NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

The NCERT Solutions for Class 9 Science Chapter 4, "Structure of the Atom," are created by subject experts from Physics Wallah. This chapter focuses on explaining the basic parts of atoms, such as protons, neutrons, and electrons. It also covers the historical models proposed by scientists like J.J. Thomson and Niels Bohr to describe how atoms are organized. These solutions provide clear explanations and diagrams to help students understand concepts like atomic number, mass number, and electron arrangement in atoms.

NCERT Solutions for Class 9 Science Chapter 4 PDF

You can find the PDF link for NCERT Solutions for Class 9 Science Chapter 4 "Structure of the Atom" below. This PDF contains detailed solutions and explanations to help students understand the concepts related to atomic structure, including the discovery of subatomic particles and the development of atomic models by scientists. It is a valuable resource for students looking to deepen their understanding of chemistry and physics concepts.

NCERT Solutions for Class 9 Science Chapter 4 PDF

Structure of the Atom

The structure of an atom refers to its fundamental composition and organization. Atoms are the basic building blocks of matter and consist of three main subatomic particles:

Protons: Positively charged particles found in the nucleus of the atom. Each proton has a relative mass of approximately 1 atomic mass unit (amu).

Neutrons: Neutrally charged particles also located in the nucleus. Neutrons have a mass similar to protons, approximately 1 amu, but they do not carry an electrical charge.

Electrons: Negatively charged particles that orbit the nucleus in specific energy levels or shells. Electrons have negligible mass compared to protons and neutrons.

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

Below we have provided NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom for the ease of the students –

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.1 Page: 47

1. What are the canal rays?

Solution: The radiations that are positively charged are canal rays. This discovery was crucial in the discovery of another subatomic particle that was positively charged – the proton.

2. If an atom contains one electron and one proton, will it carry any charge or not?

Solution: Since a proton is a positively charged particle and an electron is a negatively charged particle, the net charge becomes neutral as both particles neutralise each other.

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.2 Page: 49

1. On the basis of Thompson’s model of an atom, explain how the atom is neutral as a whole.

Solution: Thomson's model of the atom, proposed in the late 19th century, suggested the following: (i) An atom is envisioned as a uniform sphere filled with positive charge, with negatively charged electrons embedded within it like raisins in a pudding. (ii) According to Thomson's model, electrons and the positive charge are distributed uniformly throughout the atom, making the atom electrically neutral overall. This concept of equal positive and negative charges within the atom led to it being referred to as the "plum pudding model."

2. On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?

Solution:

3. Draw a sketch of Bohr’s model of an atom with three shells.

Solution:

4. What do you think would be the observation if the ∝– particle scattering experiment is carried out using a foil of a metal other than gold?

Solution: In the alpha particle scattering experiment, if any other metal foil is used instead of gold, the observations would generally remain consistent. This is because the fundamental structure of atoms, regardless of the element used, involves a dense nucleus surrounded by electrons in orbits or energy levels. Rutherford's experiment demonstrated that alpha particles, when directed at a thin foil of gold, sometimes bounced back or were deflected at large angles, indicating a concentrated positive charge (the nucleus) in the center of the atom. This basic atomic structure applies universally to all elements: a positively charged nucleus containing protons and neutrons, with electrons orbiting around it. Therefore, while the degree of scattering may vary depending on the atomic number and density of the foil used, the overall observation regarding the atom's structure would remain consistent across different metals.

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.2.4 Page: 49

1. Name the three subatomic particles of an atom.

Solution: An atom consists of three subatomic particles:

• Protons  – Positively charged
• Electrons – Negatively charged
• Neutrons – Neutral in nature (no charge)

2. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

Solution: Given: Atomic mass of helium atom = 4u, 2 protons in helium nucleus Atomic mass = number of protons + number of neutrons 4 = 2 + number of neutrons Number of neutrons = 4 – 2 = 2 Hence, Helium has 2 neutrons.

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.3 Page: 50

1. Write the distribution of electrons in Carbon and Sodium atoms.

Solution:

2. If the K and L shells of an atom are full, then what would be the total number of electrons in the atom?

Solution: K shell can hold 2 electrons. L shell can hold 8 electrons. Hence, when both the shells are full, the total number of electrons present in the atom = 2+8 = 10 electrons.

