Chapter 8 Journey Inside the Atom Class 9th Science (Exploration) ncert solution

May 29, 2026 • Admin ETeam • Class 9 NCERT Solution

Chapter 8 — Journey Inside the Atom · Complete Solutions

Grade 9 · Science · Curiosity

Journey Inside the Atom

Chapter 8 — full, step-by-step solutions for every worked calculation, In-Text “Pause & Ponder” question, and end-of-chapter “Revise, Reflect, Refine” exercise, with clean diagrams and maths.

Thomson · Rutherford · Bohr Atomic & Mass Number Electronic Configuration Isotopes & Isobars

Worked Examples

The two quantitative calculations worked through in the chapter text.

How tiny is an atom? Estimate how many atoms are needed to make up the thickness of a sheet of paper that is 0.1 mm thick, given that one atom is about $10^{-10}\ \text{m}$ across.

Solution

Convert the thickness to metres$0.1\ \text{mm} = 0.1\times10^{-3}\ \text{m} = 10^{-4}\ \text{m}$
Divide sheet thickness by atomic diameter$\text{Number of atoms} \approx \dfrac{10^{-4}\ \text{m}}{10^{-10}\ \text{m}} = 10^{6}$

So about one million atoms stacked together make up the thickness of a single sheet of paper — a vivid sense of how unimaginably small a single atom is.

Answer$\approx 10^{6}$ atoms (about one million).

Average atomic mass of chlorine. Chlorine occurs as two isotopes: $^{35}\text{Cl}$ (about 75%) and $^{37}\text{Cl}$ (about 25%). Find its average atomic mass and explain why a simple average is not used.

Solution

A simple average ignores how common each isotope is:

$$\text{Simple average}=\frac{35+37}{2}=36\ \text{u}$$

But the two isotopes are not equally abundant, so we use a weighted average — each mass is multiplied by its fractional abundance:

Multiply each mass by its abundance$\left(35\times\dfrac{75}{100}\right)+\left(37\times\dfrac{25}{100}\right)$
Simplify$=\dfrac{105}{4}+\dfrac{37}{4}=\dfrac{142}{4}$
Result$=35.5\ \text{u}$

No single chlorine atom weighs 35.5 u. The value means that in a large sample, the heavier and lighter atoms average out to 35.5 u — the weighted average reflects chlorine as it actually occurs in nature.

AnswerAverage atomic mass of chlorine $= 35.5\ \text{u}$.

In-Text — Pause & Ponder

Every “Pause and Ponder” question that appears within the chapter.

In a clay-and-beads version of Thomson’s atom (clay = positive charge, beads = electrons), what happens if (i) the positive charge on the clay is less than the total negative charge of the beads? (ii) the clay itself accidentally carries a little negative charge — would it still be a neutral atom?

Thomson’s model: electrons spread through a positive sphere

Solution

(i) The negative charge now outweighs the positive charge, so the whole model carries a net negative charge. It no longer represents a neutral atom — it behaves like a negative ion (anion).
(ii) No. If the clay is also slightly negative, then both parts are negative and nothing supplies positive charge to balance the beads. The total charge is negative, so the model cannot represent a neutral atom — neutrality needs the positive and negative charges to be exactly equal.

Answer(i) Net negative — a negative ion. • (ii) Not neutral; with no balancing positive charge it stays negatively charged.

Could an orange or lemon (seeds inside soft pulp) be a good comparison for Thomson’s atom? Where does it match the idea, and where does it fall short?

A watermelon — the textbook’s analogy for Thomson’s model

Solution

Where it matches: the soft pulp plays the role of the spread-out positive charge, and the seeds embedded within it act like electrons scattered through that positive matter — exactly the ‘plum-pudding’ picture.

Where it falls short:

Seeds in a fruit are clustered in segments, not evenly spread, while Thomson imagined electrons distributed throughout.
The pulp is not actually electrically charged, and the number of seeds has nothing to do with balancing charge, so the analogy can’t show neutrality.
Seeds are huge compared to the fruit, whereas electrons are vanishingly small compared with the atom.

