Structure of the Atom

(1) J. J Thomson discovered negatively charged sub-atomic particle electron.

It is represented as “e “.

In 1886, Eugen Goldstein discovered the presence of positively charged new radiation in the gas discharge and named them canal ray (anode ray).

The positively charged ion or radiation produced in a gas discharge tube is called canal ray (anode ray).

(2) In 1917 Rutherford discovered positively charged protons present in nucleus of atoms. It is represented as “p+”.

(3) Properties of the electron and protons are tabulated below –

SN. Property Electron Proton
1. Charge -1 +1
2. Mass Negligible One unit
3. Found Outer of the atom Interior of the atom
4. Removal from atom Easily removed Can not remove

Mass of proton = 2000(mass of electron) approx.

(4) The structures of atom

(i) Thomson’s Model (Plum – pudding model) of atom –

(a) An atom consists of a positively charged sphere and the electrons are embedded in it.

(b) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral. | Structure of the atom | Thomson's model of an atom | Diagram | NCERT | Class 9

Limitations of J.J.Thomson’s Model

(a) Thomson’s model can not explain stability of an atom. Therefore, it is fail to explain how protons hold on the electrons.

(b) Thomson’s model cannot explain nucleus.

(c) Distribution of electrons in the atom is not explained by this model.

(d)The experiments carried out by other scientists could not be explained by this model.

(ii) Rutherford’s Model of an atom-

Eranest Rutherford setup an experiment to know the arrangement of electrons within the atom. This experiment is popularly known as gold foil experiment.

Experiment –

(a) He selected a very thin gold foil layer about 1000 atoms thick.

(b) α– particles are fast moving doubly – charged helium ions that were made to fall on the thin layer of gold foil. α – particles have mass equal to 4u. Therefore, they have considerable amount of energy.

(c) Rutherford assumed that all charged particles and mass of the atom uniformly distributed in an atom (Thomson Model). Therefore, he was expecting that α – particles would pass through the gold atoms with little deflection or no deflection. | Structure of the atom | Rutherford's gold foil experiment | Diagram | NCERT | Class 9

Observation –

(a) Most of the fast moving α – particles passed straight through the gold foil.

(b) Some of the α– particles were deflected by the foil by small angles.

(c) One out of every 12000 particles appeared to rebound or bounce back (deflection by 1800) | Structure of the atom | Observation of the Rutherford's model of an atom | Diagram | NCERT | Class 9

Conclusion of α  – particle scattering experiment –

SN Observation Conclusion
1. Most of the α – particles passed through the foil without deflated Most of the space inside the atom is empty
2. Very few particles were deflected from the path Positive charge of the atom occupies very little space.


3. Very small fraction of  α-particles were deflected by 1800. All positive charge and mass of the gold atom were concentrated in a small volume within the atom. | Structure of the atom | Observation and result of the rutherford's model of an atom | Diagram | NCERT | Class 9

Proposal of Rutherford’s model –

(a) The centre of the atom, ie, nucleus is positively charged and all mass of the atom resides in it.

(b) The electrons revolve around the nucleus in circular paths.

(c) The size of the nucleus is very small in comparison of the size of the atom. | Structure of the atom | Rutherford's model of an atom | Diagram | NCERT | Class 9

Drawback of Rutherford’s model

(a) The revolution of the electrons around the nucleus in a circular path is not stable. The electrons radiate energy due to acceleration while moving on circular orbits. Therefore, revolving electron will lose energy and fall into the nucleus. As a result atoms will be unstable.

(b) He did not specify orbits and number of electrons in the orbits.

(iii) Bohr’s model of atom-

Neils Bohr’s postulations for atom structure are discuss below-

(a) Electrons can revolve in certain orbits around the nucleus. These orbits are known as discrete orbits.

(b) Electrons do not radiate energy while moving in discrete orbits.

(c)These orbits or shells are called energy levels. Smaller the size of orbit smaller the energy level of that orbit.  The orbits or shells are presented by letters K,L,M,N… or by numbers 1,2,3….. . | Structure of the atom | Bohr's model of an atom | Diagram | NCERT | Class 9

(5) Neutron is discovered by J. Chadwick in 1932. A sub – atomic particle with no charge and mass equal to proton which is found inside nucleus is known as neutron. It is represented by “n”.

