NCERT-structure of atom class 9 notes pdf download

ATOMIC STRUCTURE
4.1 INTRODUCTION
All matter is composed of tiny particles called atom.An atom is the
smallest particle of an element that can exist and still have properties
of an element.
The first theory regarding atom is given by Dalton but later on it
discarded by scientists. After that J. J. Thomson, Rutherford,
Bohr and many other scientist gave their atomic model.
4.2 DALTON’S ATOMIC THEORY
The main postulates of this theory are
(1) Each element is composed of extremely small particles called
atoms.
(2) Atom is indestructile.
(3) Atoms of a particular element are all alike but differ fromatoms
of another element.
Merits
(1) It explains the law of conservation of mass.
(2) It explains the law of definite proportion.
Demerits
(1) It fails to explain that why should atoms of an element differ in
their masses.
(2) Discovery of isotopes and isobars, proved that atoms of same
element may have different atomic mass (isotopes) and atoms of
different kinds may have same atomic masses (isobars).
(3) The discovery of various sub atomic particles like electron,
proton, neutron etc. lead to the idea that the atom was no longer the
smallest, in divisble particle of matter.
4.3 CHARGED PARTICLES IN MATTER
4.3.1 Electron
Properties of electron
(a) Electron was disovered by Sir J. J. Thomson
(b) Charge is 1.6 × 10–19
coulomb/gm
(c) The molar mass is 5.48 × 10–4
g/mole
4.1 Introduction
4.2 Dalton’s Atomic Theory
4.3 Charged particles in
matter
4.4 Thomson Atomic Model
4.5 Rutherford model of an
atom
4.6 Bohr model of atom
4.7 Atomic number and
mass number
4.8 Electronic configuration
principles
4.9 Valence electrons and
valency
“IIT-JEE FOUNDATION”
*4.10Quantum numbers
*4.11 Some Important Terms
*4.12Radioactivity
VAVA CLASSES/CHEM/9TH
All right copy reserved. No part of the material can be produced without prior permission

(e) In 1897, J. J. Thomson determined the e/m value of the electron –1.7588 × 108
coulomb/gram.
(f) Robert A. Millikan, gave the value of charge by oil drop experiment –1.6022 × 1019
Coulomb =
4.8 × 1010
esu. (g) mass of electron 9.1096 × 10–31
Kg = 0.000548 amu.
Cathode Rays
Scientists William Crookes 1879, Julius Pleuckar 1889 studied the electrical conduction through
gases at low pressures. Sir William Crookes designed discharge Tube.
(a) The electron was discovered at extremely low pressure and high voltage of the order of 10,000 volts.
J. J. Thomson’s Experiment and The Discovery of Electron
Properties of the cathode rays shows that these consist of streams of negatively charged particles having
very small mass. Sir J. J. Thomson called these particles corpuscles of negative electricity. Later,
G. J. Stoney called these particles electrons.
In 1897, an English physcist, J. J. Thomson studied the effect of electric field on cathode rays.
The discharge tube used by Sir J. J. Thomson consisted of the following parts:
 a cathode (C),
 a cylindrical metal disc having a fine hole at its centre, acts as anode (A) – this accelerates the
particles of the cathode rays,
 another metal disc (D) having a fine hole at its centre in line with the hole in the anode.
 the cathode rays (consisting of electrons) pass through these holes and strike the fluoresecnt screen
at the point E.
 Two flatplates P1 and P2, which can be connected to a source of high voltage.
 an electromagnet, generating field opposite to the field generated by the plates P1 and P2.
In the absence of any electrical or magnetic field, the cathode rays strike the fluorescent screen at point
E, and can be seen as a bright spot there.
When a high electric field is applied across the plates P1 and P2, the spot on the screen moves towards the
positively charged plate. Here, in figure the plate P2 is positively charged.
The bending of cathode rays towards the positive plate showed that the cathode rays consist of negatively
charged particles.The deflection suffered by the beam of cathode rays (or beamof electrons) inThomson’s
experiment depends upon the strength of the electric field applied across the electrodes.
Cathode ray have the following properties
(i) Travel in straight line
(ii) Rays are deflected by electric and magnetic field
(iii) Produce X-rays
(iv) Produce heating effect
(v) Produce green glow