AnswerIt captures ‘particles embedded in a body’ but fails on even distribution, charge balance and relative size.

Why did Thomson conclude that electrons are present in all atoms?

Solution

In his cathode-ray experiments, the rays were always streams of the same negatively charged particles (electrons) — and crucially, their nature did not depend on the metal used for the cathode or on the gas filling the tube.

Since the same electrons appeared no matter which material was used, they could not belong to one special substance. Thomson therefore concluded that electrons are a fundamental constituent of every atom.

AnswerCathode rays (electrons) were identical regardless of the cathode material or gas, so electrons must be common to all atoms.

What would happen if the α-particles in Rutherford’s gold-foil experiment were replaced with negatively charged particles?

Paths of positive α-particles in the gold-foil experiment

Solution

A positive nucleus repels positive α-particles, which is why some bounced back. Negative particles would instead be attracted towards the positive nucleus.

Most would still travel through the empty space almost undeflected.
Those passing close to a nucleus would be pulled toward it (their paths curving inward), rather than being pushed back.
We would therefore not see the sharp backward bounce that repulsion produced; the deflection pattern would reflect attraction instead.

AnswerNegative particles are attracted to the nucleus rather than repelled, so paths curve toward it and large-angle back-scattering is not seen.

Rutherford saw a few α-particles bounce straight back. How does this single result completely rule out Thomson’s plum-pudding model?

Solution

In the plum-pudding model the positive charge is spread thinly and evenly across the whole atom. Such a diffuse charge could only ever nudge a fast α-particle slightly — it could never stop one and throw it backward.

A bounce of more than 90° needs a near head-on encounter with something tiny, very dense and strongly positive. That can only exist if the positive charge and mass are concentrated in a minute central nucleus — the opposite of a spread-out sphere.

AnswerA backward bounce requires a concentrated dense positive centre; a uniformly spread positive sphere can’t produce it, so Thomson’s model is ruled out.

If you could ask Rutherford one question about his work, what would it be?

Solution

This is an open, reflective question — there is no single ‘correct’ answer. A thoughtful example:

“When you first saw a few α-particles bounce straight back, what convinced you to trust such a startling result instead of dismissing it as an error — and how did you make the leap to imagining a tiny central nucleus?”

A good question shows curiosity about either the reasoning behind the nuclear idea or the open puzzles he left (such as what keeps the electrons from collapsing in).

AnswerOpen-ended. Any genuine, scientifically curious question is acceptable — the sample above is one good option.

Assertion (A): Rutherford concluded that most of an atom’s mass is concentrated in a small central region, the nucleus.
Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive-charge sphere.

Both A and R true, R is the correct explanation of A.
Both A and R true, R is not the correct explanation of A.
A true, R false.
A false, R true.

Solution

A is true: the large-angle scattering of α-particles led Rutherford to the dense central nucleus.

R is true: that is a correct description of Thomson’s model.

But R describes a different model and says nothing about why Rutherford concluded the mass is central — that came from his own scattering experiment. So R does not explain A.

AnswerCorrect option: (ii) — both true, but R is not the explanation of A.

Imagine you discover a new element and name it after yourself. Justify a symbol that follows the IUPAC rules.

Solution

Sample: name the element ‘Aaravium’ after Aarav. A valid symbol would be Av.

Why it follows IUPAC norms:

The first letter is a capital (A) and the second is lowercase (v) — e.g. Av, never AV.
It is built from letters of the element’s name.
It is not already taken by an existing element, so there is no clash.

Any name works as long as the symbol is one or two letters, starts with a capital, uses lowercase for the second letter, comes from the name, and is unique.

AnswerOpen-ended — e.g. ‘Aaravium’, symbol Av: capital + lowercase, drawn from the name, and unused.

What problems could arise if every scientist used different symbols for the same element?

