Class 9 Science Chapter 8: Journey Inside the Atom — Solutions

These Class 9 Science Exploration Chapter 8 solutions cover Journey Inside the Atom from NCERT’s Exploration, first edition April 2026 (printed pages 140–161).

Class: 9 Subject: Science Book: Exploration Chapter: 8 Exercise: Revise, Reflect, Refine (15 Qs) Session: 2026–27
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Class 9 Science Exploration Chapter 8 Solutions – Overview

Chapter 8 of Exploration, Journey Inside the Atom, traces how scientists discovered that the atom is divisible. It covers the subatomic particles (electrons, protons and neutrons), the atomic models of Dalton, Thomson, Rutherford and Bohr (including the gold foil experiment), atomic number and mass number, electronic configuration and valency, and the ideas of isotopes and isobars. The worked answers distinguish historical atomic models, shell capacities, mass number and isotope masses.

Key Concepts & Definitions

Subatomic particles: electron (negative, around the nucleus), proton (positive, in the nucleus), neutron (no charge, in the nucleus).

Atomic number (Z): number of protons. Mass number (A): protons + neutrons.

Number of neutrons = A − Z; in a neutral atom, electrons = protons = Z.

Electronic configuration: electrons fill shells K, L, M… (shell capacities given by 2n2: K = 2, L = 8, M = 18…; the first 18 elements occupy shells up to 2, 8, 8). Valency = electrons gained, lost or shared (often 8 − valence electrons for non-metals).

Isotopes: same Z, different A (different neutrons). Isobars: same A, different Z.

Bohr’s model: electrons revolve in fixed energy orbits without losing energy, so the atom is stable.

“Think It Over” — Answers

Are atoms the smallest particles, or can they be divided further?

ANSWERAtoms are not the smallest — they are divisible. Atoms contain protons and electrons; most also contain neutrons. Protium (hydrogen-1) has no neutron. In a modern description electrons occupy quantum states rather than definite planetary paths.

Why do the electrons not fall into the protons in the nucleus?

ANSWERIn Bohr’s model the electrons move in fixed energy orbits and do not lose energy while doing so, so they do not spiral inward. This is a postulate of Bohr’s historical model, not a consequence of electrostatic attraction alone; modern quantum theory uses allowed electron states.

Atomic notation and model limitations

ZAX,A=p+n,Z=p{}^{A}_{Z}\mathrm X,\qquad A=p+n,\qquad Z=pn=A−Z,e=Zfor a neutral atomn=A-Z,\qquad e=Z\quad\text{for a neutral atom}Nmax⁡=2nshell2N_{\max}=2n_{\mathrm{shell}}^2

Shell capacity is not a universal rule to fill every shell completely before starting the next. The chapter’s simple distributions apply to its first 18 elements. Bohr diagrams are useful counting models; modern electrons do not follow these drawn circular tracks.

Hydrogen isotopes: protium has one proton and zero neutrons, deuterium one proton and one neutron, tritium one proton and two neutrons; each neutral atom has one electron.
Hydrogen-1 is the no-neutron exception. The three isotopes share Z = 1; their mass numbers are 1, 2 and 3. Schematic, not to scale. Open full-size diagram.

Weighted average atomic mass

Using the textbook’s simplified chlorine masses and 3:1 abundance:

mˉ=75100(35 u)+25100(37 u)\bar m=\frac{75}{100}(35\,\mathrm u)+\frac{25}{100}(37\,\mathrm u)mˉ=35.5 u\bar m=35.5\,\mathrm u

This is an average across atoms, not the mass number of one chlorine atom. Actual isotope masses and abundances give a slightly different standard atomic weight.

Think It Over: Why change atomic models?

New evidence exposed limitations: electrons contradicted an indivisible atom, alpha scattering required a small nucleus, and atomic stability required more than a classical planetary picture. Models are revised when experiments or explanatory limits demand it.