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.4 Page: 52

1. How will you find the valency of chlorine, sulphur and magnesium?

Solution: Valency of an element refers to its tendency to gain, lose, or share electrons in order to achieve a stable electronic configuration, typically with a full outer shell. Mathematically, the valency can be determined based on the number of electrons in the outermost shell (also known as the valence shell):

• If the outermost shell contains 4 or fewer electrons, the valency is equal to the number of electrons in that shell.
• If the outermost shell contains more than 4 electrons, the valency is calculated by subtracting the number of electrons in the outermost shell from 8.

Calculation of valency:

Valency of chlorine:

The electronic configuration of chlorine = 2, 8, 7 Chlorine has 7 (more than 4) electrons in its outermost shell. Therefore, the valency of chlorine = 8 – the number of electrons in the outermost shell = 8−7 = 1 Valency of Sulphur: The electronic configuration of Sulphur = 2, 8,6 Sulphur has 6 (more than 4) electrons in its outermost shell. Therefore, the valency of chlorine = 8 – the number of electrons in the outermost shell = 8−6 = 2 Valency of magnesium: The electronic configuration of Magnesium = 2, 8, 2 Magnesium has 2 (less than 4) electrons in its outermost shell. Therefore, the valency of magnesium= Number of electrons in its outermost shell = 2

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.5 Page: 52

1. If the number of electrons in an atom is 8 and the 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?

Solution: Given: Number of electrons = 8 Number of protons = 8 (i) The atomic number of an atom is the same as the number of protons in that atom; hence, its atomic number is 8. (ii) In an atom, the number of protons is equal to the number of electrons. Hence, both the charges – positive and negative – neutralise each other. Therefore, the atom does not possess any charge.

2. With the help of the given table, find out the mass number of oxygen and sulphur atom.

Table: Composition of Atoms of the First Eighteen Elements with Electron Distribution in Various Shells.

Name of Element	Symbol	Atomic number	Number of Protons	Number of Neutrons	Number of electrons	Distribution of electrons K L   M   N				Valency
Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium	H He Li Be B C N O F Ne Na Mg	1 2 3 4 5 6 7 8 9 10 11 12	1 2 3 4 5 6 7 8 9 10 11 12	– 2 4 5 6 6 7 8 10 10 12 12	1 2 3 4 5 6 7 8 9 10 11 12	1 2 2 2 2 2 2 2 2 2 2 2	– – 1 2 3 4 5 6 7 8 8 8	– – – – – – – – – – 1 2	– – – – – – – – – — – –	1 0 1 2 3 4 3 2 1 0 1 2
Aluminium Silicon Phosphorus Sulphur Chlorine Argon	Al Si P S Cl Ar	13 14 15 16 17 18	13 14 15 16 17 18	14 14 16 16 18 22	13 14 15 16 17 18	2 2 2 2 2 2	8 8 8 8 8 8	3 4 5 6 7 8	– – – – –	3 4 3,5 2 1 0

Solution: (a) To find the mass number of Oxygen, Number of protons = 8 Number of neutrons = 8 Atomic number = 8 Atomic mass number = Number of protons + number of neutrons = 8 + 8 = 16 Therefore, the mass number of oxygen = 16 (b) To find the mass number of Sulphur, Number of protons = 16 Number of neutrons = 16 Atomic number = 16 Atomic mass number = Number of protons + number of neutrons = 16 + 16 = 32

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom Exercise-4.6 Page: 53

1. For the symbols H, D and T, tabulate three subatomic particles found in each of them.

Solution: The following table depicts the subatomic particles in Hydrogen (H), Deuterium (D), and Tritium(T).

Isotope	Symbol	Mass no.	Atomic no.	No. of electrons	No. of protons	No. of neutrons
Hydrogen	H	1	1	1	1	0
Deuterium	D	2	1	1	1	1
Tritium	T	3	1	1	1	2

2. Write the electronic configuration of any one pair of isotopes and isobar.

Solution:

(a) Isotopes: Isotopes are atoms of the same element that have the same number of protons (and therefore the same atomic number) but differ in the number of neutrons. This difference in neutrons results in isotopes having different mass numbers. Isotopes of an element behave similarly in chemical reactions because they have the same number of electrons and hence the same electronic configuration.