Solution

Constant confusion and miscommunication — the same symbol could mean different elements to different people.
Errors in writing chemical formulae and equations, which could be dangerous in medicine or industry.
Scientists across countries and languages could no longer share results or reproduce each other’s work reliably.

A single, internationally agreed set of symbols (now maintained by IUPAC) removes all this ambiguity.

AnswerWidespread confusion, formula/equation errors, and a breakdown of clear global scientific communication.

An atom has atomic number 26 and 56 nucleons. Find its number of electrons, protons and neutrons.

Solution

Protons = atomic number$p = Z = 26$
Electrons = protons (neutral atom)$e = 26$
Neutrons = mass number − protons$n = A – Z = 56 – 26 = 30$

AnswerProtons = 26, Electrons = 26, Neutrons = 30 (the element is iron, Fe).

A nucleus has 20 protons and a mass number of 41. Find the number of neutrons.

Solution

Neutrons = mass number − protons$n = A – p = 41 – 20 = 21$

AnswerNumber of neutrons = 21.

An atom has 18 neutrons and an atomic number of 17. What is its mass number?

Solution

Mass number = protons + neutrons$A = Z + n = 17 + 18 = 35$

AnswerMass number = 35 (this is chlorine, $^{35}\text{Cl}$).

An atom $^{23}\text{A}$ has 11 electrons. Find the number of neutrons in it.

Solution

Protons = electrons (neutral atom)$p = e = 11$, so $Z = 11$
Neutrons = mass number − protons$n = A – Z = 23 – 11 = 12$

AnswerNumber of neutrons = 12 (the element is sodium, Na).

Identify the number of electrons in the outermost shell of (i) $^{12}_{\;6}\text{C}$ (ii) $^{19}_{\;9}\text{F}$ (iii) $^{28}_{14}\text{Si}$.

Solution

(i) Carbon ($Z=6$): configuration 2, 4 → outermost shell has 4 electrons.
(ii) Fluorine ($Z=9$): configuration 2, 7 → outermost shell has 7 electrons.
(iii) Silicon ($Z=14$): configuration 2, 8, 4 → outermost shell has 4 electrons.

Answer(i) 4 • (ii) 7 • (iii) 4 valence electrons.

Write the electronic configuration of the elements with atomic numbers 12, 16 and 18.

Z = 12 → 2, 8, 2 (Mg)

Z = 16 → 2, 8, 6 (S)

Z = 18 → 2, 8, 8 (Ar)

Fig. 8.17 — electron arrangements (a)–(d)

Solution

Fill shells in order K (max 2), L (max 8), M (max 8 for these):

$Z=12$ (Magnesium): 2, 8, 2
$Z=16$ (Sulfur): 2, 8, 6
$Z=18$ (Argon): 2, 8, 8

Answer12 → 2, 8, 2 • 16 → 2, 8, 6 • 18 → 2, 8, 8.

Riddle: “I have a mass number of 23 and 11 protons. I am a soft metal that reacts vigorously with water. Who am I, and how many neutrons do I have?”

Solution

11 protons means atomic number 11 — the element is sodium (Na), a soft metal that reacts vigorously with water.

Neutrons = mass number − protons$n = A – p = 23 – 11 = 12$

AnswerThe atom is sodium (Na) with 12 neutrons.

Two atoms each have 11 protons; one has 12 neutrons, the other 13 neutrons. Compare their atomic and mass numbers. Same element or different?

Solution

Atomic numbersBoth have 11 protons, so both have $Z = 11$ — same atomic number.
Mass numbers$A_1 = 11+12 = 23$ and $A_2 = 11+13 = 24$ — different mass numbers.

Same atomic number means they are the same element (sodium). Same element but different mass numbers → they are isotopes.

AnswerSame atomic number (11), different mass numbers (23 and 24). They are the same element (sodium) — isotopes of each other.

Bromine exists as two isotopes, $^{79}_{35}\text{Br}$ (49.7%) and $^{81}_{35}\text{Br}$ (50.3%). Calculate its average atomic mass.

