Atomic Structure

ATOMIC STRUCTURE

• Planck’s Quantum theory gave a new dimension to the study of atoms.

• It introduced this new concept of Quantisation on which all the future
models are based.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Before going further and study Bohr’s model we will take a look at
what are atomic spectra?

What is spectra?

CHEMISTRY
Atomic Structure

What is spectra? ATOMIC STRUCTURE

SPECTRA

• Spectra is an array of entities (such as light waves or particle) ordered in accordance with the
magnitudes of a common physical property (such as wavelength).

• Often the band of colours produced when sunlight is passed through a prism comprising
VIBGYOR.

CHEMISTRY
Atomic Structure
What is the classification of Spectra?
ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

What is the classification ATOMIC

of Spectra? STRUCTURE

CHEMISTRY
Atomic Structure
What is the continuous Emission
ATOMIC Spectra? STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC
What is the continuous Emission Spectra? STRUCTURE

CHEMISTRY
Atomic Structure

What is the Discontinuous ATOMIC STRUCTURE

Emission Spectra?

CHEMISTRY
Atomic Structure

ATOMIC
What is the Discontinuous Emission Spectra?STRUCTURE

CHEMISTRY
Atomic Structure

What is the DiscontinuousATOMIC STRUCTURE

Absorption Spectra?

CHEMISTRY
Atomic Structure
What is the Discontinuous Absorption Spectra?
ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Hydrogen spectrum in detail.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What is Atomic Spectra of Hydrogen ?

 Atomic spectra are a characteristic of a given element.

 Also known as Finger Print of an element.
 It was observed that atomic spectrum of hydrogen is a discontinuous line spectrum
 where some wavelengths are missing in between.
 Rydberg found that the wavelengths of the spectra follow the following relation

7 –1
 R = 1.097 × 10 m
 Bohr proposed a model based on these findings. Next we will discuss that
CHEMISTRY
Atomic Structure
Define Hydrogen spectrum in detail.
ATOMIC STRUCTURE
Hydrogen spectrum
n=8 γ
n=7 γ β
n=6
δ γ β α
n=5 Pfund series I.R region (far zone)
δ γ β α
n=4 I.R region (mid zone)
γ β α Brackett series
n=3
α Paschen series I.R region (near zone)
β
n=2 1
Balmer series visible region 1 1
α = R Z2 –
λ 22 n2
n=1
Lyman series U.V.
CHEMISTRY Rydberg Balmer equation
Atomic Structure

What is Atomic Spectra ofATOMIC

Hydrogen ? STRUCTURE

CHEMISTRY
Atomic Structure

Define Hydrogen spectrumATOMIC

in detail. STRUCTURE
Hydrogen spectrum

Series n1 n2 Spectral Region

Lyman 1 2, 3… Ultraviolet
Balmer 2 3, 4… Visible
Paschen 3 4, 5… near infrared
Brackett 4 5, 6… mid infrared
Pfund 5 6, 7… far infrared

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the wavelength for α- line of the Lyman series for H-atom?

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the wavelength for α- line of the Lyman series for H-atom?

Solution Rydberg equation

For α- line of Lyman series n1 = 1 and n2 = 2
1 = 2 1 – 1 n2 > n1
RZ n 2
λ 1 n22

1 1 3
⇒ = RZ2 – 1 = RZ2 For H-atom Z = 1
λ 1 4 4
4
⇒ λ = 911.5  1 
3 R = 1.097  107 m–1
1 = 911.5 A
λ = 1216 A R

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

The Lyman series of H-spectrum can be represented by the equation

υ = 3.28  1015 1 –1 S –1.

12 n2
Calculate the minimum and maximum frequency.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

The Lyman series of H-spectrum can be represented by the equation

υ = 3.28  1015 1 –1 S–1.

2
12 n
Calculate the minimum and maximum frequency.
For minimum value of frequency ‘n’
Solution has to be minimum
υmin = 3.28  1015 1 –1 S–1
4 For maximum value of frequency
υmin ‘n’ has to be maximum
= 2.46  1015 Hz
υmax 1– 1
= 3.28  1015 ∞
S–1 For any series maximum value of
‘n’ is ∞
υmax = 3.28  1015 Hz For Lyman series
minimum value of ‘n’ is 2
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the wave number for the longest wavelength transition in the
Balmer series of atomic hydrogen.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the wave number for the longest wavelength transition in the
Balmer series of atomic hydrogen.
Solution and for longest wavelength n2 = 3
1 1 1
= RZ2 –
λ n12 n22
For H-atom Z = 1
1 1 1
⇒ =1.097  107  –
λ 4 9
5 1 = υ
⇒ υ = 1.097  106  m–1
36 λ
Wave number = υ = 1.57  106 m–1
For Balmer series n1 = 2
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Estimate the λ of second line (also called β -line) in Balmer series of H-atom.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Estimate the λ of second line (also called β -line) in Balmer series of H-atom.
Solution
1 1 1 For second line in Balmer series
= RZ2 – n1 = 2 and n2 = 4
λ n12 n22

1 3
⇒ = RZ2 1 – 1 = RZ2 For H-atom Z = 1
λ 4 16 16

⇒ λ = 911.5  1  16 R = 1.097  107 m–1

3 1 = 911.5 A
R
λ = 4861 A

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy released when 2.0 mol of atoms of Hydrogen undergo transition
giving spectral line of lowest energy in visible region of its atomic spectra (in kJ).

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy released when 2.0 mol of atoms of Hydrogen undergo transition
giving spectral line of lowest energy in visible region of its atomic spectra (in kJ).

Solution –RH Z2
E =
n2 For lowest energy transition in Balmer series
ΔE = E3 – E2 n1 = 2 and n2 = 3
1– 1 = RHZ2 1 – 1
ΔE = RHZ2 RH = 1312 kJ.mol–1
n22 n32 4 9
5 visible region indicates Balmer series
ΔE = 1312 × 36 = 182.22 kJ.mol–1
For 2 moles,
ΔE = 182.22 × 2 = 364.4 kJ
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy required to excite 0.041 moles of H2 gas to the first excited state
of atomic hydrogen. The energy required for the dissociated of H– H bond is 436
kJ/mol. Also calculate the minimum frequency of a photon to break this bond.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy required to excite 0.041 moles of H2 gas to the first excited state
of atomic hydrogen. The energy required for the dissociated of H– H bond is 436
kJ/mol. Also calculate the minimum frequency of a photon to break this bond.
Solution E1
H2 2H
0.041 mol 2 × 0.041 moles of atom
Energy required for dissociation of 0.041 moles of H2 = 0.041 × 436
1
ΔE = E2 – E1 = RH – 41 = 1312 × 3 kJ/mol
1 4
RH = 1312 kJ.mol–1
∴ For 0.082 moles of H-atom = 0.082 × 1312 × 3 kJ
4

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy required to excite 0.041 moles of H2 gas to the first excited state
of atomic hydrogen. The energy required for the dissociated of H– H bond is 436
kJ/mol. Also calculate the minimum frequency of a photon to break this bond.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Find the energy required to excite 0.041 moles of H2 gas to the first excited state
of atomic hydrogen. The energy required for the dissociated of H– H bond is 436
kJ/mol. Also calculate the minimum frequency of a photon to break this bond.
Solution NA = 6.623  1023
Total energy required for excitation
3
= 0.041 × 436 + 0.082 × 1312 ×
4
= 98.564 kJ
h = Planck’s constant
NA  hυ = 436  103 J = 6.627  10-34 J.Sec
436  103
⇒ υ = Sec-1
6.623  1023  6.627  10–34
υ = 9. 93  1014 Hz
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the frequency of revolution for electron of He+ ion, when it is at n = 3.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the frequency of revolution for electron of He+ ion, when it is at n = 3.
2πrn
Solution Time period of revolution, T = vn
Frequency of revolution,
1 vn
υ = =
T 2πrn n2
2
rn = 0.529  A
r3 = 0.529  3 Z
A
2
v3 = 2.165  106  2 m/s
3 vn = 2.15  106 Z m/s
2.165 10 6 2 2 -1 n
υ = Sec
2 3.14 0.529 9 10–10 3
14 For He+ ion Z = 1
υ = 9.6 10 Hz
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

What is Maxwell’s EM wave Theory?

CHEMISTRY
Atomic Structure

What is Maxwell’s EM waveATOMIC

Theory? STRUCTURE
A source emits energy in form of radiations continuously i.e. there is no change
in wavelength or frequency of the emitted radiations even on increasing the
energy radiated

If we increase the energy emitted then the intensity of the radiation should
increase without having any effect on the wavelength or the frequency.

A moving charged particle under an electric field should lose energy

continuously in form of EM radiations.

Continuous emission => no change in Wave length or Frequency

CHEMISTRY
Atomic Structure

ATOMIC
Was this theory applicable STRUCTURE
in all cases?
Blackbody radiations and Photoelectric Effect.

CHEMISTRY
Atomic Structure

ATOMIC
Was this theory applicable in all cases? STRUCTURE
Blackbody radiations and Photoelectric Effect.

These experiments proved that

The theory of Experiments like energy emission in form of
continuous energy BB Radiations radiations is a discontinuous
emission failed. and PE Effect. process in which frequency plays
a very important role.

CHEMISTRY
Atomic Structure

ATOMIC
What are black body radiations STRUCTURE
and how were they studied?

CHEMISTRY
Atomic Structure
What are black body radiations and how were they studied?
ATOMIC STRUCTURE
Before understanding black body radiations we will define what is a Black body:-

A black body is an ideal object which can absorb all the radiation incident upon it and
once absorbed it can emit them as well. It is ideal absorber or ideal emitter.

CHEMISTRY
Atomic Structure
What are black body radiations and how were
ATOMIC they studied?
STRUCTURE
Blackbody Radiation Experiment
As the temperature of the object increased, the wave
length corresponding to which observed intensity at
maximum, decreases.

Higher temperatures, though the intensity rises as

predicted by Maxwell’s theory but the wavelength
decreases.

These observations suggest that

somehow the energy radiated is
related to frequency and it is not
the case of continuous emission.
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

What is Wein’s displacement law?

CHEMISTRY
Atomic Structure
What is Wein’s displacement law?
ATOMIC STRUCTURE
Wein’s Displacement Law states that the wavelength distribution peaks at a value that is inversely
proportional a to the temperature i.e Wavelength X Frequency = Constant

CHEMISTRY
Atomic Structure

ATOMIC
What is photo electric effect? STRUCTURE

CHEMISTRY
Atomic Structure

What is photo electric effect?