Pause and Ponder: Answers 1–18

  1. (i) Excess negative charge makes the model negatively charged. (ii) Adding negative charge to the clay also prevents neutrality unless balanced by positive charge elsewhere.
  2. An orange or lemon can illustrate inclusions inside a larger body, but seeds and pulp do not reproduce electric charge, atomic scale or the distribution of electrons. It is an analogy.
  3. Cathode-ray properties did not depend on the electrode material or gas, suggesting electrons are components common to atoms.
  4. A negative projectile is attracted toward the positive nucleus, rather than repelled like an alpha particle. Its mass and energy also affect scattering, so changing only the charge does not fully specify an experiment.
  5. A few very large deflections require a strong, concentrated repulsive interaction; diffuse positive charge in Thomson’s model could not account for them.
  6. Sample question: How did you distinguish nuclear scattering from other effects in the foil?
  7. Option (ii): both statements are true; Thomson’s description is not the reason for Rutherford’s nuclear conclusion.
  8. Sample: a student named Anaya might propose “Anayium”, Ay. A capital first letter and lower-case second letter follow symbol style, but a classroom proposal is not an officially approved element name or symbol.
  9. Different symbols for the same element would make scientific communication and formula interpretation unreliable.
  10. 26 electrons, 26 protons and 30 neutrons.
  11. 41 − 20 = 21 neutrons.
  12. 17 + 18 = 35.
  13. 23 − 11 = 12 neutrons, assuming a neutral atom.
  14. C: 4; F: 7; Si: 4 valence electrons.
  15. Z = 12: 2, 8, 2; Z = 16: 2, 8, 6; Z = 18: 2, 8, 8.
  16. Sodium-23, with 12 neutrons. Sample riddle: I have Z = 8 and A = 16; my neutral atom has eight electrons and eight neutrons. I am oxygen-16.
  17. Both are sodium, Z = 11. Their mass numbers are 23 and 24, so they are isotopes.
  18. Use the given hypothetical abundances, rather than substituting remembered natural values:
mˉ=79(49.7100)+81(50.3100)\bar m=79\left(\frac{49.7}{100}\right)+81\left(\frac{50.3}{100}\right)mˉ=80.006 u≈80.01 u\bar m=80.006\,\mathrm u\approx80.01\,\mathrm u

Explorations and sample activities

Thicker foil: multiple scattering and energy loss become more likely; fewer particles pass straight through, and a sufficiently thick foil can stop them. A thicker foil is less suitable for attributing a trajectory to a single scattering. Radioactive-source experiments require specialist facilities and must not be reproduced as classroom handling activities.

10−4 m10−10 m=106\frac{10^{-4}\,\mathrm m}{10^{-10}\,\mathrm m}=10^6

The paper-thickness estimate is about one million atomic diameters, not an exact atom count for real paper. For Table 8.4, common introductory valencies H to Ar are: 1, 0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 3, 2, 1, 0. Some elements exhibit other valencies in compounds.

Journey Beyond sample: build a prediction board using Z = 12, A = 24, p = e = 12 and n = 12; plot K/L/M electron counts for sodium (2/8/1), magnesium (2/8/2) and aluminium (2/8/3). A documentary response must name the film actually watched and separate its claims from questions left open; do not invent viewing or observations.

Class 9 Science Exploration Chapter 8 Solutions — Revise, Reflect, Refine

1. Choose the correct options and explain, in the context of Rutherford’s gold foil experiment. (i) The experiment clearly showed the existence of neutrons in the nucleus. (ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom. (iii) The large deflection of a few alpha particles indicated that most of the mass and positive charge are packed into a tiny centre. (iv) The way alpha particles were deflected showed that electrons move around the nucleus.

ANSWER Correct: (ii) and (iii). (ii) is correct — most alpha particles passed straight through, so the atom is mostly empty space, disproving the plum pudding model and pointing to a small central nucleus. (iii) is correct — only a few particles bounced back, so the positive charge and most of the mass are concentrated in a tiny, dense nucleus. (i) is wrong — the experiment did not show neutrons (discovered later by Chadwick).(iv) is wrong — it revealed the nucleus, not the motion of electrons.

2. Which statements are correct or incorrect according to Bohr’s atomic model? Give a reason for each. (i) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus. (ii) Electrons can exist anywhere around the nucleus with no fixed energy. (iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. (iv) Electrons can be found between energy levels as they move around the nucleus.