Solution

Weight each mass by its abundance$\left(79\times\dfrac{49.7}{100}\right)+\left(81\times\dfrac{50.3}{100}\right)$
Compute each term$=\dfrac{3926.3}{100}+\dfrac{4074.3}{100}$
Add$=\dfrac{8000.6}{100}=80.006\ \text{u}$

AnswerAverage atomic mass of bromine $\approx 80.0\ \text{u}$.

Exercise — Revise, Reflect, Refine

Full solutions to the end-of-chapter questions.

In the context of Rutherford’s gold-foil experiment, choose the correct option(s) and give reasons:

It clearly showed the existence of neutrons in the nucleus.
The results disproved the plum-pudding model and led to the idea of a nucleus.
The large deflection of a few α-particles showed most of the mass and positive charge are packed into a tiny centre.
The deflections showed how electrons move around the nucleus.

Solution

(i) Incorrect. Neutrons were discovered much later (Chadwick, 1932); this experiment said nothing about them.
(ii) Correct. Back-scattering could not fit a spread-out positive sphere, so it overturned plum-pudding and pointed to a central nucleus.
(iii) Correct. Only a tiny, dense, highly positive centre could deflect heavy α-particles through large angles.
(iv) Incorrect. The experiment revealed the nucleus, not the motion of electrons.

AnswerCorrect options: (ii) and (iii).

Which statements are correct/incorrect according to Bohr’s model? Give a reason for each.

Electrons lose energy in fixed orbits and slowly fall into the nucleus.
Electrons can exist anywhere around the nucleus with no fixed energy.
Electrons revolve in orbits of fixed energy without losing energy.
Electrons can be found between energy levels.

Bohr’s fixed energy levels (shells)

Solution

(i) Incorrect. The whole point of Bohr’s model is that in a fixed orbit (stationary state) an electron does not lose energy.
(ii) Incorrect. Electrons may only occupy fixed orbits, each with a definite energy — not just anywhere.
(iii) Correct. This is exactly Bohr’s key postulate, which explains atomic stability.
(iv) Incorrect. Electrons cannot exist between allowed shells.

AnswerOnly (iii) is correct; (i), (ii) and (iv) are incorrect.

For three species — X (p = 18, n = 19), Y (p = 17, n = 18), Z (p = 17, n = 20) — explain the relation between (i) Y and Z and (ii) Z and X.

Solution

First find each mass number $A = p + n$:

X: $A = 18+19 = 37$, $Z = 18$
Y: $A = 17+18 = 35$, $Z = 17$
Z: $A = 17+20 = 37$, $Z = 17$

(i) Y and Z: same number of protons (17) but different neutrons → same element with different mass numbers (35 and 37) → they are isotopes.

(ii) Z and X: different atomic numbers (17 vs 18) but the same mass number (37) → they are isobars.

Answer(i) Y and Z are isotopes. • (ii) Z and X are isobars.

What did Rutherford conclude about the position and characteristics of the atom’s positive charge from the few α-particles that bounced back or were deflected through large angles?

Rutherford’s nuclear (planetary) model

Solution

Because only a very few α-particles were strongly deflected while most passed straight through, Rutherford concluded that the positive charge is:

Concentrated in an extremely small central region — the nucleus — not spread out across the atom.
Dense and massive, carrying most of the atom’s mass, so it could throw back heavy α-particles.
Positively charged, since it repelled the positive α-particles.
Surrounded by mostly empty space, which is why most particles went through undeflected.

AnswerThe positive charge and nearly all the mass sit in a tiny, dense, positively charged central nucleus, with the rest of the atom mostly empty space.

Arrange the atomic models in correct chronological order: (i) Bohr (fixed orbits with definite energy); (ii) Thomson (plum-pudding); (iii) Rutherford (dense central nucleus); (iv) Dalton (indivisible particles).