ATOMIC STRUCTURE
• The emission of electrons from a metal surface on the
incidence of electromagnetic radiations is called
“Photoelectric Effect”.
• Basic principle-the radiations transfer the energy to
electrons inside the metal and if this energy is sufficient then
the electrons are ejected from the surface.
• According to Maxwell’s Theory the ejection of electrons
should depend on intensity of radiation that is if electrons
are not being ejected, then on increasing the intensity they
can be ejected and so on.
• i.e. the energy of the electrons should have been
proportional to the intensity of the light, not the frequency.
CHEMISTRY
Atomic Structure

What were the observations

ATOMICof photo electric
STRUCTUREeffect?

CHEMISTRY
Atomic Structure
What were the observations of photo electric effect?
ATOMIC STRUCTURE
1. No electron is ejected if the incident frequency, n, is less than
certain minimum value n0.

2. When n < n0 then even on increasing the intensity here is no

ejection of electrons. Minimum energy required to exhibit
photoemission of surface electrons.This is called as Threshold
Frequency

3. When n > n0, howsoever low the intensity is, ejection is observed.

4. When n > n0, on increasing the intensity the number of

photoelectrons keep on increasing.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Were these findings justifying Maxwell’s theory?

Quantum Theory

CHEMISTRY
Atomic Structure

Were these ATOMIC

findings justifying Maxwell’s theory?
STRUCTURE

These findings again contradict Maxwell’s theory. To justify these findings Max Von Planck gave
his Quantum Theory which is one of the most important theories of modern science.

CHEMISTRY
Atomic Structure

ATOMIC
Now, Let’s discuss a problem STRUCTURE

Problem:

At 1500K the maximum wavelength of radiation that an object emits out is 7000A°. What will
be the maximum wavelength if radiation if the T is increased to 3000K.

CHEMISTRY
Atomic Structure

ATOMIC
Now, Let’s discuss a problem STRUCTURE
Problem
At 1500K the maximum wavelength of radiation that an object emits out is 7000A°. What will
be the maximum wavelength if radiation if the T is increased to 3000K.

Solution
λ max T = constant
λ 1 T1 = λ 2 T2

=> 7000 X 1500 = λ 2 X 3000

λ 2 = 3500 A°

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define Structure of atom.

CHEMISTRY
Atomic Structure
Define Structure of atom.
ATOMIC STRUCTURE
What is matter ?
Anything that has mass, occupies space and which can be perceived by any of our senses is
called matter. e.g- Chalk, Book, Table..

What is matter made up of ?

It is made up of tiny, indivisible unit particles which are known as

ATOMS.

CHEMISTRY
Atomic Structure
Define Structure of atom.
ATOMIC STRUCTURE
What is an atom
An atom is considered as the smallest indivisible particle of matter
which can participate in a chemical reaction.
First explanation was given by
Based upon an important property
“atomic mass”

Gave a theory known as Dalton’s

atomic theory 1805

John Dalton CHEMISTRY

Atomic Structure

What are the Postulates

ATOMIC of Dalton’s Atomic Theory ?
STRUCTURE
Dalton’s Atomic Theory

CHEMISTRY
Atomic Structure
What are the Postulates of Dalton’s Atomic Theory ?
ATOMIC STRUCTURE
Postulates of Dalton’s atomic theory
1) Atom is considered as hard, dense and smallest indivisible particle of
matter

2) Each element consists of a unique kind of atoms and have identical mass
H2 molecule

Each H atom has an atomic mass of 1 amu

So if you observe each atom of H2 has same atomic

mass therefore their properties are similar
CHEMISTRY
Atomic Structure
What are the Postulates of Dalton’s Atomic Theory ?
ATOMIC STRUCTURE
Postulates of Dalton’s atomic theory
3) Atoms of different elements have different properties
H - atom (Mass - 1amu)
C - atom (Mass - 12amu)
Hydrogen is a gas

Carbon is found in solid state

Moreover there is a difference

in atomic masses of C and H
Hence, properties of C and H
are different
CHEMISTRY
Atomic Structure
What are the Postulates of Dalton’s Atomic Theory ?
ATOMIC
Postulates of Dalton’s atomic theory STRUCTURE
Consider the reaction between Na and H2O

we add sodium pellets to it

It’s a vigorous reaction in which effervescence is produced

This proves that

The same reaction will take place for all the sodium atoms

4) All the atoms of a given element having same atomic mass have Identical
Properties,
CHEMISTRY
Atomic Structure
What are the Postulates of Dalton’s Atomic Theory ?
ATOMIC STRUCTURE
Postulates Dalton’s of atomic theory
5). Compounds of different elements are formed when atoms of
different elements combine in a fixed ratio of whole numbers

Chlorine Sodium

Let’s take a crystal of Chlorine and Sodium

They chemically combine to form NaCl

6) Chemical reactions involve reorganization of atom
CHEMISTRY
Atomic Structure
What are the Postulates of Dalton’s Atomic Theory ?
ATOMIC STRUCTURE
Postulates of Dalton’s atomic theory

Laws of chemical
combination

The law of The law of The law of

definite multiple reciprocal
proportion proportion proportion

7) Dalton successfully explained all the laws of chemical combination

CHEMISTRY
Atomic Structure

Was this the correct explanation

ATOMIC for structure of atom?
STRUCTURE

Define Limitations of Dalton’s Atomic theory.

CHEMISTRY
Atomic Structure
Define Limitations of Dalton’s Atomic theory.
ATOMIC STRUCTURE
Limitations of Dalton’s Atomic theory

1. According to Dalton’s theory all Hydrogen atoms must have same atomic masses but the
discovery of isotopes discarded his theory.

What are isotopes???

2. Let’s recollect postulate number 3.

Compounds of different elements are formed when atoms of different elements combine in a
fixed ratio of whole numbers but this is not true in the case of “Non Stoichiometric
Compounds”
e.g- Fe0.93O
CHEMISTRY
Atomic Structure
Define Limitations of Dalton’s Atomic theory.
ATOMIC STRUCTURE
Limitations of Dalton’s Atomic theory

3. According to Dalton’s atomic theory atom was Indivisible but atom was found to be
divisible when sub atomic particles were discovered.

With advance in technology as of today we know that atom is further subdivided. We have
35 sub atomic particles. Among which fundamental particles are:
A………….
B………….
C………….

CHEMISTRY
Atomic Structure
Define Limitations of Dalton’s Atomic theory.
ATOMIC STRUCTURE
Limitations of Dalton’s Atomic theory

• Isotopes: Elements with same atomic number(Z) but with different mass number (A) are
called isotopes.

• Fe0.93O. It is a mixture of FeO and Fe2O3. In Fe0.93O we may have small proportion of FeO
and Fe2O3which leads to the empirical formula of Fe0.93O. FeO and Fe2O3 individually do
not violate the theory individually.

• Sub atomic particles A.Electrons B. Protons C. Neutrons

CHEMISTRY
Atomic Structure

ATOMIC
How were Sub Atomic particlesSTRUCTURE
discovered?

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Explain discovery of electron- study of

cathode rays?

CHEMISTRY
Atomic Structure
Explain discovery of electron- study of cathode rays?
ATOMIC
Discovery of Electron
STRUCTURE
Atom was found to be divisible. Limitation of Dalton’s Theory

CATHODE RAY EXPERIMENT

William Crookes J.J Thomson

CHEMISTRY
Atomic Structure
Explain discovery of electron- study of cathode rays?
ATOMIC
Cathode Ray Experiment
STRUCTURE

Vacuum pump
Cathode (–) Anode (+)
Cathode rays
z

Gas at low pressure

(10-2 – 10-3 mm)

Battery
CHEMISTRY
Atomic Structure
Explain discovery of electron- study of cathode rays?
ATOMIC
Cathode Ray Experiment STRUCTURE
 The apparatus used in this experiment cylindrical glass tube, William
Crookes designed the tube so it is called as Crooke’s tube

Cathode ray tube Crooke’s tube Discharge tube

 Fitted with two metal electrodes, Metal electrode are connected to
high voltage power supply. (5000-10000 V)
 It is fitted with a vacuum pump, It helps to maintain a low pressure.
 A ZnS screen was placed behind anode, Metal electrode are connected
to high voltage power supply
 At this high voltage & low pressure it was observed ,the ZnS behind the
anode started glowing
 The property of ZnS is, it starts glowing when a charged particle hits
the screen
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define Properties of Cathode Rays and

Explain them.

CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays and Explain them.
ATOMIC STRUCTURE
Properties of cathode rays
 Cathode rays travels in a straight line
 Cathode rays are made up of particle
 Cathode rays are made of negative charged particle
 Cathode rays are deflected by electric field and magnetic field
 Cathode rays show heating effect
 Cathode rays ionize the gases through which they travel
 Cathode rays produces x rays on hitting a metallic target
Source of cathode rays

 Metal of electrode (cathode)

 Gas atoms in the medium
CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays and Explain them.

Properties of Cathode Rays

ATOMIC STRUCTURE
1) CATHODE RAYS TRAVEL IN A STRAIGHT LINE
An opaque object was placed in the path of cathode rays Adjacent to Vacuum pump
ZnS screen is placed Example : Wooden block, Stone

It is observed that the Cathode (–) Anode (+)

shadow of the object is Cathode rays
formed on ZnS screen
z

Gas at low pressure

Battery
(10-2 – 10-3 mm)
CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays and Explain them.
ATOMIC
Properties of Cathode Rays STRUCTURE
2) CATHODE RAYS ARE MADE UP OF PARTICLE
 This time a pin wheel is kept in the path of cathode Vacuum pump
rays It was observed that the pin wheel started
rotating
Cathode (–) Anode (+)
 It means that cathode rays
Cathode rays
posses energy which was
transferred to the wheel. The z
energy associated with motion
is K.E. As we know mass is
possessed by particle Gas at low pressure
Battery
(10-2 – 10-3 mm)
CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays and Explain them.
ATOMIC
Properties of Cathode Rays STRUCTURE
3) CATHODE RAYS ARE MADE OF NEGATIVE CHARGED PARTICLE
Vacuum pump

 This time an external Anode(+)

electric field was applied to Cathode (–)
it, Cathode rays were
Cathode rays
deflected towards the
positive end. So, they carry z
negative charge

Gas at low pressure

Battery
(10-2 – 10-3 mm)

CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays and Explain them.
ATOMIC
Properties of Cathode Rays STRUCTURE
4) Cathode rays are deflected towards magnet

Cathode (+) Anode (+)

If magnetic field is applied
using a horse shoe magnet
z

Gas at low pressure

(10-2 – 10-3 mm)

Battery
CHEMISTRY
Atomic Structure
Define Properties of Cathode Rays.
ATOMIC STRUCTURE
Properties of Cathode Rays

5) Cathode rays show heating effect Anode (+)

If a metal foil is placed in
the path of cathode rays Metal foil
Cathode (-)

It’s temperature increases

z
and it becomes hot

CHEMISTRY Battery
Atomic Structure
Define Properties of Cathode Rays and Explain them.
ATOMIC STRUCTURE
6) Cathode rays ionize the gases through which they travel.