ANSWER Only (iii) is correct — Bohr proposed that electrons move in fixed orbits (energy levels) without radiating energy, which keeps the atom stable. (i) is incorrect — in a fixed orbit the electron does not lose energy, so it does not fall in. (ii) is incorrect — electrons have definite (quantised) energies, not any value. (iv) is incorrect — electrons are found only in allowed orbits, not between them.

3. The nuclei of three species are: X (18 protons, 19 neutrons), Y (17 protons, 18 neutrons), Z (17 protons, 20 neutrons). Explain the relation between: (i) Y and Z (ii) Z and X

ANSWER (i) Y and Z both have 17 protons (same element) but different neutrons (18 and 20), so they are isotopes. (ii) Z has mass number 17 + 20 = 37 and X has mass number 18 + 19 = 37 — same mass number but different atomic numbers (17 and 18), so they are isobars.

4. What conclusion did Rutherford draw about the position and characteristics of the atom’s positively charged part, based on the few alpha particles that bounced back or were deflected at large angles?

ANSWER Because only a few alpha particles bounced back, the positive charge cannot be spread out — it must be concentrated in a very small, dense central region called the nucleus. This tiny nucleus carries almost all the mass of the atom and the entire positive charge, while the rest of the atom is mostly empty space.

5. Arrange these in the correct chronological order to show how atomic models evolved. (i) Bohr’s model — electrons move in fixed orbits, each with a definite energy. (ii) Thomson’s model — the ‘plum pudding’ atom with electrons embedded in positive charge. (iii) Rutherford’s model — a dense central nucleus. (iv) Dalton’s model — atoms as indivisible particles.

ANSWER Correct order: Dalton → Thomson → Rutherford → Bohr (options iv, ii, iii, i respectively). Dalton (atoms indivisible) → Thomson (plum pudding) → Rutherford (nucleus) → Bohr (fixed energy orbits).

6. Electrons move around the nucleus in orbits. Why do they not fly away from the atom? Explain what keeps them attracted to the nucleus.

ANSWER The negatively charged electrons are held by the electrostatic force of attraction from the positively charged protons in the nucleus. In the historical orbit model, this attraction provides the centripetal force. Attraction alone does not explain stability: classical orbiting charges would radiate energy. Bohr added non-radiating stationary states; modern quantum theory gives a fuller explanation.

7. Assertion (A): The discovery of subatomic particles helped in understanding atomic structure. Reason (R): The number of electrons is equal to the number of protons in an atom. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

ANSWER Correct option: (ii) — both A and R are true, but R is not the correct explanation of A. A is true: knowing the subatomic particles did explain atomic structure. R is also true (a neutral atom has equal protons and electrons), but that fact does not explain why the discovery helped — so R is not the explanation of A.

8. For a magnesium atom with mass number 24 and atomic number 12, find the number of (i) protons, (ii) neutrons, (iii) electrons, and illustrate the arrangement of electrons.

ANSWER

For the neutral magnesium-24 atom, Z = 12 and A = 24:

p=12,n=A−Z=24−12=12p=12,\qquad n=A-Z=24-12=12e=12,(K,L,M)=(2,8,2)e=12,\qquad(K,L,M)=(2,8,2)
Magnesium-24 shell model with 12 protons and 12 neutrons in the nucleus and 2, 8 and 2 electrons in K, L and M shells.
Bohr-style shell model for counting electrons; schematic, not to scale. Colours are a visual key, not actual particle colours. Open full-size diagram.

9. For the elements shown in Fig. 8.17, find: (i) name, (ii) symbol, (iii) total electrons, (iv) valence electrons, (v) valency, (vi) protons, (vii) atomic number.

ANSWER Reading the shell diagrams (number of electrons in each shell):
Fig.ConfigNameSymbolTotal e−Valence e−ValencyProtonsAtomic no.
(a)2, 1LithiumLi31133
(b)2, 5NitrogenN75377
(c)2, 8, 3AluminiumAl13331313
(d)2, 7FluorineF97199

10. Both Rutherford’s and Bohr’s models have electrons orbiting the nucleus. Why did Rutherford’s model fail to explain atomic stability, while Bohr’s model succeeded?

