Complete handwritten notes • every diagram, table & flow chart
with all “Pause and Ponder” questions and
all “Revise, Reflect, Refine” exercises answered, topic by topic
Chapter at a glance — the five big ideas
Think it over (before you begin)
Are atoms really the smallest, indivisible particles?
Why do electrons not fall into the nucleus even though protons attract them?
Why did scientists keep changing the model of the atom again and again?
Keep these three questions in mind — the whole chapter is the answer to them.
8.1 Rediscovering the Roots of Atomic Theory
The question “What is everything made up of?” is more than 2000 years old. Two
civilisations — ancient India and ancient Greece — arrived at almost the same answer, purely by thinking.
Table A — The earliest ideas about matter
Thinker
Place / Time
What was proposed
Acharya Kanada
India, ~6th cent. BCE
Divide matter (dravya) again and again → you reach the smallest indivisible particle,
the parmanu. It is infinitely small and cannot be sensed. Parmanus join to form
dyads (2) and triads (3), and these build the whole universe. Recorded in the
Vaisesika Sutras. Limitation: it did not state in what proportion parmanus combine.
Leucippus & Democritus
Greece, ~5th cent. BCE
Same idea — indivisible particles called atomos
(Greek atomos = “that which cannot be cut”). This is where the word atom comes from.
John Dalton
England, 1808
The first scientific (experiment-based) atomic theory: all matter is made of
tiny indivisible particles called atoms, which are the fundamental building blocks of matter.
KEY The idea of the atom began as an imaginative / philosophical idea,
not from experiments. Dalton was the first to base it on scientific experiments — that is why his
theory became the starting point of modern atomic structure.
After Dalton, three questions drove the next 100 years of science:
What are atoms made up of?
What would an atom look like if we could see it?
What makes the atoms of one element different from another?
8.2 A Short Historical Journey Through Atomic Models
Till the late 19th century everyone believed the atom was the smallest, indivisible unit.
Then came the discovery of radioactivity — certain elements emit invisible energy and
particles called radiation. If something is coming out of the atom, the atom must have parts inside!
MODEL A model is a simple picture scientists build to explain
their observations. When a new experiment does not fit, the model is changed or replaced. Old models are not
“useless” — they show how science moves forward, one step at a time.
Discovery of the electron — Thomson’s cathode ray experiment (1897)
J. J. Thomson studied the conduction of electricity through gases at very low pressure. He took a glass
tube with two electrodes and applied a high voltage.
Fig. 8.1 — Line diagram of a cathode ray tube
Observation: rays travelled from the cathode (–) to the
anode (+). These were named cathode rays. On applying electric and magnetic fields: the rays bent — proving they are streams of
negatively charged particles with a mass far smaller than an atom.
These particles were later named electrons.
WHY IT MATTERS The nature of cathode rays did not depend on the
material of the cathode or on the gas filled in the tube. Same rays, every time.
⇒ Electrons are a fundamental part of every atom of every element.
Charge of an electron = –1.602 × 10–19 C, taken as –1 by convention for
convenience.
Meet a Scientist — J. J. Thomson
Discovered the electron — the first subatomic particle ever identified, and a part of every atom.
Nobel Prize in Physics, 1906, for his study of electrical conduction in gases. As head of the famous
Cavendish Laboratory, Cambridge, he guided many scientists — including Ernest Rutherford.
8.2.1 Thomson’s Model of an Atom
Thomson faced a puzzle: electrons are negative, but atoms are neutral — so where is the positive charge?
His answer: the atom is a sphere of positive charge with electrons stuck all through it.
Fig. 8.2 — Thomson’s “plum pudding” model of the atom
The two famous analogies
Analogy
Positive charge
Electrons
Plum pudding
the pudding
the plums embedded in it
Watermelon (Fig. 8.3)
the red pulp
the seeds spread throughout
Note Atoms have no colour. The colours in all these diagrams
(red nucleus, blue electrons) are only for illustration.
Score card of Thomson’s model✔ Explained: the atom is electrically neutral — total (+) charge = total (–) charge. ✘ Failed: could not explain the results of the gold foil experiment (next topic).
Pause and Ponder Q1 – Q3
1Suppose you made your own ‘atom’ as Thomson described,
using clay for the positive charge and small beads for the electrons spread through it. What will happen if
(i) the positive charge on the clay is less than the total negative charge of the beads?
(ii) by mistake, the clay itself carries a bit of negative charge — would your model still be a neutral atom?
(i) The negative charge would no longer be cancelled. The model would carry a
net negative charge — so it would represent a negative ion (anion), not a neutral atom.
(ii) No. If the clay is also negative, there is nothing positive left to balance the beads. The whole
model becomes negatively charged and breaks the basic requirement of Thomson’s model —
that a positive sphere must exactly balance the embedded electrons.
2Could an orange or a lemon, which also contain seeds
inside soft pulp, be a good comparison? In what ways does it match Thomson’s idea and where does it fall short?
Where it matches: it is roughly spherical, the seeds (electrons) are inside a soft bulk
(positive matter), and the seeds are much smaller than the fruit — just as electrons are far lighter than the atom.
Where it falls short:
The pulp is divided into segments, but Thomson’s positive charge is continuous and uniform.
The seeds sit near the centre / in rows, not spread evenly throughout.
The peel forms a definite boundary; an atom has no such “skin”.
The seeds carry no charge, so the neutrality idea is not modelled at all.
3Why did Thomson conclude that electrons are present in
all atoms?
Because the properties of cathode rays were always the same — the same
negative charge and the same (very small) mass — no matter which metal was used as the cathode or
which gas was filled in the tube. If a particle can be pulled out of every material, it must already be
present in every atom. Hence the electron is a universal constituent of all atoms.
8.2.2 Testing Thomson’s Model — The Gold Foil Experiment (1911)
Geiger and Marsden, working under Ernest Rutherford, fired a narrow beam of
α-particles at an extremely thin sheet of gold foil.