When cathode rays hits the gas atoms,

e– moves out and atom becomes +ve Ion

Cathode Anode
holes in cathode

e+–
e–
+ z
e+–

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Determine the value of e/m.

CHEMISTRY
Atomic Structure
Determine the value of e/m.
ATOMIC STRUCTURE
Determination of e/m value

7) When this experiment was conducted in a dark room cathode

rays produces X rays on hitting a metallic target
e/m = 1.76  108 C/g
Cathode Anode
X
Y

-CHEMISTRY
+ Fluorescent screen
Atomic Structure
Determine the value of e/m.
ATOMIC STRUCTURE
Determination of e/m value
7) When this experiment was conducted in a dark room cathode
rays produces X rays on hitting a metallic target
 The ray hits at a point ‘Y’ and its starts glowing

 Cathode rays deflect towards the positive electrodeThus hitting the screen
at a point say ‘X’
 Cathode rays deflect towards magnet ,Thus hitting at a point say ‘Z’

 Using these three points X,Y,Z

J.J Thomson determined the value of e/m

e/m = 1.76  108 C/g
CHEMISTRY
Atomic Structure
Determine the value of e/m.
ATOMIC STRUCTURE
Conclusion
Cathode rays are made of negative charge particles “e”
Cathode rays are made up of particles “m”

Determination of e/m value

e Charge

m mass

J.J Thomson conducted an

experiment to find the value of e/m
CHEMISTRY
Atomic Structure
Determine the value of e/m.

Determination of e/m value

ATOMIC STRUCTURE
This ratio was found to be constant irrespective of the nature of
cathode or the nature of gas taken in discharge tube
Since it is constant for all gases

That means this (–) ve charged particle is

Electron
universal and it is common to all atoms
The charge of electron
was found to be e = –1.602  10–19c
This was found with the help of an experiment conducted by
Scientist : R.A. Millikan Name of electron was given by
Oil drop experiment
CHEMISTRY the scientist Stoney
Atomic Structure
Determine the value of e/m.
ATOMIC STRUCTURE
Determination of mass of electron

By knowing charge (e) and (e/m) ‘m’ was determined

Mass of electron
e 1.6022  10–19C
me = =
e/m 1.758  1011 CKg–1
me = 9.1 10-31 Kg

1
Note : Mass of electron is th
1837
the mass of an H atom.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Explain discovery of electron- study of

Anode rays?

CHEMISTRY
Atomic Structure
Explain discovery of electron- study of Anode rays?
ATOMIC STRUCTURE
Discovery of Anode Rays

Atom Neutral Discovery of (–)ve charge particle

Electron
Positively charged particle ??

??

CHEMISTRY
Atomic Structure
Explain discovery of electron- study of Anode rays?

Discovery of Proton ATOMIC STRUCTURE

ZnS screen is placed behind the cathode, It was found that ZnS started glowing

Perforated Vacuum pump

Cathode (-)
Anode (+)

ZnS screen

CHEMISTRY Battery
Atomic Structure
Explain discovery of electron- study of Anode rays?
ATOMIC STRUCTURE
But the question is are they actually coming from anode ??
Anode rays are not produced by the anode rather they are produced by
the Ions of the gas, which are produced when cathode rays hits the gaseous atom.

cathode Vacuum pump

Perforated
+ e– Anode
Cathode + e–
+ e–
z

ZnS screen

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define Properties of Anode Rays and explain

them.

CHEMISTRY
Atomic Structure
Define Properties of Anode Rays and explain them.
ATOMIC STRUCTURE
Properties of anode rays
 Anode rays travels in a straight line
 Anode rays are made of particle
 Anode rays are made of positive charged particle

 Anode rays are deflected towards negative electrode

 Anode rays are deflected towards magnet
 Anode rays show heating effect

CHEMISTRY
Atomic Structure
Define Properties of Anode Rays and explain them.

Properties of anode rays

ATOMIC STRUCTURE
1) Anode rays travels in a straight line.
Like cathode rays, an object is placed in thepath of anode rays
Vacuum pump
Anode (+)
Cathode (-)

Shadow
Gas at low pressure
(10-2 – 10-3 mm)

CHEMISTRY Battery
Atomic Structure
Define Properties of Anode Rays and explain them.
Properties of anode raysATOMIC STRUCTURE

2) Anode rays are made of particle

Cathode (-) Anode (+)

Gas at low pressure

(10-2 – 10-3 mm)

Battery
CHEMISTRY
Atomic Structure
Define Properties of Anode Rays and explain them.

Properties of anode rays

ATOMIC STRUCTURE
3) Anode rays are made of positive charged particle

Anode (+)
Cathode (-)

CHEMISTRY
Atomic Structure
Define Properties of Anode Rays and explain them.
ATOMIC STRUCTURE
4) Anode rays are deflected towards magnet

Anode rays are deflected

Vacuum pump
towards the magnet
Anode Cathode

Gas at low pressure Glowing glass wall

(10-2 – 10-3 mm)

Battery
CHEMISTRY
Atomic Structure
Define Properties of Anode Rays and explain them.
ATOMIC
5) Anode rays show heating effect STRUCTURE
If a metal foil is placed in It’s temperature increases
the path of anode rays and it becomes hot

Metal foil Anode

Cathode

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Discovery of Proton and Neutron

CHEMISTRY
Atomic Structure
Discovery of Proton and Neutron
ATOMIC STRUCTURE
Conclusion
Anode rays are made of positive charge particles “e”

Anode rays are made up of particles “m”

For different gases different
e/m values were found

 For H2 gas the value of e/m was found to be minimum

 These particles possessed minimum mass and a unit positive charge

 These unit positive charged particles were termed as Protons

CHEMISTRY
Atomic Structure
Discovery of Proton and Neutron
ATOMIC STRUCTURE
For protons the values are
found to be

e/m = 9.58 × 104 C/g

e = + 1.6 × 10-19 C
1.6 × 10-19 C
mass =
9.58 × 104 C/g

= 1.67 × 10-24 g

(nearly same as that of hydrogen atom)

CHEMISTRY
Atomic Structure
Discovery of Proton and Neutron
ATOMIC STRUCTURE
Discovery of Proton
 When the gas taken inside the tube is hydrogen then charge and mass
of the ions generated are eH+ and mH+.
 But for other gases charge and mass were found to be different from
that of hydrogen and these are eion and mion.
It was found that
p and q are some constants
eion = p eH+ and mion = q mH+
and they are integer
H+ is called proton
Charge on proton = + 1.6 10–19 c
Mass of proton = 1.67  10–27 kg. protons are simplest and fundamental
particle of atom just like electron
CHEMISTRY
Atomic Structure
Discovery of Proton and Neutron
ATOMIC STRUCTURE
Discovery of Neutron
eion = p eH+ , mion = q mH+
If all mass contributing particle contribute charge also; q = p
But it was found that p q
There must be another particle that contributes mass not charge
This particular particle was named neutron
Neutron is 1837 times heavier than electron

CHEMISTRY
Atomic Structure
Discovery of Proton and Neutron
ATOMIC STRUCTURE
Discovery of Neutron
Discovered by James Chadwick
9 4 12 1
Be + He C + n
4 2 6 0
neutron
Summarization

Fundamental Absolute Relative

Particles Discovery Charge Charge Mass Symbol
Electron Thomson – 1.6 10–19c –1 9.1 10–31kg e
Proton Goldstein + 1.6 10–19c +1 1.6726 10–27kg p
Neutron Chadwick 0 0 1.6729 10–27kg n
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define Atomic Number, Mass Number, Isotopes , Isobars and

isotones .

CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .

ATOMIC STRUCTURE
Atomic number (Z) and mass number (A)
Z = No. of proton = No. of electron in a neutral atom
A = Sum of proton and neutron Mass number Y
A Element
Z =p = e n = A–P Atomic Number X
A = P+n = A–Z

CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .

ATOMIC STRUCTURE
Isotopes
Elements with same atomic number(Z) but with different mass number (A)
are called isotopes. due to difference in the no of neutron

35 37
e.g. Cl Cl
17 17

Isotopes
35
37
Relative abundance of Cl and Cl are 75% and 25% respectively.
17 17
75 25
Average atomic mass =  35 + 100  37 = 35.5
100

CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .

ATOMIC STRUCTURE
Is there any difference between mass number
and atomic mass ??

 Mass number is a whole number which is sum of protons and neutrons

 Atomic mass is a fractional which is the sum of actual mass of electrons,
protons and neutrons
1
E.g. H Protium(P)
1
2
H Deuterium(D)
1
3
H Tritium (T)  Radioactive
1
CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .

ATOMIC STRUCTURE
Isotopes
E.g. 16 18 Used for investigating
O and O Radioactive
8 8 reaction mechanism
12 14 14
C C and C Radioactive
6 , 6 6

Isobars Used for dating rocks and fossils

Elements with same mass number(A) but with different atomic number (Z)
are called isobars.
E.g. 40 40
Ar and Ca
18 20
CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .

ATOMIC STRUCTURE
Isotones
Elements with same number of neutrons are called isotones.
e.g. 40 37
Ca and Cl
20 17
Number of
= A - Z = 20 20
neutrons
e.g.
23 24
Na and Mg
11 12
Number of
= A - Z = 12 12
neutrons
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Isoelectronic Species

CHEMISTRY
Atomic Structure
Define Isoelectronic Species
ATOMIC STRUCTURE
Isoelectronic Species
Elements or ions which have same number of electrons are called isoelectronic species.

e.g. Species O2 F Ne Na  Mg 2 Al 3

Number of
8 9 10 11 12 13
protons
Number of 10
electrons
Species P3  S2 Cl Ar K Ca2
e.g.
Number of
15 16 17 18 19 20
protons
Number of
18
electrons CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Problems Based on e/m, A & Z

CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .
Problem 1 ATOMIC STRUCTURE
The mass to charge ratio for A+ is 1.97  10–7 kg /C. Find the mass of A.

CHEMISTRY
Atomic Structure
Define Atomic Number, Mass Number, Isotopes ,Isobars and isotones .
Problem 1 ATOMIC STRUCTURE
The mass to charge ratio for A+ is 1.97  10–7 kg /C. Find the mass of A.
Given
m
= 1.97  10–7 kg /C
e
 m = 1.97 10–7  1.6 10–19 kg
 m = 1.97 1.6 10–26 kg
m = 3.152 10–26 kg

CHEMISTRY
Atomic Structure

Problem 2 ATOMIC STRUCTURE

Find the ratio of e/m value for electron to proton.

CHEMISTRY
Atomic Structure

Problem 2 ATOMIC STRUCTURE

Find the ratio of e/m value for electron to proton.