(vi) Penetrate through metal likeAl (Aluminium)
(vii) Effect the photographic plates
(viii) The ratio of charge to mass i.e. charge/mass is same for all the cathode rays irrespective of the gas
used in the tube.
4.3.2 Proton
Properties of proton: (a) Discovered by Goldstein (b) Charge of +1.602 × 10–19
coulomb (c) Mass is
1.672 × 10–27
kg or 1.0072 amu
Positive Rays-Discovery of Proton
Perforated
cathode
Cathode ray
Positive
rays
Cathode
+
(a) E. Goldstein in 1886 discovered proton by repeating the same discharge tube experiments by using
a perforated cathode.
(b) When a high potential difference was applied, not only cathode rays were produced but also a new
type of rays were produced simultaneously from anode moving towards cathode and passed through the
holes of the cathode. These termed as canal ray or anode ray.
(c) Characteristics of Anode rays are as follows
(i) Travel in straight line and cast shadow
(ii) Deflected by the magnetic and electric fields
(iii) These rays have kinetic energy and produces heating effect also.
(iv) Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube
(v) These rays can pass through thin metal foils.
(vi) They are capable to produce ionization in gases.
4.3.3 Neutron
Properties of Neutron (n): (a) Chadwick (1932), discovered neutral particles which was called neutron.
Nuclear reaction is as follows 1
0
12
6
4
2
9
4 n
C
He
Be 



(b) A neutron is a subatomic particle which has a mass 1.675 × 10–24
g, approximately 1 amu.
4.3.4 Meson or Pions ()
(a) Discovered by Yukawa in 1935.
(b) On the basis of charge, the meson is of three types, -meson, µ-meson and neutral meson(°).
(c) -meson are called pions.
(d) It tells about the stability of nucleus.
4.4 THOMSON ATOMIC MODEL(THE RAISIN PUDDING MODEL) (PLUM
PUDDING MODEL): FIRST ATOMIC MODEL
J. J. Thomson suggested atom is a sphere (Pudding) of positive electricity with a electrons (raisins)

distributed within a sphere, sufficient to neutralize the +ve charge the stability of an atom was explained
as a result of the balance between the repulsive forces between the electron and their attraction towards
the centre of the positive sphere.
Electrons
Sphere of
positive
electricity
4.4.1 Limitations
Thomson’s model of an atom could explain the main characteristics of the atom at that time. But it did not
have any experimental support. Therefore, it was opposed by his coscientists and was rejected.
Rutherford scattering experiment raised an objection against the model.
4.4.2Acceptance
The prediction that an atom is electrically neutral and has no net charge, is still accepted. This was indeed
a big contribution towards the structure of the atom.
4.5 RUTHERFORD MODEL OF AN ATOM
Rutherford designed an experiment. In this experiment, a fast moving -particle is made to fall on a thin
goldfoil.
He took gold foil that was about 100 nm of 10–5
cm thick.by -particle. -particle are +2 charged helium
ions. Since they have mass of 4u, the fast moving -particle have a considerable amount of energy.
Atom of
metal foil
+
Beam of 
particles
Very low
Nucleus
Majority of rays
Majority of rays
Few
Few
4.5.1 Scattering of -particles by a gold foil
The -particle scattering experiment gave totally unexpected result. It has following observation:-
(i) Most of the fast moving -particles passed straight through the gold foil.
(ii) Some of them were deflected by small angles.
(iii) One out of every 12000 particles approved to rebound.
Rutherford gave following conclusion.
(i) Most of the space inside the atom is empty because most of the -particles passed through the gold
foil with out getting deflected.
(ii) Very few particles were deflected from their path, indicating that the positive charge of an atom
occupies very little space.
(iii) A very small fraction of -particle were deflected by 180°, indicating the all the positive charge and
mass of gold atom were concentrated in a very small volume within the atom.
From above data he also calculated that the radius of the nucleus is about 105
times less

than the radius of the atom.
On the basis of his experiment, Rutherford put forward the nuclear model of an atom, which
had the following features.
(a) The size of an atom is 10-8
cm. Most of the space of the atom is empty called extra nuclear space.
(b) The centre is positively charged body called nucleus and its size is very small i.e. 10–13
cm. Nearly all
the mass of an atom resides in the nucleus.
(c) The electrons revolve around the nucleus in well-defined orbits.
4.5.2 Drawbacks of Rutherford’s model
(a) Stability of the atom is not defined.
(b) The observed spectrum should be continuous but found to be discontinuous
The orbital revolution of the electron is not expected to the stable.Any particle in the circular orbit would
undergo acceleration.
According to classic science, whenever any charge particle revolve in circular orbit and undergoes
acceleration, it emits radiation and loses energy. As a result of this, the orbit will become smaller and the
electrons will drop into the nucleus. This however does not happen.
Illustration 1
What is ratio of mass proton and electron?
Solution
1837
Kg
10
1
.
9
Kg
10
67
.
1
31
27





Illustration 2
The mass-charge ratio for A+
ion is 1.97 × 10–7
Kg C–1
. Calculate the mass of A atom?
Solution
Given:
7
10
97
.
1
e
m 