DEFα (alpha) particle — a tiny, fast, positively charged
particle emitted by radioactive elements. It is actually the nucleus of a helium atom: 2 protons + 2 neutrons
(charge +2, mass 4 u). Scattering — deflection of a particle from its straight path. That is why
this is also called the α-ray scattering experiment.
Fig. 8.4 — Schematic view of the gold foil (α-scattering) experiment
Table B — Observation → Conclusion (the heart of this chapter)
Observation
Expected (Thomson)
Conclusion drawn
Most α-particles passed straight through, undeflected
same
Most of the atom is empty space
Some were deflected through small angles
only very slight deflection
There is a positive charge inside the atom that repels them,
but it occupies a very small volume
Very few (about 1 in 12000) bounced straight back
never expected!
All the positive charge and almost all the mass are packed into an
extremely small, dense centre — the nucleus
FAMOUS LINE Rutherford said it was “as incredible as if you fired a
15-inch shell at a piece of tissue paper and it came back and hit you.”
Thomson’s model failed here: if the positive charge were spread thinly over the whole atom,
the repulsive push at any point would be far too weak to turn back a heavy, fast α-particle.
Think as a ScientistWhat if the gold foil were made thicker?
A thicker foil = many more layers of atoms = many more nuclei in the path. So the chance of a close encounter rises:
fewer α-particles would pass straight through, many more would be deflected, and more would be
scattered through large angles or bounce back. Multiple scattering would also blur the pattern on the screen, so the
result would be harder to interpret — this is exactly why Rutherford insisted on an extremely thin foil.
Exercise Q1 Revise, Reflect, Refine
1Choose the correct options and explain the reason for the
correct and incorrect options in the context of Rutherford’s gold foil experiment.
Correct: (ii) and (iii). Incorrect: (i) and (iv).
(i) Showed the existence of neutrons — INCORRECT. α-particles are positive
and are deflected by charge. A neutral particle produces no such effect. The neutron was discovered much later,
by Chadwick in 1932.
(ii) Disproved the plum pudding model and led to the nucleus — CORRECT. A uniformly
spread positive charge can never turn a fast α-particle back; a concentrated positive centre can.
(iii) Large deflection of a few α-particles ⇒ mass and positive charge packed in a tiny centre —
CORRECT. Only something very heavy and highly positively charged can reverse the
α-particle; and since only a few were affected, that region must be very small.
(iv) Showed how electrons move — INCORRECT. Electrons are ~7300 times lighter than
an α-particle, so they cannot deflect it noticeably. The experiment gave no information at all about electron
motion; that came later from Bohr.
A. Rutherford’s Model of the Atom (the nuclear model)
Fig. 8.5 — The planetary (nuclear) model suggested by Rutherford
Most of the atom is empty space — because most α-particles went straight through.
The nucleus is a very small, dense centre carrying all the positive charge and
almost all the mass of the atom.
Electrons revolve around the nucleus, like planets around the Sun ⇒ the
planetary model.
How ridiculously small the nucleus is
Ready to Go Beyond — a nice calculation
How many atoms are stacked across a sheet of paper 0.1 mm thick?
Thickness = 0.1 mm = 10–4 m, diameter of one atom ≈ 10–10 m
Number of atoms = 10–4 ÷ 10–10 = 106 = about one million atoms!
Pause and Ponder Q4 – Q6
4What do you think would happen if α-particles were
replaced with negatively charged particles in Rutherford’s gold foil experiment?
The force would flip from repulsion to attraction. Negative particles (e.g. electrons /
β-particles) would be pulled towards the positive nucleus instead of being pushed away, so
they would bend inwards and none would bounce straight back the way α-particles did.
Also, being about 7300 times lighter than an α-particle, they would be knocked about easily — even by the
electrons of the gold atoms — giving scattering in all directions. The clean “most go straight, a few rebound”
pattern would be lost, and the nucleus would be much harder to detect.
5Rutherford found that a few α-particles bounced back
sharply. How does this single surprising result completely rule out Thomson’s ‘plum pudding model’?
To reverse a fast, heavy, positive α-particle you need a huge repulsive force acting over a
very short distance, i.e. a target that is (a) highly positively charged, (b) very massive and
(c) extremely concentrated.
In the plum pudding model the positive charge is smeared thinly over the whole atom, so at
any point the charge — and hence the repulsion — is tiny. Such an atom could never push an α-particle back;
at most it could nudge it slightly.
So even one rebound is fatal to the model: in science, a single reproducible observation that a theory
cannot explain is enough to reject it.
6If you could ask Rutherford one question about his work,
what would it be?
(Open-ended — your own question is valid. Sample answers:)
“When the first α-particle bounced back, did you think it was a mistake in the apparatus rather than a
discovery?”
“If the nucleus is so tiny, what holds it together against the repulsion of all those protons?”
“Your model could not explain why the atom does not collapse — did that trouble you?”
Tip: a good scientific question asks about evidence, limitation or
next step — not just a fact.
Exercise Q4 Revise, Reflect, Refine
4What 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?
Position: the positive charge is not spread over the whole atom — it is
concentrated at the centre, in a region he named the nucleus.
Characteristics:
Extremely small — about 105 (one lakh) times smaller than the atom
(dnucleus ≈ 10–15 m vs datom ≈ 10–10 m).
Positively charged — it repelled the positive α-particles.
Very dense and heavy — it carries almost the entire mass of the atom, which is why it could turn
a heavy α-particle back instead of being pushed aside.
The rest of the atom, where the electrons move, is essentially empty space.
B. Limitation of Rutherford’s Model — the stability problem
Fig. 8.6 — Spiral path of a charged particle that keeps losing energy
THE PROBLEM A particle moving in a circle is constantly changing direction
⇒ it is accelerating (Chapter 4, Describing Motion Around Us). An accelerating charged particle
must radiate energy. An electron that keeps losing energy would spiral inward and crash into the nucleus —
so every atom would collapse in a fraction of a second. But atoms are stable! Hence Rutherford’s
model was incomplete.
C. Discovery of the Proton
Rutherford showed the nucleus is positive; that charge comes from particles called protons.