(e/m) p (ep/mp)
 =
(e/m) e (ee/me)
ee = ep
me 1
 =
mp 1836

CHEMISTRY
Atomic Structure

Problem 3 ATOMIC STRUCTURE

Find the ratio of (e/m) value for proton to  – particle ? (He2+)

CHEMISTRY
Atomic Structure

Problem 3 ATOMIC STRUCTURE

Find the ratio of (e/m) value for proton to  – particle ? (He2+)

(e/m) p (ep /mp)

 =
(e/m) (e /m)
(e/m) p ep m
 =
(e/m) e  mp
ep 4mp 2
= =
2ep mp 1

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

80 .
a) Calculate the number of electrons, protons and neutrons in
35 Br
b) The number of electrons, protons and neutrons in a species are
equal to 18, 16 and 16 respectively. Assign with proper Symbol.

CHEMISTRY
Atomic Structure
Problem a) Calculate the number of electrons, protons and neutrons in 80 .
ATOMIC STRUCTURE 35 Br
b) The number of electrons, protons and neutrons in a species are
equal to 18, 16 and 16 respectively. Assign with proper Symbol.

Solution (a) Number of protons = Number of electrons = 35

Number of neutron = 80 – 35 = 45

∵ number of protons  number of electrons; So it is a charged species

Solution (b)
Mass number(A) = neutrons +
Number of protons = 16  Z = 16 32 2– protons = 16+16 = 32
S
Number of neutrons = 16 16
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

How many neutrons and protons are there in the following nuclei ?
13 16 24 56 88
O O Mg Fe Sr
6 8 12 26 38

Atomic number = number of protons

Mass number = neutrons + protons

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

How many neutrons and protons are there in the following nuclei ?
13 16 24 56 88
O O Mg Fe Sr
6 8 12 26 38
Solution Mass number Y
A Element
Atomic Number X
13 16 24 56 88
Species O O Fe Sr
6 8 12 Mg 26 38
Number of
protons 6 8 12 26 38
Number of 7 8 12 30 50
neutrons
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Naturally occurring boron consists of 80% 11B (nuclide mass = 11.01) and 20%
another isotope. To account for the atomic weight 10.81, what must be the
nuclide mass of the isotope.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Naturally occurring boron consists of 80% 11B (nuclide mass = 11.01) and 20%
another isotope. To account for the atomic weight 10.81, what must be the
nuclide mass of the isotope.
Solution Let p be the nuclide mass of the isotope
80 20
100  11.01 + 100  p = 10.81

 80  11.01 + 20p = 1081

1081 – (80  11.01)
 p = = 10.01
20
 Nuclide mass (p)= 10.01

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Complete the following table
Particle Mass no. Atomic no. P n e

N - atom 7 7

Ca - ion 20 20

O - atom 16 8

Bromide ion 35 45 36

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Complete the following table
Particle Mass no. Atomic no. P n e

N - atom 14 7 7 7 7

Ca - ion 40 20 20 20 18

O - atom 16 8 8 8 8

Bromide ion 80 35 35 45 36

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

After the discovery of these particles inside the atom, many models
were proposed based on different set of assumptions. We will take a
journey through the history of atom.

CHEMISTRY
Atomic Structure

ATOMIC
Define Thomson's Model STRUCTURE
and Its Limitations.

CHEMISTRY
Atomic Structure

ATOMIC
Define Thomson's Model STRUCTURE
and Its Limitations.
In 1898 Thomson gave his model of atom known as
Thomson’s Plum Pudding Model of Atom which is also
known as Watermelon model, Raisin pudding model,
Chocochip cookie model .

According to Thomson the atom is considered to be a

large sphere of positive charge with electrons being
embedded on it.

Number of electrons = Number of protons Atom is neutral

CHEMISTRY
Atomic Structure

ATOMIC
Define Thomson's Model STRUCTURE
and Its Limitations.

Limitations of Thomson’s model of atom

• Thomson’s model of atom is entirely theoretical and it has no practical approach.

• This model of atom could not explain Rutherford’s experiment and atomic spectrum
which will be discussed later

• Thomson’s model of atom was challenged by his student Rutherford

CHEMISTRY
Atomic Structure

Define Rutherford’s ATOMIC

Model of an STRUCTURE
Atom and its observations .

Ernest Rutherford
CHEMISTRY
Atomic Structure
Define Rutherford’s Model of an Atom and its observations .
ATOMIC STRUCTURE
Rutherford’s alpha particles Scattering Experiment

CHEMISTRY
Atomic Structure
Define Rutherford’s Model of an Atom and its observations .
ATOMIC STRUCTURE
Rutherford’s alpha particles Scattering Experiment
• Alpha-scattering experiment -the first definitive
experiment to establish the basic structure of atom,
specially the discovery of nucleus containing protons
and neutrons.
• To check the conclusions of Thomson’s model
Rutherford bombarded a thin sheet of Gold by fast
moving alpha particles coming from a radioactive
source and observed their deviations after passing
through the foil.
• Alpha-particles are positively charged helium nuclei
with atomic mass 4 a.m.u.

CHEMISTRY
Atomic Structure
Define Rutherford’s Model of an Atom and its observations .
ATOMIC STRUCTURE

• Most of the α–particles

passed through the foil
non-deflected
• A few a-particles gets
deflected by certain
angles
• A very few a-particles
were deflected through
an angle of 1800

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Why only Gold foil was used Why not any other Element ?

CHEMISTRY
 Atomic Structure
Why only Gold foil was used Why not any other Element ?
ATOMIC STRUCTURE
• Gold is a heavy metal
• Gold is malleable i.e. can change shape easily on
hammering
• When fast moving ∝ particles strikes on gold foil, scattering
can be observed
• easily due to its heavy mass
• Deflection of ∝ particles observed will be very less due to its
light mass

CHEMISTRY
Atomic Structure

What were theATOMIC STRUCTURE

conclusions from Rutherford Model?

CHEMISTRY
Atomic Structure
What were the conclusions from Rutherford Model?
ATOMIC STRUCTURE
• Most of the space inside an atom is empty.
• As some particles deviated and alpha particles being positively charged
helium nuclei, can be deviated by a positive charge, so there must be some
positively charged region in the atom.
• As only a few particles deviated, it was concluded that the positively charged
region was very small as compared to the atom that is the positive charge of
the atom is concentrated in a very small space called “The Nucleus”. The
order of diameter of a nucleus is 10-15 m and that of an atom is about 10-10m.
• The biggest achievement of the model was the discovery of the nucleus.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What were the conclusions from Rutherford Model?

On the bases of these observation and conclusions, Rutherford gave

his model of atom. What was it?

“Like planets revolved around the sun in solar system Similarly e– revolves around
nucleus”

CHEMISTRY
Atomic Structure
What were the conclusions from Rutherford Model?
ATOMIC STRUCTURE
Rutherford’s atomic model
• An atom consists of the tiny positively charged nucleus at
its centre.

• Nucleus contain positively charged particles. Almost

entire mass of an atom and positive charge is
concentrated in the nucleus.

• Electron revolve around the nucleus in circular paths

called as orbits.This model resembles the solar system
thus called as planetary model

• The electron and nucleus are held together by

electrostatic force of attraction.
CHEMISTRY
Atomic Structure

Was Rutherford’s model adopted?

ATOMIC STRUCTURE
Define drawbacks of Rutherford’s Model

CHEMISTRY
Atomic Structure
Define drawbacks of Rutherford’s Model.
ATOMIC STRUCTURE
Drawbacks of Rutherford’s Model

• According to classical theory of electromagnetic waves, (Maxwell’s theory) when a charge is

subjected to a continuous motion around an opposite charge it radiates energy continuously
in form of radiations and thus will lose energy. That’s why the electron in Rutherford’s Model
should lose energy and follow a spiral path and collapse into the nucleus in 10–8s. Although
this does not happen but Rutherford had no explanations for the same.

• It says nothing about the structure of atom , i.e. how the electrons are distributed
around the nucleus and their energies

CHEMISTRY
Atomic Structure

Due to the drawbacks ofATOMIC STRUCTURE

Rutherford’s model, Other Scientists gave forth their
Atomic models.
• Bohr’s Theory

• Quantum Theory

• Electromagnetic Radiations

To go further into the atomic mysteries,

we will have to understand the nature of electromagnetic radiations and
study “Maxwell’s Electromagnetic Wave Theory”.

CHEMISTRY
Atomic Structure

ATOMIC
What are Electro magnetic STRUCTURE
Radiations?

CHEMISTRY
Atomic Structure
What are Electro magnetic Radiations?
ATOMIC STRUCTURE
Electromagnetic Radiation:

Radiations which consist of electric (E) and magnetic

field (B) oscillating perpendicular to each other and
both perpendicular to the direction of propagation.

The direction of propagation is given by :

Cross product of electric and magnetic field

(E) ×(B)

CHEMISTRY
Atomic Structure

What are Characteristics of Transverse

ATOMIC Waves?
STRUCTURE

CHEMISTRY
 Atomic Structure
What are Characteristics of Transverse Waves?
ATOMIC STRUCTURE
• Electromagnetic Radiations are waves which are formed as a result of oscillating
magnetic and electric fields which are perpendicular to each other and both are
perpendicular to direction of motion.
• Light has wave characteristics with maxima and minima as shown in the figure.
• These waves can be produced by a charged particle moving in magnetic field.

CHEMISTRY
Atomic Structure

What are CharacteristicsATOMIC

of TransverseSTRUCTURE
Waves?
1. Wavelength, l : The distance between two consecutive crests or troughs in m, nm, A0 etc.

2. Frequency, v : The number of waves passing through a point in space in 1 second. Its units
are s-1 or Hertz (Hz). λ

3. Speed, c: The speed of electromagnetic radiations in vacuum is found to be c = 3 × 108 m/s.

The relation between λ, ν, c
c = νλ

4. Amplitude: The maximum displacement from the mean position or the axis.

5. Wave Number: The number of waves per unit length. This is reciprocal of wavelength (1/λ)
with unit m-1.
CHEMISTRY
Atomic Structure

Now, Let’s discuss problems;

ATOMIC STRUCTURE
Problem :
Vividh Bharti station of All India Radio, Delhi broadcasts on a frequency of 1368 kHz. What is
the wavelength (λ) of the radiation ?
c = 3 × 108 m/s

CHEMISTRY
Atomic Structure

Now, Let’s discuss problems;

ATOMIC STRUCTURE
Problem:
Vividh Bharti station of All India Radio, Delhi broadcasts on a frequency of 1368 kHz. What is
the wavelength (λ) of the radiation ?
c = 3 × 108 m/s
Solution:
Formula ν × λ = C
c/ ν =λ 3 × 108m ≅ 2.2 × 102 m
1368 × 103

λ = 220 m

CHEMISTRY
Atomic Structure

Now, Let’s discuss problems;

ATOMIC STRUCTURE
Problem :

Calculate wave number of a radiation having wavelength 5800 Å .