 (since, e = 1.602 × 10–19
C)
 m = 1.97 × 10–7
× 1.602 × 10–19
Kg
= 3.16 × 10–26
Kg
4.6 BOHR’S MODEL OF ATOM
NEIL BOHR [1885-1962]
4.6.1 The important postulates of Bohr’s model of atom
(a) Electron revolves around the nucleus in a fixed circular orbit of definite energy. These orbits or shells
are called energy levels.
(b) mvr = n(h/2)
where: m = mass of the electron, v = velocity of electron

n = number of orbit in which electron revolves i.e. n = 1,2,3.....r = radius of the orbit
(c) As long as the electron occupy a definite energy level, it is said to be in stationary state.
(d) The energy is emitted or absorbed only when the electron jumps from one energy level to another.
This amount of energy emitted or absorbed is given by the difference of the energies of the two energy
levels concerned.
energy
Emitted
E
____
1
n
energy
Absorbed
E
_____
1
n
E
E
E
E
_____
2
n
E
E
E
E
_____
2
n
1
1
2
1
2
1
2
2













4.6.2 Failures/Limitations of Bohr’s Theory
(i) He could not explain the line spectra of atoms containing more than one electron.
(ii) Could not explain the presence of multiple spectral lines
(iii) In 1923, de Broglie, the French physicist, suggested that electron, like light has a dual character. It
has particle as well as wave nature. Bohr had treated electron only as a particle.
(iv) The main objection to Bohr’s theory was raised by Heisenberg’s uncertainty principle.According to
this principle, it is impossible to determine simultaneously the exact position and the momentumof a small
moving particle like an electron.
4.7 ATOMIC NUMBER AND MASS NUMBER
Atomic number (Z) : Number of protons = Number of electrons.
Mass number or Nucleon number (A) :
The mass number being the sum of the number of protons and neutrons in the nucleus, which is always a
whole number. A = Z + N where : N = Number of neutrons

Nuclear composition of the first eighteen elements
Element Atomic
number (Z)
Mass
number (A)
Number of
Symbol* Name Neutrons
(N)
Protons
(P)
Electrons
(e)
H
1
1
Hydrogen 1 1 0 1 1
He
4
2
Helium 2 4 2 2 2
Li
7
3
Lithium 3 7 4 3 3
Be
9
4
Beryllium 4 9 5 4 4
B
11
5
Boron 5 11 6 5 5
C
12
6
Carbon 6 12 6 6 6
N
14
7
Nitrogen 7 14 7 7 7
O
16
8
Oxygen 8 16 8 8 8
F
19
9
Fluorine 9 19 10 9 9
Ne
20
10
Neon 10 20 10 10 10
Na
23
11
Sodium 11 23 12 11 11
Mg
24
12
Magnesium 12 24 12 12 12
Al
27
13
Aluminium 13 27 14 13 13
Si
28
14
Silicon 14 28 14 14 14
P
31
15
Phosphorus 15 31 16 15 15
S
32
16
Sulphur 16 32 16 16 16
Cl
35
17
Chlorine 17 35 18 17 17
Ar
40
18
Argon 18 40 18 18 18
* Mass number is written at the top-left of the symbol, and the atomic number at the bottom-left.
Illustration 6
How many protons, electrons and neutrons are present in P
30
15 ?
Solution
Number of protons in one atom of Proton = Number of electrons in one atom of proton = 15
Number of neutrons in one atom of P = (A–Z) = 30–15 = 15
4.8 ELECTRONIC CONFIGURATION PRINCIPLES
The distribution of electrons in different orbitals is known as electronic configuration of the atoms.
It is based on certain guide-lines or rules given by Bohr and Bury. This is known as Bohr-Bury Scheme
According to this scheme.
1. The maximum number of electrons which can be present in a particular energy shell of an
atom is given by 2n2
. Here ‘n’ is the number of the energy shells or energy levels.
Let us find the maximum number of electrons in the first three energy shells. These are K, L and M as
given by Bohr.
(a) In first energy shell or K-shell (n = 1):
Maximum no. of electrons = 2n2
= 2 x (1)2
= 2
(b) In seconds energy shell or L-shell (n = 2)
= 2 x (2)2
= 8

(c) In third energy shell or M-shell (n = 3)
= 2 (3)2
= 18
Starting from the nucleus of the atom, the order of energy of these energy shells is: K < L < M.
2. The outermost energy shell in an atom cannot have more than eight electrons even if it has
a capacity to take up more electrons according to first rule.

4.9 VALENCE ELECTRONS AND VALENCY
Valence electrons: the electrons in the outermost shell of an atom are called its valence electrons.
The number of valence electron is sodium is one. The number of valence electrons in chlorine is seven.
Valency of an element: The number of valence electrons of an element which actually take part in
chemical reactions is called the valency of that element or the number of hydrogen atoms, or chlorine
atoms, or double the number of oxygen atoms which combine with one atom of the element is called its
valency.
How can the electronic configuration of an element explain its chemical reactivity
The elements having a completely filled outermost orbit (or shell) will be chemically inert (non-reactive).
These elements do not form compounds with other elements
Elem ent Total num ber of
electrons
D istribution of electrons in
various shells
K L M
H e
N e
A r
2
10
18
2
2
2
8
8 8
Because of this chemical inactivity, these gases are called noble gases (earlier these were called inert
gases.
The elements containing only one, or seven electrons in their outermost shell show greater
chemical reactivity, i.e., such elements react very fast with other elements. For example, sodium and
chlorine having the follwing electronic configurations are highly reactive.
Sodium 2, 8, 1 - Here, the outermost shell has only one electron: one more than the completely filled shell.
Chlorine 2, 8, 7 - Here, the outermost shell has seven electrons: one less than that required to fill the shell
completely.
4.9.1 Ion Formation
electron
e
Na
Na
8
,
2
1
,
8
,
2