A proton is much heavier than an electron, and carries a charge equal and opposite to it (+1).
For an atom to be electrically neutral: number of protons = number of electrons.
no. of p+ = no. of e– ⇒ atom is NEUTRAL
Examples — helium: 2 p+, 2 e– | sodium: 11 p+,
11 e–. In each case total (+) = total (–), so the atom is neutral. This is true for all atoms.
Meet a Scientist — Ernest Rutherford
Born in New Zealand; came to Cambridge to work with J. J. Thomson and later became the
“Father of Nuclear Physics”. He discovered the atomic nucleus, explained radioactive decay
(Nobel Prize in Chemistry, 1908) and proposed the nuclear model in 1911. His portrait appears on
New Zealand’s $100 note.
Pause and Ponder Q7 — Assertion & Reason
7Assertion (A): Rutherford concluded that most of the
mass of an atom is concentrated in a small region at the centre called the nucleus. Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive
charge sphere.
Correct option: (ii) — Both A and R are true, but R is not the correct explanation of A.
A is true: the large-angle deflection and rebound of α-particles proved that the mass and positive
charge sit in a tiny central nucleus.
R is true: that is a correct statement of Thomson’s model.
But R does not explain A: R only describes an older, different model. Rutherford’s conclusion came
from the observations of the gold foil experiment, not from Thomson’s description. In fact the experiment
contradicted R.
8.2.3 Bohr’s Model of the Atom (1913)
To rescue the atom from collapsing, Niels Bohr made a bold proposal.
Fig. 8.7 — Energy levels (shells) in an atom, and jumps between them
Bohr’s postulates
Electrons do not move randomly. They revolve in fixed circular paths called
stationary states / orbits / shells.
Each shell has a definite amount of energy ⇒ shells are also called energy levels.
Shells are named K, L, M, N … or numbered n = 1, 2, 3, 4 …
Electrons can exist only in these allowed shells — never in between them.
While moving in a fixed shell, an electron does not lose energy. ← this is the
whole point of the model.
K (n = 1) is nearest the nucleus and has the least energy. Energy increases as you move
outward: EK < EL < EM < EN.
An electron moves from one shell to another only by absorbing (jump outward) or releasing
(fall inward) a fixed amount of energy = the difference between the two levels.
Each shell can hold only a certain maximum number of electrons (see 8.7).
HOW STABILITY IS EXPLAINED In Bohr’s model the electron still moves in a
circle — but he postulated that in a stationary state its energy stays constant even though it is
moving. No energy is radiated ⇒ no spiralling in ⇒ the atom is stable.
Threads of Curiosity — why K, L, M, N and not A, B, C, D?
The names come from early X-ray work by Charles Barkla, who called the first X-ray line he saw “K”.
He deliberately did not start at A, leaving room in case an earlier series was discovered — none ever was.
Bohr borrowed the same lettering for atomic shells.
Meet a Scientist — Niels Bohr
Professor of physics at Copenhagen University, Denmark. He was troubled that older models could not explain
why electrons stay around the nucleus without collapsing into it. His explanation of atomic structure won him the
Nobel Prize in Physics, 1922.
Next level up Even Bohr’s model was later found to have limitations.
It was replaced by the quantum mechanical model, in which electrons do not follow neat fixed paths at all but
exist as “electron clouds” — regions where an electron is most likely to be found. You will study this
in higher classes.
Exercise Q2, Q6 & Q10 on Bohr’s model
2Which of the following statements are correct or incorrect
according to Bohr’s atomic model? Give a reason for each.
(i) Electrons lose energy in fixed orbits and slowly fall into the nucleus —
INCORRECT. This is exactly the flaw of Rutherford’s model. Bohr postulated that
in a stationary state the electron does not radiate energy, so it never spirals in.
(ii) Electrons can exist anywhere around the nucleus with no fixed energy —
INCORRECT. Energy is quantised: an electron in a given shell has one definite
energy value, not any arbitrary value.
(iii) Electrons revolve in orbits of fixed energy without losing energy —
CORRECT. This is Bohr’s central postulate of stationary states, and it is what
explains the stability of the atom.
(iv) Electrons can be found between energy levels — INCORRECT. Only certain
orbits are permitted. The space between two shells is a forbidden region; an electron crossing over must
jump in one go by absorbing or emitting exactly the energy difference.
6Electrons move around the nucleus in orbits. Why do they
not fly away from the atom? Explain what keeps them attracted to the nucleus.
Electrons carry a negative charge and the nucleus carries a positive charge (protons).
The electrostatic force of attraction between opposite charges pulls the electron towards the
nucleus.
This inward pull acts as the centripetal force that keeps the electron moving in its circular shell —
exactly as gravity keeps a planet in orbit around the Sun. The electron’s motion balances the pull: it neither flies
off nor falls in.
To actually escape, an electron must be supplied energy at least equal to its binding (ionisation) energy —
which is why atoms hold on to their electrons and matter stays intact.
10Both 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?
Rutherford failed because he applied ordinary (classical) physics: an electron revolving in a
circle is continuously accelerating; an accelerating charge must radiate energy; losing energy it would
spiral inward and hit the nucleus almost instantly (~10–8 s) — the atom would collapse. His model gave
no reason why this does not happen.
Bohr succeeded because he added a new rule: only certain fixed orbits (stationary states) are
allowed, and in these orbits the electron does not radiate energy at all. Energy can be lost
or gained only in fixed jumps between levels — never continuously. With no continuous energy loss, there is no
inward spiral, and the atom is stable.
In short: same picture, but Bohr quantised it.
Fig. 8.16 — Journey of the development of atomic models
Exercise Q5 chronological order
5Explain and arrange the statements in the correct
chronological order to show how atomic models evolved.
Correct order: (iv) → (ii) → (iii) → (i)
(iv) Dalton (1808) — the atom is an indivisible, solid particle; the first scientific theory of
matter.
(ii) Thomson (1904) — after discovering the electron, the atom became a ‘plum pudding’: electrons embedded
in a sphere of positive charge. The atom is no longer indivisible.