CHEMISTRY
Atomic Structure

Now, Let’s discuss problems;

ATOMIC STRUCTURE
Problem
5800 Å
Calculate wave number of a radiation having wavelength 5800 Å .
Solution
We already know 1 Å=10–10m

ν = 1/λ
ν =1/5800 × 10–10 = 1/58 × 10–8
ν = 1.7 × 106 m–1

CHEMISTRY
Atomic Structure

ATOMIC
So What is a Electromagnetic STRUCTURE
Spectrum?

CHEMISTRY
Atomic Structure

ATOMIC
So What is a Electromagnetic STRUCTURE
Spectrum?

When all the electromagnetic radiations are arranged in increasing order of wavelength or
decreasing frequency the band of radiations obtained is termed as electromagnetic
spectrum.
The wavelength increase in the order
Gamma rays < X – rays < UV rays < Visible < Infrared < Microwaves < Radio waves
< Long RadioWaves

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Now , We will discuss Quantum Theory

What is Planck’s Quantum Theory?

CHEMISTRY
Atomic Structure
What is Planck’s Quantum Theory?
ATOMIC STRUCTURE
Planck’s Quantum Theory

 Energy is emitted in form of radiations from a source in a discontinuous manner that is in form
of packets of energy.
 And energy of each packet depends on the frequency of radiation. These packets are called
“Quantum of energy”.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What are the properties of Quantum of Energy?

CHEMISTRY
Atomic Structure
What are the properties of Quantum of Energy?
ATOMIC STRUCTURE
Properties of Quantum of Energy

 At a time, only one quantum can be supplied to one electron or any other particle.
 One Quantum cannot be divided or distributed.
 An electron can absorb one packet at a time and get excited. And the energy of a
Quantum increases with frequency.

CHEMISTRY
Atomic Structure
What are the properties of Quantum of Energy?
ATOMIC STRUCTURE
Properties of Quantum of Energy

 Energy of Quantum Frequency

E∝ν
 The constant of proportionality is called Planck’s constant represented by “h”
=> E = h ν
 E is in J and the units of h are Js (Joule-seconds).
h = 6.626 × 10-34 Js.
 If it contains “n” Quanta of the same frequency then the total energy will become:
E = nhν
CHEMISTRY
Atomic Structure
What are the properties of Quantum of Energy?
ATOMIC STRUCTURE
Properties of Quantum of Energy
 This is also called “Quantisation of Energy”. This can also be expressed in
terms of wavelength:
E = nhc/λ
 We can also conclude that energy transferred by an electromagnetic radiation
is an integral multiple of hν
Note: Quantum of light is called “Photon”.

 We can also conclude that energy transferred by an electromagnetic radiation

is an integral multiple of hν.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Now Let’s solve some problems

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Solution

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

How can we explain Blackbody radiations and

Photoelectric Effect on the basis of this theory?

CHEMISTRY
 Atomic Structure
How can we explain Blackbody radiations and PE Effect on the basis of this theory?
ATOMIC STRUCTURE
Blackbody radiations

 Temperature increases => Energy emitted increases => Increase in the frequency of
the emitted radiations

 Frequency increases => Wavelength shifts to lower values.

 Basically frequency affects the energy of one quantum or photon and due to
increase in number of Quanta emitted the intensity increases.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Explanation of Photoelectric Effect

CHEMISTRY
Atomic Structure

ATOMIC
Explanation of Photoelectric Effect STRUCTURE

As one electron absorbs one quantum that is

why if the energy of one quantum is insufficient
for the ejection of electrons then there is no use
of increasing the number by increasing the
intensity.

To make ejection possible the energy per Quantum should be increased by increasing
frequency. And for every metal there is a minimum value of frequency at which ejection
starts which is called “Threshold Frequency” represented by νo

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What are the observations?

CHEMISTRY
Atomic Structure

WhatATOMIC STRUCTURE
are the observations?

 When ν < ν0 then energy of one quantum is insufficient, ejection does not take place

 When ν = ν0 then ejection just takes place and electrons don’t have any kinetic energy

 When ν > ν0 then every ejected electron possesses kinetic energy that is the excess energy of a
quantum gets converted into kinetic energy of the electron.

 When ν > ν0, then on increasing the intensity the number of quanta incident increases thereby
increasing the number of photoelectrons ejected.

The Threshold energy required for emission is called “Work Function” (f) that is “hνo”.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
How can we explain Photoelectric effect mathematically?

CHEMISTRY
 Atomic Structure
How can we explain Photoelectric effect mathematically?
ATOMIC STRUCTURE
By energy balance:

Energy of one quantum = Threshold Energy + Kinetic Energy

hν = hν0 + (1/2)mev2

 The Kinetic energy of electron varies linearly with the incident frequency .
 Einstein gave this equation of photoelectric effect and he concluded that Light has particle like
properties as well.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What was the conclusion from this theory?

CHEMISTRY
Atomic Structure
What was the conclusion from this theory?
ATOMIC STRUCTURE

We can conclude that according to Maxwell, Light is a wave but according to modern
theory it may also have particle like properties, as it contains photons each with
discrete amount of energy.
Therefore, Light has both the properties: Wave and Particle.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

What ATOMIC
are the postulates of Bohr’s Atomic Model?
STRUCTURE

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model
(1)
 Electron revolves around the nucleus in circular
orbits called as shells or energy levels.
 These are numbered as 1, 2, 3, 4….. etc., or
designated as K, L, M, N…
 These orbits are associated with definite energies.

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model

2. Electron revolves around the nucleus due to the

centripetal force provided by the electrostatic
force of attraction (kq1q2/r2).

This centripetal force is balanced by

centrifugal force of circular motion (mv2/r).

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model
3. Energy of the shells is fixed for a given atom and these shells are called stationary energy
level i.e., an atom will not lose or gain energy when it is present in a shell.

4. When electrons gain energy it gets excited to higher energy level and when it gets de-excited
it loses energy in the form of electromagnetic radiation and comes down to lower energy level.

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model

5. Only those orbits are permitted in which the angular momentum of the electron is a whole
number multiple of h/2π.
Like energy, angular momentum of the electrons in an atom is also quantized.

Angular momentum = moment of inertia (I) × angular velocity (w) .

Since I = mer2 and w = v/r where v is the linear velocity

∴ Angular momentum = mer2 × v/r = mevr

mvr = nh/2π

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model

6) As we move away from the nucleus, energy of the shell/orbit increases.

Energy of K < L < M < N

i.e. if the energy of the 1st orbit is E1 and that of 2nd orbit is E2, then => E2 > E1

So the difference in energies can be calculated as : E2 – E1 = hν

CHEMISTRY
 Atomic Structure
What are the postulates of Bohr’s Atomic Model?
ATOMIC STRUCTURE
Postulates of Bohr’s Atomic Model

7. When energy from external source is supplied to the electron it

jumps to higher energy level.
When this electron jumps back to the lower level it radiates
same amount of energy in the form
of photons.
E2 – E1 = hν
(E2 – E1)/ h = ν

The energy required is equal to the difference in energies of the two

levels

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Study of Bohr’s Model Mathematically

Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure
Define Study of Bohr’s Model Mathematically
ATOMIC STRUCTURE
Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Study of Bohr’s Model Mathematically

Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure
Define Study of Bohr’s Model Mathematically
ATOMIC STRUCTURE
Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure
Define Study of Bohr’s Model Mathematically
ATOMIC STRUCTURE
Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure
Define Study of Bohr’s Model Mathematically
ATOMIC STRUCTURE
Study of Bohr’s Model Mathematically

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Study of Bohr’s Model Mathematically

Bohr Radius, Velocity and energy Value of H-atom

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Ionization Energy.

CHEMISTRY
Atomic Structure

Define Ionization Energy. ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Concept of Induction radiation .

CHEMISTRY
Atomic Structure
Define Concept of Induction radiation .
ATOMIC STRUCTURE
Concept of Induced Radiation

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

CHEMISTRY
Atomic Structure
What were the limitations of Bohr’s Model?
ATOMIC STRUCTURE
Limitation of Bohr’s Model of H-atom

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What were the limitations of Bohr’s Model?

Limitations/ Demerits in Bohr’s Model

 Bohr’s model is combined concepts of classical physics and quantum physics
 He could not explain the spectra of multi-electronic species
 No justification was given for the quantisation of angular momentum,
although this was a correct assumption
 Bohr could not explain formation of chemical bond

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
What were the limitations of Bohr’s Model?

Limitations/ Demerits in Bohr’s Model

 Bohr could not explain splitting of spectral lines in presence of electrical field
 (Stark effect) and magnetic field (Zeeman effect)

 Similarly fine spectrum can be explained by magnetic quantum no.

and doublets in hydrogen spectra by spin quantum no.
In spectroscopy fine structure is the splitting of the main spectral lines of an atom into two
or more components, each representing a slightly different wavelength

these can be explained by

azimuthal/ angular quantum no.

CHEMISTRY
Atomic Structure

Define de Broglie’s Matter-Wave

ATOMIC STRUCTURE Duality Hypothesis.

h
λdb = p
p= Linear momentum = mv

CHEMISTRY
Atomic Structure
Define de Broglie’s Matter-Wave Duality Hypothesis.
ATOMIC STRUCTURE
de Broglie’s Matter-Wave Duality Hypothesis
de Broglie found wavelength of a particle is associated with the linear momentum of that
particle.
h p = Linear momentum = mv
λdb =
p
According to Planck light has both Similarly matter
wave and particle nature can also have wave-particle duality

de Broglie’s hypothesis of wave-

particle If we consider rotation of a fan we can not
duality was found to be useful here determine
the exact position of the blades
CHEMISTRY
Atomic Structure
Define de Broglie’s Matter-Wave Duality Hypothesis.
ATOMIC STRUCTURE
Let’s consider the case of an electron me = 9.1  10–31 kg
h 6.627  10–34 J.Sec ve = 2.165  106 m/s
λdb = mv =
9.1  10–31 kg  2.165  106 m/s
= 6.627  10–10 m
λdb = 6.627 A0
4
Radius of 2nd Bohr’s orbit = 0.529  2.1 A
1 =

It means radius of 2nd Bohr’s orbit is smaller

How is it possible ??
than the wavelength of the electron

Let’s discuss an another case……

CHEMISTRY
Atomic Structure
Define de Broglie’s Matter-Wave Duality Hypothesis.
ATOMIC STRUCTURE
Let’s take the case of a cricket ball hit bay a batsman
Typical mass of a cricket ball is
h 6.627  10–34 J.Sec 200g = 0.2 kg
λdb = mv =
0.2 kg  50 m/s

λdb = 6.627  10–35 m And if the velocity of the ball is

180 km/hr = 50 m/s

Typical radius of a cricket ball is 4 cm

Late Davisson and Germer did electron diffraction experiment on Ni. Crystal and
they proved that electron can have wave nature
It means de Broglie wavelength for that ball For that reason we cannot see
is negligible to the radius ofCHEMISTRY
the ball the wavelength of the ball
Atomic Structure

ATOMIC STRUCTURE
Define DeBroglie's Model and What are applications of
DeBroglie's Model ?

