Class 9 Science Ch 4 Structure Of Atom

This chapter Class 9 Science Ch 4 Structure of Atom explores the internal structure of atoms, the basic building blocks of matter. It explains how scientists developed different models of the atom over time, including

Thomson’s model, Rutherford’s model, and Bohr’s model. The chapter introduces the three main subatomic particles — electrons, protons, and neutrons — and their arrangement within the atom. Students also learn about atomic number, mass number, isotopes, and isobars, which are essential for understanding the behavior of elements and their placement in the periodic table.

⚡ Discovery of Cathode Rays (Detailed Explanation)

The discovery of cathode rays was one of the most important steps in understanding the structure of the atom. It helped scientists realize that atoms are made up of smaller particles — a finding that completely changed the way people thought about matter.

🔹 Background

In the mid-1800s, scientists were trying to understand how electricity passes through gases. They used special glass tubes called discharge tubes (or cathode ray tubes), which had two metal plates at the ends — the cathode (negative electrode) and the anode (positive electrode).. Most of the air inside these tubes was removed using a vacuum pump, leaving only a small amount of gas. Then, a high voltage was applied between the two electrodes.

🔹 What Scientists Observed

When the voltage was increased and the gas pressure was very low, scientists noticed a faint greenish glow on the glass wall of the tube, just behind the anode. This glow was not coming from the gas but from invisible rays that started at the cathode (negative plate) and moved toward the anode (positive plate). These invisible rays were called Cathode Rays.

🔹 Experiments by J.J. Thomson

The real understanding of these rays came from the experiments of Sir J.J. Thomson, a British physicist, in 1897.
Thomson studied the behavior of cathode rays using different materials and electric and magnetic fields.

Cathode rays always travel in a straight line from the cathode to the anode.

They can cast shadows of objects placed in their path.

When a light paddle wheel was placed in their path, it started to move, showing that the rays have mass and momentum.

When an electric or magnetic field was applied, the rays bent towards the positive side, showing that they carry negative charge.

From these observations, Thomson concluded that cathode rays are made up of very tiny negatively charged particles, which were later called electrons.

🔹 Key Findings of the Discovery

1. Cathode rays are streams of electrons. 2. Electrons are negatively charged particles. 3. Electrons are present in all atoms, no matter what gas or material the cathode is made of.

This means that atoms are not indivisible (as once thought in Dalton’s Atomic Theory) — they are made of smaller subatomic particles.

🔹 Conclusion

The discovery of cathode rays by J.J. Thomson in 1897 led to the discovery of the electron, the first known subatomic particle.
This discovery proved that atoms are made up of smaller particles, leading to the development of Thomson’s model of the atom and later atomic theories.

Class 9 Science Ch 4 Structure of Atom – Textbook

Question 1. Compare the properties of electrons, protons and neutrons.
Answer:

Question 2. What are the limitations of J.J. Thomson’s model of the atom?
Answer: According to J.J. Thomson’s model of an atom, the electrons are embedded all over in the positively charged spheres. But experiments done by other scientists showed that protons are present only in the centre of the atom and electrons are distributed around it.

Question 3. What are the limitations of Rutherford’s model of the atom?
Answer: According to Rutherford’s model of an atom the electrons are revolving in a circular orbit around the nucleus. Any such particle that revolves would undergo acceleration and radiate energy. The revolving electron would lose its energy and finally fall into the nucleus, the atom would be highly unstable. But we know that atoms are quite stable.

Question 4. Describe Bohr’s model of the atom.
Answer: Bohr’s model of the atom
(1) Atom has nucleus in the centre.
(2) Electrons revolve around the nucleus.
(3) Certain special orbits known as discrete orbits of electrons are allowed inside the atom.
(4) While revolving in discrete orbits the electrons do not radiate energy.
(5) These orbits or shells are called energy levels.
(6) These orbits or shells are represented by the letters K, L, M, N or the numbers n = 1, 2, 3, 4

Question 5. Compare all the proposed Bohr’s models of an atom given in this chapter.
Answer:

Question 6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer: The rules for writing of distribution of electrons in various shells for the first eighteen elements are:
(i) The maximum number of electrons present in a shell is given by the formula-2 n²
∵ n = orbit number i.e., 1, 2, 3
∵ Maximum number of electrons in different shells are:
K shell n = 1 2n² => 2(1)² = 2
L shell n = 2 2n² => 2(2)² = 8
M shell n = 3 2n² => 2(3)² = 18
N shell n = 4 2n² => 2(4)² = 32
(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.
(iii) Electrons are not accommodated in a given shell unless the inner shells are filled. (Shells are filled step-wise).