Example: Carbon has two major isotopes, Carbon-12 (6C12) and Carbon-14 (6C14). Both isotopes have the same atomic number (6) and thus the same number of protons and electrons, but Carbon-12 has 6 neutrons and Carbon-14 has 8 neutrons, leading to different mass numbers (12 and 14, respectively).

(b) Isobars: Isobars are atoms of different elements that have the same mass number but different atomic numbers. Unlike isotopes, which are of the same element, isobars are different elements altogether. This means isobars have different numbers of protons and electrons and therefore different atomic numbers.

Example: Calcium-40 (20Ca40) and Argon-40 (18Ar40) are isobars because they both have a mass number of 40. Calcium-40 has an atomic number of 20 (20 protons), while Argon-40 has an atomic number of 18 (18 protons). Despite the different atomic numbers, they can have similar electron configurations in their respective shells due to having the same number of electrons (20 for Calcium-40 and 18 for Argon-40) in their neutral state.

NCERT Solutions Class 9 Science Chapter 4 Exercise Page: 54

1. Compare the properties of electrons, protons and neutrons.

Solution:

Property	Electrons	Protons	Neutrons
Charge	Negatively charged	Positively charged	No charge.
Location	Located outside the nucleus	Located within the nucleus	Located inside the nucleus of an atom
Weight	Mass is negligible	1 a.m.u	1 a.m.u
Affinity	Attracted towards positively charged	Attracted towards negatively charged	Do not get attracted to any charged particle

2. What are the limitations of J.J.Thomson’s model of the atom?

Solution: The limitations of J.J. Thomson's model of the atom include:

1. Failure to Explain Alpha Particle Scattering: Thomson's model couldn't explain the results of Rutherford's alpha particle scattering experiment. This experiment showed that while most alpha particles passed through the gold foil, some were deflected at large angles and a few even bounced back. Thomson's model, which envisioned electrons embedded in a uniform positive sphere, did not account for the concentrated positive charge and empty space observed in Rutherford's experiment.
2. Lack of Experimental Evidence: Thomson's model was largely theoretical and based on the concept of the "plum pudding" model, where electrons were dispersed within a positively charged sphere. However, it lacked direct experimental evidence to support its claims about the structure and behavior of atoms.

3. What are the limitations of Rutherford’s model of the atom?

Solution: The limitations of Rutherford's model of the atom include:

1. Lack of Expected Stability in Electron Orbits: According to classical electromagnetic theory, an accelerating charged particle (like an electron orbiting a nucleus) should continuously emit energy in the form of electromagnetic radiation. This would cause the electron to lose energy and spiral into the nucleus, leading to instability rather than the observed stable atomic structures.
2. Energy Loss and Atom Stability: The model suggests that atoms, if governed purely by classical physics, would be highly unstable because electrons should continuously lose energy and collapse into the nucleus. However, atoms are observed to be stable entities, which contradicts this prediction

4. Describe Bohr’s model of the atom.

Solution:

• An atom holds the nucleus at the centre.
• Negatively charged electrons revolve around the nucleus.
• The atoms in it contain distinct orbits of electrons.
• Electrons do not radiate energy when they are in their orbits.
• The distinct orbits are named K, L, M, and N orbits. Numbers used to denote them are n=1, 2, 3, 4

5. Compare all the proposed models of an atom given in this chapter.

Solution:

Thomson	Rutherford	Bohr
● Sphere is positively charged. ● Electrons are negatively charged and scattered all through the inside of the sphere. ● Positively charged = negatively charged ● The net charge in the atom is zero.	● The nucleus is at the centre and is positively charged, holding the entire mass. ● Electrons are negatively charged, revolving in a well-defined path ● In comparison with the nucleus, the size of the atom is very large. ● Force of attraction of the electrons towards the nucleus is balanced by centrifugal force acting away from it. As a result, electrons are not drawn close to the nucleus.	● Nucleus is present at the centre and is positively charged ● Electrons are negatively charged, revolving around but do not radiate energy. ● The distinct orbits are labelled as K, L, M, and N

6. Thomson’s Model of Atom.

7. Rutherford’s Model of Atoms.

8. Bohr’s model of the atom.

Summarise the rules for the writing of the distribution of electrons in various shells for the first eighteen elements.

