Quantum Number ( n )

Set of four
numbers required 2 Azimuthal Quantum number ( l )
to define an
electron in an 3 Magnetic Quantum Number ( ml )
atom completely

4 Spin Quantum Number ( s )

 Principal Quantum Number (n)

Designates the shell to which

1
the electron belongs

Signifies energy level for single

2
electron species

Accounts for the main lines in

Proposed by
3
the atomic spectrum
Niels Bohr
 Principal Quantum Number (n)

n = 1, 2, 3...
Describes the size
of electron wave Represented as K, L, M, N,...
& the total energy
of the electron Angular momentum in any nh
shell = 2π
 Azimuthal Quantum Number (l )

Designates the subshell to

1
which the electron belongs

Energy of the orbital in

2 multielectron species (both n&l)

Accounts for the fine lines in

Proposed by
3 atomic spectrum
Sommerfeld
 Azimuthal Quantum Number (l )
Also known as

Describes the 3-D Subsidiary Quantum Number

shape of the
orbital or the Orbital Angular momentum Quantum
electron cloud Number
For a given value of Principal Quantum Number (n)

l = 0 to (n - 1)
=
 Boundary Surface Diagram

Encloses the 3D region where probability

of finding electrons is maximum

Example

Shape : Spherical
 Classification of Orbitals

p
Orbital
d

f
 Shape of Orbitals

s - orbital p - orbital Shape : Double

Shape : Spherical Shape : Dumb bell dumb bell
 Probability Density Plots

s orbital p orbital d orbital

Shapes of Orbitals

Orbital Shape

s Spherical

p Dumb bell

d Double dumb bell

Leaf like /
f
Complicated
 Subshell

Collection of
similar shaped
orbitals of same n.
 Subshell

l Subshell Description

0 s Sharp

1 p Principal

2 d Diffused

3 f Fundamental

4 g Generalised
 Subshell Representations

Number of subshells in the nth shell

Subshell
n l
notation
1 0 1s

2 0, 1 2s, 2p

3 0, 1, 2 3s, 3p, 3d

4 0, 1, 2, 3 4s, 4p, 4d, 4f

 Azimuthal Quantum Number (l )

Orbital angular h
momentum (L) = l (l + 1) ħ ħ=
2π

Orbital angular
Subshell
momentum
s 0

p 2 ħ

d 6 ħ
 Magnetic Quantum Number (ml )

Designates the orbital to

1
which the electron belongs

Describes the orientation of

2
orbitals

Accounts for the splitting of lines of

Proposed by 3 atomic spectrum in magnetic field
Linde
Magnetic Quantum Number (ml )

Can have values from - l to + l

including zero

Each value corresponds to an orbital

For d
subshel, ml = -2, -1, 0, 1, 2
l=2
Magnetic Quantum Number (ml )

Maximum number of
orbitals in a subshell
= 2l + 1

Subshell Number of orbitals

s 1

p 3 (px, py, pz)

d 5 (dxy, dyz, dzx, dx - y , dz )

2 2 2

f 7
 s - orbital

Shape : Spherical
Non-directional in nature

x
 p - orbital

Shape : Dumb bell

Directional in nature

z z z

y y y

x x x

pz py px
 d - orbital
z z z
y y y

x
Non axial
x x
d-orbitals

dyz dxy dxz

Shape : Dumb bell

z z
Directional in nature
y y Axial
x d-orbitals
x
dz2
dx2- y2
 Remember!

An orbitals can accommodate maximum of 2 electrons.

Maximum number of
electrons in a subshell = 2 (2l + 1)

Subshell s p d f

l 0 1 2 3

Number of electrons 2 6 10 14
 Spin Quantum Number (s or ms)

Presence of two closely spaced lines in

atomic spectrum

Spin of an electron

1 1
s= + 2 s= 2
Proposed by George Uhlenbeck (left)
and Samuel Goudsmit (Right)
 Spin Quantum Number (s)

Spin magnetic
moment (μ) = √n(n+2) B.M.

n = number of unpaired electron

 Spin Quantum Number (s)

Spin angular h
momentum = 2π
√s(s+1)

Maximum Spin 1 n
of an atom (S) = 2

Spin
multiplicity = 2 |S| + 1
 Orbit and Orbital

Orbit Orbital

Well defined circular path 3D region around the nucleus

around the nucleus where where electrons are most likely
electrons revolve to be found

Maximum number of electrons Cannot accommodate more

in nth orbit is 2n2 than two electrons
 Orbit and Orbital

Orbit Orbital

Not in accordance with In accordance with

Heisenberg’s Uncertainty Heisenberg’s Uncertainty
Principle Principle

Designated as K, L, M, N, ... Designated as s, p, d, f, ...

