11C02
Structure of Atom
Matter
Atoms
Smallest Particle
atomio
Indivisible
Introduction of Atom
Greek word
John Dalton
Dalton’s Atomic Theory
made of
John Dalton
Law of conservation of mass
Law of constant composition
Law of multiple proportion
Generation of electrical charge while rubbing glass by silk
Unexplained:
Neutral Atom
Dalton’s Atomic Theory
+
+
Electron
Proton
Neutron
Daltons Assumption that atoms are indivisible is wrong
divided
11C02.1
Discovery of Subatomic Particles
Learning Objectives
Discovery of Electron
Discovery of Proton & Neutron
11C02.1 Discovery of Subatomic Particles
11C02.1
CV1
Discovery of Electron
Like charges repel each other
Unlike charges attract each other
Experiment Setup: Cathode Ray Discharge Tube
Cathode
Anode
Vacuum Pump
Battery
Cathode Rays
Modifications:
1. Completely evacuated tube
2. Hole was created in Anode
3. ZnS screen was placed behind the Anode
J.J. Thomson
Result of these Experiment:
Cathode Rays Negative charge particle Electron
Deflection depends on:
1. Mass of electron
2. Charge of electron
3. Strength of the Field
J.J. Thomson
Magnetic Force
Electric Force
Static
Millikan‘s oil drop Method:
Millikan ‘s oil drop Apparatus
Electric Field
Gravity
Static
Atomizer
Source of ionizing radiation
Telescope
Negatively charged plate
Positively charged plate
11C02.1 CV2
Discovery of Proton & Neutron
Atom
Neutral
Positive + Negative
Modification:
Discovery of Proton
Observations:
1. Glow was seen: indicate charged particle
3. In presence of electric field, it deflects toward negative plate: Indicate positively charged particle
4. Mass of Particle depends on Nature of Gas
Smallest & Lightest positive ion Hydrogen Proton
Discovery of Proton
Discovery of Neutron
Radioactive element
Beryllium sheet
Chadwick
Neutron
Neutral Particle
Properties of Sub-atomic Particles:
Name | Symbol | Absolute charge/C | Relative charge | Mass/kg | Mass/u | Approx. mass/u |
Electron
Proton Neutron | | | | | | |
11C02.1
PSV1
Q. What is the decreasing order of e/m ratio of electron, proton and neutron?
Sol:
Electron
Proton
Neutron
>
>
Electron > Proton > Neutron
Summary
Reference Questions
NCERT Exercises: 1
Workbook Question: 4, 6, 16
11C02.1 Discovery of Subatomic Particles
11C02.2
Atomic Models
Learning Objectives
Thomson Model of Atom & Rutherford Model
Atomic number , Mass number , Isotopes, Isobars & Isotones
11C02.2 Atomic Models
11C02.2
CV1
Thomson Model of Atom &
Rutherford Model
J.J. Thomson
Thomson model of Atom
Watermelon
Seed = electron
Red juicy part = Positive charge
Neutrality of Atom
Explained:
Atom consists of Positively charged sphere with uniform distribution of electrons
Rutherford
Rutherford Model
Set-up
Observations:
Rutherford Model
Conclusions drawn by Rutherford:
Nuclear Model of Atom:
Limitations:
Rutherford’s model could not explain
11C02.2 CV2
Atomic Number & Atomic Mass
Atomic Number
Atomic Mass
Representation of Element
Element Symbol
Neutral Atom
No. of Electrons in Neutral Atom
Positive Charge
Negative Charge
Protons
Electrons
Positively charged
Charged Species
No. of Protons > No. of electrons
Negatively charged
e = 12-2 = 10
No. of Protons < No. of electrons
e = 7+3 = 10
Isotopes, Isotones & Isobars
Isotopes
17
Isotones
Isobars
12
14
Equal no of protons
Equal no of Neutrons
Same atomic mass
11C02.2
PSV 1
Q. Two nuclides A and B are isoneutronic. Their atomic mass are 76 and 77 respectively. If the atomic number of A is 32 , find the atomic number of B .
Sol:
Given:
A & B are isoneutronic, AA = 76 AB = 77 , ZA = 32 , ZB = ?