The atomic number of sodium (Na) is 11. Its electronic configuration is 2, 8, 1. Na+
ion is obtained when
one electron is lost from sodium atom.
So, the electronic configuration of Na+
is 2,8. The electronic configuration 2,8 is the electronic configuration
of neon (Ne). Thus, Na+
resembles neon (Ne) in its electronic configuration
8
,
8
,
2
7
,
8
,
2



 Cl
e
Cl
The atomic number of chlorine (Cl) is 17. Its electronic configuration is 2, 8, 7. Cl—
ion is obtained when
one electron is gain from chlorine atom.
So, the electronic configuration of Cl—
is 2,8,8. The electronic configuration 2,8,8 is the electronic
configuration of argon (Ar). Thus, Cl–
resembles argon (Ar) in its electronic configuration
*4.10 QUANTUM NUMBERS
(i) The measurement scale by which the orbitals are distinguished, can be represented by sets of numbers
called as quantum number.
(ii) Each orbital in an atom is specified by a set of three quantum numbers and each electron is designated
by a set of four quantum numbers.

4.10.1 These quantum numbers are as follows
(i) Principal quantum numbers (Bohr) denoted by ‘n’
(a) It determines the size of atom and energy of the electron in an orbit.
(b) The maximum number of an electron in an orbit is calculated by 2n2
(c) If value of n is 1,2,3,.....it can be denoted by K, L, M.....
(ii) Azimuthal quantum number or angular quantum number, (Sommerfeld) denoted by ‘l ’
(a) It determines the number of subshells or sublevels to which the electron belongs
(b) It tells about the shape of subshells.
(c) It also expresses the energies of subshells s < p < d < f (increasing energy)
(d) The value of l = (n–1) always where ‘n’ is the number of principle shell.
(e) Value l = 0 1 2 3 ...(n–1)
Name of subshell= s p d f
Shape of subshell= spherical dumbbell double complex
dumbbell
(f) It represent theorbital angular momentum, which isequal to )
1
(
2
h


l
l
(g) The number of electrons in subshell = 2(2l + 1)
(iii) Magnetic quantum number (Zeeman) denoted by ‘m’
(a) It gives the number of permitted orientation of subshells.
(b) The value of m varies from –1 to +1 through zero.
(c) It tells about the splitting of spectral lines in the magnetic field i.e. it proves the Zeeman effect.
Zeeman Effect : In the presence of a magnetic field, each spectral line gets split up into closely spaced
lines. This phenomenon, known as Zeeman effect.
(d) For a given value of ‘I’ the total value of ‘m’ is equal to (2l + 1)
(e) Degenerate orbitals - Orbitals having the same energy are known as degenerate orbitals.
eg. for P subshell Px , Py, Pz
(iv) Spin Quantum number (s) Goldshmidt & UlenBack and denoted by the symbol of ‘S’
(a) The value of ‘s’ is +½ and –½, which is signified the spin or rotation or direction of electron on it’s
axis during the movement and the spin may be clockwise and anticlockwise.
(b) It represents the value of spin angular momentum is equal to )
1
(
2
h


s
s
4.10.2 Shape of Orbitals
Orbital: Orbital is the three dimensional region around the nucleus where there is a maximum tendency
of finding an electron of definite energy.
Shape of orbitals on the basis of quantum number

Y
Y
Y Y Y
Z
Z
Z Z Z
X
X
X X X
+
+
+ +
–
2s
1s
2 px 2 py 2 pz
–
–
+
–
Shape of ‘s’ orbital:
(a) For ‘s’ orbital l = 0 and m = 0, so ‘s’ orbital have only one unidirectional orientation i.e. the probability
of finding the electron is same in all directions.
(b) Spherical shape and does not have directional property.
Shape of ‘p’ orbital:
(a) For ‘p’orbital l = 1 and m = +1, 0, –1 means there are three ‘p’orbitals, which is symbolised as Px, Py,
Pz.
(b) Shape of ‘p’orbital is dumbbel and it has directional property.
Shape of ‘d’ orbital:
(a) For the ‘d’ orbital l = 2 then the values of ‘m’ are –2, –1, 0, +1, +2. It shows that the ‘d’ orbitals has
five orbitals as dxy, dyz, dxz, 2
2
2
z
y
x
d
,
d  .
(b) The ‘d’ orbital is double dumb belled and it has directional properties.
4.10.3 Electronic Configuration Principles
The distribution of electrons in different orbitals is known as electronic configuration of the atoms. Filling
up of orbitals in the ground state of atom is governed by the following rules
(i) Aufbau Principle
(a) It is a German word, meaning ‘building up’
(b) According to this principle, “In the ground state, the atomic orbitals are filled in order of increasing
energies”, i.e. in the ground state the electrons occupy the lowest orbitals available to them.
(c) In fact the energy of an orbital is determined by the quantum number n and l with the help of (n+1)
rule or Bohr Bury rule.