Rules for Filling Electrons in Orbitals

Aufbau Principle

Pauli’s Exclusion
Rules
Principle

Hund’s Rule of
Maximum Multiplicity
Aufbau Principle

Electrons are
filled in various
orbitals in order of
their increasing
energies
Energies of Subshells of H-like Species

Energy of single electron species depends only on the

Principal Quantum Number

Order of
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < …
Energy
 Energies of Subshells of H-like Species

4s 4p 4d 4f

Degenerate Orbitals:
3s 3p 3d
Energy

Orbitals that have same

energy.
2s 2p
Example: In the given
diagram 3s, 3p and 3d
1s are degenerate orbitals.
Energy of Subshells of Multi-electron Species

Different subshells have different energy

which depends on:

Principal Quantum Azimuthal

Number Quantum Number
 ( n + l ) rule or Bohr-Bury’s Rule

Lower value of Lower will be

(n+l) energy of subshell

Two subshells with Subshell with lower ‘n’

same ( n + l ) value value has lower energy
Comparison of orbital energy

1s < 2s

2s < 2p

5p > 4d

2p < 3p

3d < 4p
Energies of Subshells of Multi electron Species

4p

3d
4s
3p

3s
Energy

2p
2s

1s
Energies of Subshells of Multi electron Species
H-like v/s Multi-electron species

H-like species Multi-electron species

Energy of a subshell Energy of a subshell

depends on ‘n’ only. depends on (n + l).

Electrons experience
Only attractive forces are attractive forces towards
present between the the nucleus as well as
nucleus and the electron. repulsive forces from other
electrons.
 One-electron species
There is only attractive
force here

n=2
n=1

Attraction
 Multiple-electron species
Outer Electron
Attractive Force

n=2 Repulsive Force

n=1

Shielding Electron

Repulsion

Attraction
 Pauli’s Exclusion Principle

No two electrons
in an atom can
have the same
set of all four
quantum
numbers
Wolfgang Pauli
Pauli’s Exclusion Principle

Restrict the filling of number of electrons

in an orbital

Wrong

Right
 Subshell electron capacity

s2 ⥮

p6 ⥮ ⥮ ⥮

d10 ⥮ ⥮ ⥮ ⥮ ⥮

f14 ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮
Hund’s Rule of Maximum Multiplicity

Friedrich Hund
 Hund’s Rule of Maximum Multiplicity

No electron pairing takes place

Until each orbital is occupied
in
by 1 electron with parallel spin
the orbitals in a subshell

Hund’s rule is an empirical rule

Determines the lowest energy

arrangement of electrons
 Why Maximum multiplicity?

Electrons with parallel spins

Maximum spin of 1
an atom (S) = 2
xn
Repel each other
Spin Multiplicity
(S.M.) = 2S + 1
Have a tendency to stay apart

Spin Multiplicity Stability Atom shrink slightly

(S.M.)

Electron-nucleus interaction is
improved
 Electronic Configuration

H ↿
1s1

Distribution
He ⥮
of electrons in
1s2
orbitals of
an atom
Li ⥮ ↿
1s2 2s1
Electronic Configuration of Various Elements

Be ⥮ ⥮ N ⥮ ⥮ ↿ ↿ ↿
1s2 2s2 1s2 2s2 2p3

B ⥮ ⥮ ↿ O ⥮ ⥮ ⥮ ↿ ↿
1s2 2s2 2p1 1s2 2s2 2p4

C ⥮ ⥮ ↿ ↿ F ⥮ ⥮ ⥮ ⥮ ↿
1s2 2s2 2p2 1s2 2s2 2p5
Electronic Configuration of Various Elements

