2. Session Objectives
1. Dalton’s theory
2. Discovery of fundamental particles
3. Thomson’s model of an atom
4. Rutherford’s model
5. Concept of atomic number and mass number
6. Drawback of Rutherford’s model
7. Electromagnetic waves
8. Planck’s quantum theory
9. Bohr’s model
3. • All matter is composed of atoms.
• All atoms of a given element
are identical.
Atom → Can not be cut.
Indivisible and indestructible
Dalton’s Theory – Atom Is
Fundamental Particle
Pre 1897
5. Properties of Cathode Rays
They are deflected from their path by electric and
magnetic fields
6. Properties of cathode rays
They are material particles as they produce mechanical motion
in a small paddle wheel
e/m ratio of cathode rays :1.758820 x 10 11
C kg -1
7. Anode rays
+
–
Z n S
C o a t in g
P e r fo r a te d
c a t h o d e
H g a s in s id e
a t lo w p r e s s u r e
2
Unlike cathode rays, the charge to mass
ratio of canal rays on the nature of gas
taken in the discharge tube.
It is maximum when lightest gas hydrogen is
taken which is equal to 9.58 x 107
Ckg-1
8. Properties of anode Rays
They travel in straight line
They are deflected by electric and magnetic field
The nature of anode rays depends upon the nature of gas
e/m ratio for anode rays is not constant
It is maximum when lightest gas hydrogen
is taken which is equal to 9.58 x 107
Ckg-1
9. Sub – atomic
particles
Subatomic
particles
Symbol Unit charge Unit mass
Charge in
Coulomb
Mass in amu
Proton
Neutron
Electron Negligible
1
1H
1
0n
0
1e−
-19
+1.60 × 10
-19
+1.60 × 10 -4
5.489×10
1.008665
1.007825
-1
0
+1 1
1 0
12. Rutherford’s Experiment -
Results
A beam of α particles aimed
at thin gold foil.
• Most of the particles
passed through. Most of the space is empty
• A few came back Presence of concentrated mass
at the centre
• Others deflected at
various angles
Repulsion between two +vely
charged particles
“ Like firing shells at paper handkerchief with few
of them coming back.” - Ernst Rutherford
13. Rutherford’s Model
Atom consist of two parts:
(a)Nucleus:Almost the whole mass of the atom is
concentrated in this small region
(b)Extra nuclear part:this is the space around the
nucleus
in which electrons are revolving at high speeds in
fixed path
14. Concept of atomic mass and
atomic number
Atomic number(Z)=number of protons
Mass number(A)=number of protons+number of neutrons
Entire mass of the atom
is concentrated at the
centre
15. Concept of atomic number
and mass number
aN23
11
For example:
Mass number=number of protons+number of neutrons
=23
Atomic number =number of protons
=11
16. Concept of atomic number
and mass numberWe express weight of an atom in
terms of atomic mass unit (a.m.u).
Mass of a proton=Mass of neutron
=1 a.m.u(approx)
∴ Mass number=Atomic weight (expressed in a.m.u)
19. Electromagnetic waves
Light is an oscillating
electro-magnetic field.
Oscillating electric field
generates the magnetic
field and vice-versa.
Electric and magnetic fields are perpendicular to each other
21. . λ
(i) Wavelength: It is represented by
Units:
m, cm(10-2
m), nm(10-9
m), pm(10-12
m) or A0
(10-10
m).
Characteristics of a wave
λ
Direction of
Propogation of wave
22. (iv) Wave number: The number of
waves present in 1 cm length.
It is represented by . Its unit is cm-1
.
1
λ
(iii) Velocity: The linear distance travelled by a crest or a
trough in one second. Its unit is cm s-1
.
(ii) Frequency: The number of waves
passes through a given point
in 1 second. It is represented by .ν
Its unit is Hertz or second-1 .
Characteristics of a wave
24. Radio city broadcasts on a frequency
of 5,090 KHz.What is the wavelength
of electromagnetic radiation emitted
by the transmitter?
Illustrative problem 1
c
λ =
ν
8
3
3 10
5090 10
×
λ =
×
2
0.589 10= ×
58.9 m=
s/m103isc 8
×
25. Radiant energy is emitted or
absorbed discontinuously in the
form of quanta.
Planck’s quantum theory :
34
c
E h h hc
Wavelength
Frequency
Wave number
h Plank 's cons tant
6.626 10 Js−
= ν = = ν
λ
λ =
ν =
ν =
=
= ×
27. The ratio of the energy of a photon
of 2000 wavelength radiation to
that of 6000 radiation is
(a) ¼ (b) 4
(b)½ (d) 3
0
A
0
A
Illustrative Problem
2
hc
E h= ν =
λ 1 2
1 2
hc hc
E E= =
λ λ
1 2
2 1
E
E
λ
∴ =
λ
0
0
6000A
3
2000A
= =
Hence, answer is (d).
Solution:
29. Bohr’s Postulates
Retained key
features of
Rutherford’s
model.
Concept of
stationary
circular orbits.