(ii) Pauli’s Exclusion Principle
According this principle, “no two electrons in an atom can have all the four quantum numbers n, l, m and
s identical.
(iii) Hund’s Rule of Maximum Multiplicity
(a) This rule governs the filling up of degenerate orbitals of the same sub-shell.
(b) According to this rule “As far as possible electron in the degenerate orbitals remain single, pairing will
not take place unless and untill each degenerate (equal energy) orbital has got a single electron.
7N  1s2
, 2s2
, 2p3
Wrong
Correct
1s2
2s2 1
2 x
p ,
1
2 y
p , 1
2 z
p
Thus,
2 2 1 1 1
7 1 ,2 ,2 , 2 , 2
x y z
N s s p p p

(iv) (n+l) Rule
This rule states that electrons are filled in orbitals according to their n+1 values. When (n+l) is same for
sub energy levels, the electrons first occupy the sublevels with lowest ‘n’ value.
Thus, order of filling up of orbitals is as follows
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d
(v) Half Filled and Completely Filled Orbitals
Half-filled and completely filled sub-shells have extra stability on the basis of Exchange Energy.
Extra stability is shown, when p, d, f orbitals are half filled or completely filled.
For example,
24Cr  1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s2
, 3d4
1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s2
, 3d5
29Cu  1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s2
, 3d9
1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s1
, 3d10
4.11 SOME IMPORTANT TERMS
4.11.1 Isotopes
(i) First proposed by soddy.
(ii) The isotopes have same atomic number but different atomic weight.
e.g. 1H1
(Hydrogen) 1H2
(Deuterium) 1H3
(Tritium)
Atomic no. Z = 1 1 1
Mass no. A = 1 2 3

Other such examples
(i) Carbon C
12
6 and C
14
6 (ii) Chlorine Cl
35
17 and Cl
37
17
The chemical properties of isotopes are similar but their physical properties are different.
If an element has no isotopes then the mass of its atom would be the same as sum of proton and neutrons
in it. But if an element occurs in isotopic forms then the we have to know the percentage of each isotopic
form and then the average mass is calculated.
How to calculate the atomic mass of an element from the mass numbers of its isotopes
Atomic mass of the element
X =
100
)
X
of
number
Mass
X
of
Percentage
(
)
X
of
number
Mass
X
of
Percentage
( 2
2
1
1 


Chlorine has two isotopes Cl
35
17 and Cl
35
17 ; these are found in 3 : 1 ratio or 75% : 25% respectively.
.
Isotopic mass may be calculated as
Isotopic mass of chlorine
37
37
35
35
Cl
of
mass
100
Cl
of
%
Cl
of
mass
100
Cl
of
%




5
.
35
37
100
25
35
100
75





OR
Isotopic mass of chlorine
ratio
of
Sum
Cl
of
Mass
Cl
of
Ratio
Cl
of
mass
Cl
of
Ratio 37
37
35
35




4
37
1
35
3 


 = 35.5
Illustration 12
If bromine occurs is in the form of say two isotopes %)
7
.
49
(
Br
79
35 and %)
3
.
50
(
Br
81
35 , then calculate
the atomic mass of bromine atom.
Solution
Atomic mass of bromine
79u 49.7 81u 50.3
49.7 50.3
  


= 80 u
So, the atomic mass of bromine is 80.
Applications of isotopes or radioisotopes
(a) For estimating the age of old archaeological samples.
(b) For the treatment of diseases
cobalt-60: destroying malignant cells in patients suffering from cancer
iodine-131: studing disorders of the thyroid gland
sodium-24: circulation of blood
(c) For estimating the age of glaciers
Radiocarbon dating: Estimating the age of a carbon-containing object by measuring the concentration (or
activity) of C
14
6 in it is called radiocarbon dating.