Ne ⥮ ⥮ ⥮ ⥮ ⥮
1s2 2s2 2p6

Na ⥮ ⥮ ⥮ ⥮ ⥮ ↿
1s2 2s2 2p6 3s1

Mg ⥮ ⥮ ⥮ ⥮ ⥮ ⥮
1s2 2s2 2p6 3s2
 Simplified Electronic Configuration
Configuration of Sodium:

Na ⥮ ⥮ ⥮ ⥮ ⥮ ↿
1s2 2s2 2p6 3s1

Ne

Simplified configuration: Na Ne ↿

3s1
 Electronic Configuration

a) 21
Sc 1s2 2s2 2p6 3s2 3p6 4s2 3d1
[Ar] 4s2 3d1 45
[Ar] 3d1 4s2 21 Sc
Scandium

↿ ⥮
3d1 4s2

Number of unpaired electrons 1

1 1
Total spin + or -
2 2
 Electronic Configuration
b) Fe
26
1s2 2s2 2p6 3s2 3p6 4s2 3d6
[Ar] 4s2 3d6 56
[Ar] 3d6 4s2 26 Fe
Iron

⥮ ↿ ↿ ↿ ↿ ⥮
3d6 4s2

Number of unpaired electrons 4

4 4
Total spin + or -
2 2
 Exceptions

Not
Cr [Ar] 4s2 3d4
24
Correct
52
Cr
24
Cr
24
[Ar] 4s1 3d5 Correct

d5 is more stable than d4 configuration

 Exceptions

Not
29
Cu [Ar] 4s2 3d9
Correct
63
Cu
29
Cu
29
[Ar] 4s1 3d10 Correct

d10 is more stable than d9 configuration

 Half-filled & Fully Filled Orbitals

Exactly half filled & Stability of half filled & fully filled
fully filled orbitals make the orbitals
configuration more stable

p3, p6, d5, d10, f7 & f14

Exchange
configurations are stable Symmetry
Energy
 Symmetry

Consequently, their
shielding of one another is
relatively small
Symmetrical Symmetry
distribution of leads to
electrons stability Electrons are more strongly
attracted by the nucleus

Electrons in Different
Equal
the same spatial Have less energy
energy
subshell distribution and more stability
 Exchange Energy

Energy released when two or more

electrons with the same spin in the
degenerate orbitals

Tends to exchange their positions

Number of exchanges that

When subshell is either half
can take place is maximum
filled or fully filled.
 Exchange Energy

a b

↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿

4 exchange by electron ‘a’ 3 exchange by electron ‘b’

c d
↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿ ↿

2 exchange by electron ‘c’ 1 exchange by electron ‘d’

Electronic Configuration of Ions

Cations

Ions

Anions
Electronic Configuration of Cations

Formed by removing outermost

electron from a neutral atom

Na ⥮ ⥮ ⥮ ⥮ ⥮ ↿
1s2 2s2 2p6 3s1

Na+ ⥮ ⥮ ⥮ ⥮ ⥮
1s2 2s2 2p6 3s0
 Electronic Configuration of Cations

electrons are first removed

In d-block metals from ns orbital, then from the
penultimate (n-1)d orbital
 Examples

Fe: [Ar] 3d6 4s2 or Ar ⥮ ↿ ↿ ↿ ↿ ⥮

3d6 4s2

Fe2+: [Ar] 3d6 4s0 or Ar ⥮ ↿ ↿ ↿ ↿

3d6 4s0

Fe3+: [Ar] 3d5 4s0 or Ar ↿ ↿ ↿ ↿ ↿

3d5 4s0
 Examples

29
Cu : Ar ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ↿
3s2 3p6 3d10 4s1

Cu+ :
29
Ar ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮ ⥮
3s2 3p6 3d10 4s0

18 electrons Pseudo inert gas configuration

 Electronic Configuration of Anions

Formed by adding electrons to a neutral atom

according to the 3 rules (Pauli’s, Aufbau & Hund’s rule)

F ⥮ ⥮ ⥮ ⥮ ↿
1s2 2s2 2p5

_
F ⥮ ⥮ ⥮ ⥮ ⥮
1s2 2s2 2p6
Schrodinger Wave Equation (SWE)

