RBSE Class 9 Science Notes Chapter 4 Structure of the Atom

These comprehensive RBSE Class 9 Science Notes Chapter 4 Structure of the Atom will give a brief overview of all the concepts.

RBSE Class 9 Science Chapter 4 Notes Structure of the Atom

→ Charged Atomic Particles of Matter : Atom and molecules are the fundamental building block of matter. Atom can be broken into electron, proton and neutron.

→ Discovery of Electron : J.J. Thomson discovered the electron with negative charge in 1897. Thomson carried a cathode ray experiment in which he observed a flow of negatively charged particles coming out of cathode towards the anode. These particles were named as electrons.

→ Discovery of Proton : Ernest Goldstein discovered proton in 1886. Goldstein observed in the same gas discharge tube that the anode emitted positive particles which he named as Canal Rays. This experiment led to the discovery of proton.

→ Discovery of Neutron : J. Chadwick discovered neutron in 1932. Neutron is present in the nucleus of all atoms.

→ Structure of an Atom : An atom consists three basic particles of an atom named as electron, proton and neutron.

→ Dalton’s Atomic Theory : John Dalton proposed atomic theory in 1808 on the basis of laws of chemical combination.

→ Thomson’s Model of Atom : This model was given by Joseph James Thomson in 1904.

→ Thomson’s model of an atom proposed that :

(i) An atom consists of a positively charged sphere and the electrons are embedded in it.
(ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.

This model is also known as watermelon model or plum pudding model or apple pie model.

→ Limitations of Thomson’s Model : Thomson’s model failed to explain the arrangement of protons and electrons in atom as shown by Rutherford.

→ Rutherford’s Model of Atom : This model was given by Ernest Rutherford in 1909. Rutherford’s model was based on an experiment in which a-particles were bombarded on a thin gold foil.

→ Conclusion of Rutherford’s experiment:

• Most of the part in an atom is empty.
• Very few. particles deflected from their path because of positively charged centre in atom. This centre is called nucleus.
• Very few particles bounced back at their original path because of the size of .. nucleus is very small compared to an atom.
• Electrons revolve round the nucleus.

→ Valence Electrons : The electrons present in the outermost shell of an atom are known as valence electrons.

→ Valency: Valency is the combining capacity of an atom or number of electrons lost or gained by an atom to acquire noble gas configuration.

→ Atomic Number (Z) : Atomic number of an element is equal to the number of protons present in the nucleus of an atom.

Atomic Number (Z) = no. of protons = no. of electrons.
According to Moseley, atomic number is a fundamental property of the atom. No two elements have the same atomic number.

→ Atomic Mass (A) : The total number of the protons and neutrons present in the nucleus of an atom is called mass number. It is denoted by A.
Mass number (A) = Number of protons + Number of Neutron
[∵ Number of protons = Atomic number (Z)]
Mass number (A) = Atomic number (Z) + Number of Neutrons
∴ Number of Neutrons = Mass number (A) – Atomic number (Z)

→ Nucleon : The protons and neutrons together are called nucleon. It can be represented by N.

→ Isotopes : Atoms of the same element having same atomic number but different mass number are known as isotopes.

→ Isobars : Atoms of different elements having same atomic mass are known as isobars.

→ Bohr-Bury Scheme : Bohr-Bury, in 1921, independently gave identical scheme for the distribution of electrons in various orbits or shells.

The maximum number of electrons that can be accommodated in an orbit or shell is equal to 2n², where n is the number of the shell.

• For first shell (K shell), n = 1 ∴ max e^– = 2 × 1² = 2
• For second shell (L shell), n = 2 ∴ max e^– = 2 × 2² = 8
• In the third shell (M shell), n = 3 ∴ max e^– = 2 × 3² = 18
• In the fourth shell (N shell), n = 4 ∴ ii max e^– = 2 × 4² = 32

→ Calculation of Average Atomic mass of two Isotopic Forms :