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define DeBroglie's Model and What are applications of DeBroglie's Model ?

de Broglie's model of Atom de Broglie thought the same way

L He took a string confined between two ends

and did the following experiment
L = 2 nodes

L =   2 The energy which is given to the string by vibration

2
is confined within this region

L =   3
2
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define DeBroglie's Model and What are applications of DeBroglie's Model ?

de Broglie's model of Atom

the energy of the electron is confined
n = 1 the an orbit like a standing wave

One wavelength

n = 2

Two wavelengths
Likewise electrons rotate in an orbit like wave
n = 3 and the wave ends where it begins

Three wavelengths
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define DeBroglie's Model and What are applications of DeBroglie's Model ?

If the wavelength does not fit into the circumference,

the electron interferes destructively;
Bohr’s orbit
n=3 it cannot exist in such an orbit

2πr = nλ 2πr′ ≠ nλ′

n = integer n = integer
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define DeBroglie's Model and What are applications of DeBroglie's Model ?
de Broglie wavelength,
λdb = h
P
de Broglie considered p = √ 2meV0
h electron as wave
λdb =
√ 2meV0

de Broglie's model of Atom

 Bohr introduced the concept of quantisation of angular momentum but
justification of quantisation was given by de Broglie

 In a guitar the strings are confined between points guitar. When guitarist plays
guitar he holds the string at some point and the strings start vibrating
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

An electron is accelerated through a potential difference of Vo volts. Find the
de Broglie wavelengths associated with the electron.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

An electron is accelerated through a potential difference of Vo volts. Find the
de Broglie wavelengths associated with the electron.

Solution
here we are accelerating an electron
We know p = mv
so here the charge is ‘e’

Now if we accelerate the electron The energy attained by the electron

through V0 potential difference in the form of K.E.

CHEMISTRY
Atomic Structure
Problem ATOMIC STRUCTURE
An electron is accelerated through a potential difference of Vo volts. Find the
de Broglie wavelengths associated with the electron.

Solution
Energy gained by an e– on being accelerated through a potential difference of V
volts
eV0 = 1 mv
2
2
2 2
⇒ eV0 = m 2m
v

p2
⇒ eV0 =
2m
⇒ p=
√ 2meV0
CHEMISTRY
Atomic Structure

Define HeisenBerg’s
ATOMIC STRUCTURE
Uncertainty Principle.

“Particle Waves”

CHEMISTRY
Atomic Structure
Heisenberg’s Uncertainty Principle
ATOMIC STRUCTURE
“It is impossible to determine the position and momentum of an electron
simultaneously and accurately.”

 Viewing electrons and other particles as “Particle Waves” which are highly
de-localised changes the way we see universe.
 Instead of things having exact location and motion they are distributed in some region in space.
 Heisenberg proposed that locating an electron and determining its velocity,
both cannot be done simultaneously and accurately because of the wave characteristic of electron.

CHEMISTRY
Atomic Structure

Calculate the uncertainty inATOMIC STRUCTURE

position assuming uncertainty in momentum within
0.1% for:
(a) a tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s.

(b) a electron moving in an atom with a velocity of 2 × 106 m/s.

CHEMISTRY
Atomic Structure
Calculate the uncertainty in position assuming uncertainty in momentum within
ATOMIC STRUCTURE
0.1% for: (a) a tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s.

(b) a electron moving in an atom with a velocity of 2 × 106 m/s.

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Problems on Heisenberg Uncertainty Principle

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

A cricketer hits a ball of mass 0.2 kg which moves with a velocity of 100 m/s.
Error in velocity measurement is 1%. Then determine the uncertainty in
measurement of position of the ball.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

A cricketer hits a ball of mass 0.2 kg which moves with a velocity of 100 m/s.
Error in velocity measurement is 1%. Then determine the uncertainty in
measurement of position of the ball.
This error is less negligible
Solution
in comparison to λ of the cricket ball
⇒ Δ v  100 = 1 ⇒ Δ v = 1 m/s
100
h Which means Heisenberg uncertainty principle
Δv.Δx ≥
4πm has no significance in the macroscopic world
h 6.626  10–34
⇒ Δx ≥ =
4πm Δv 4  3.14  0.2  1
Δv
Δ x = 2.49  10–34 m ≈ 10–34 m v  100 = 1 %
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

A microscope using suitable photons is employed to locate an electron
in an atom within a distance of 0.1 A0. What is the uncertainty involves
in the measurement of its velocity.

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

A microscope using suitable photons is employed to locate an electron
in an atom within a distance of 0.1 A0. What is the uncertainty involves
in the measurement of its velocity.
Solution Δ x = 0.1 A ° = 0.1  10 –10m = 10 –11 m
h
Δv.Δx ≥
4πm

Δv ≥
h = 6.626  10–34
⇒
4πm Δx 4  3.14  10–30  10–11
6.626
12.28 
= 107 m/s

∴Uncertainty in velocity = 0.5  107 m/s

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the uncertainty in position assuming uncertainty in momentum
within 0.1% for

(a) A tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s
(b) An electron moving in an atom with a velocity of 2 106

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the uncertainty in position assuming uncertainty in momentum
within 0.1% for
(a) A tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s
(b) An electron moving in an atom with a velocity of 2 106
Solution (a)
Δv
⇒ 10  100 = 0.1 ⇒ Δ v = 0.01 m/s
h
Δv.Δx ≥
4πm

Δx ≥
h
= 6.626  10–34
⇒ 4πm Δv 4  3.14  0.2  0.01
∴Uncertainty in position = 2.5  10-32 m
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

Calculate the uncertainty in position assuming uncertainty in momentum
within 0.1% for
(a) A tennis ball weighing 0.2 kg and moving with a velocity of 10 m/s
(b) An electron moving in an atom with a velocity of 2 106
Δv
Solution (b)
 100 = 0.1 %
v
Δv
⇒ 6  100 = 0.1 ⇒ Δ v = 2 103 m/s
2 10
h
Δv.Δx ≥
4πm

⇒ Δx ≥
h 6.626  10–34
=
4πm Δv 4  3.14  10–30  2 103
Δ x = 250 A
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define modern quantum mechanical model of atom.

CHEMISTRY
Atomic Structure
Define modern quantum mechanical model of atom.
ATOMIC STRUCTURE
Modern Quantum Mechanical Model of Atom
P (x,y)
Schrodinger : Wave equation Treat electron as

r sin θ
r
⇓ wave. θ
What is ψ O
To Analyse electron r cos θ
ψ (psi) :- It’s a representation of electron as a wave Representation of
ψ (x,y,z) Spherical coordinate ψ (r,θ, φ) a point in terms
of r and θ
ψ :- Gives Full (probable)information about the
electron like: For x = cos θ
y = sin θ
 Energy of electron
 Angular momentum of electron
 Position of electronCHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define Orbit and difference between Orbit and Orbital.

CHEMISTRY
Atomic Structure
Define Orbit and difference between Orbit and Orbital.
ATOMIC STRUCTURE
Orbital
 An orbital is region around the nucleus where maximum probability of
finding the electron. If the electron density is
Higher in some region, we say
 Probability should be 95% or 0.95 Probability of finding e– will be higher
By solving Schrodinger equation we get :-
Quantum number
n – Principal quantum numbers
l – Azimuthal / Angular quantum number.
ml – Magnetic quantum number.

Probability density :
|ψ|2 orbital can be related like electron
CHEMISTRY
Atomic Structure
Define Orbit and difference between Orbit and Orbital.
ATOMIC STRUCTURE
What is the difference between Orbit and Orbital
Orbit Orbital
 Circular path around the  It is the 3D region around the nucleus in
nucleus in which the Electron which e– is present as wave.
moves.
 Characterise by the orbital  It characterise by ‘n’ ‘l’ ‘m’
No ‘n’
 No real meaning  Represents probability of
finding electron (|ψ|2)

Schrodinger wave equation :- Hψ = Eψ

H :- Hamiltonian operator
E :- Energy of e– (KE + PE)
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define Quantum Numbers.

CHEMISTRY
Atomic Structure

Define Quantum Numbers. ATOMIC STRUCTURE

What are Quantum Numbers?

A number which occurs in the theoretical expression for the value of

some quantized property of a subatomic particle, atom, or molecule
and can only have certain integral or half-integral values.
 Location of electron

 Energy of electron

 Magnetic nature of electron

 Momentum of electron

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Define types of Quantum Numbers.

CHEMISTRY
Atomic Structure

Define types of Quantum Numbers.

ATOMIC STRUCTURE
1. Principle Quantum Number (n)

 It Represent the distance from the nucleus and also called as

Shell of electron.
n = 1 K
2 L
3 M
4 N
 If higher the value of ‘n’ ,higher will be the
energy of electron, lower will be its stability
and it will be more diffused out in the space.
 It represent Total energy of electron.
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.

2. Azimuthal/ Angular momentum quantum (l)

 As we know, e is moving around the nucleus so it must have K.E. associated
with it.
 As it is in motion and have attractive force with nucleus it must have angular
 momentum,
So we say, Azimuthal quantum no represents energy of e associated with motion

 It represent the energy of motion of electron and also called as

Sub shell of electron.
l - 0,1,2…( n – 1)
if n = 1 l=0
n = 2 l = 0,1
n = 3 l = 0,1,2
n = 4 l = 0,1,2,3 CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.

Shape of the orbital

l = 0 1 2 3 4
s – Spherical
Designation = s p d f g
p – Dumb-bell
d – Double dumb-bell
Value of ‘n’ will be equal f – Complex
to total possible value of l

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.

3. Magnetic Quantum Number (ml)

 Let’s say if we consider electron as a particle and the electron is rotating,
It is forming a current conducting loop as the electron is charged particle,
Which implies that a magnetic field is associated with it.

 We assign this magnetic field with the magnetic

quantum number

 We are considering the movement of electron

in all possible directions of the magnetic field

 As the magnetic quantum number (m) is dependent

on angular momentum (l)
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.
3. Magnetic Quantum Number (ml)
 We calculate all the possible magnetic field associated to electron
 Total values = 2l + 1
ml = – l to +l
Sub shell (l) ml
s(0) 0
p(1) –1, 0, +1
d(2) –2,–1, 0, +1, +2
f(3) –3, –2,–1, 0, +1, +2, +3

 It will show possible orientation of orbitals in

a sub shell CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.