Based on the dual nature of matter

Describes the behavior of electron

around the nucleus
 Schrodinger Wave Equation

∂2𝚿 ∂2𝚿 ∂2𝚿 8𝝅2m ( E - V ) 𝚿

∂x2
+
∂y2
+
∂z2
+
h2 = 0

● x, y, z = Cartesian coordinates

𝚿 = Amplitude of the electron wave or Wave function

● h = Planck’s constant

● V = Potential energy of the electron

● E = Total energy of the electron

 Wavefunction

𝚿 corresponds to the allowed

A2 solutions of SWE
A1

x1 𝚿 contains all the information related to the

motion of an electron in an atom
x2

Amplitude of a standing wave is a function of x

Similarly, 𝚿 is a function of coordinates

 Wavefunction

Wave Radial part of Angular part of

function wave function wave function

𝚿 (r, 𝜃, ɸ) = 𝚿 (r) × 𝚿 (𝜃, ɸ)

n,
n, ll l,l, m
ml
l
Physical Significance of 𝚿

Amplitude of the
electron wave
Physical
significance Can be +ve or -ve
of 𝚿

As such there is no
significance of 𝚿
 Physical significance of 𝚿2

Maxwell’s wave theory

Intensity of wave ∝ Square of amplitude

Max Born suggested that

Probability of finding an
𝚿2 electron per unit volume or
probability density
 Probability density

𝚿(r,𝜃,ɸ) = 𝚿(r) × 𝚿(𝜃,ɸ)

𝚿2(r,𝜃,ɸ) = 𝚿2(r) × 𝚿2(𝜃,ɸ)

Probability Radial Angular

density probability probability
density density
 Radial Distribution Function

Radial probability of finding an

electron in a shell of thickness ‘dr’ = 𝚿R2 × dV
at a radial distance ‘r’

This probability which is

independent of direction is called
radial probability and is equal to
[4𝜋r2drR2].

It gives the probability of finding

the electron at a distance r from
the nucleus regardless of direction.
 Radial Distribution Function

4 4 3
dV = 3
𝜋 (r + dr)3 -
3
𝜋r = 4𝜋r2 . dr

Radial probability
density on a layer
× Volume of that layer = 𝚿R2 . 4𝜋r2 . dr

Radial distribution function = 𝚿R2 . 4𝜋r2

Radial Distribution Function
 Summary

Radial wave function 𝚿R

Radial probability density 𝚿R2

Radial probability distribution 4𝜋r2𝚿R2

 Nodes

Region where the probability density is zero i.e.

where the probability of finding an electron is zero

𝚿2.dV = 0

∵ dV can’t be zero

⟹ 𝚿2 = 0
 Nodes

Radial Node

Nodes

Angular Node
 Node

∵ 𝚿(r,𝜃,ɸ) = 𝚿(r) × 𝚿(𝜃,ɸ) = 0

⟹ 𝚿(r) = 0 or 𝚿(𝜃,ɸ) = 0
 Radial Node

Spherical region 𝚿R or 𝚿R2 is zero

around nucleus

Number of radial nodes in an orbital = n-l-1

 Angular Node

Plane or a surface
passing through the 𝚿𝜃, Φ or 𝚿2𝜃, Φ is zero
nucleus

Number of angular nodes in an orbital = l

 Nodes

Total number Radial nodes Angular nodes

of nodes = (n - l - 1) + (l) = n-1
 What are these radial and angular nodes?

Dark coloured regions

show increased Radial node
electron density

Angular node
Radial Node vs Angular Node

Radial node Angular node

Spherical regions where the Flat planes or cones where

probability of finding an the probability of finding an
electron is zero. electron is zero.

Have fixed radii. Have fixed angles.

Number of radial nodes is Number of angular nodes is

given by (n - l - 1) given by (l)
 Nodes of ‘s’ orbitals
Node Nodes

1s 2s 3s
 Angular Nodes of ‘p’ orbitals
z z

y y
z

x
x
y

px pz
x

py
 Angular Nodes of ‘d’ orbitals
z z

y
y x

z
x

dzx dxy
x
y

dyz
Angular Nodes of ‘d’ orbitals

z
z

y
y

x
x

dx2- 2 dz 2
y
 Comparison of Penetration Power

Measure of orbital’s closeness to the nucleus

𝚿R2 4𝜋r2

3d
3p
3s

r
Comparison of Penetration Power

Additional maximas in 3s curve

Electron in 3s spends maximum time

near the nucleus compared to 3p and 3d.