Question 7. Define valency by taking examples of silicon and oxygen.
Answer: Valency is the combining capacity of an atom.
Atomic number of oxygen = 8 Atomic number of silicon = 14 K L M
Electronic configuration of oxygen = 2 6 –
Electronic configuration of silicon =2 8 4
In the atoms of oxygen the valence electrons are 6 (i.e., electrons in the outermost shell). To fill the orbit, 2 electrons are required. In the atom of silicon, the valence electrons are 4. To fill this orbit 4 electrons are required.
Hence, the combining capacity of oxygen is 2 and of silicon is 4.
i.e., Valency of oxygen = 2
Valency of silicon = 4

Question 8. Explain with examples:
(i) Atomic number (ii) Mass number,
(iii) Isotopes and (iv) Isobars.
Give any two uses of isotopes.
Answer: (i) Atomic number: The atomic number of an element is equal to the number of protons in the nucleus of its atom. e.g., Oxygen has 6 protons hence atomic no. = 6.
(ii) Mass number: The mass number of an atom is equal to the number of protons and neutrons in its nucleus.
Nucleons = number of protons + number of neutrons Example: Protons + Neutrons = Nucleus = Mass number 6 + 6 = 12

(iii) Isotopes: Isotopes are atoms of the same element which have different mass number but same atomic number.

(iv) Isobars: Isobars are atoms having the same mass number but different atomic numbers.

Both calcium and argon have same mass number but different atomic number.
Two uses of isotopes are:
(i) An isotope of iodine is used in the treatment of goitre.
(ii) An isotope of uranium is used as a fuel in nuclear reactors.

Question 9. Na⁺ has completely filled K and L shells. Explain.
Answer: Sodium atom (Na), has atomic number =11
Number of protons =11
Number of electrons = 11
Electronic configuration of Na = K L M – 2 8 1
Sodium atom (Na) looses 1 electron to become stable and form Na⁺ ion. Hence it has completely filled K and L shells.

Question 10. If bromine atom is available in the form of say, two isotopes ⁷⁹₃₅Br (49.7%) and ⁸¹₃₅Br (50.3%), calculate the average atomic mass of bromine atom.
Answer:
Average Atomic Mass=(79 u×0.497)+(81 u×0.503)

Average Atomic Mass=39.263 u+40.743 u

Average Atomic Mass=80.006 u

Question 11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes ¹⁶₈X and ¹⁸₈X in the sample?
Answer: Let the percentage of ¹⁶₈X be x and the percentage of ¹⁶₈X be 100 – x.
To find the percentages of the two isotopes, you can use the formula for average atomic mass. Let the fractional abundance of the isotope 16X be represented by x. Then the fractional abundance of the isotope 18X will be (1 – x).

The average atomic mass is calculated as:

Average Atomic Mass=(Mass of Isotope 1×Abundance of Isotope 1)+(Mass of Isotope 2×Abundance of Isotope 2)

16.2=(16×x)+(18×(1−x)) = x=0.9

Since x is the fractional abundance of the 16X isotope, its percentage is 0.9×100%=90%.

The percentage of the 18X isotope is (1−x)×100%=(1−0.9)×100%=0.1×100%=10%.

Question 12. If Z = 3, what would be the valency of the element? Also, name the element.
Answer: Z = 3, (i.e, atomic number —> z)
∴ Electronic configuration = 2, 1
Valency = 1
Name of the element is lithium.

Question 13. Composition of the nuclei of two atomic species X and Y are given as under
X – Y
Protons =6 6
Neutrons = 6 8
Give the mass number of X and Y. What is the relation between the two species?
Answer: Mass number of X = Protons + Neutrons
= 6 + 6 = 12
Mass number of Y = Protons + Neutrons = 6 + 8 = 14
As the atomic number is same i.e., = 6.
[atomic number = number of protons].
Both X and Y are isotopes of same element.

Question 14. For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore,it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer: (a) False (b) False
(c) True (d) False

Put tick against correct choice and cross (x) against wrong choice in questions 15, 16 and 17.
Question 15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus (c) Proton
(b)Electron (d)neutron
Answer: (a) Atomic nucleus

Question 16.Isotopes of an element have
(a) the same physical properties (c) different number of neutrons
(b)different number of neutrons (d) different atomic numbers.
Answer: (c) different number of neutrons

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NCERT Class 10 Mathematics Solutions

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