(i) Every atom except hydrogen atom has neutron in their nucleus.

(ii) Mass of the atom = Mass of the proton + mass of the neutron.

(6)Distribution of electrons in orbits or shells was suggested by Bohr and Bury.

Rules of distribution of electrons in orbits-

Structure of the atom, Bohr's model of an atom, NCERT, class 9(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.

(iii)The shells are filled in a step by step. Therefore, inner shells or orbits should be filled first.

Structure of the atom, atomic structures of some elements, diagram, NCERT, class 9

(7) Electrons present in outermost shell of the any atom is known as valency electrons.

(i) According to Bohr – Bury the maximum number of 8 electrons can be accommodated in the outer shell. The atoms that have 8 electrons in outermost orbit are chemical inactive or very little active. Therefore, valency of these atoms is zero.

(ii) An outermost – shell which had 8 electrons was said to posses an octet.

(iii) The inert elements have 8 electrons in outermost shell of the atom except helium which have two electrons in outermost shell.

(iv) The atoms react to achieve octet in outermost – shell by sharing, gaining or losing electrons.

Rules to calculate valency

(a) If number of electron in outermost shell is four or less than four, then valency of that atom is number of electron present in outermost shell.

If number of electrons in outermost shell is 4or less than 4,

Valency = Number of electron in outermost shell

Eg:- Hydrogen has one electron in its outermost shell.

Valency = Number of electron in outermost shell

= 1

Hence, valency of hydrogen is 1.

(b) If number of electron in outermost shell is more than 4, then valency of that atom can be calculated by subtracted number of electron from octet.

If number of electron in outermost shell is more than 4,

Valency = Octet – number of electron in outermost shell

Eg:- Chlorine has 7 electrons in its outermost shell.

Valency = Octet – number of electron in outermost shell

= 8 – 7

= 1

Hence, valency of chlorine is one.

Composition fo Atoms of the first eighten elements with electronb distribution in various shells are tabulated below – | Structure of the atom | Electronic distribution in various shells | Table | NCERT | Class 9

(8) Atomic number is the number of protons present in the nucleus of an atom.

(i) All atoms of an element have same atomic number.

(ii) Atomic number is represented by “Z”.

Eg:- Hydrogen has one proton in its nucleus.

Hence, its atomic number Z = 1.

(8) The mass number is defined as the sum of the total number of protons and neutrons in the nucleus of an atom.

Mass number is represented by “A”.

Eg:- Find the mass number of carbon atom.

Number of protons in carbon nucleus = 6 p

Number of neutrons in carbon nucleus = 6 n

Mass number of carbon atom = 6 + 6

= 12 u

The mass number “A”, atomic number ”Z” and symbol of an element X can be written as

Structure of the atom, diagram, NCERT, class 9

The protons and neutrons in the nucleus are called nucleons.

(9) Atoms of the same element that have same atomic number but different mass number are called isotopes of the element.

Structure of the atom, NCERT, class 9

Properties of isobars –

(i) Many elements consist of mixture of isotopes.

(ii) Each isotope of an element is a pure substance.

(iii) The chemical properties of the isotopes are similar.

(iv) The physical properties of the isotopes are different.

(v) If an element occurs in isotopic form then average mass can be calculated with the percentage of each isotopic form.

Average atomic mass of element

= (Fractional abundance of isotope 1 ×  mass of isotope 1) + (Fractional abundance of isotope 2 ×  mass of isotope 2)…

If abundance in percentage, divide it by 100

Structure of the atom, example of average atomic mass of the element, NCERT, class 9Application –

Some isotopes have special properties. Some of them are given below –

(i) An isotope of uranium is used as a fuel in nuclear reactors.

(ii) An isotope of cobalt is used in the treatment of cancer.

(iii) An isotope of iodine is used in the treatment of goiter.

(10) Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars.

Eg:- Atomic number of calcium = 20

Mass number of calcium = 40

Atomic number of argon = 18

Mass number of argon = 40

Therefore, both elements have same mass number equal to 40 but different atomic numbers.

Helping Topics

NCERT solutions, class 9

Practice sheet, class 9

Atoms and Molecules


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