3. Magnetic Quantum Number (ml)

 Total values = 2l + 1
ml = – l to +l
Sub shell (l) ml
s(0) 0
p(1) –1, 0, +1
d(2) –2,–1, 0, +1, +2
f(3) –3, –2,–1, 0, +1, +2, +3
 It will show possible orientation of orbitals in
a sub shell
CHEMISTRY
Atomic Structure

Define Quantum Numbers. ATOMIC STRUCTURE

4. Spin Quantum number (ms)

 Electrons motion about its axis  Hypothetical quantum concept

 Can’t be related with classical spinning motion

of a particle about its own axis
1
s = + clockwise spin/ up-spin
2

1
s = – anticlockwise spin/ down-spin
2

CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE
Define types of Quantum Numbers.

4. Spin Quantum number (ms)

Representation of blocks:
Example :-
s
n=4 l=1 4p - subshell
ml = 0 4pz - Orbital p

1 d
If ms = + 
2
or
1 
ms = –
2
An orbital can have a maximum of two electrons
CHEMISTRY
Atomic Structure

ATOMIC
Define types of Quantum Numbers. STRUCTURE
Capacity of Subshell Orbital
s 2 1
p 6 3
d 10 5
f 14 7
Q. No. Range
n 1, 2, 3, …
l 0 to n–1 n values
ml –l to +l 2l + 1 values (No. of orbitals)
ms –1/2 to +1/2 2 values
l=0 s - subshell S - orbital
l=1 p - subshell Px, Py & Pz - orbital
CHEMISTRY
Atomic Structure

What is Orientation ofATOMIC

subshells STRUCTURE

CHEMISTRY
Atomic Structure

What is Orientation ofATOMIC

subshells STRUCTURE
For
1. s – spherical (ml = 0)
 Only one orientation is possible

y
z

Spherical Subshell

CHEMISTRY
Atomic Structure

What is Orientation ofATOMIC

subshells STRUCTURE
z
y
2. p – dumb-bell (ml = +1, 0, -1) x
 It can be represented Px
in x, y, z axis
y Elipsoid
z

x y
Pz z

Py x

CHEMISTRY
Atomic Structure

ATOMIC
What is Orientation of subshells 3. d –double dumbell (ml = +2, +1, 0, -1, -2)
STRUCTURE
y z  0 is conventionally
y
explained as dz2
 If x, y directions are
considered
x z x

dxy dyz dxz

ml = +1 y ml = –1 z ml = –2

x y
dx2 – dy2 dz2
ml = +2 ml = 0
CHEMISTRY
Atomic Structure

ATOMIC STRUCTURE

Problems on QUANTUM NUMBERS

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2

Solution
a) 3p :- n = 3 (3, 1, 1, +1/2) or (3, 1, 1, –1/2)
l=1
( 3, 1, 0, +1/2) or ( 3, 1, 0, –1/2)
ml = -1, o, +1
1 1 (3, 1, –1, +1/2) or (3, 1, –1, –1/2)
ms = + or –
2 2
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2
Solution
l ml ml =(2l + 1) = No. of orbital
b) 5s 0 1 = 1 + 3 +5 + 7 + 9 = 25
5p 1 3 No. of orbital in nth shell = n2
5d 2 5
5f 3 7 No. of electrons in a shell = 2n2
5g 4 9
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2, l = 1 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) Give the set of quantum numbers that describe an electron in a 3p orbital
b) How many orbitals are there in a shell with n = 5
c) Designate the orbitals using quantum numbers s, p, d, f for the following
combination
i) n = 2, l = 1 ii) n = 4, l = 0 iii) n = 5, l = 3 iv) n = 3, l = 2
Solution
c) i) n = 2 l=1 iii) n = 5 l = 3
 2p  5f

ii) n = 4 l = 0 ii) n = 3 l = 2
 4s  3d
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) How many valid sets of quantum numbers are there for 4d orbitals.
b) Which shell would be the first shell to have a ‘g’ sub-shell

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

a) How many valid sets of quantum numbers are there for 4d orbitals.
b) Which shell would be the first shell to have a ‘g’ sub-shell
Solution
a) 10 values (4, 2, –2, –1/2), (4, 2, –2, +1/2)
n=4&l=2 b) g l=4
ml = –2 (4, 2, –1, –1/2), (4, 2, –1, +1/2)
= –1
=0
= +1
= +2
1 1
ms = + or –
2 2
CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

What is the maximum no. of electron which can be accommodated
a) In a shell with n = 4
b) In the 4f sub shell
c) In a 4f orbital

CHEMISTRY
Atomic Structure

Problem ATOMIC STRUCTURE

What is the maximum no. of electron which can be accommodated
a) In a shell with n = 4
b) In the 4f sub shell
c) In a 4f orbital
Solution

a) 2  42 = 32
b) 14

c) 2

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Atomic Structure

ATOMIC STRUCTURE
Define AufBau’s Principle.

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Atomic Structure

Define AufBau’s Principle. ATOMIC STRUCTURE

Aufbau’s Principle

 Sub shell :- For a given shell the energy of different sub shell increases as the
value of ‘l’ increases.
ns  np  nd  nf (for multi electronic species)

 In the absence of external field (electric/magnetic),

energies of orbitals in the sub shell are equal.
 Degenerate atomic orbitals
Px = Py = Pz in a p sub shell

 For a hydrogen like species :-

1s  2s = 2p  3s = 3p = 3d  4s = 4p = 4d =4f
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Atomic Structure

Define AufBau’s Principle.

ATOMIC
AufBau principle STRUCTURE
: (Building – up)
1s = 1+0= 1
1s 2s = 2+0= 2
1s will be filled first

2s 2p 2p = 2+1= 3
Values are same
3s = 3+0= 3
3s 3p 3d 3p = 3+1= 4
3d = 3+2= 5 4s will be filled
4s 4p 4d 4f before 3d
4s = 4+0= 4
5s 5p 5d 5f 5g 4p = 4+1= 5
4d = 4+2= 6 5s will be filled
6s 6p 6d 6f 4f = 4+3= 7 after 4p and then 4d
7s 7p 5s = 5+0= 5 After 4d, 5p will
5p = 5+1= 6 be filled
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Atomic Structure

AufBauATOMIC STRUCTURE
Define AufBau’s Principle.
principle : (Building – up)
How the configuration
of an atom is built
1s
 We fill the electron
2s 2p on the basis of (n + l)
3s 3p 3d  Lower the value of ‘n’
electrons are filled first
4s 4p 4d 4f
 Values of n + l is less, energy
5s 5p 5d 5f 5g is lower, will fill e– first

6s 6p Shell which is close to nucleus will

6d 6f
have lower energy and the shell which
7s 7p is far from nucleus will have high energy
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Atomic Structure

Define AufBau’s Principle.

ATOMIC STRUCTURE
Auf – Bau principle : (Building – up)

1s 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2

4d10 5p6 6s2
2s 2p
4f14 5d10 6p6
3s 3p 3d 7s2 5f14 6d10
4s 4p 7p6
4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
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Atomic Structure

Define AufBau’s Principle.

ATOMIC STRUCTURE
AufBau principle : (Building – up)

 According to this principle electron are added progressively to the various orbitals
is their order of increasing energy starting with the lowest energy.

 As a working rule a new electron enters in empty

orbital for which the value of (n + l) is minimum.

 If the value of (n + l) is same for two or more

orbital, the new electrons enters in the orbital
with lower ‘n’ value.

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Atomic Structure

ATOMIC STRUCTURE
Important Points

 Orbital angular momentum

h
=  l (l + 1)
2

 Spin angular momentum

h
=  s (s + 1) s : Total unpaired spin
2

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Atomic Structure

ATOMIC STRUCTURE
Define Pauli’s Rule.

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Atomic Structure

Define Pauli’s Rule. ATOMIC STRUCTURE

Pauli's Exclusion Rule
 No two electrons in atom can have the set of all four quantum numbers.

 Let’s take one orbital 2px and filled with 2 electrons.

1
A) n = 2, l = 1, ml = + 1, ms = + 2

1
B) n = 2, l = 1, ml = + 1 ms = – 2

ml – can be any one value -1, 0, +1

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Atomic Structure

ATOMIC STRUCTURE
Define Hund’s Rule of maximum multiplicity.

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Atomic Structure

Define Pauli’s Rule. ATOMIC STRUCTURE

Hund’s Rule of Maximum Multiplicity
 In a sub shell, electrons never pair until no available empty degenerated orbitals
are left to them.
or
 Electron do not start to pair up in a sub shell until
all orbitals are filled at least once.

 Also, all the singly occupied orbitals will have

electron with parallel spin.

 Electrons do not pair until all the orbitals are filled at least once, so that
electrons stay as far way from each other as possible and minimise inter
electronic repulsion.
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Atomic Structure

ATOMIC STRUCTURE
Define Hund’s Rule of maximum multiplicity.

Hund’s Rule of Maximum Multiplicity

 Further, the electrons is singly occupied orbitals have parallel spins, as this
configuration has lesser inter electronic repulsion than when the electrons
are present with anti - parallel spins
An electron from 4s orbital is
For Eg: promoted into 3d orbital.
2 4
24Cr [Ar] 4s 3d
[ Expected]

[Actual]
 Half Filled Stability
It is feasible, because difference in
 Exchange energy energy level of 3d and 4s is small
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Atomic Structure

ATOMIC STRUCTURE
Define Hund’s Rule of maximum multiplicity.

Hund’s Rule of Maximum Multiplicity

Exchange energy More the exchange more amount
of energy released that
provide more stability to atom
Total exchange
d4
= 6 (4C2)

With every exchange

Total exchange Small amount of
d5
= 10 (5C2) energy is released

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Atomic Structure

ATOMIC STRUCTURE
Define Hund’s Rule of maximum multiplicity.

Hund’s Rule of Maximum Multiplicity n(n – 1)

No. of
=
exchange 2
24Cr [Ar] 4s1 3d5
no. of e– with
n =
same spin

25Mn [Ar] 4s2 3d5

26Fe [Ar] 4s2 3d6

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Atomic Structure

Define Hund’s Rule of maximum multiplicity.

ATOMIC STRUCTURE
Hund’s Rule of Maximum Multiplicity We always exchange same spin
27Co [Ar] 4s2 3d7 we don’t exchange opposite spins.

28Ni [Ar] 4s2 3d8

29Cu [Ar] 4s1 3d10

30Zn [Ar] 4s2 3d10

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Atomic Structure

ATOMIC STRUCTURE
Define Configuration of Ions.
.

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Atomic Structure

ATOMIC STRUCTURE
O is gaining two electrons
Configuration of Ions
those will add up in outermost p-orbital
 Two types of ions
1) Cation :- By the loss of e in atom
2) Anion :- By the gain of e in atom
 Both loss / gain will take place in valence shell.