(iii) Rutherford (1911) — the gold foil experiment showed a dense central nucleus with empty space
around it.
(i) Bohr (1913) — electrons move in fixed orbits of definite energy, which finally explained why
the atom does not collapse.
Each model kept what was right in the previous one and repaired what it could not explain.
8.3 What Components Contribute to the Mass of an Atom?
Rutherford showed that nearly all the mass sits in the nucleus; electrons are so light that their mass can be
ignored. But a puzzle remained:
THE PUZZLE Hydrogen has 1 proton; helium has 2 protons.
So helium should be twice as heavy as hydrogen — but it is actually about four times
heavier! Is there something else in the nucleus that adds mass without adding charge?
8.3.1 Discovery of the Neutron (1932)
James Chadwick, a student of Rutherford, found a new subatomic particle with a mass nearly equal to that of
a proton but with no charge — the neutron, symbol n.
Neutrons are present in the nucleus of all atoms except ordinary hydrogen (11H).
Mass of an atom comes mainly from protons + neutrons packed in the nucleus.
This is why atoms are heavier than the total mass of their protons alone.
The three subatomic particles and where they live
Table 8.1 — Symbols and relative charges of subatomic particles
S. No.
Subatomic particle
Symbol
Relative charge
Relative mass (u)
Location
1.
Electron
e–
–1
≈ 1/1836 (negligible)
outside nucleus, in shells
2.
Proton
p+
+1
1
inside nucleus
3.
Neutron
n0
0
1
inside nucleus
The last two columns are extra — they are not in the NCERT table but make the table far more useful
for questions.
Why do heavier atoms need more neutrons?
Element
Protons
Neutrons
Pattern
Carbon
6
6
light atoms: p+ ≈ n0
Oxygen
8
8
Iron
26
30
heavy atoms: n0 > p+
Uranium
92
146
Threads of Curiosity — why don’t the protons push each other apart?
Every proton repels every other proton (all are positive). Neutrons help in two ways:
being neutral, they sit between protons and increase the distance between them, weakening the
repulsion;
they strengthen the nuclear force — the very strong, very short-range force that binds all nucleons
together.
That is why heavy nuclei need many extra neutrons to stay bound.
Meet a Scientist — James Chadwick
Working under Rutherford at the Cavendish Laboratory, Cambridge, he discovered the neutron in 1932 —
solving the atomic-mass puzzle. Nobel Prize in Physics, 1935. Because neutrons are uncharged they can slip
into nuclei easily, which led to artificial radioactive elements and to the splitting of uranium — the beginning of
the ‘atomic age’, giving both nuclear power and nuclear weapons.
What if … an atom had no empty space? Matter would become
unimaginably dense — the entire Earth would shrink to a ball a few hundred metres across, and everyday objects would
be billions of times heavier. (This actually happens in neutron stars: a teaspoon would weigh ~1 billion
tonnes.)
India’s Scientific Contributions
The Bhabha Atomic Research Centre (BARC), Mumbai (Fig. 8.8), runs advanced neutron-scattering
experiments using reactors such as Dhruva. These reveal the inner structure of superconductors, battery
electrodes and drug molecules — helping build better medicines, energy storage and industrial alloys in India.
Exercise Q7 Assertion & Reason
7Assertion (A): The discovery of subatomic particles
helped in understanding the atomic structure. Reason (R): The number of electrons is equal to the number
of protons in an atom.
Correct option: (ii) — Both A and R are true, but R is not the correct explanation of A.
A is true: each discovery — electron (1897), proton, neutron (1932) — revealed one more piece of the
atom and forced a better model.
R is true: in a neutral atom the number of electrons does equal the number of protons.
R does not explain A: R states only one particular fact about a neutral atom (its charge balance).
It does not explain why discovering the particles improved our understanding of atomic structure. It is a
consequence of the discoveries, not the reason for them.
Quick recap — who discovered what
Particle
Discovered by
Year
Key experiment / idea
Electron
J. J. Thomson
1897
Cathode ray tube — rays bent by fields
Nucleus & Proton
E. Rutherford
1911
α-particle scattering from gold foil
Neutron
J. Chadwick
1932
Neutral particle explaining the extra mass
8.4 Symbols of Elements
By 1869 scientists knew about 69 elements. Today 118 elements are known — some of them
made artificially — and the search continues.
1803 — John Dalton gave the first pictorial symbols (circles with dots, lines, letters) for the
known elements (Fig. 8.9). They were hard to draw and remember.
1813 — J. J. Berzelius suggested using letters from the Latin names → alphabetic symbols.
Today — IUPAC (International Union of Pure and Applied Chemistry) approves the names and symbols of all
elements.
Rules for writing symbols
Table C — IUPAC rules with examples
Rule
Examples
Many symbols are the first letter or the first two letters of the name
Hydrogen → H, Carbon → C, Aluminium → Al
The first letter is always CAPITAL, the second (if any) is always small
Al not AL; Co not CO; Cl not CL
(CO would mean carbon + oxygen!)
Sometimes the second letter is taken from elsewhere in the name, not the 2nd letter
Chlorine → Cl, Zinc → Zn, Magnesium → Mg
Some come from Latin, Greek or German names
Iron → Fe (ferrum), Mercury → Hg (hydrargyros), Tungsten → W (wolfram)
Table 8.2 — Common elements and their symbols (learn these by heart)
Element
Symbol
Element
Symbol
Element
Symbol
Aluminium
Al
Copper (Cuprum)
Cu
Nitrogen
N
Argon
Ar
Fluorine
F
Oxygen
O
Barium
Ba
Gold (Aurum)
Au
Potassium (Kalium)
K
Boron
B
Hydrogen
H
Silicon
Si
Bromine
Br
Iodine
I
Silver (Argentum)
Ag
Calcium
Ca
Iron (Ferrum)
Fe
Sodium (Natrium)
Na
Carbon
C
Lead (Plumbum)
Pb
Sulfur
S
Chlorine
Cl
Magnesium
Mg
Uranium
U
Cobalt
Co
Neon
Ne
Zinc
Zn
Red symbols = the ones that do not match the English name — these are
the most commonly asked in exams.