Quantization of
angular
momentum.nh
mvr
2
=
π
Energy emitted/absorbed when
electrons jump from one orbit
to another. f iE E E
h
∆ = −
= ∆ν
32. centrifugal force
2
c
mu
F
r
= (ii)
F Fa c=
2 2
2
kZe mu
r r
∴ =
2
2 kZe
u
mr
=
Calculation of radius of Bohr orbit
nh
mur
2
=
π
Bohr’s postulate
nh
u
2 mr
=
π
2 2
2
2 2 2
n h
u
4 m r
=
π
33. For hydrogen Z=1
= ×
2
0n
r 0.529 A
Z
2
2 kZe
u
mr
=
2 2
2
2 2 2
n h
u
4 m r
=
π
222
222
rm4
hn
mr
kZe
π
= 22
22
mkZe4
hn
r
π
=
For n=1,Z=1 , k = 9
109× Nm2
/C2
Calculation of radius of Bohr orbit
34. u is the velocity with which the electron revolves in
an orbit
nh
mur
2
=
π
(1)
2 2
2
kZe mu
rr
= (2)
Dividing (1) by (2),we get:
Calculation of velocity of
electron
nh
kZe2
u
2
π
= u is in m/s
35. Number of revolutions per second
velocity of electron
circumference of an orbit
=
Calculation of number of
revolutions
36. Total energy(T.E) P.E K.E= +
21
K.E mu
2
=
2
kZe
P.E
r
= −
2
2
2
1 kZe
T.E mu
2 r
= −
Calculation of energy of an
electron
37. We know that
2 2
2
mu kZe
r r
=
2
2 kZe
mu
r
=
2 2
kZe kZe
E
2r r
= −
2
kZe
2r
= −
Substituting the value of r we get
22
2422
n
hn
mkeZ2
E
π
−=
P.E. = 2K.E.
K.E. = -Total energy
38. Bohr’s model
Bohr’s postulates
8
n
Z
v 2.18 10 cm /sec
n
= ×
2 0
n
n
r 0.529 A
Z
= ×
2
n 2
2
19
2
Z
E 13.6 eVper atom
n
Z
21.8 10 J per atom
n
−
= − ×
= − × ×
39. Quantum Mechanical Model of an Atom
Schrodinger’s Equation:
∂2
/∂xψ 2
+ ∂2
/∂yψ 2
+ ∂2
/∂zψ 2
+ 8π2
m/h2
( E - v ) =ψ
o
Where x, y, z are certain coordinates of the electron
m = mass of the electron
E = total energy of the electron.
V = potential energy of the electron;
h = Planck’s constant and
ψ (psi) = wave function of the electron.
Significance of ψ2
: ψ2
is a probability factor. It describes the
probability of finding an electron within a small space. The
space in which there is maximum probability of finding an
electron is termed as orbital. The important point of the
solution of the wave equation is that it provides a set of
numbers called quantum numbers which describe energies
of the electron in atoms, information about the shapes
and orientations of the most probable distribution of
electrons around nucleus.
40. Nodal point:
The point where there is zero probability of finding
the electron is called nodal point.
No. of radial nodes = n – l – 1
No. of angular nodes = l
Total number of nodes = n – 1
Nodal plane:
Nodal planes are the planes of zero probability of
finding the electron.
The number of such planes is also equal to l.
41. λ = h/mv ……. De – Broglie’s Equation
Where mv = p, momentum of the particle
De- Broglie’s hypothesis
Dual nature of Particles
(PARTICLE AND WAVE CHARACTER OF MATTER
AND RADIATION)
E = hυ
(According to the Planck’s quantum theory)
E = mc2
(according to Einstein’s equation)
42. HEISENBERG’S UNCERTAINTY PRINCIPLE
“It is impossible to measure simultaneously
the position and momentum of a small
microscopic moving particle with
absolute accuracy or certainty”
43. Quantum
Numbers
Quantum numbers may be defined as a set of four
numbers with the help of which we can get complete
information about all the electrons in an atom. It
tells us the address of the electron i.e., location,
energy, the type of orbital occupied and orientation
of that orbital.
Principal quantum number (n): • main shell in which the electron
resides
• also tells the maximum number of
electrons that a shell can
accommodate is 2n2
, where n is the
principal quantum number
44. Azimuthal
(or)
angular momentum quantum number
(l):
represents the number of subshells present in the
main shell
The orbital angular momentum of the electron is given
as √l (l +1) h/2π
or
√l (l+1) nh/2Π for a particular value of ‘n’
(where h = Planck’s constant).
For a given value of n values of possible l vary from
0 to n – 1
45. Magnetic Quantum Number
(m):
The magnetic quantum number determines the
number of preferred orientations of the
electron present in a subshell.
m can assume all integral values between
–l to +l including zero.
Thus m can be –1, 0, +1 for l = 1.
Total values of m associated with a particular value of l
is given by 2l+ 1.
46. Spin Quantum Number (s):
spin quantum number can have two
values, i.e., +1/2 and –1/2 or these
are represented by two arrows
pointing in the opposite directions,
i.e., - and .↑ ↓
It helps to explain the magnetic
properties of the substances.
