Case Study Questions for Class 9 Science Chapter 8 Journey Inside the Atom (Exploration Book) 2026-27

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This page provides Case Study Questions for Class 9 Science Chapter 8 – Journey Inside the Atom from the latest NCERT Exploration textbook. Each set contains a reading passage followed by 4 objective/short-answer questions, exactly as expected in the CBSE Board examination pattern.

1

Case Study: The Gold Foil Experiment

Read the passage carefully, then answer all four questions
Ch 8 · Journey Inside the Atom

In 1911, Geiger and Marsden, working under Ernest Rutherford, tested Thomson’s model of the atom through what became famous as the gold foil experiment. They aimed a narrow beam of alpha particles at an extremely thin sheet of gold foil. Alpha particles are tiny, positively charged particles emitted from certain radioactive elements. According to Thomson’s model, the positive charge in the atom was spread out evenly. So they expected the alpha particles to pass straight through the gold foil or be deflected only slightly.

But to their surprise, while most particles passed through undeflected, some were sharply deflected, and a few even bounced back. This deflection from the straight path is called scattering. This single surprising result completely ruled out Thomson’s ‘plum pudding model’ of the atom and led Rutherford to propose the nuclear model of the atom.

1
What are alpha particles as described in the context of Rutherford’s experiment?
a Negatively charged electrons
b Uncharged neutrons
c Positively charged nuclei of helium atoms
d High-energy light waves
Correct Answer (c) Positively charged nuclei of helium atoms
Explanation

Alpha particles are tiny, positively charged particles emitted from radioactive elements. They are essentially the nucleus of a helium atom, consisting of two protons and two neutrons.

2
What did the observation that the vast majority of alpha particles passed straight through the gold foil indicate?
a That most of an atom is simply empty space.
b That the nucleus is negatively charged.
c That electrons are extremely heavy.
d That the positive charge is spread evenly across the atom.
Correct Answer (a) That most of an atom is simply empty space.
Explanation

Because most alpha particles went straight through the foil without any deflection, Rutherford concluded that the atoms in the gold foil were largely composed of empty space rather than solid, uniformly distributed matter.

3
Which specific finding directly contradicted Thomson’s “plum pudding” model?
a Alpha particles being positively charged.
b Electrons revolving in circular orbits.
c A few alpha particles bouncing sharply backwards.
d The continuous emission of radiation.
Correct Answer (c) A few alpha particles bouncing sharply backwards.
Explanation

Thomson’s model suggested a diffuse, spread-out positive charge that wouldn’t exert enough force to repel a fast-moving, heavy alpha particle. The sharp bouncing back of a few particles proved the existence of a dense, concentrated positive center.

4
Based on the conclusions of the gold foil experiment, where is the entire positive charge and most of the mass of an atom located?
a Spread uniformly throughout the atom.
b Dispersed among the revolving electrons.
c In a tiny, dense central region called the nucleus.
d In the empty space between the electron orbits.
Correct Answer (c) In a tiny, dense central region called the nucleus.
Explanation

To explain the sharp deflections, Rutherford concluded that all the positive charge and nearly all the mass of the atom are tightly packed into an extremely small central core, which he named the nucleus.

2

Case Study: Isotopes and Average Mass

Read the passage carefully, then answer all four questions
Ch 8 · Journey Inside the Atom

Scientists discovered that atoms of the same element can have the same number of protons (atomic number, Z) yet have different numbers of neutrons, resulting in different mass numbers. These “twin atoms” with the same atomic number but different mass numbers are called isotopes. For example, carbon has three isotopes: Carbon-12, Carbon-13, and Carbon-14. Each of these has six protons and six electrons, but they differ in their number of neutrons.

Because these isotopes occur in different proportions in nature, the atomic mass of a natural element is calculated as the weighted average of the masses of its naturally occurring isotopes. For instance, chlorine occurs in nature in two isotopic forms, 35u and 37u, in a ratio of 3:1 (or 75% to 25%). Calculating the weighted average gives us the accurate atomic mass of 35.5u for chlorine as it exists in nature.

1
What fundamentally causes isotopes of the same element to have different mass numbers?
a They have a different number of protons.
b They have a different number of electrons.
c They have a different number of neutrons.
d They have a different number of valence shells.
Correct Answer (c) They have a different number of neutrons.
Explanation

Isotopes of an element have the exact same number of protons (which defines the element) but differ in the number of neutrons present in their nucleus, which changes their total mass number.

2
Why do all isotopes of a given element exhibit identical chemical properties?
a Because they have the same mass number.
b Because they have the same number of neutrons.
c Because they have the same electronic configuration.
d Because they are found in the same physical state.
Correct Answer (c) Because they have the same electronic configuration.
Explanation

Chemical properties depend primarily on the number of valence electrons. Since all isotopes of an element have the same number of protons and electrons, their electronic configuration and chemical behavior remain identical.

3
A sample of chlorine consists of 75% Chlorine-35 and 25% Chlorine-37. How is its average atomic mass calculated?
a By taking a simple average: (35 + 37) / 2
b By adding their percentages together: 75 + 25
c By multiplying their mass by their abundance: (35 × 0.75) + (37 × 0.25)
d By subtracting the mass of neutrons from protons.
Correct Answer (c) By multiplying their mass by their abundance: (35 × 0.75) + (37 × 0.25)
Explanation

Because isotopes do not occur in equal ratios, the accurate atomic mass is calculated using a weighted average. This is done by multiplying the mass of each isotope by its percent relative abundance and adding the values together.

4
Which specific isotope of carbon is widely used by scientists in archaeology to determine the age of ancient fossils?
a Carbon-12
b Carbon-13
c Carbon-14
d Carbon-16
Correct Answer (c) Carbon-14
Explanation

Carbon-14 is an isotope of carbon that is utilized in carbon dating. It helps archaeologists and geologists estimate the age of ancient organic materials and artefacts.

Chapters covered in CBSE Class 9 Science Latest Book – Exploration

  • Chapter 1: Exploration: Entering the World of Secondary Science
  • Chapter 2: Cell: The Building Block of Life
  • Chapter 3: Tissues in Action
  • Chapter 4: Describing Motion Around Us
  • Chapter 5: Exploring Mixtures and their Separation
  • Chapter 6: How Forces Affect Motion
  • Chapter 7: Work, Energy, and Simple Machines
  • Chapter 8: Journey Inside the Atom
  • Chapter 9: Atomic Foundations of Matter
  • Chapter 10: Sound Waves: Characteristics and Applications
  • Chapter 11: Reproduction: How Life Continues
  • Chapter 12: Patterns in Life: Diversity and Classification
  • Chapter 13: Earth as a System: Energy, Matter, and Life

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Case Study Questions for Class 9 Science Chapter 8 Journey Inside the Atom (Exploration Book) 2026-27

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