ANSWER In Rutherford’s model, an orbiting electron is constantly accelerating, so by classical physics it should continuously radiate energy, spiral inward and fall into the nucleus — the atom would be unstable. Bohr removed this problem by postulating that electrons revolve only in certain fixed orbits (stationary states) of definite energy and do not radiate energy while in them, so the atom is stable.

11. An atom 70X has 31 electrons. How many neutrons are there in its nucleus?

ANSWER

“Atom” implies a neutral species, so protons equal electrons:

Z=p=e=31Z=p=e=31n=A−Z=70−31=39n=A-Z=70-31=39

There are 39 neutrons.

12. An atom has 79 protons and a mass number of 197. Calculate (i) the number of neutrons, and (ii) the number of electrons.

ANSWER

(i) Neutrons:

n=A−Z=197−79=118n=A-Z=197-79=118

(ii) A neutral atom has 79 electrons, equal to its 79 protons. The element is gold (Au).

13. Complete Table 8.5.

ANSWER Using protons = atomic number = electrons, and neutrons = mass number − atomic number:
Atomic numberMass numberNeutronsProtonsElectronsElement
511655Boron (B)
714777Nitrogen (N)
1224121212Magnesium (Mg)
1531161515Phosphorus (P)
11011Hydrogen (H)

14. Element X has a mass number of 35 and contains 18 neutrons. (i) How many electrons and protons does X have? (ii) What is its atomic number? (iii) Identify element X. (iv) Write its electronic configuration. (v) How many valence electrons does it have? (vi) What will be the mass number if two neutrons are added? (vii) What will be the relation of X with the new atom?

ANSWER (i) Protons = 35 − 18 = 17, so electrons = 17 and protons = 17. (ii) Atomic number = 17. (iii) Element X is chlorine (Cl). (iv) Electronic configuration = 2, 8, 7. (v) Valence electrons = 7. (vi) Adding 2 neutrons: new mass number = 35 + 2 = 37. (vii) The new atom has the same number of protons (17) but more neutrons, so it is an isotope of X.

15. An atom has 12 protons and 12 neutrons. Imagine all the electrons are replaced by hypothetical particles with the same charge as electrons but 500 times heavier. What is the effect on the atom’s: (i) Atomic number (ii) Atomic mass (iii) Mass number (iv) Overall charge

ANSWER

(i) Atomic number remains 12, because proton number is unchanged. (ii) The atom’s mass increases. Ignoring binding-energy changes in this hypothetical model:

Δm=12(500−1)me=5988me\Delta m=12(500-1)m_e=5988m_eΔm≈59881823 u≈3.28 u\Delta m\approx\frac{5988}{1823}\,\mathrm u\approx3.28\,\mathrm u

(iii) Mass number remains 24: it counts nucleons, not the new negatively charged particles. (iv) Twelve replacement particles have the same total negative charge as the original electrons, so the species remains electrically neutral. This is a hypothetical counting model; it does not establish the stability or detailed structure of such an exotic atom.

Common Mistakes to Avoid

Watch out for these

  • Confusing atomic number (protons) with mass number (protons + neutrons).
  • Forgetting that neutrons = mass number − atomic number.
  • Mixing up isotopes (same protons, different neutrons) with isobars (same mass number, different protons).
  • Confusing shell capacity with occupancy: K, L and M can hold 2, 8 and 18 electrons, respectively; 2, 8, 8 describes argon, not the maximum M-shell capacity.
  • Thinking the gold foil experiment discovered neutrons (it revealed the nucleus; neutrons came later).
  • Saying valency equals valence electrons for every element — for many non-metals valency = 8 − valence electrons.

Extra Practice Questions

Very Short Answer Type Questions

Q1. Who discovered the electron?

ANSWERJ. J. Thomson.

Q2. What is the maximum number of electrons in the L shell?

ANSWER8 electrons.

Q3. Name the particle with no charge in the nucleus.

ANSWERThe neutron.

Short Answer Type Questions

Q1. Write the electronic configuration of sodium (Z = 11) and state its valency.

ANSWERConfiguration = 2, 8, 1. It has 1 valence electron, so its valency is 1.

Q2. State two uses of isotopes.

ANSWERUranium-235 is used as nuclear fuel; carbon-14 is used to date suitable once-living materials. These are examples of particular isotopes, not uses of every isotope of each element.