4.11.2 Isobar
The two different atoms which have same atomic masses but different atomic number is called as isobar
e.g. 18Ar40
19K40
20Ca40
Atomic mass 40 40 40
Atomic number 18 19 20
4.11.3 Isoelectronic
Ion or atom or molecule or species which have the same number of electron is called isoelectronic
species.
*4.11.4 Isosters
Substance which have same number of electron and atoms called Isosters.
e.g. CO2 N2O
22 22
*4.11.5 Isodiapheres
The elements which have same value of (n–p) is called isodiapheres.
e.g. 7N14
8O16
value of (n–p) 0 0
4.11.6 Isotones
Elements which contain same number of neutron is called isotones.
e.g. 14Si30
15P31
16S32
number of neutrons 16 16
4.11.7 Kernel
Orbit which present after removing the outer most orbit of that atom is called kernel and electrons which
is present that orbit called kernel electrons e.g. Mg = 1s2
2s2
2p6
3s2
. Total kernel electron = 2 + 2 + 6 = 10
4.11.8 Core
(i) The outer most shell of an any atom called Core and the number of electron present of that shell is
called Core electron. e.g. Cl = 1s2
2s2
2p6
3s2
3p5
Core electron = 2 + 5 = 7
(ii) If the core is unstable for an atom then that atom shows variable valency.
4.12 RADIOACTIVITY
The elements like uranium, thorium, radium and polonium emit some invisible, ionizable rays which effect
a photographic plate just like ordinary light rays.
This spontaneous emission of invisible, ionizable radiation by some elements like uranium,
thorium, radium & polonium due to splitting of their nucleus is called radioacivity & the element
which emit such radiations are called radioactive elements.
Radioactivity was discovered by a (French physicist Henri Bacquerl in 1896).

4.12.1 Nature of Radioactive radiations
The invisible rays emitted by radioactive elements consist of three different type of radiations
1.Alpha particles 2. Beta particles 3. Gamma rays
Comparision of ,  &  rays:
Property -particle -particle -rays
1. Charge
2. Nature
3. Relative
Penetration
power
4. Velocity
5. Ionising
nature
6. Mass
7. Energy
8. Effect on
ZnS plate
+ve
He
4
2
1
1 – 1/10% of light
Very high
4 a.m.u.
4.9 mev
High
- ve
Electron
100
33 – 99%
Medium
9.1 x 10-29
gm
5 – 2 mev
Medium
Neutral
Energetic rays
100 x 100
Same as light
Very low
Zero
E = hv
Very low
4.12.2 Cause of Radioactivity: unstable nuclei
When the nucleus of an atom contains relatively more neutrons than what is normally found in nature, it
becomes unstable & shows radioactivity.
no. of protons no. of neutrons Stability
C -12 atom 6 6 very stable
C - 14 atom 6 8 radioactive
Examples of -decay
C
14
6 
 
 decay
-
â
N
14
7 + e
0
1
-
+ energy
Examples of -decay
U
238
92 
 
 decay
-
á Th
234
90 + He
4
2
+ energy
4.12.3 Applications of Radioactivity
(i) The radioactive isotopes are used to estimate the age of fossils like dead plants and animals,
and rocks called as radio-chemical dating or radio-isotope dating. An important case is (carbon-
dating). Which is a method of estimating the age of old carbon containing objects like dead plants and
animals by measuring the levels of C-14 radioactivity in them.
(ii) The radioacive isotopes are used as ‘tracers’ in medicine to detect the presence of tumors
and blood clots etc in the human body.
Important Information:
(a) Arsenic - 74 tracer is used to detect the presence of (tumors)
(b) Sodium-24 tracer is used to detect the presence of (blood clot)
(c) Iodine-131 radioisotope is used to determine the activity of (thyroid gland)

3. A controlled nuclear fission of radioactive substance is used in nuclear reaction to generrte
electricity.
4. Radioactive isotopes are used in industry to detect the leakage in underground all pipe
lines, gas pipe lines & water pipes.
Solved Examples
Example 1
Complete the following table
Number of
Element Atomic no.
Protons Electrons Neutrons
Mass
number
A 17 - - 18 -
B - - 14 14 -
C - 9 - - 19
Solution
For an atom, we know that,
Atomic number = Number of protons
Number of electrons = Number of protons = Atomic number
Mass number = Number of protons + Number of neutrons
Using these relationships the gaps in the above table are filled up as follws. The filled number are show in
bold.
Number of
Element Atomic no.
Protons Electrons Neutrons
Mass
number
A 17 17 17 18 35
B 14 14 14 14 28
C 9 9 9 10 19
Example 2
Fill in the blanks
Atom Mass no. Atomic no.
No. of
neutrons
Co
60
27
- - -
Na


24 11 -
Cl
37 - - 20
Solution
We know that,
Number on the top-left of the symbol is Mass number and
Number at the bottom-left of the symbol is Atomic number
Using the relationship
Mass number = Atomic number + Number of neutrons
One can calculate value of the missing quantity (shown in bold). The completed table is