Pause and Ponder Q8 – Q9
8Imagine you are a scientist who has discovered a new
element. Name this element after yourself and justify that the symbol you have chosen follows the IUPAC rules.
(Sample answer — use your own name.) Name: Mayankium → Symbol: My Justification against each IUPAC rule:
The symbol is short — two letters, taken from the element’s name.
The first letter M is capital and the second letter y is small — written My, never MY or my.
It is not already used by any of the 118 known elements (Mg, Mn, Mo, Mt, Md are taken — My is free), so
there is no confusion.
The name ends in “-ium”, the ending IUPAC uses for newly discovered metallic elements.
It is easy to pronounce and can be written the same way in every language.
Note: in real practice IUPAC does not allow an element to be named after a living discoverer —
names usually honour a scientist, place, mineral, mythological figure or property.
9What problems could arise if every scientist used
different symbols for the same element?
Confusion in formulae and equations — the same formula would mean different compounds to different people.
Experiments could not be repeated, because nobody would be sure which substance was used.
Language barriers would return — a symbol is understood worldwide, a name is not.
Dangerous errors in medicine, industry and trade — e.g. mixing up a harmless and a poisonous substance.
Textbooks, research papers and databases from different countries could not be compared, slowing science
down.
That is why the symbols are standardised by IUPAC and used identically all over the world.
8.5 Atomic Number (Z)
DEFAtomic number (Z) = the number of protons in the
nucleus of an atom. It determines the identity of the element and its chemical behaviour.
Z = number of protons = number of electrons (in a neutral atom)
Hydrogen: 1 p+, 1 e– ⇒ Z = 1 | Helium: 2 p+,
2 e– ⇒ Z = 2 | Lithium: 3 p+, 4 n0 ⇒ Z = 3
Two elements can never have the same atomic number. Change Z and you change the element
itself.
8.6 Mass Number (A)
DEFMass number (A) = total number of protons + neutrons in
the nucleus. Protons and neutrons together are called nucleons.
A = p+ + n0 ⇒ n0 = A – Z
The electron’s mass is almost negligible, so it is never counted in the mass number.
Table 8.3 — Mass number of some elements
Element
Protons (p+)
Neutrons (n0)
Mass number (A)
Hydrogen
1
0
1
Helium
2
2
4
Lithium
3
4
7
Standard notation of an atom
How to read AZX — mass number on top, atomic number below
Memory trickA is Above and it is the
Atomic mass-ish number. Z is at the bottom (like the last letter of the alphabet) and counts
protons. Neutrons are never written — you always subtract to get them.
Pause and Ponder Q10 – Q13 (numericals)
10An atom with an atomic number of 26 has 56 nucleons.
Find its number of electrons, protons and neutrons.
Given:Z = 26, nucleons (A) = 56
Protons = Z = 26 Electrons = protons (neutral atom) = 26 Neutrons = A – Z = 56 – 26 = 30 The element is iron (Fe), written as 5626Fe.
11The nucleus of an atom contains 20 protons. If its mass
number is 41, find the number of neutrons in it.
Given:Z = 20, A = 41
Neutrons = A – Z = 41 – 20 = 21 (Z = 20 ⇒ the element is calcium, so this is 4120Ca.)
12An atom has 18 neutrons and an atomic number of 17. What
is its mass number?
Given: n0 = 18, Z = 17
A = p+ + n0 = 17 + 18 = 35 The element is chlorine — 3517Cl.
13An atom 23A has 11 electrons. Find the number
of neutrons in it.
Given: A = 23, electrons = 11
In a neutral atom, protons = electrons ⇒ Z = 11 Neutrons = A – Z = 23 – 11 = 12 The element is sodium (Na) — 2311Na.
Exercise Q11, Q12 & Q15 numericals on Z and A
11An atom 70X has 31 electrons. How many
neutrons are there in its nucleus?
Electrons = 31 ⇒ protons = 31 ⇒ Z = 31; A = 70
Neutrons = A – Z = 70 – 31 = 39 Z = 31 ⇒ the element is gallium (Ga).
12An atom has 79 protons and a mass number of 197.
Calculate (i) the number of neutrons and (ii) the number of electrons.
(i) Neutrons = A – Z = 197 – 79 = 118 (ii) Electrons = protons = 79 (the atom is neutral)
Z = 79 ⇒ the element is gold (Au), 19779Au.
15In an atom there are 12 protons and 12 neutrons.
Now imagine all the electrons are replaced by hypothetical particles with the same charge as an electron but
500 times heavier. What effect will this have on the atom’s (i) atomic number, (ii) atomic mass,
(iii) mass number, (iv) overall charge?
The atom is magnesium: Z = 12, A = 24, 12 electrons.
(i) Atomic number — no change, Z = 12.Z counts only protons, and the
protons were not touched.
(ii) Atomic mass — increases noticeably. Normally 1 electron ≈ 1/1836 u, so
12 electrons ≈ 0.0065 u (ignorable). Now each weighs 500 × 1/1836 ≈ 0.27 u, so 12 of them ≈ 3.3 u.
New atomic mass ≈ 24 + 3.3 ≈ 27.3 u — electron mass can no longer be ignored.
(iii) Mass number — no change, A = 24. Mass number counts only nucleons
(protons + neutrons), never electrons.
(iv) Overall charge — no change, the atom stays neutral (0). The new particles carry
the same charge (–1 each), and 12 negative charges still cancel 12 protons. Mass changed; charge did
not.
8.7 How Are Electrons Distributed in Different Energy Levels?
The rules were given by Bohr and Bury:
RULE 1 Maximum number of electrons in a shell =
2n², where n is the shell number.
Shell
n
2n²
Maximum electrons
K
1
2 × 1²
2
L
2
2 × 2²
8
M
3
2 × 3²
18
N
4
2 × 4²
32
RULE 2 The outermost shell can hold a maximum of
8 electrons — no matter what 2n² allows. (The first shell can hold only 2.)