Neutrons in A = Neutrons in B
Neutrons = Atomic mass – Atomic Number
AA- ZA = AB - ZB
76 – 32 = 77 - ZB
ZB = 33
Concept Test
Ready for Challenge
Q. Calculate the number of electrons which will together weigh one gram.
Pause the video
Time duration: 2 minutes
Sol:
Q. Calculate the number of electrons which will together weigh one gram.
Summary
= Number of electrons in a neutral Atom
Reference Questions
NCERT Exercises: 2,3,4,22, 26(i), 27, 42, 43, 44
Workbook Question: 5, 10, 11, 13
11C02.2 Atomic Models
11C02.3
Developments leading to Bohr’s Model
Learning Objectives
Electromagnetic Wave
Plank quantum theory
Photoelectric effect
Atomic spectra
11C02.3 Developments leading to Bohr’s Model
11C02.3
CV1
Electromagnetic Wave
Drawbacks of Rutherford Model
Stability of Electron
Atomic Spectra
Distribution of electrons
Rutherford’s Model
Drawbacks of Rutherford Model
1. Stability of Electron
Drawbacks of Rutherford Model
2. Atomic Spectra
Acc. to electromagnetic theory
Continuous Spectrum
Line Spectrum
Real
Theoretical
1. Stability of Electron
3. Distribution of electrons
Distribution of electron
Drawbacks of Rutherford Model
2. Atomic Spectra
1. Stability of Electron
Bohr’s Model
Rutherford’s Model
Next
Electromagnetic Radiation :
Earth
Sun
Electromagnetic Radiation
Nature of Electromagnetic Radiation
Wave Nature
Particle Nature
Electromagnetic Radiation
James Maxwell
Electric field
Magnetic field
Wave Nature
Properties of electromagnetic Waves
1. Oscillating Magnetic field Electric field
Propagation of waves Magnetic field & Electric field
3. Speed
2. EM Waves can travel in vacuum
Properties of electromagnetic Waves
Visible spectrum
4 . Electromagnetic spectrum
Representation of Electromagnetic waves
Water waves
Representation of Electromagnetic waves
SI Unit = m
Representation of Electromagnetic waves
SI Unit = Sec-1 or Hertz
Relation b/w speed , frequency and wavelength
Representation of Electromagnetic waves
SI Unit = m-1
1 Unit
11C02.3
PSV 1
Q. Calculate the wave number and frequency of violet radiation having
wavelength 4000 .
Sol:
Given :
A
0
A
0
1A
0
ConcepTest
Ready for challenge
Q. Calculate the wavelength, frequency of a light wave whose period is
2.0 × 10–10 s.
Pause the video
Time duration: 2 minutes
Sol:
Given :
T = 2.0 × 10–10 s
Q. Calculate the wavelength, frequency of a light wave whose period is
2.0 × 10–10 s.
11C02.3
CV2
Planck ‘s Quantum Theory
Diffraction
Wave nature of Electromagnetic radiation
Diffraction
Interference
Wave nature of Electromagnetic radiation
Wave nature of Electromagnetic radiation
Black body radiation
Photoelectric effect
Line spectrum (H)
Variation of Heat capacity of solids as f (T)
Black Shirt
White Shirt
Comfortable in Hot
Comfortable in Hot
Black body radiation
Black body
Emits and Absorbs
Black body radiation
Black body
Intensity
Intensity depends :
Unexplained by Wave nature
Black body radiation
Black body
Packets of Energy
Continuous
Max. Planck
Planck ’s quantum Theory
Black body
Planck ’s quantum Theory
Packet of Energy
Quantum
h = Planck ‘s constant
Max. Planck
11C02.3
PSV 2
Q. 100 Watt bulb emits the monochromatic light of 400nm . Calculate the
number of photons emitted per second by the bulb .