Atom Mass no. Atomic no.
No. of
neutrons
Co
60
27
60 27 33
Na
24
11
24 11 13
Cl
37
17
37 17 20
Example 3
Give the electronic configuration of the element having atomic number 17.
Solution
Atomic number (17): 1s2
2s2
2p6
3s2
3p5
Example 4
Write the electronic configurations of Cr2+
and Mn4+
and give the number of unpaired electrons
present in each case (Atomic number Cr = 24, Mn = 25)
Solution
Cr: 1s2
2s2
2p6
3s2
3p6
3d5
4s1
Cr2+
: 1s2
2s2
2p6
3s2
3p6
3d4
4s0
Mn: 1s2
2s2
2p6
3s2
3p6
3d5
4s2
Mn4+
: 1s2
2s2
2p6
3s2
3p6
3d3
The number of unpaired electron is 4 in Cr2+
and is 3 in Mn4+
Example 5
How many unpaired electrons are there in each of the following in the ground state?
(i) O (ii) O+
(iii) O–
(iv) Fe
(v) Mn (vi) S (vii) F (viii) Ar
Solution
(i) 2 (ii) 3 (iii)1 (iv) 4
(v) 5 (vi) 2 (vii) 1 (viii)0

EXERCISE-I
1. Who determined the charge-to-mass ratio of the positive rays ? [1]
2. Describe the essential differences between cathode rays and the canal rays. [2]
3. Which problem of atomic structure was solved by the discovery of neutron ? [1]
4. How did the thickness of the gold foil affect the scattering of -particle in Rutherford’s scattering experi-
ment ? What conclusions can be drawn from it ? [3]
5. Which of the Na+
and He has completely filled K and L shells ? [1]
6. Which electrons of an atom are involved in the bond formation between different atoms ? [1]
7. Is it possible to write down the electronic configuration of an atom if we know its atomic number ?[1]
8. Write the electronic configuration of a positively charged sodium ion (Na+
). What is the atomic number of
Na+
? (atomic number of sodium (Na) is 11) [2]
9. The atomic number of two elements A and B are 18 and 16 respectively. Which of the two should be
chemically more reactive ? [2]
10. Is there any relationship between the valency of an element and the number of electrons its atom has in its
outermost shell ? Predict the valencies of helium (He), phosphorus (P), sulphur (S) and neon (Ne). The
atomic numbers of these elements are 2, 15, 16 adn 10 respectively. [3]
11. For the following statements, write T for true and F for False : [5]
(a) J. J. Thomson proposed that the nucleus of an atom contains only nucleons
(b) Aneutron is formed by an electron and proton combining together, therefore, it is neutral.
(c) The mass of an electron is about 1840 times that of proton
(d) Matter is electrical in nature and negatively charged
(e) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
EXERCISE-II
1. If an atom contains one electron and one proton, will it carry any charge or not ? [1]
2. On the basis of Thomson’s model atom is neutral as a whole comment. [1]
3. On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an
atom ? [1]
4. Draw a sketch of Bohr’s model of an atom with three shells. [1]
5. Name the three sub-atomic particles of an atom. [1]
6. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it has?
[1]
7. If K and L shells of an atom are full, then what would be the total number of electrons in the atom ?[1]
8. Write the distribution of electrons in carbon and sodium atoms. [2]

9. If number of electrons in an atom is 8 and number of protons is also 8, then (i) what is the atomic number
of the atom? and (ii) what is the charge on the atom ? [2]
10. Define valency by taking examples of silicon and oxygen. [2]
11. Na+
has completely filled K and L shells. Explain. [2]
12. If Z = 3, what would be the valency of the element ? Also, name the element. [2]
13. Composition of the nuclei of two atomic species X and Y are given as under [2]
X Y
Protons = 6 6
Neutrons =6 8
Give the mass numbers of X and Y. What is the relation between the two species ?
14. The average atomic mass of a sample of an element X is 16.2 u. What are the percentage of isotopes
X
and
X 18
8
16
8 in the sample ? [3]
15. Write the electronic configuration of any one pair of isotopes and isobars. [3]
16. Compare the properties of electrons, protons and neutrons. [3]
17. What are the limitations of J.J. Thomson’s model of the atom ? [3]
18. What are the limitations of Rutherford’s model of the atom ? [3]
19. Describe Bohr’s model of the atom. [5]
20. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two
uses of isotopes. [5]
EXERCISE-III
SECTION-A
 Multiple Choice question with one correct answers
1. An atom with atomic number 18 and mass number 40, has the following arrangement.
(A) 18p, 18e, 22n (B) 18p, 18e, 40n (C) 22p,18e, 18n (D) 22p,22e,18n
2. The isotope of carbon used in radiocarbon dating is
(A) C
12
6 (B) C
13
6 (C) C
14
6 (D) C
15
6
3. Which one of the following is correct electronic configuration of sodium ?
(A) 2,8 (B) 8,2,1 (C) 8,2 (D) 2,8,1
4. Particle in cathode rays have same charge to mass ratio as
(A)  -particles (B)  -rays (C) -rays (D) Protons
5. Which of the following particles has maximum charge to mass ratio ?
(A) Electrons (B) Protons (C)  -particle (D) Neutrons
6. Which of the following pairs of species are iso-electronic ?
(A) H2
O, H2
S (B) 
 2
3
3 ,CO
NO (C) H3
O+
, K+
(D) HF, HCI
7. Which of the following traids represents isotones ?
(A) 6
C12
, 6
C13
, 6
C14
(B) 18
Ar40
, 20
Ca42
, 21
Sc43
(C) 18
Ar40
, 20
Ca40
, 21
Sc41
(D) 7
N14
, 8
O16
, 9
F19
8. Which of the following statements is correct ?

