According to this model :
The positive charge and most of the mass of the atom is densely
concentrated in the extremely small region. This portion of atom is called
nucleus.
The nucleus is surrounded by electrons that move around the nucleus with
a very high speed in circular paths called orbits.
Electrons and the nucleus are held together by electrostatic force of
attraction.
The volume of the nucleus is very small and is only a minute fraction of
the total volume of the atom.
(VA = 1015 VN)
The radius of a nucleus is proportional to the cube root of the number of
nucleons within it.
R A1/3 R = R0A1/3
where, R0 = 1.33 × 10–15 m (constant),
A = mass number (p + n) ,
R = radius of the nucleus.
R = 1.33 × 10–15 × A1/3 m
N TE
Order of radius of nucleus is 10-15 m and Order of radius of atom is
10-10 m.
According to Maxwell, an electron loose its energy continuously in the
form of electromagnetic radiations. As a result, the e– should loose
energy at every turn and move closer and closer to the nucleus following
a spiral path. The ultimate result will be that it will fall into the nucleus,
thereby making the atom unstable.
e
Example
Calculate the radius of a nucleus of an atom that has 216 nucleons in it.
Solution
Example
Prove that the density of nucleus is independent of the element which we
are taking
Solution
Example
Calculate the radius of nucleus of 13Al
27 ?
Solution
Part -2
Atomic number and
Mass number
(1) Atomic Number (Z)
The number of protons present in the nucleus is called atomic number of
an element.
Number of proton = Atomic number(Z)
For neutral atom : Number of e– = Number of proton
For charged atom : Number of e– = Z – (charge on atom)
(2) Mass Number (A)
The sum of number of Neutrons and protons is called the mass number of
element.
It is also known as number of nucleons.
It is always a whole number.
Formula
A = number of protons + number of Neutrons
Mass Number A
Atomic Number Z
X Element Symbol
4
2He
Mass Number = No. of protons + No. of neutrons
= Atomic number (Z) + No. of neutrons
Number of neutrons =A–Z
(A) ISOTOPES
They are atoms of a given element which have the same atomic number
but differ in their mass number.
1H , 1H , 1H
1 2 3
Example
1H 1H 1H
1 2 3
Protium (H) Deuterium (D) Tritium (T)
Proton →
Electron →
Neutron →
Neutron is not available in Protium
N TE
Isotopes have the same nuclear charge but differ in the number of neutrons
in the nucleus.
(B) ATOMIC WEIGHT
The atomic weight of an element is the average of mass of all the
isotopes of that element.
If an element have three isotopes
Isotopes (1) Isotopes (2) Isotopes (3)
Isotopes y1 y2 y3
Weights w1 w2 w3
Percentage
x1 x2 x3
Occurrence
𝐰𝟏 𝐱 𝟏 + 𝐰𝟐 𝐱 𝟐 + 𝐰𝟑 𝐱 𝟑
Average atomic weight =
𝐱𝟏 + 𝐱𝟐 + 𝐱𝟑
Example
Cl atom has two isotopes so average atomic mass is calculated as :
35Cl 37Cl
Percentage 75% 25%
Ratio 3 1
𝟑𝟓×𝟑 + 𝟑𝟕×𝟏
Average Atomic weight = = 35.5
𝟑+𝟏
(C) ISOBARS
Isobars are the atoms of different element which have the same mass
number but different Atomic number.
Example
20Ca
40
19K
40
p= p=
e= e=
n= n=
n+p= n+p=
(D) ISODIAPHERS
They are the atoms of different element which have the same difference
of the number of Neutrons & protons.
Example
90Th 92U
234 238
p= p=
n= n=
n–p= n–p=
(E) ISOTONE/ISONEUTRONIC SPECIES
They are the atoms of different element which have the same number of
neutrons.
Example
19K 20Ca
39 40
n= n=
(F) ISOSTERS
They are the molecules which have the same number of atoms &
electrons.
Example
CO2 N2O
Atoms = =
Electrons = =
(G) ISOELECTRONIC SPECIES
They are the atoms, molecules or ions which have the same number of
electrons.
Example
N3- NH3 Ne
e=
Example
Find the number of neutrons, protons and electrons in:
Species Protons Neutrons Electrons
(i) 9F19
16 2–
(ii) 8O
(iii) 24
12Mg
2+
Solution
Example
If in A–2 numbers of electrons are 18 and mass number is 34, then
determine the number of protons, number of neutrons & number of
electrons in A+3.
Solution
Example
The mass number of two atoms X & Y, having the same atomic number are
108 & 110 respectively. If X has 58 neutrons in its nucleus then find number
of neutrons in Y & its atomic number.
Solution
Example
Average atomic weight of an element 'Y' = 41.8. If it exists in the form of
2 isotopes 40Y and 42Y, then find their percentage occurrence in nature.
Solution