Sol:
Given :
100 watt = 100 j / sec
100 j / sec = Energy of n number photons / sec
ConcepTest
Ready for challenge
Pause the video
Time duration: 2 minutes
Sol:
Given :
According to Planck’s quantum theory :
11C02.3
CV3
Photoelectric Effect
Photoelectric Effect
H. Hertz
Metal Plate
For instance :
Potassium , Rubidium etc
Photoelectric Effect
Brightness
Result observed :
No time lag b/w light striking & Electron ejection
Threshold Frequency
Frequency
<
Photoelectric effect
Observations could not be explained by wave nature of light
Photoelectric Effect
No. of Electron ejected depends
Intensity of light
Energy of electron
Potassium Plate
-
-
-
Red light
Yellow light
Frequency of light
If light is a wave or continuous form of energy than
Photoelectric Effect
Albert Einstein
Quantum Theory of
Electromagnetic Radiation
Work function
Max. Planck
Photoelectric Effect
Albert Einstein
Mass of electron
Velocity of ejected electron
Dual Behaviour of Electromagnetic Radiation
Photoelectric Effect
Interference and diffraction
Particle nature
Wave nature
Albert Einstein
Dilemma??
Dual nature of light
11C02.3
PSV 3
Sol:
Given :
11C02.3
CV4
Line spectrum of Hydrogen
Spectrum
Continuous Band
of Colour
Spectrum
Red
Violet
Light
Ground state
Excited state
Atom
Emission spectrum
Spectrum
Absorption spectrum
Absorption of energy
Emission of energy
Continuous Spectrum
Spectrum
Line Spectrum
Hydrogen spectrum
Balmer
Several series of Line
n = 3, 4, 5……..
Rydberg Constant
Balmer formula
Line Spectrum of Hydrogen
Hydrogen spectrum
All series of Line
n1 =1, 2……..
n2 = n1+1, n1+2 ……..
Rydberg
formula
Lyman Series
Balmer Series
Paschen Series
Brackett Series
Pfund Series
Line Spectrum of Hydrogen
Line Spectrum of Hydrogen
n1 < n2
Line Spectrum of Hydrogen
11C02.3
PSV 4
Q. What is the wavelength of light emitted when the electron in a
hydrogen atom undergoes transition from an energy level with n = 4
to an energy level with n = 2?
Sol:
Given :
Rydberg
formula
Summary
Reference Questions
NCERT Exercises: 5, 6, 7, 8, 9, 11, 12, 13, 17, 33, 51, 52, 53, 54
Workbook Question: 3, 8, 9, 14, 15, 19
11C02.3 Developments leading to Bohr’s Model
11C02.4
Bohr’s Model for Hydrogen Atom
11C02.4 Bohr’s Model for Hydrogen Atom
Learning Objectives
Postulates of Bohr’s Model
Results obtained from Bohr’s Model
Explanation of line spectrum of Hydrogen
Limitations of Bohr’s Model
11C02.4
CV 1
Postulates of Bohr’s Model
Introduction to Bohr’s Model
Neils Bohr
General features of the structure of hydrogen atom and its spectrum is explained
Postulates of Bohr’s Model
1. Electron revolve around nucleus in fixed circular paths(Orbits). These Orbits have constant energy and radius
It removed the limitation of collapsing of electron in Rutherford Atomic Model
+
n = 1
n = 3
n = 2
2. The orbit of electron does not change unless energy is added or removed
Gain energy
Lose/ emit energy
Jumps from lower to higher n
Jumps from higher to lower n
Postulates of Bohr’s Model
depends upon the energy difference of the initial & final orbits
Postulates of Bohr’s Model
4. The angular momentum of an electron is quantized
v
r
n = Integer
Postulates of Bohr’s Model
ConcepTest
Ready for Challenge
Q. Which of the following cannot be the value of angular momentum of electron in Hydrogen atom ?
Pause the video
Time duration : 1 minute
Ans. C
Not an Integer
Q. Which of the following cannot be the value of angular momentum of electron in Hydrogen atom ?
Sol.
11C02.4
CV 2
Results obtained from Bohr’s Model
Results obtained from Bohr’s model
1. The stationary states(orbits) for electron are numbered n = 1,2,3..........
These integral numbers are known as Principal quantum numbers
2. The radii of the stationary states are expressed as :
+
3. The energy associated with electron in its stationary state/orbit/shell is given by :
Note : Energy associated with free electron at rest is zero
Energies & radii of the stationary states associated with these kinds of ions are given by :
Energy
Radius
Z = Atomic Number
5. Velocities of an electron moving in a stationary orbit can be calculated as :
11C02.4
PSV 1
Q. What is the radius of third orbit of the hydrogen atom ?
Pause the video
Time duration : 1 minute
Q. What is the radius of third orbit of the hydrogen atom ?
Sol.