Penetration power : 3s > 3p > 3d

 Probability Curves

Radial Wave
function (𝚿) against r

Radial Probability
Plots density (𝚿2) against r

Radial Probability distribution

(4𝜋r2𝚿R2) against r
 s-orbital

Density of electron cloud is spherical

Probability density is maximum at the

nucleus and decreases at large distance
Radial Probability of 1s

𝚿 ΨR2

r r
4𝜋r2ΨR2

r
 Radial Probability of 2s

Node
𝚿 ΨR2

r r
4𝜋r2ΨR2

r
 Radial Probability of 3s

𝚿 ΨR2

r
r
4𝜋r2ΨR2

r
Radial Probability of 2p

𝚿 ΨR2

r r
4𝜋r2ΨR2

r
Radial Probability of 3p

3p
𝚿 ΨR2

r r
4𝜋r2ΨR2

r
 Radial Probability of 3d

3d
𝚿 ΨR2

r r
4𝜋r2ΨR2

3d

r
 Analysis: 𝚿 Plots

1s 2p 3p 3d

𝚿 𝚿 𝚿
𝚿

r r r r

2s 3s
𝚿 𝚿

r r
 Analysis: 𝚿2 Plots

1s 2p 3p 3d

𝚿R2 𝚿R2 𝚿R2 𝚿R2

r r r r

2s 3s

𝚿R2 𝚿R2

r r
 4𝜋r2ΨR2 4𝜋r2ΨR2

r
r
1s

2s
2
4𝜋r2ΨR2
4𝜋r ΨR2

r
2p

3s

4𝜋r2ΨR2
r
3p
Analysis: 𝚿2.4𝜋r2 Plots

4𝜋r2ΨR2
r
3d
 (ΨR2) 4𝜋r2

1s
r

2s
(ΨR2) 4𝜋r2
Node
r

3s
Nodes

(ΨR2) 4𝜋r2
r
 Nodes
Ψ2 Ψ2 Node Ψ2

r r r
1s 2s 3s
 Radial Wavefunction of Hydrogenic Atoms

3/2 1/2 l
2Z (n - l - 1)! -Zr/na0 2Zr 2l + 1 2Zr
Rnl(r) = na0 2n [(n + l)!]3 e na0
Ln - l - 1
na0
 Laguerre Polynomial

n-l-1 (-1)i [ (n + l)! ]2 𝜌i

2l + 1
Ln - l - 1 (𝜌) = ⅀
i=0 i! (n - l - 1 - i)! (2l + 1 + i)!

Z
where 𝜌 = 2kr and kn =
a0n
 1s orbital

2Zr
R (1s) = K1σ e- σ/2 where, σ = n a0

Number of radial nodes = n-l-1 = 1-0-1 = 0

 Radial Probability of 1s

4𝜋r2𝚿R2
𝚿R 𝚿R2

r r r
 2s orbital

- σ/2
R (2s) = K2(2 - σ) e

Number of radial nodes = n-l-1 = 2-0-1 = 1

 Radial Probability of 2s

4𝜋r2𝚿R2
𝚿R Node
2
𝚿R

r r r
 p-orbitals

p-orbitals consist of lobes

z

Dumb bell Shaped

 2p orbital

- σ/2
R (2p) = K3σ e

Number of radial nodes = n-l-1 = 2-1-1 = 0

 Radial Probability of 2p

4𝜋r2𝚿R2
𝚿R 𝚿R2

r r r
 3p orbital

- σ/2
R (3p) = K4(4 - σ) σ e

Number of radial nodes = n-l-1 = 3-1-1 = 1

 Radial Probability of 3p

4𝜋r2𝚿R2
𝚿R 𝚿R2

r r
r

Radial node
xy nodal plane
 Nodes in p-orbital
3pz orbital
z

2pz orbital
z

Radial node
xy nodal plane
xy nodal plane (Angular node)
(Angular node)
Nodes in d-orbitals

x y

Double Dumb bell Shaped

 3d orbital

- σ/2
R (3d) = K5σ 2 e

Number of radial nodes = n-l-1 = 3-2-1 = 0