(A) 7
N14
and 6
C13
are isotones (B) 7
N14
and 6
C14
are isotopes
(C) 7
N14
and 6
C12
are isobars (D) 7
N14
and 6
C15
are isotones
9. Which of the followings statement is not correct ?
(A) Isotones are atoms of different elements having same number of neutrons
(B) Isotopes are atoms of differenet elements having same number of protons
(C) Isobars are atoms of different elements having same number of neutrons
(D) Isotones and isobars are atoms of different elements.
10. When atoms of the gold sheet are bombarded by a beam of -particles, only a few -particles get
deflected whereas most of them go straight undeflected. This is because:
(A) the force of attraction on -particles by the oppositely charged electron is not sufficient.
(B) the nucleus occupies much smaller volume as compared to the volume of atom
(C) the force of repulsion on fast moving -particles is very small.
(D) the neutrons in the nucleus do not have any effect on -particles
11. A region in space around the nucleus of an atom where the probability of finding the electron is maximum
is called
(A) sub-level (B) orbit (C) orbital (D) electron shell
12. The possible sub shells in n = 3 energy shell are
(A) s, p, d (B) s, p, d, f (C) s, p (D) s only
13. Which of the following orbital does not make sense?
(A) 3d (B) 3f (C) 5p (D) 7s
EXERCISE-IV
SECTION-A
 Multiple Choice question with one correct answers
1. The maximum number of electrons with clockwise spin that can be accomodated in a f-subshell is
(A) 10 (B) 3 (C) 5 (D) 7
2. Indicate which electronic configuration amongst the following correctly represent SULPHUR atom ?
(A) 1s2
2s2
2p6
3s2
3p2
3d2
(B) 1s2
2s2
2p6
3s2
3p2
4s2
(C) 1s2
2s2
2p6
3s2
3p6
4s1
4p1
(D) 1s2
2s2
2p6
3s2
3p4
3. As we move away from nucleus, the energy of orbit.
(A) decreases (B) increases (C) remains unchanged (D) none of these
4. Electronic configuration of calcium atom can be written as
(A) [Ne] 4p2
(B) [Ar] 4s2
(C) [Ne] 4s2
(D) [Kr] 4p2
5. Which one of the following pairs of atoms/atom-ion have identical ground state configuration ?
(A) Li+
and He+
(B) Cl–
and Ar (C) Na and K (D) F+
and Ne
6. The correct electronic configuration of Cu(29) is
(A) 1s2
2s2
2p6
3s3
3p6
3d10
4s1
(B) 1s2
2s2
2p6
3s2
3p5
3d11
4s1
(C) 1s2
2s2
2p6
3s2
3p6
3d9
4s2
(D) 1s2
2s2
2p6
3s2
3p5
3d10
4s2
7. The number of d-electrons in Fe2+
(atomic number = 26) is equal to that of
(A) p-electrons in 10
Ne (B) s-electrons in 12
Mg
(C) d-electrons in Fe (D) p-electrons in Cl

8. Which of the following is the electronic configuration of Cu2+
(Z = 29) ?
(A) [Ar]4s1
3d8
(B) [Ar]4s2
3d10
4p1
(C) [Ar]4s1
3d10
(D) [Ar]3d9
9. Unpaired electrons in Ni++
, (Z = 28) is
(A) 0 (B) 2 (C) 4 (D) 8
10. Which of the following electronic configurations is not possible according to Hund’s rule ?0
(A) 1s2
2s2
(B) 1s2
2s1
(C) 1s2
2s2
2p1
x
2p1
y
2p1
z
(D) 1s2
2s2
2p2
x
2p1
y
2po
z
SECTION-B
 Multiple choice questions with one or more than one correct answers
1. Which of the following statement are wrong ?
(A) An atom is electrically neutral
(B) An atom & its ion have an unequal number of protons
(C) The size of a cation is smaller than that of corresponding atom
(D) An atom & its corrosponding anion have equal number of electrons
2. Which of the following sub-atomic particles is/are present in the nucleus of an atom?
(A) proton (B) electron (C) positron (D) meson
*******

Answers
Exercise-III
SECTION-A
1. (A) 2. (C) 3. (D) 4. (B) 5. (A)
6. (B) 7. (B) 8. (A) 9. (C) 10. (B)
11. (C) 12. (A) 13. (B)
Exercise-IV
SECTION-A
1. (D) 2. (D) 3. (B) 4. (B) 5. (B)
6. (A) 7. (A) 8. (D) 9. (B) 10. (D)
SECTION-B
1. (B,D) 2. (A,C,D)

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