We know that :
Radius of orbit of Hydrogen is given by :
Here, n = 3, Z = 1
11C02.4
PSV 2
Pause the video
Time duration : 1 minute
Q. Calculate the difference in energy of electron, between 4th and 2nd orbit of hydrogen atom.
Sol.
Q. Calculate the difference in energy of electron, between 4th and 2nd orbit of hydrogen atom.
11C02.4
CV 3
Explanation of Line Spectrum of Hydrogen
Explanation of Line Spectrum of Hydrogen
This expression is similar with Rydberg expression
Each spectral line, whether in absorption/ emission spectrum, can be associated to the particular transition in hydrogen atom
Energy :
The intensity of spectral lines depends upon the number of photons of same wavelength or frequency absorbed/ emitted
In case of large number of hydrogen atoms, different possible transitions can be observed and thus leading to large number of spectral lines
Explanation of Line Spectrum of Hydrogen
11C02.4
CV 4
Limitations of Bohr’s Model
Limitations of Bohr’s Model
Bohr’s model was a big improvement over Rutherford’s nuclear model, as
it could account for :
of hydrogen atom and hydrogen like ions
But...
Limitations of Bohr’s Model
Bohr’s model was unable to explain the following points :
Summary
11C02.4 Reference questions
Workbook questions : Q7, Q12, Q17, Q18, Q20
NCERT Exercise questions : 2.13, 2.14, 2.16, 2.18, 2.19, 2.33
11C02.5
Towards Quantum Mechanical Model of the Atom
Learning Objectives
Dual Behaviour of Matter
Heisenberg’s Uncertainty Principle
11C02.5 Towards Quantum Mechanical Model of the Atom
11C02.5
CV1
Dual Behaviour of Matter
Dual Behaviour of Matter
Matter
Particle properties
Wavelike properties
Louis de Broglie
Wave like
Wave and particle
Particle like
Dual Behaviour of Matter
Louis de Broglie
Electron beam
Particle nature
Diffraction
Wave nature
Develop electron microscope
Dual Behaviour of Matter
Louis de Broglie
Wavelength
Short
For ordinary object like ball etc.
Because large masses
Can be detected experimentally.
For electrons and other
subatomic particles
Because very small masses
11C02.5
PSV 1
Q. Similar to electron diffraction, neutron diffraction microscope is also used
for the determination of the structure of molecules. If the wavelength used
here is 800 pm, calculate the characteristic velocity associated with the
neutron.
Q. Similar to electron diffraction, neutron diffraction microscope is also used
for the determination of the structure of molecules. If the wavelength used
here is 800 pm, calculate the characteristic velocity associated with the
neutron.
Sol
ConcepTest
Ready for Challenge
Q. Find the wavelength of a particle of 100 g moving with velocity 100 m s-1
Pause the video
Time duration: 2 minutes
Sol.
Q. Find the wavelength of a particle of 100 g moving with velocity 100 m s-1
11C02.5
CV2
Heisenberg’s Uncertainty Principle
Heisenberg’s Uncertainty Principle
It is impossible to determine simultaneously, the exact position and exact momentum of an electron.
Werner Heisenberg
Significance of uncertainty principle
11C02.5
PSV 2
Sol.
ConcepTest
Ready for Challenge
Pause the video
Time duration: 2 minutes
Q. An electron is confined to a region of width 5.00x 10-11 m, which
is its uncertainty in position. Estimate the minimum uncertainty
in its momentum.
Q. An electron is confined to a region of width 5.00x 10-11 m, which
is its uncertainty in position. Estimate the minimum uncertainty
in its momentum.
Sol.
Summary
Dual Behaviour of Matter
Louis de Broglie
Wavelength
Heisenberg’s Uncertainty Principle
It is impossible to determine simultaneously, the exact position and exact momentum of an electron.
Significance of uncertainty principle
11C02.5 Towards Quantum Mechanical Model of the Atom
Reference question:
NCERT exercise question: 2.20, 2.21, 2.57, 2.58, 2.59, 2.60, 2.61
Workbook question: 1
11C02.6
Quantum Mechanical Model of an Atom
11C02.6 Quantum Mechanical Model of an Atom
Learning objectives
Quantum Mechanics
Quantum numbers
Shape of atomic orbitals
Effective Nuclear Charge
Electronic configuration
11C02.6
CV 1
Quantum Mechanics
Classical mechanics V/s Quantum mechanics
Classical mechanics
Based on wave properties of matter
Behavior of microscopic particles such as protons and electrons
Quantum mechanics
Dual behavior
of matter
Explain behavior of microscopic particles
Quantum mechanics
It deals with dual behavior of matter both wave like and particle like properties.