How the model of the atom evolved over time

Solution

The models developed in this order:

(iv) DaltonAtoms are tiny indivisible particles — the first scientific atomic theory (1808).
(ii) ThomsonElectrons embedded in a positive sphere — the plum-pudding model (1897–1904).
(iii) RutherfordGold-foil scattering reveals a dense central nucleus (1911).
(i) BohrElectrons occupy fixed energy levels, explaining stability (1913).

AnswerCorrect order: (iv) → (ii) → (iii) → (i) [Dalton → Thomson → Rutherford → Bohr].

Electrons orbit the nucleus — so why don’t they fly away from the atom? What keeps them attracted to the nucleus?

Solution

The nucleus is positively charged (protons) and electrons are negatively charged. The electrostatic force of attraction between them pulls the electrons toward the nucleus.

For an electron moving in a circular path, this inward pull provides exactly the centripetal force needed to keep it in orbit — balancing its tendency to fly off in a straight line. In Bohr’s picture the electron stays in a fixed energy level without losing energy, so it neither escapes nor falls in.

AnswerThe electrostatic attraction between the negative electrons and the positive nucleus supplies the centripetal force that holds electrons in their orbits.

Assertion (A): The discovery of subatomic particles helped in understanding atomic structure.
Reason (R): The number of electrons equals the number of protons in an atom.

Both A and R true, R is the correct explanation of A.
Both A and R true, R is not the correct explanation of A.
A true, R false.
A false, R true.

Solution

A is true: finding electrons, protons and neutrons is precisely what let scientists build models of the atom.

R is true: in a neutral atom electrons do equal protons.

But R is just one fact about neutrality; it does not explain why discovering the particles advanced our understanding. So R is not the correct explanation of A.

AnswerCorrect option: (ii) — both true, but R is not the explanation of A.

Magnesium has mass number 24 and atomic number 12. Find its (i) protons, (ii) neutrons, (iii) electrons, and illustrate the electron arrangement.

Solution

(i) Protons = atomic number$p = Z = 12$
(ii) Neutrons = mass number − protons$n = A – Z = 24 – 12 = 12$
(iii) Electrons = protons (neutral atom)$e = 12$
Electron arrangementFill K, L, M → configuration 2, 8, 2

Magnesium (Z = 12): electron arrangement 2, 8, 2

AnswerProtons = 12, Neutrons = 12, Electrons = 12; configuration 2, 8, 2.

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

(a) 2, 1

(b) 2, 5

(d) 2, 7

Fig. 8.17 — electron arrangements (a)–(d)

Solution

Reading the shells of each atom:

(a) 2, 1 → Lithium (Li): total e = 3, valence e = 1, valency = 1, protons = 3, $Z$ = 3.
(b) 2, 5 → Nitrogen (N): total e = 7, valence e = 5, valency = 3 (gains/shares $8-5=3$), protons = 7, $Z$ = 7.
(c) 2, 8, 3 → Aluminium (Al): total e = 13, valence e = 3, valency = 3 (loses 3), protons = 13, $Z$ = 13.
(d) 2, 7 → Fluorine (F): total e = 9, valence e = 7, valency = 1 (gains $8-7=1$), protons = 9, $Z$ = 9.

Answer(a) Li, Z=3, valency 1 • (b) N, Z=7, valency 3 • (c) Al, Z=13, valency 3 • (d) F, Z=9, valency 1.

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

Rutherford’s problem: an orbiting electron should spiral inward

Solution

Rutherford’s problem: an electron moving in a circle is constantly accelerating. By classical physics an accelerating charge must continuously radiate energy; losing energy, it would spiral inward and crash into the nucleus in a fraction of a second. His model therefore predicted that atoms could not be stable.

Bohr’s fix: Bohr postulated special stationary states (fixed orbits) in which an electron revolves without radiating any energy. As long as the electron stays in such a level its energy is constant, so it never spirals in — which explains why atoms are stable.

AnswerRutherford’s orbiting electrons should radiate energy and collapse inward; Bohr’s postulate of non-radiating fixed energy levels removed this, explaining stability.

An atom $^{70}\text{X}$ has 31 electrons. How many neutrons are in its nucleus?