RULE 3 Electrons are filled step by step from the inside out:
K → L → M → N. A shell is filled only after the one before it is complete.
Why sodium is 2,8,1 and not 2,9 The L shell is full at 8 (2n² = 8), so the
11th electron must start a new shell M. And why not 2,8,1 → 2,8,1 only? Because Rule 2 caps the outermost
shell at 8, and Rule 3 forbids skipping a shell.
8.7.1 Building Up Atoms — the first eighteen elements
DEFElectronic configuration = the distribution of electrons among
the various shells of an atom, written as K, L, M, N (e.g. sodium = 2, 8, 1).
H(1)
1
He(2)
2
Li(3)
2,1
Be(4)
2,2
B(5)
2,3
C(6)
2,4
N(7)
2,5
O(8)
2,6
F(9)
2,7
Ne(10)
2,8
Na(11)
2,8,1
Mg(12)
2,8,2
Al(13)
2,8,3
Si(14)
2,8,4
P(15)
2,8,5
S(16)
2,8,6
Cl(17)
2,8,7
Ar(18)
2,8,8
Fig. 8.11 — Schematic atomic structures of the first 18 elements
(red centre = nucleus, blue dots = electrons; the number in brackets is Z)
Table 8.4 — The first eighteen elements (with the valency column added, as the book asks)
Element
Symbol
Z
p+
n0
e–
K
L
M
N
Valency
Hydrogen
H
1
1
–
1
1
–
–
–
1
Helium
He
2
2
2
2
2
–
–
–
0
Lithium
Li
3
3
4
3
2
1
–
–
1
Beryllium
Be
4
4
5
4
2
2
–
–
2
Boron
B
5
5
6
5
2
3
–
–
3
Carbon
C
6
6
6
6
2
4
–
–
4
Nitrogen
N
7
7
7
7
2
5
–
–
3
Oxygen
O
8
8
8
8
2
6
–
–
2
Fluorine
F
9
9
10
9
2
7
–
–
1
Neon
Ne
10
10
10
10
2
8
–
–
0
Sodium
Na
11
11
12
11
2
8
1
–
1
Magnesium
Mg
12
12
12
12
2
8
2
–
2
Aluminium
Al
13
13
14
13
2
8
3
–
3
Silicon
Si
14
14
14
14
2
8
4
–
4
Phosphorus
P
15
15
16
15
2
8
5
–
3
Sulfur
S
16
16
16
16
2
8
6
–
2
Chlorine
Cl
17
17
18
17
2
8
7
–
1
Argon
Ar
18
18
22
18
2
8
8
–
0
Pause and Ponder Q14 – Q16
14Identify the number of electrons in the outermost shell
of: (i) 126C (ii) 199F (iii) 2814Si
(i) 126C — Z = 6 ⇒ 6 electrons ⇒ configuration 2, 4 ⇒
4 electrons in the outermost shell.
(ii) 199F — Z = 9 ⇒ 9 electrons ⇒ configuration 2, 7 ⇒
7 electrons in the outermost shell.
(iii) 2814Si — Z = 14 ⇒ 14 electrons ⇒ configuration 2, 8, 4 ⇒
4 electrons in the outermost shell.
Remember: use the lower number (Z), not the mass number, to count electrons.
15Write the electronic configuration of the elements
having atomic numbers 12, 16 and 18.
Z = 12 (Magnesium, Mg): K = 2, L = 8, M = 2 ⇒ 2, 8, 2
Z = 16 (Sulfur, S): K = 2, L = 8, M = 6 ⇒ 2, 8, 6
Z = 18 (Argon, Ar): K = 2, L = 8, M = 8 ⇒ 2, 8, 8 — a complete octet, so argon is
unreactive (a noble gas).
16Solve this riddle: I am an atom with a mass number of 23
and 11 protons. I am a soft metal and react vigorously with water. Who am I and how many neutrons do I have?
11 protons ⇒ Z = 11 ⇒ I am sodium (Na), written 2311Na.
Neutrons = A – Z = 23 – 11 = 12 (Configuration 2, 8, 1 — one loose valence electron is exactly why sodium is so reactive.) Make your own riddle: “I have 17 protons and 18 neutrons, I am a greenish-yellow gas and I disinfect your
drinking water. Who am I?” → Chlorine, 3517Cl.
Exercise Q8 & Q13 configuration practice
8Magnesium is essential for many biological processes,
including muscle contraction. For an atom of magnesium with mass number 24 and atomic number 12, determine
(i) protons, (ii) neutrons, (iii) electrons, and illustrate the arrangement of electrons.
Given: A = 24, Z = 12
(i) Protons = Z = 12
(ii) Neutrons = A – Z = 24 – 12 = 12
(iii) Electrons = protons = 12
Arrangement: K = 2, L = 8, M = 2 ⇒ 2, 8, 2
2 valence electrons ⇒ Mg loses them easily ⇒ valency 2 ⇒ forms Mg²⁺ (as in MgCl2).
13Complete Table 8.5.
Use Z = p+ = e– and A = p+ + n0. Filled-in
values are shown in green.
Atomic number
Mass number
Neutrons
Protons
Electrons
Name of 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
Working, row by row:
Row 1: Z = 5 ⇒ p = e = 5; A = 5 + 6 = 11 ⇒ boron.
Row 2: e = 7 ⇒ Z = p = 7; n = 14 – 7 = 7 ⇒ nitrogen ✓.
Row 3: p = 12 ⇒ Z = e = 12; n = 24 – 12 = 12 ⇒ magnesium.
Row 4: Z = 15 ⇒ p = e = 15; A = 15 + 16 = 31 ⇒ phosphorus.
Row 5: A = 1, n = 0 ⇒ p = 1 ⇒ Z = e = 1 ⇒ hydrogen (the only atom with no neutron).
8.8 Combining Capacity of an Atom: Valency
DEFCombining capacity = the number of atoms of hydrogen or
chlorine with which one atom of an element combines. (H and Cl are used as the yardstick because both have a
combining capacity of 1.)
