DK015
Chapter 2 :
ATOMIC
STRUCTURE
2.1 : Bohr’s Atomic Models
CHEMISTRY UNIT
KOLEJ MATRIKULASI KEDAH
OBJECTIVES
At the end of this lesson, students should be able to:
describe Bohr’s atomic model.
explain the existence of energy levels in an atom.
calculate the energy of an electron using:
1
E n  R H ( 2 ),
n
R H  2.18 x 10 -18 J
calculate the energy change of an electron during transition.
1
ΔE  R H ( 2
ni
1
- 2 ),
nf
where R H  2.18 x 10-18 J
calculate the photon emitted by an electron that produces a
particular wavelength during transition.
History of Atomic Models
The Development of Modern Atomic Theory
2.1 : Bohr’s Atomic Models
4 Bohr’s atomic postulate :
1
Electron moves in circular orbit about the nucleus.
Each orbit has a different energy level/energy state.
2
In the specific energy level, the energy of electron is
fixed in value or is quantised.
The electron does not release or absorb any
energy when it is in the ground state or at any
particular energy level/orbit.
This means that an electron moving in an orbit can
have a certain amount of energy.
(quantised = fixed/specific/definite)
n=2
electron
PYQ:
‘The energy of an electron in an atom
is quantized.’
Describe the above statement.
[3M]
n=1
nucleus
(proton)
energy of electron at n=1 ≠ energy of electron at n=2
The energy of an electron in its level is given by:
 1 
E n   RH  2 
n 
where,
RH (Rydberg constant)= 2.1810-18J.
n (orbit/ energy level) = 1, 2, 3,…, (integer)
NOTES:
 Energy is zero (E = 0) if electron is located infinitely
(n = )far from nucleus.
 Energy associated with forces of attraction are
taken to be negative (thus, negative sign)
3
At ordinary condition, the electron is at the ground
state (lowest energy state).
If energy is supplied, electron absorbed the energy
and is promoted from lower energy level to a higher e
nergy level (electron is excited)
4
Energy at excited state is unstable.
It will fall back to lower energy level and released
a specific amount of energy in the form of light or
photon.
The light have specific wavelength.
The line spectrum is formed.
The energy absorb / released during transition can
be calculated by:
 1
1

ΔE  R H 

2
2
 ni
n
f





Where,
RH = Rydberg constant (2.18 x 10-18 J)
n= principal quantum number (1, 2, 3, …, )
ni = position(n) of electron initially
nf = position(n) of electron finally
The amount of energy released by the electron
during transition is called a photon of energy.
A photon of energy is released in the form of light
with appropriate frequency and wavelength.
Therefore, the energy change (released/absorbed)
can be related as:
ΔE  hυ
where,
h (Planck’s constant) =6.63  10-34 Js
= frequency;
c (speed of light) = 3.00108 ms-1
Photon = a packet of light energy equals to h
Which of the following is an
example when an electron
absorbs energy?
A.When it goes from the fourth energy
level to the first.
B.When an electron releases a photon.
C.When it maintains constant energy and
doesn't change his energy level.
D.When it goes from the second energy
level to the fourth.
1.
The energy of an electron at particular orbit:
 1 
En   RH  2 
n 
2.
RH = 2.18 x 10-18 J
The energy absorb / released during transition:
 1

1

ΔE  R H 

 ni 2 n f 2 


ΔE  hυ
Rydberg Equation Calculator
ΔE  h υ
c
υ

A photon of energy is emitted in the form of
radiation with appropriate frequency and wavelength.
Where:
c
υ

Thus:
hc
ΔE 

Where:
v = frequency
λ = wavelength
c = speed of light (3.00x108 ms-1)
h = Plank’s constant (6.63 x 10-34 Js)
3.
Find wavelength, λ :
 1 1 
ΔE  R H  2  2 
 n1 n2 
RH = Rydberg constant,
=1.097 x 107 m-1
Using the Bohr Rydberg equation
Calculation 1
Energy of electron at particular orbit
 1 
E n   RH  2 
n 
where,
RH (Rydberg constant)= 2.1810-18J.
n (orbit/ energy level) = 1, 2, 3,…, (integer)
Example 1
Calculate the energy of electron that occupies n=2
Answer:
Example 2
The energy of an electron at its orbit (n) is -1.36x10-19J.
Determine the value of n.
Answer:
1. Calculate the energy of electron that
occupies n=6
2. The energy of an electron at its orbit
(n) is -8.72x10-18J.
Determine the value of n.
Calculation 2
Energy released/absorbed by electron during transition
 1
1
ΔE  R H 