Solution

Protons = electrons (neutral atom)$p = e = 31$
Neutrons = mass number − protons$n = A – p = 70 – 31 = 39$

AnswerNumber of neutrons = 39 (the element is gallium, Ga).

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

Solution

(i) Neutrons = mass number − protons$n = 197 – 79 = 118$
(ii) Electrons = protons (neutral atom)$e = 79$

Answer(i) Neutrons = 118 • (ii) Electrons = 79 (the element is gold, Au).

Complete Table 8.5. (Bold values are the ones you must fill in.)

Solution

Use $A = p + n$, $Z = p = e$ (neutral atom):

Atomic no. (Z)	Mass no. (A)	Neutrons	Protons	Electrons	Element
5	11	6	5	5	Boron
7	14	7	7	7	Nitrogen
12	24	12	12	12	Magnesium
15	31	16	15	15	Phosphorus
1	1	0	1	1	Hydrogen

Row 1: $Z=5,\ n=6 \Rightarrow p=5,\ e=5,\ A=11$ → Boron.
Row 2: $A=14,\ e=7 \Rightarrow p=7,\ Z=7,\ n=7$ → Nitrogen.
Row 3: $A=24,\ p=12 \Rightarrow Z=12,\ e=12,\ n=12$ → Magnesium.
Row 4: $Z=15,\ n=16 \Rightarrow p=15,\ e=15,\ A=31$ → Phosphorus.
Row 5: $A=1,\ n=0 \Rightarrow p=1,\ Z=1,\ e=1$ → Hydrogen.

AnswerBoron, Nitrogen, Magnesium, Phosphorus, Hydrogen — completed values shown bold above.

Element X has mass number 35 and contains 18 neutrons. Answer: (i) electrons & protons, (ii) atomic number, (iii) identity, (iv) electronic configuration, (v) valence electrons, (vi) new mass number if 2 neutrons are added, (vii) relation of X to the new atom.

Solution

(i) Protons & electrons$p = A – n = 35 – 18 = 17$, so $p = e = 17$
(ii) Atomic number$Z = p = 17$
(iii) Identify the element$Z=17$ → Chlorine (Cl)
(iv) Electronic configuration2, 8, 7
(v) Valence electrons7 (in the outermost shell)
(vi) Add 2 neutronsnew $A = 35 + 2 = 37$ (protons unchanged)
(vii) RelationSame $Z$ (17), different mass numbers (35 and 37) → isotopes

AnswerX is chlorine: p = e = 17, Z = 17, config 2, 8, 7, valence 7; adding 2 neutrons gives mass 37 — an isotope of X.

An atom has 12 protons and 12 neutrons. Imagine all its electrons are replaced by hypothetical particles with the same charge as electrons but 500× heavier. What is the effect on the atom’s (i) atomic number, (ii) atomic mass, (iii) mass number, (iv) overall charge?

Solution

(i) Atomic number — unchanged (12). Atomic number depends only on the number of protons, which is untouched.
(ii) Atomic mass — increases. Normal electrons are so light their mass is ignored. Particles 500× heavier are no longer negligible (each is roughly $500/1836 \approx 0.27$ of a proton’s mass), so 12 of them add a noticeable amount of mass.
(iii) Mass number — unchanged (24). Mass number counts only nucleons: $p + n = 12 + 12 = 24$. Electrons (or their replacements) are not included.
(iv) Overall charge — unchanged (neutral). The new particles still carry one negative charge each, so 12 of them still balance the 12 protons.

Answer(i) same (12) • (ii) increases • (iii) same (24) • (iv) still neutral.

The quest continues…

From Dalton’s indivisible sphere to Bohr’s energy levels and beyond — the story of the atom is still being written.

Chapter 8 Journey Inside the Atom Class 9th Science (Exploration) ncert solution

Worked Examples

In-Text — Pause & Ponder

Exercise — Revise, Reflect, Refine

The quest continues…

Leave a Reply Cancel reply