Compound
Element
Combines with
Combining capacity
H2O
Oxygen
2 H atoms
2
NH3
Nitrogen
3 H atoms
3
MgCl2
Magnesium
2 Cl atoms
2
CH4
Carbon
4 H atoms
4
DEFValence shell = the outermost shell of an atom that contains electrons. Valence electrons = the electrons present in the valence shell. Octet = a valence shell containing 8 electrons. Valency = the number of electrons gained, lost or shared by an atom to complete its octet.
THE OCTET RULE Elements with a complete octet (8 valence
electrons) — or 2 in the case of helium — are largely unreactive and stable (the noble gases).
Atoms with incomplete valence shells are reactive: they lose, gain or share electrons to reach an octet.
Flow chart — how to find the valency of any element from its configuration
Worked examples of valency
Element
Configuration
Valence e–
What it does
Valency
Sodium (Na)
2, 8, 1
1
loses 1 e– → 2, 8 (octet)
1
Magnesium (Mg)
2, 8, 2
2
loses 2 e–
2
Carbon (C)
2, 4
4
shares 4 e– (cannot easily lose/gain 4)
4
Oxygen (O)
2, 6
6
gains 2 e– → 2, 8
2
Chlorine (Cl)
2, 8, 7
7
gains 1 e– → 2, 8, 8
1
Neon (Ne)
2, 8
8
nothing — already an octet
0
Can an atom that already has 8 valence electrons still lose or gain electrons?No — it has no reason to. Its octet is already complete, so it is stable and does not
normally react. Its valency is 0. That is why neon, argon, etc., are called noble (inert) gases.
Some compounds appear to break the usual valency rule — you will study these in higher
classes.
Exercise Q9 read the diagrams
9Find the following information for the elements shown in
Fig. 8.17: (i) name, (ii) symbol, (iii) total electrons, (iv) valence electrons, (v) valency, (vi) protons,
(vii) atomic number.
First count the dots shell by shell, then everything else follows.
(a)
(b)
(c)
(d)
(i) Name
Lithium
Nitrogen
Aluminium
Fluorine
(ii) Symbol
Li
N
Al
F
(iii) Total electrons
3
7
13
9
Configuration
2, 1
2, 5
2, 8, 3
2, 7
(iv) Valence electrons
1
5
3
7
(v) Valency
1
3
3
1
(vi) Protons
3
7
13
9
(vii) Atomic number (Z)
3
7
13
9
Why those valencies: Li has 1 valence e– (< 4) → loses 1 → valency 1. N has 5 (> 4) →
gains 3 → valency 3. Al has 3 (< 4) → loses 3 → valency 3. F has 7 (> 4) → gains 1 → valency 1.
8.9 A Deeper Look into Atomic Structure
8.9.1 Isotopes
Dalton said all atoms of an element are identical and equally heavy. Scientists later found this is
not true: atoms of the same element can carry different numbers of neutrons.
DEFIsotopes = atoms of the same element having the
same atomic number (Z) but different mass numbers (A) — i.e. the same number of protons but a
different number of neutrons. Think of them as “twin atoms”.
Isotopes of hydrogen
Fig. 8.12 — The three isotopes of hydrogen (all have 1 proton and 1 electron)
Isotope
Symbol
p+
n0
e–
A
Natural abundance
Protium
11H
1
0
1
1
≈ 99.98 %
Deuterium
21H
1
1
1
2
≈ 0.015 %
Tritium
31H
1
2
1
3
traces only
All three have 1 electron — because all three have 1 proton and are neutral.
Isotopes of carbon
Fig. 8.13 — Isotopes of carbon; each has 6 protons and 6 electrons, only the neutrons differ
VERY IMPORTANT
Isotopes have the SAME chemical properties — because chemical behaviour depends on the number of
valence electrons, and all isotopes have the same number of electrons and the same electronic configuration.
They have DIFFERENT physical properties (density, melting point, boiling point, rate of
diffusion) — because these depend on mass, and their masses differ.
Uses of isotopes — Bridging Science and Society
Isotope
Element
Use
23592U
Uranium
Fuel in nuclear reactors to generate electricity in nuclear power plants (Fig. 8.14)
6027Co
Cobalt (radioactive)
Radiation therapy for the treatment of cancer
13153I
Iodine
Treatment of goitre and thyroid cancer
146C
Carbon
Carbon dating — finding the age of ancient fossils and artefacts in archaeology and geology
Ready to Go Beyond — the unit ‘u’
Atoms are far too tiny to weigh in kilograms, just as a grain of wheat is measured in milligrams, not kilograms.
So scientists use the unified atomic mass unit (u). (The older name was amu, atomic mass unit.)
Exercise Q14 a full case-study question
14An 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 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?
Given: A = 35, n0 = 18 ⇒ p+ = A – n0 = 35 – 18 = 17
(i) Protons = 17, electrons = 17 (neutral atom).
(ii) Atomic number Z = 17.
(iii)Z = 17 ⇒ X is chlorine (Cl), i.e. 3517Cl.
(iv) Configuration: K = 2, L = 8, M = 7 ⇒ 2, 8, 7.
(vi) Adding 2 neutrons: n0 = 20, protons still 17 ⇒ new A = 17 + 20 = 37,
i.e. 3717Cl.
(vii) Same Z (17) but different A (35 and 37) ⇒ the two atoms are isotopes of each other.
They are both chlorine and behave identically in chemical reactions.
A. Average Atomic Mass
Chlorine occurs in nature as two isotopes — one of mass 35 u, the other 37 u, in the ratio
3 : 1. So is the mass of a chlorine atom 35 u or 37 u?
Average atomic mass = Σ (mass of isotope × its % abundance) ÷ 100
DON’T MISUNDERSTANDNo single chlorine atom weighs 35.5 u.
It means that in, say, 10,00,000 chlorine atoms there are about 7,50,000 atoms of
3517Cl and 2,50,000 atoms of 3717Cl, and 35.5 u is their
weighted average.