 ni 2 n f 2





RH = Rydberg constant (2.18 x 10-18 J)
n= principal quantum number (1, 2, 3, …, )
ni = position(n) of electron initially
nf = position(n) of electron finally
Example 1
Calculate the energy released when an electron falls
from n=3 to n=2.
Answer:
Example 2
Calculate the energy required to promote an electron from
the first energy level to the third energy level of a
hydrogen atom.
Answer:
Calculate the energy released
when an electron excited
from n=1 to n=4.
Calculation 3
Energy released/absorbed by electron during transition
ΔE  hυ
h (Planck’s constant) =6.63  10-34 Js
= frequency;
c (speed of light) = 3.00108 ms-1
Example 1
Calculate the energy absorbed by an electron that produces a
wavelength of 9.38 x 10-8 m.
Answer:
Example 2
An electron released 4.58x10-19 J of energy when falls from
n=5 to n=2. Calculate the frequency and wavelength
produced.
Answer:
1.
Calculate the energy released by an electron that
produces a wavelength of 9.49 x 10-8 m.
2.
An electron absorbed 2.09x10-18 J of energy
when excited from n=1 to n=5. Calculate the
frequency and wavelength produced.
1.
2.
Calculate the energy released by an electron that
produces a wavelength of 9.49 x 10-8 m.
An electron absorbed 2.09x10-18 J of energy
when excited from n=1 to n=5. Calculate the
frequency and wavelength produced.
ans:
v
λ
=
=
Calculation 4
Calculate:
(a) The energy (in J) of an electron has when it occupies a
level equivalent to the quantum number of n = 3 and n = 4.
(b) The energy (in kJ/mol) of photon emitted when one mole of
electron drops from the 4th energy level to the 3rd energy
level.
(c) The frequency (in s-1) and wavelength (in nm) of this
photon.
Answer:
(a)
Energy of an electron at n = 3 :
Energy of an electron at n = 4 :
b) For one electron:
For 1 mol of electrons:
1 mol electrons = 6.02 x 1023 electron
c)
∆E = – 1.06 x 10–19 J
Calculate the energy of an electron in the hydrogen atom
when n = 2, and when n = 6.
Calculate the wavelength released when an electron
moves from n = 6 to n = 2.
If this line in the visible region of electromagnetic
spectrum?
(Note: visible light: 400 ~ 700 nm)
 1 
Calculate the energy of an electron in the
En   RH  2 
hydrogen atom when n = 2, and when n = 6.
n 
OBJECTIVES
At the end of this lesson, students should be able to:
describe the formation of line spectrum of hydrogen atom.
perform calculations involving the Rydberg equation for
Lyman, Balmer, Paschen, Brackett and Pfund series:
1