Solution:

• The maximum number of electrons that can be accommodated in a shell is given by the formula: 2n ² , where n= 1, 2, 3…
• The maximum number of electrons in different shells are:

K shell – n=1 ; 2n ² = 2(1) ² = 2 L shell – n=2 ; 2n ² = 2(2) ² = 8 M shell – n=3 ; 2n ² = 2(3) ² = 18 N shell- n=4 ; 2n ² = 2(4) ² = 32

• The outermost orbit can be accommodated with 8 electrons at the maximum.
• The electrons are not taken in unless the inner shells are filled, which are filled step-wise; hence, the highest element has K-2; L-8; M-8 distribution of electrons.

9. Define valency by taking examples of silicon and oxygen.

Solution: Valency refers to the combining capacity of an atom, specifically how many electrons an atom needs to gain, lose, or share to achieve a stable electron configuration, typically with a full outer shell (octet rule).

Example of Silicon (Atomic Number 14):

• Silicon has 14 electrons.
• Electron distribution: K-shell (2 electrons), L-shell (8 electrons), M-shell (4 electrons).
• To complete its outermost shell (M-shell), silicon needs 4 more electrons.
• Therefore, the valency of silicon is 4.

Example of Oxygen (Atomic Number 8):

• Oxygen has 8 electrons.
• Electron distribution: K-shell (2 electrons), L-shell (6 electrons).
• To complete its outermost shell (L-shell), oxygen needs 2 more electrons.
• Therefore, the valency of oxygen is 2.

10. Explain with examples

(i) Atomic number,

(ii) Mass number,

(iii) Isotopes and

(iv) Isobars.

Give any two uses of isotopes.

Solution: (i) The number of positively charged protons present in the nucleus of an atom is defined as the atomic number and is denoted by Z. Example: Hydrogen has one proton in its nucleus; hence, its atomic number is one. (ii) The total number of protons and neutrons present in the nucleus of an atom is known as the mass number. It is denoted by A. ₂₀ Ca ⁴⁰ . The mass number is 40. The atomic number is 20. (iii) The atoms which have the same number of protons but a different number of neutrons are referred to as isotopes. Hence, the mass number varies. Example: The most simple example is the Carbon molecule which exists as ₆ C ¹² and ₆ C ¹⁴ (iv) Isobars: Isobars are atoms which have the same mass number but differ in atomic number. Examples are, ₂₀ Ca ⁴⁰ and ₁₈ Ar ⁴⁰ Uses of isotopes

• The isotope of the Iodine atom is used to treat goitre, an iodine-deficient disease.
• In the treatment of cancer, an isotope of cobalt is used.
• Fuel for nuclear reactors is derived from the isotopes of the Uranium atom.

11. Na+ has completely filled K and L shells. Explain.

Solution: Sodium (Na) has an atomic number of 11, which means it has 11 electrons.

• Electronic Configuration before Ionization: K-shell (2 electrons), L-shell (8 electrons), M-shell (1 electron)
• Sodium tends to lose one electron to achieve a stable electron configuration, as elements tend to gain or lose electrons to reach a noble gas configuration.
• Formation of Sodium Ion (Na^+): When sodium loses one electron, it forms a sodium ion (Na^+).
• Electronic Configuration after Ionization: K-shell (2 electrons), L-shell (8 electrons).
• The resulting sodium ion (Na^+) now has a full outer shell (noble gas configuration), making it more stable.
• It is indeed difficult to remove an electron from a filled shell, as atoms and ions tend to achieve stability by filling or emptying their outermost electron shells.

12. If the bromine atom is available in the form of, say, two isotopes ₃₅ Br ⁷⁹ (49.7%) and ₃₅ Br ⁸¹ (50.3%), calculate the average atomic mass of the Bromine atom.