O2– :- 8O = 1s2 2s2 2p4

O2– = 1s2 2s2 2p6 [Ne]

Electron will be removed

Na+ :- 11Na = 1s2 2s2 2p6 3s1
Na+ = 1s2 2s2 2p6 [Ne]
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Atomic Structure

Problem ATOMIC STRUCTURE

Mark the configuration of S–2, Al+3, Cr+3, Sc+, Sc+3, Fe+2, Fe+3, Ni+2, N–3, P–3

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Atomic Structure

Problem ATOMIC STRUCTURE

Mark the configuration of S–2, Al+3, Cr+3, Sc+, Sc+3, Fe+2, Fe+3, Ni+2, N–3, P–3
Solution Valence shell is 4s so first,
electrons will be removed
S = 1s2 2s2 2P6 3s2 3p4
from 4th shell
S2– = 1s2 2s2 2P6 3s2 3p6 OR [Ar]

Al3+ = [Ne] Fe2+ = [Ar] 3d6

Cr = [Ar] 4s1 3d5 Fe3+ = [Ar] 3d5
Cr+3 = [Ar] 4s0 3d3 Ni2+ = [Ar] 3d8
Sc+ = [Ar] 4s1 3d1 N3– = [Ne]
Sc3+ = [Ar] P3– = [Ar]
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Atomic Structure

ATOMIC STRUCTURE
Define Magnetic moment of the ions.

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Atomic Structure

Magnetic moment of the ions

ATOMIC STRUCTURE
If an ion is placed in presence of external magnetic field,
it will get attracted or repelled by the field.

μB = √ n (n+2) B.M n = No. of unpaired electrons

No. of unpaired Magnetic moment B.M = Bohr Magneton
electrons (n) (μB)
Species with μB = 0,
0 0 Diamagnetic are called as diamagnetic
1 √3
2 √8 Paramagnetic Species with μB ≠ 0,
are called as paramagnetic
3 √15
Paramagnetic substances will get attracted to the external magnetic field and
Diamagnetic substances will getCHEMISTRY
repelled from external magnetic field
Atomic Structure

ATOMIC STRUCTURE
Problem
What is the magnetic moment for Fe2+, Fe3+ and Ni2+ ?

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Atomic Structure

Problem ATOMIC STRUCTURE

What is the magnetic moment for Fe2+, Fe3+ and Ni2+ ?
Solution
For Fe2+ ⇒ [Ar] 3d6 ⇒

μB = √ 4 (4+2) = √ 24 ∼ 5 BM

For Fe3+ ⇒ [Ar] 3d5 ⇒

μB = √ 5 (5+2) = √ 35 ∼ 6 BM

For Ni2+ ⇒ [Ar] 3d8 ⇒

μB = √ 2 (2+2) = √ 8 ∼ 2.83 BM

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Atomic Structure

ATOMIC Curves
Define Radial Probability STRUCTURE
and probability of
finding a electron.
Let’s see where the electrons
are present inside the atom

A single wave function (ψ )

describes the entire system

electrons are present within orbitals

We will get the probability of finding

the electrons within certain region of space
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Atomic Structure

ATOMIC STRUCTURE
Define Radial Probability Curves and probability of finding a electron.

Analysis of ψ : Orbital picture and probability distribution curves

ψ(x,y,z) can be determined in terms of Polar coordinate (r,θ ,φ )

ψ(r,θ ,φ ) can be further written in terms of ψ(r). ψ(θ ,φ)

ψ(r,θ ,φ ) = ψ(r). ψ(θ ,φ)

Where ψ(r) is the radial component & ψ(θ ,φ) is the angular component

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Atomic Structure

ATOMIC STRUCTURE
Define Radial Probability Curves and probability of finding a electron.

Analysis of ψ : Orbital picture and probability distribution curves

How ψ changes with
distance from the nucleus

ψ2s (r) ψ2 : gives probability

ψ1s(r) density of electron
+ NODE
+
r – r
NODE : Point where
|ψ1s (r)|2 |ψ2s (r)2|
probability of finding e– is Zero
+ + +
r r
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Atomic Structure

Define Radial Probability Curves and probability of finding a electron.

ATOMIC STRUCTURE
Probability of finding electron
If we consider a spherical shell inside the
sphere of radius ‘r’ and thickness ‘dr’

If we multiply this the probability density

with volume of the region
we will get probability of finding the electron

now we know the probability density

P(r) = radial probability

distribution function
P(r) =|ψ|2 4π r2
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Atomic Structure

ATOMIC STRUCTURE
Define Radial Probability Curves and probability of finding a electron.

Probability of finding electron

dr
r
dv = (4π r2) dr
Probability of finding an electron in a small shell of thickness ‘dr’ is defined as
dP = |ψ|2 dv
= |ψ|2 4π r2 dr
Total probability will be
p R2
P =o ∫ dP = R1∫ |ψ|2 4π r2 dr
R2
P = 1∫ P(r) dr
R
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Atomic Structure

ATOMIC STRUCTURE
Define Probability Distribution Curves.

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Atomic Structure
Define Probability Distribution Curves.
ATOMIC STRUCTURE
Radial Probability Distribution Function Curve
2 2
P1s (r) r = 0.529 nz = a0 nz

r
r = ROMP = radius of maximum probability
= 0.529 A0 = a0 (Bohr’s radius)

Radial node
P2s (r) Here probability of finding
electron is zero

r
r = ROMP = 4a0 CHEMISTRY
Atomic Structure
Define Probability Distribution Curves.
ATOMIC STRUCTURE
Radial Probability Distribution Function Curve
Initially | | 2 is decreasing bur r2 is increasing At nucleus r =0, so P(r) = o
but as r increases | | 2 decreases exponentially so graph will start form the origin
so the overall curve tends to become zero
(ROMP)1s < (ROMP)2s

From the graph is evident that probability

of
r2 is increasing function but | |2 is
finding electron at the nucleus and at
decreasing function
infinite
distance from the nucleus is zero

That means electron in 2s orbital can be found

at a larger distance from the nucleus
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than the electron in 1s orbital
Atomic Structure
Define Probability Distribution Curves.
ATOMIC STRUCTURE
Radial Probability Distribution Function Curve
Radial node Radial node
P2s (r) P2P(r) P3P (r)

r r r
Radial node
P3d(r) P4d (r)

r r

 Total number of radial nodes = n – l – 1

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Atomic Structure

ATOMIC STRUCTURE
Define Nodes and Nodal Planes

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Atomic Structure

ATOMIC STRUCTURE
Define Nodes and Nodal Planes
Nodes Radial node S orbitals do not have any nodal plane
Nodes
Angular node / planar node
Angular nodes:
No of nodal planes = l (azimuthal Q.N.)
xz- plane
The plane is passing between the lobes of py orbital
py-orbital where the probability of finding the electron is zero

For py  orbital xz is the nodal plane

Angular node is dependent on the
For px orbital yz will be nodal plane shape of the orbital
For pz orbital xy will be nodal plane
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Atomic Structure

ATOMIC STRUCTURE
Define Nodes and Nodal Planes
Nodes d orbital For d orbital l = 2 planes
So for dwill
xy, dhave
yz andtwo
dxz nodal
x are combination of the remaining axis
with the axes of the orbital
y

dxy orbital
For dxy orbital nodal planes are xz and yz

For dyz orbital nodal planes are xy and xz

For dxz orbital nodal planes are xy and yz

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Atomic Structure

ATOMIC STRUCTURE
Define Nodes and Nodal Planes

Nodes
x
x=y
Nodal plane Nodal Cone
y
Nodal plane
Nodal Cone
x = -y
d x2 – y2 orbital d z2 orbital
2 nodal planes 2 nodal planes

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Atomic Structure

ATOMIC STRUCTURE
Define Nodes and Nodal Planes

No. of angular nodes =l

No. of radial nodes = n–l–1
Total no. of nodes = (n – l – 1) + l = n – 1

Ex. total nodes in 3d orbital?

Angular nodes =l =2
Radial nodes = n – l – 1 = 3–2–1=0
Total no. of nodes = 2

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Atomic Structure

ATOMIC STRUCTURE

Problems on Nodes & Nodal planes

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Atomic Structure

Problem ATOMIC STRUCTURE

How many nodal planes are present in atomic orbital for quantum no. n = 3 ?

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Atomic Structure

Problem ATOMIC STRUCTURE

How many nodal planes are present in atomic orbital for quantum no. n = 3 ?
Solution For n = 3 l = 0, 1, 2
So the possible atomic orbitals are 3s, 3p, 3d
1) For 3s orbital l=0  no nodal plane
2) For 3p orbital l=1
3p 1 1 1
For three p- orbital, 3 1=3 nodal planes
3) For 3d orbital l=2
3d 2 2 2 2 2
For five d- orbital, 5  2 = 10 nodal planes Total no. of nodal planes = 13

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Atomic Structure

Problem ATOMIC STRUCTURE

How many nodal planes are present in atomic orbital for quantum no. n = 4 ?

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Atomic Structure

Problem ATOMIC STRUCTURE

How many nodal planes are present in atomic orbital for quantum no. n = 4 ?
Solution For n = 4 l = 0, 1, 2, 3
So the possible atomic orbitals are 4s, 4p, 4d, 4f
For n = 4 total no. of nodal planes = 13 + nodal planes due to 4f orbital

4f 3 3 3 3 3 3 3

For seven f- orbital, 7  3 = 21 nodal planes

Total no. of nodal planes = 13 +21 = 34

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Atomic Structure
3
Problem 1 ATOMIC
1 2 r STRUCTURE
–
r
a
ψ2s = a0 2– a e 0
4 √ 2π 0

Where a0 is the Bohr radius (r0 : radius / distance from nucleus). Find the
location of radial node ?

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Atomic Structure
3
Problem 1 ATOMIC
1 2 r STRUCTURE
–
r
a
ψ2s = a0 2– a e 0
4 √ 2π 0

Where a0 is the Bohr radius (r0 : radius / distance from nucleus). Find the
location of radial node ?
Solution To have the radial node, ψ2s = 0
r
∴ 2– = 0 ⇒ r = 2a0
a0

– r
e a0 Can not be zero , because it is zero when r = ∞

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Atomic Structure

Problem: ATOMIC
For a quadratic equation STRUCTURE
7r 2
12 – r
a0 + a0 = 0

Find the location of radial nodes ?

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Atomic Structure

Problem: ATOMIC
For a quadratic equation STRUCTURE
7r 2
12 – r
a0 + a0 = 0 Find the location of radial nodes ?
r
Solution Let’s a = x
0
2
x – 7x + 12 = 0
⇒ (x – 4)(x – 3) = 0
 (x – 4) = 0 and (x – 3) = 0
r ⇒ r
⇒
a0 = 4 a0 = 3
⇒ r = 4a0 ⇒ r = 3a0

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