 RH (
1
n12
-n
1
2
)
2
where RH  1.079 x 107 m-1 and n1  n2
calculate the ionisation energy of hydrogen atom from Ly
man series.
OBJECTIVES
State the weakness of Bohr’s atomic model.
State the dual nature of electron using de Broglie’s
postulate and Heisenberg’s uncertainty principle.
(calculation is not required)
FORMATION LINE SPECTRUM
HYDROGEN ATOMS
1
2
Spectrum is a series of lines or a set of colours.
They are two types of spectrum: line spectrum, and c
ontinuous spectrum.
Line spectrum
Continuous spectrum
A spectrum that contains a series of diA spectrum that contains continuous
screte lines. Each line corresponds to band of light radiation of all wavelen
a specific wavelength.
gths.
Each line separated by blank area
Continuous. No blank area in between.
Source: emission spectrum of atom ( Source: white light, sunlight
from gas discharge tube of hydrogen
, etc.)
How is the formation of line spectrum
of hydrogen atom?
When the electron of a hydrogen
atom at its ground state absorbs s
ufficient amount of energy, it will
move to a higher energy level
At higher energy evel, electron is
unstable. It will falls back to a
lower energy level.
During the transition (falls form
higher to lower energy level),
energy will be released in a form
of light at certain wavelength and
frequency.
Since energy is quantised (fix in
value), line spectrum is produced.
n=4
n=3
n=2
n=1
nucleus
3 The electromagnetic spectrum:
4
There is several emission series of lines obtained
during the transition of electrons in hydrogen spectrum
5
6
The series in the hydrogen line spectrum are found
in the visible region, infrared region and ultraviolet
region.
These spectral lines are the result of the transition of
excited electron to the lower energy level.
series
Spectrum region
Lyman
Ultraviolet
Balmer
Visible
Paschen
Infrared
Brackett
Infrared
Pfund
Infrared
7
Balmer series is the only one visible to the
unaided eye. It contains four coloured line against
a black background.
Line
number
1st line
2nd line
3rd line
4th line
Colour in Balmer
series
Wavelength
(nm)
red
656
Blue-green
486
indigo
434
violet
410
7
The possible energy levels in the hydrogen atom
and the transitions of electrons that produce the
lines in the hydrogen spectrum:
7
Lines produced caused by the transition of electrons
from higher energy level to lower energy level.
eeee-
e-
n=
n=7
n=6
Series
Line
Lyman
Transition of e-
n=5
1st
n=2 to n=1
n=4
2nd
n=3 to n=1
3rd
n=4 to n=1
4th
n=5 to n=1
5th
n=6 to n=1
n=3
n=2
n=1
Continuum
limit
ΔE  h
h
c

E,,
eeee-
n=
n=7
n=6
n=5
Series
Line
Balmer
Transition of e-
1st
n=3 to n=2
n=4
2nd
n=4 to n=2
n=3
3rd
n=5 to n=2
4th
n=6 to n=2
n=2
n=1
Continuum
limit
ΔE  h
h
c

E,,
eeee-
n=
n=7
n=6
n=5
n=4
n=3
n=2
n=1
Series
Paschen
Line
Transition of e-
1st
n=4 to n=3
2nd
n=5 to n=3
3rd
n=6 to n=3
4th
n=7 to n=3
Continuum
limit ΔE  h
h
c

E,,
eee-
n=
n=7
n=6
n=5
n=4
n=3
Series
Brackett
Line
Transition of e-
1st
n=5 to n=4
2nd
n=6 to n=4
3rd
n=7 to n=4
n=2
n=1
Continuum
limit
ΔE  h
h
c

E,,
ee-
n=
n=7
n=6
Series
n=5
Line
Transition of e-
n=4
1st
n=6 to n=5
n=3
2nd
n=7 to n=5
Pfund
n=2
n=1
Continuum
limit
ΔE  h
h
c

E,,
8
Each line in the hydrogen spectrum has a specific
wavelength.
9
The wavelength can be calculated by using the
Rydberg equation:
 1
1 
 RH  2  2 

n2 
 n1
1
Where

= wavelength
RH
= Rydberg constant = 1.097 x 107 m-1
n1 and n2 = 1, 2, 3, …,  (principal quantum numbers)
Since  should have a positive value thus n1
< n2
Example
Calculate the wavelength of the indigo line in the
Balmer series.
Solution:
Indigo line  3rd line in Balmer series
 n=5
 1

1
 RH  2  2 
n


n
2 
 1
1
; n1  n2
to n=2
 1 x 10-9
m
nm
Try!!!
1.
Calculate the wavelength of the continuum limit in
the Paschen series.
2.
For the Lyman series, calculate the wavelength
when electron falls from n=4 to n=1.
Changing the Energy
Ultraviolet
Infrared
Visible
• The further they fall, more energy is releas
ed and the higher the frequency.
• The orbitals also have different energies in
side energy levels.
ΔE 
hc

ΔE  hv
Example
The following diagram depicts the line spectrum of
hydrogen atom. Line A is the first line of the Lyman series.
A
B
C
D E
Line
spectrum
∆E


Specify the increasing order of the radiant energy,
frequency and wavelength of the emitted photon.
Which of the line that corresponds to
i) the shortest wavelength?
ii) the lowest frequency?
53
Example
E D
C
B
Line
spectrum
A
Paschen series
Which of the line in the Paschen series corresponds to the
longest wavelength of photon?
Describe the transition that gives rise to the line.
Example
W
Line
spectrum
Y
Balmer series
Describe the transitions of electrons that lead to the
lines W, and Y, respectively.
Compare the difference:
1
n2i
E = RH
1