Werner Heisenberg
Erwin Schrodinger
Schrodinger equation
Hamiltonian operator
Total energy
Wave function
>
Boundary Surface diagram
Darker the region greater the probability.
Drawn for orbital for which is constant.
Encloses the region with probability density of more than 90%.
Why do we not draw a boundary surface diagram, which bounds a region in which the probability of finding the electron is 100%??
At any distance from the nucleus, the probability density of finding an electron is never zero.
11C02.6
CV 2
Quantum numbers
Quantum numbers
Principle Quantum number
Azimuthal Quantum number
Magnetic Quantum number
Spin Quantum number
Quantum numbers
It determines size, shape and orientation of orbital
Principle Quantum number(n)
K
L
M
N
n=1
n=2
n=3
n=4
Azimuthal quantum number(l)
n | l | Subshell notation |
1 | 0 | 1s |
2 | 0 | 2s |
2 | 1 | 2p |
Subshell
| |
| |
| |
Orientation of the orbital with respect to standard set of co-ordinate axis.
11C02.6
PSV 1
ConcepTest
Ready for challenge
Pause the video
Time duration : 2 minutes
11C02.6
CV 3
Effective Nuclear Charge
Degenerate orbital
For single electron species energy of orbitals having same quantum number have same energy.
For multi-electron atom orbitals with same quantum number posses different energy
Orbitals having same energy are called degenerate orbitals
For single electron system,
Higher the value of n, higher the energy
For multi electron system,
Higher the value of (n+l), higher the energy
+
Ze
Attraction by nucleus and repulsion by electron in inner shell
+
Ze
ConcepTest
Ready for challenge
Q. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
Pause the video
Time duration:1 min
Q. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
Sol.
4p will experience lowest effective nuclear charge as it is farthest from the nucleus.
2p
3p
4p
11C02.6
CV 4
Shape of Atomic Orbital
Boundary surface diagram of 1s and 2s orbital.
Boundary surface diagram of the three 2p orbital
Boundary surface diagram of the five 3d orbital.
Radial node=(n – l - 1)
Angular node/nodal plane=l
Total=(n-1)
node
11C02.6
CV 5
Electronic Configuration
Aufbau principle
In ground states of the atoms, the orbitals are filled in order of their increasing energies value of (n+l).
orbital | Value of n | Value of l | Value of (n+l) | |
1s | 1 | 0 | 1 | |
2s | 2 | 0 | 2 | |
2p | 2 | 1 | 3 | 2p(n=2)has lower energy than |
3s | 3 | 0 | 3 | 3s(n=3) |
3p | 3 | 1 | 4 | 3p(n=3)has lower energy than |
4s | 4 | 0 | 4 | 4s(n=4) |
3d | 3 | 2 | 5 | 3d(n=3)has lower energy than |
4p | 4 | 1 | 5 | 4p(n-4) |
Aufbau principle
Pauli Exclusion Principle
Wolfgang Pauli
No two electron in an atom can have the same set of four quantum number.
| | n | l | m | s |
2s | Electron 1 | 2 | 0 | 0 | |
| Electron 2 | 2 | 0 | 0 | |
It shows that only two electron may exist in the same orbital and these electrons must have opposite spin.
Hund’ s rule of maximum multiplicity
Friedrich Hund
Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.
Electronic Configuration
Electronic configuration can be represented by
Orbital diagram
Example:
Stability of half filled and completely filled subshell
Why actual electronic configuration was different from expected?
Causes of Stability of half filled and completely filled subshell
1. Symmetrical distribution of electrons
2. Exchange energy
4 exchange by electron 1
3 exchange by electron 2
2 exchange by electron 3
1 exchange by electron 4
Summary
>
11C02.6 Quantum Mechanics
Reference question:
NCERT exercise question: 2.28, 2.29, 2.30, 2.31, 2.62, 2.63, 2.64, 2.65,
2.66, 2.67