Long Answer Type Question

Q1. Describe Rutherford’s gold foil experiment and the conclusions drawn from it.

ANSWER A beam of fast-moving alpha particles was directed at a very thin gold foil and the scattering was observed on a screen. Most alpha particles passed straight through, so the atom is mostly empty space; a few were deflected, so there is a concentrated positive charge; and very few bounced back, so this positive charge and most of the mass lie in a tiny, dense nucleus. Conclusion: the atom has a small, dense, positively charged nucleus at the centre, with electrons around it — replacing Thomson’s plum pudding model.

MCQs & Assertion–Reason

1. The particle with a negative charge is the:

(a) proton    (b) neutron    (c) electron    (d) nucleus

2. The atomic number of an element is the number of:

(a) neutrons    (b) protons    (c) protons + neutrons    (d) electrons + neutrons

3. The gold foil experiment was performed by:

(a) Dalton    (b) Thomson    (c) Rutherford    (d) Bohr

4. The maximum number of electrons in the K shell is:

(a) 2    (b) 8    (c) 18    (d) 32

5. Atoms with the same atomic number but different mass numbers are called:

(a) isobars    (b) isotopes    (c) ions    (d) isomers

6. The number of neutrons in an atom with mass number 23 and atomic number 11 is:

(a) 11    (b) 12    (c) 23    (d) 34

7. The electronic configuration of an atom with 13 electrons is:

(a) 2, 8, 3    (b) 2, 8, 8    (c) 2, 3, 8    (d) 8, 2, 3

8. According to Bohr, electrons revolve in:

(a) random paths    (b) fixed energy orbits    (c) straight lines    (d) the nucleus

9. Atoms of different elements with the same mass number are:

(a) isotopes    (b) isobars    (c) isotones    (d) ions

10. The valency of an atom with configuration 2, 8, 7 is:

(a) 7    (b) 1    (c) 8    (d) 2

Answer key: 1-(c), 2-(b), 3-(c), 4-(a), 5-(b), 6-(b), 7-(a), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Isotopes have the same chemical properties.

Reason: They have the same number of protons and electrons.

A-R 2. Assertion: Most of an atom is empty space.

Reason: Most alpha particles passed straight through the gold foil.

A-R 3. Assertion: The mass of an atom is concentrated in its nucleus.

Reason: Electrons have negligible mass compared with protons and neutrons.

A-R 4. Assertion: Neutrons are present in the nucleus of every atom.

Reason: A protium (hydrogen-1) nucleus contains no neutron.

A-R 5. Assertion: Bohr’s model explains the stability of the atom.

Reason: Electrons in fixed orbits do not radiate energy.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • An atom has a tiny, dense, positive nucleus (protons + neutrons) with electrons around it.
  • Atomic models evolved: Dalton → Thomson → Rutherford → Bohr.
  • Atomic number Z = protons; mass number A = protons + neutrons; neutrons = A − Z.
  • For the first 18 elements use the textbook’s shell distributions; capacity follows 2n2. Valency is not always the number of valence electrons.
  • Isotopes: same Z, different A; isobars: same A, different Z.
  • Bohr’s fixed-energy orbits explain why atoms are stable.

Real-life Applications

Electronic structure helps explain chemical bonding and spectra. Uranium-235 can supply nuclear fuel, and carbon-14 helps date suitable once-living materials within the method’s limits. Modern quantum theory gives a fuller description than the historical Bohr model.

How to revise this chapter

Remember neutrons = mass number − atomic number, and in a neutral atom electrons = protons. For the first 18 elements, use the appropriate shell distribution; do not treat 8 as the M-shell’s full capacity. Keep isotopes (same Z) and isobars (same A) clearly separate, and link each atomic model to the scientist and its key idea.

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 8 about?

Journey inside the atom — subatomic particles, the atomic models of Dalton, Thomson, Rutherford and Bohr, atomic number and mass number, electronic configuration, valency, and isotopes and isobars.

What is the difference between atomic number and mass number?

Atomic number is the number of protons; mass number is the total number of protons and neutrons.

What are isotopes and isobars?

Isotopes have the same atomic number but different mass numbers; isobars have the same mass number but different atomic numbers.

Are these Class 9 Science Exploration Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

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