=
RH
1
n2f
1
n12
RH = 2.18 x 10–18 J
1
n22
RH = 1.097 x 107 m-1
n1 can be ni or nf as long as n1 < n2
 to get positive value for 
Both equations can be used to calculate
wavelength () and frequency () of any line
of H atom emission series
Remember! RH values and units are different
in both equations !
EXAMPLE-1
Use the Rydberg equation to calculate
the wavelength (in nm) of the forth
line in the Balmer series of Hydrogen
spectrum .
ANS:
410.2 nm
ANS.EXAMPLE-1
Balmer series, electron: higher level to n = 2
1st line: n: 3 to 2 ; So, 4th line: n: 6(n2) to 2(n1)
By using Rydberg equation:
Therefore,  = 4.102 x 10-7 m = 410.2 nm
EXAMPLE-2
An electron in the n = 5 level of an
H atom emits a photon of wavelength
1281 nm. To what energy level does
the electron move?
ANS:n = 3
ANS.EXAMPLE-2
Electron transition: ni = 5 nf = ?
By using Rydberg equation:
EXAMPLE-3
Calculate what is;
For Lyman series; n1 = 1
i ) Wavelength
& n2 = ∞
ii ) Frequency
iii ) Wave number
of the last line of
hydrogen spectrum
in Lyman series
Wave number = 1/wavelength
The minimum energy required to remove
one mole of electrons from one mole of
the gaseous atoms or ions.
M (g) → M+ (g) + e-
ΔH1= +ve kJ
Ionization Energy is the energy required to cause an electron to
escape from the atom.
The energy levels are a bit like steps of a ladder that the electrons
can climb, if an electron is given enough energy to climb to the last
energy level the electron will escape from the atom and becomes a
positive ion.
The energy needed for this jump is called IONIZATION ENERGY
EXAMPLE
Calculate the ionization energy of
H atom in kJmol-1.
Ans: EXAMPLE
Electron transition: ni = 1 (ground state)
nf = ∞ (complete removal)
 E = RH
1
n2
i
1
n2
f
= 2.18 x 10-18 J x
1
12
1
∞2
= 2.18 x 10-18 J
(for removal of one electron form one H atom)
EXAMPLE
EXAMPLE
–– 14
14 – 14
Ans:
Ans:
EXAMPLE
EXAMPLE
6.02 x 1023 electrons 6.02 x 1023 H atoms
For removal of 1 mol electrons from 1 mol H atom:
E =
2.18 x 10-18 J
x
6.02 x 1023 electrons H
1 electron
= 1.3124 x 106 J/mol
= 1312.4 kJ/mol
1 mol
Bohr’s theory has a number of weaknesses. Bohr’s atomic
model
• Unable to explain the line spectrum of atoms or ions cont
aining more than one electron (such as helium).
• Electron is restricted to move in a certain distance
around the nucleus of an atom.
• Unable to explain the extra lines formed in the hydrogen
spectrum.
• Unable to explain the dual nature of electrons.
(A) de Broglie’s postulate
• in 1924, Louis de Broglie, a French physicist, suggested
that light and matter appear to have dual natures, that is,
both light and matter are wave-like as well as particle-like.
• de Broglie suggested an equation that allows the calculation
of the wavelength of an electron or a particle with mass (m),
moving at velocity ():
h
λ
m
where,
h = Planck’s constant (J s)
m = particle mass (kg)
 = velocity (m/s)
 = wavelength of a matter wave
(B) Heisenberg’s uncertainty principle
• Bohr’s theory:
 the electron was thought of as orbiting the nucleus.
 This would mean that at any moment, we would
know both the precise position and precise
speed of the electron.
 This is not allowed in quantum mechanics.
• In 1927, Werner Heisenberg states in his principle that:
“It is impossible to know simultaneously both the
momentum, p (defined as mass times velocity) and
the position of a particle with certain”.
This principle is known as Heisenberg’s uncertainty
principle.
• Mathematically, his principle stated as:
h
x p 
4
where
x= uncertainty in measuring the position
p=uncertainty in measuring the momentum
= mv
h = Planck’s constant
Bohr’s Atomic Models
9M
ANS: https://image.ibb.co/eu8bmQ/tysc2.jpg