Solution: The atomic mass of an element is the mass of one atom of that element. Average atomic mass takes into account the isotopic abundance. Isotope of bromine with atomic mass 79 u = 49.7% Therefore, Contribution of ³⁵ Br ₇₉ to atomic mass = (79 × 49.7)/100 ⇒ 39.26 u Isotope of bromine with atomic mass 81 u = 50.3% Contribution of ³⁵ Br ₈₁ to the atomic mass of bromine = (81 × 50.3)/100

⇒ 40.64u

Hence, the average atomic mass of the bromine atom = 39.26 + 40.64 u = 79.9u

13. The average atomic mass of a sample of element X is 16.2 u. What are the percentages of isotopes ₈ X ¹⁶ and ₈ X ¹⁸ in the sample?

Solution: Let the percentage of ₈ X ¹⁶ be ‘a’ and that of ₈ X ¹⁸ be ‘100-a’. As per the given data, 16.2u = 16 a / 100 + 18 (100-a) /100 1620 = 16a + 1800 – 18a 1620 = 1800 – 2a a = 90% Hence, the percentage of the isotope in the sample ₈ X ¹⁶ is 90% and that of ₈ X ¹⁸ = 100-a = 100- 90=10%

14. If Z=3, what would be the valency of the element? Also, name the element.

Solution: Given: Atomic number, Z = 3 The electronic configuration of the element = K-2; L-1, hence its valency = 1 The element with atomic number 3 is Lithium.

15. 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 numbers of X and Y. What is the relation between the two species?

Solution: Mass number of X: Protons + neutrons = 6+6 = 12 Mass number of Y: Protons + neutrons = 6+8 = 14 They are the same element, and their atomic numbers are the same. They are isotopes, as they differ in the number of neutrons and hence their mass numbers.

16. 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 a proton. (d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. Solution: (a) The statement is False. (b)  The statement is False. (c)  The statement is True. (d)  The statement is False.

17. Put a tick(✓) against the correct choice and cross(x) against the wrong choice in questions 15, 16 and 17.

Rutherford’s alpha–particle scattering experiment was responsible for the discovery of

(a) Atomic nucleus (b) Electron (c) Proton (d) Neutron Solution: (a) Atomic nucleus

Isotopes of an element have

(a) The same physical properties (b) Different chemical properties (c) Different number of neutrons (d) Different atomic numbers Solution: (c) Different number of neutrons

18. Number of valence electrons in Cl ^– ion are

(a) 16 (b) 8 (c) 17 (d) 18 Solution: (b) 8 The electronic distribution of Cl is K-2, L-8, M-7. Valence electrons are 7; hence, chlorine gains one electron for the formation of Cl ^– . Therefore, its valency is 8.

19. Which one of the following is a correct electronic configuration of Sodium?

(a) 2, 8 (b) 8, 2, 1 (c) 2, 1, 8 (d) 2, 8, 1 Solution: (d) 2, 8, 1

Complete the following table.

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9 16 – – –	– 32 24 2 1	10 – – – 0	– – 12 1 1	– – – – 0	– Sulphur – – –

Solution: The following table depicts the missing data: Atomic number(Z) = Number of protons Mass number = Number of neutrons + atomic number (or) Mass number(A) = Number of neutrons + number of neutrons

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9 16 12 1 1	19 32 24 2 1	10 16 12 1 0	9 16 12 1 1	9 16 12 1 0	Fluorine Sulphur Magnesium Deuterium Hydrogen

Benefits of NCERT Solutions for Class 9 Science Chapter 4

• Comprehensive Understanding: These solutions provide clear explanations and detailed answers to help students understand the fundamental concepts of atomic structure, including the discovery of subatomic particles and atomic models proposed by scientists.
• Structured Learning: NCERT Solutions provide a structured approach to learning, covering each topic systematically and ensuring thorough comprehension of complex concepts related to atoms and their properties.
• Exam Preparation: By practicing with NCERT Solutions, students can effectively prepare for exams. These solutions include solved examples and exercises that help in grasping concepts and improving problem-solving skills.
• Clarity and Accuracy: The solutions are prepared by subject matter experts, ensuring accuracy and clarity in explanations, which helps students in developing a strong foundation in science.
• Enhanced Learning Experience: NCERT Solutions encourage self-study and promote deeper engagement with the subject matter through interactive and illustrative explanations.

NCERT Solutions Class 9 Science Chapter-wise List

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around Us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of The Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

Chapter 9 - Force and Laws of Motion

Chapter 10 - Gravitation

Chapter 11 - Work and Energy

Chapter 12 - Sound