47. Shapes of Orbitals
S- and p- orbitals
S- orbital –
spherical in
shape
P- orbital –
Dumb-bell in shape
50. Illustrative Problem
3
The energy of the electron in the
second and third Bohr orbits of
the hydrogen atom is
-5.42 X 10-12
and –2.41 X 10-12
respectively. Calculate the
wavelength of the emitted
radiation, when the electron
drops from third to second orbit.
51. Solution
2 1E E E∆ = −
According to Planck’s quantum theory
12 12
5.42 10 ( 2.41 10 )ergs− −
= − × − − ×
12
3.01 10 ergs.−
= − ×
E h= υ
c
E h=
λ
27
h 6.6 10 ergs−
= ×
54. Class Exercise - 1
Which of the following fundamental
particles are present in the nucleus
of an atom?
(a) Alpha particles and protons
(b) Protons and neutrons
(c) Protons and electrons
(d) Electrons, protons and neutrons
Solution
The nucleus of an atom is positively charged and almost
the entire mass of the atom is concentrated in it. Hence,
it contains protons and neutrons.
Hence, answer is (b).
55. Class Exercise - 2
The mass of the proton is
(a) 1.672 × 10–24
g
(b) 1.672 × 10–25
g
(c) 1.672 × 1025
g
(d) 1.672 × 1026
g
Solution
Hence, answer is (a).
The mass of the proton is 1.672 × 10–24
g
56. Class Exercise -3
Which of the following is not true
in case of an electron?
(a) It is a fundamental particle
(b) It has wave nature
(c) Its motion is affected by magnetic field
(d) It emits energy while moving in orbits
Solution
Hence, answer is (d).
An electron does not emit energy while moving in orbit.
This is so because if it would have done that it would
have eventually fallen into the nucleus and the atom
would have collapsed.
57. Class Exercise - 4
Positive charge of an atom is
(a) concentrated in the nucleus
(b) revolves around the nucleus
(c) scattered all over the atom
(d) None of these
Solution
Hence, answer is (a).
Positive charge of an atom is present entirely in the
nucleus.
58. Class Exercise - 6
Why only very few a-particles are
deflected back on hitting a thin
gold foil?
Solution
Due to the presence of a very small centre in
which the entire mass is concentrated.
59. Class Exercise - 5
Calculate and compare the energies
of two radiations which have
wavelengths 6000Å and 4000Å
(h = 6.6 x 10-34
J s, c = 3 x 108
m s-1
)
Solution
− −
−
× × × × ×
= =
λ ×
34 8 1
1 7
hc 6.6 10 Js 3 10 ms
E
6 10 m
= 3.3 x 10-19
J
− −
−
× × ×
=
×
34 8 1
2 7
6.6 10 Js 3 10 ms
E
4 10 m
= 4.9 x 10-19
J
19
1
19
2
E 3.3 10 J
E 4.95 10 J
−
−
×
=
×
= 0.666 : 1
60. Class Exercise - 7
Explain why cathode rays are
produced only when the pressure
in the discharge tube is very low.
Solution
This is happened because at higher pressure
no electric current flows through the tube
as gases are poor conductor of electricity.
62. Class Exercise - 8
If a neutron is introduced into the
nucleus of an atom, it would result
in the change of
(a) number of electrons
(b) atomic number
(c) atomic weight
(d) chemical nature of the atom
Solution
Hence, answer is (c).
Neutrons contribute in a major way to the weight
of the nucleus, thus addition of neutron would
result in increase in the atomic weight.
63. Class Exercise - 9
The concept of stationary orbits lies
in the fact that
(a) Electrons are stationary
(b) No change in energy takes place in
stationary orbit
(c) Electrons gain kinetic energy
(d) Energy goes on increasing
Solution
Hence, answer is (c).
When an electron revolves in a stationary orbit,
no energy change takes place. Energy is emitted
or absorbed only when the electron jumps from
one stationary orbit to another.
64. Class Exercise - 10
What is the energy possessed by
1 mole of photons of radiations
of frequency 10 × 1014
Hz?
Solution
E = h ν
E = 6.6 × 10–34
× 10 × 1014
E = 66 × 10–20
= 6.6 × 10–19
joules
∴ energy of 1 mole of photons = 6.6 × 10–19
× 6.023 × 1023
= 39.7518 × 104
= 397.518 kJ/mol
65. Class test
1.The radius of hydrogen atom in ground
state is 5.3x10–11
m. It will have a
radius of 4.77A after colliding with an
electron. The principal quantum number
of the atom in the excited state is
(a) 2 (b) 4 (c)3 (d)5
2
n o
2
10 11
2
n
Since r r x
Z
n
4.77 10 5.3 10 (for hydrogen atom)
1
n 9
n 3
− −
=
× = × ×
=
=
Solution
Hence, answer is (c).
Thompson explained emission lines by assuming his electrons were trapped in a continuous blob of positive charge, like negative plums in a positive pudding. However, there was no way he could explain the precise wavelength patterns emitted by particular elements.