Pause and Ponder Q17 – Q18
17Two different atoms have 11 protons each, but one has
12 neutrons and the other 13 neutrons. How do their atomic numbers and mass numbers compare? Are they the same
element or different elements?
Atomic numbers: both have 11 protons ⇒ Z = 11 for both — identical.
Mass numbers: A₁ = 11 + 12 = 23; A₂ = 11 + 13 = 24 — different.
Same or different element? The identity of an element is fixed by Z alone, so both are the
same element — sodium (Na). Since Z is the same but A differs, they are
isotopes: 2311Na and 2411Na. They will show identical
chemical properties.
18Bromine occurs as two isotopes,
7935Br (49.7 %) and 8135Br (50.3 %). Calculate the average atomic mass
of bromine.
Average atomic mass = (79 × 49.7/100) + (81 × 50.3/100)
= (3926.3 ÷ 100) + (4074.3 ÷ 100)
= 39.263 + 40.743
= 80.006 u ≈ 80 u Check: the two isotopes are almost equally abundant, so the answer should lie almost exactly
midway between 79 and 81 — and 80 does. Always do this sanity check.
8.9.2 Isobars
DEFIsobars = atoms of different elements having the
same mass number (A) but different atomic numbers (Z). They have the same total number of nucleons,
but the protons/neutrons split differently.
Isotopes vs Isobars — the one comparison you must never mix up
The classic isobar trio — all have A = 40
Element
Symbol
Z (p+)
n0
A
Argon
4018Ar
18
22
40
Potassium
4019K
19
21
40
Calcium
4020Ca
20
20
40
Memory hookIsotopes → “o” for the same atomic number (same element, different
weight). Isobars → “a” for the same A (mass number).
Exercise Q3 isotopes vs isobars
3The composition of the nuclei of three atomic species
X, Y and Z is given. X: 18 p, 19 n | Y: 17 p, 18 n | Z: 17 p, 20 n.
Explain the relation between (i) Y and Z, (ii) Z and X.
Step 1 — work out Z and A for each:
Species
Protons = Z
Neutrons
A = p + n
Element
X
18
19
37
Argon
Y
17
18
35
Chlorine
Z
17
20
37
Chlorine
(i) Y and Z: same atomic number (Z = 17) but different mass numbers (35 and 37) ⇒ they are
ISOTOPES. Both are chlorine, so their chemical properties are identical; only their physical
properties differ slightly.
(ii) Z and X: same mass number (A = 37) but different atomic numbers (17 and 18) ⇒ they are
ISOBARS. They are different elements (chlorine and argon) with completely different chemical
properties, but their atoms contain the same total number of nucleons.
At a Glance — the whole chapter on one page
Summary points (NCERT)
Atoms are the building blocks of matter.
J. J. Thomson — electrons are embedded in a positively charged sphere (plum pudding).
Rutherford — the atom is mostly empty space, with a dense, positively charged nucleus at the centre
and electrons orbiting it.
Niels Bohr — electrons move in fixed energy levels (shells) around the nucleus.
Shells are named K, L, M, N …
James Chadwick discovered the neutron.
The three subatomic particles are electrons, protons and neutrons.
An octet in the outermost shell (or 2 e– for helium) makes an atom stable and unreactive.
Valency = combining capacity = electrons gained, lost or shared to reach a stable configuration.
Atomic number (Z) = number of protons in the nucleus.
Mass number (A) = total number of nucleons (protons + neutrons).
Isotopes — same Z, different A. Isobars — same A, different Z.
Average atomic mass is calculated from the relative abundance of the isotopes in nature.
Formula & fact sheet (learn these cold)
Formula / fact
Meaning & typical use
Z = p+ = e–
Atomic number; true only for a neutral atom
A = p+ + n0
Mass number = nucleons
n0 = A – Z
The most used line in every numerical
Max e– in shell = 2n²
K = 2, L = 8, M = 18, N = 32
Outermost shell ≤ 8 e–
(≤ 2 if it is the K shell)
Valency
= valence e– if ≤ 4 | = 8 – valence e– if > 4
Avg. atomic mass
Σ (isotope mass × % abundance) ÷ 100
datom ≈ 10–10 m
Nucleus ≈ 10–15 m ⇒ 105 times smaller
α-particle = He nucleus
2 p+ + 2 n0, charge +2, mass 4 u
e– charge = –1.602 × 10–19 C
Taken as –1 by convention
Common exam traps
Counting electrons from the mass number instead of the atomic number. Always use Z.
Writing sodium as 2, 9 — the L shell stops at 8, so it must be 2, 8, 1.
Saying the neutron was discovered by the gold foil experiment. It was Chadwick, 1932.
Saying isotopes have different chemical properties. They do not — only physical properties differ.
Confusing “accelerating” with “speeding up”. In a circle, direction changes ⇒ it is accelerating even at
constant speed.
Forgetting that mass number is always a whole number, while average atomic mass can be a fraction
(e.g. Cl = 35.5 u).
The Quest Continues …Is it possible to completely understand everything that happens inside an atom?
The story does not end with Bohr. We now know electrons do not travel on neat fixed tracks at all — they exist as
electron clouds, and we can only predict where they are most likely to be, not exactly where they are.
Instruments such as Scanning Tunnelling Microscopes (STM) and Transmission Electron Microscopes (TEM)
now let us see individual atoms (Fig. 8.15). The journey inside the atom is far from over.
Meet a Scientist — Homi Jehangir Bhabha
Indian physicist, known as the father of the Indian nuclear programme. He founded the
Tata Institute of Fundamental Research (TIFR) and the Bhabha Atomic Research Centre (BARC), for the
peaceful use of atomic energy — generating electricity, supporting agriculture and advancing medical treatment.
The Journey Beyond — things to try
Make an ‘Atomic Prediction Board’ game: give clues (Z, A, valency) and let friends name the element.
Write a report on how atomic properties help us in healthcare, energy, agriculture and technology.
Make a role-play or story — “Journey Inside the Atom” — with Dalton, Thomson, Rutherford, Bohr and
Chadwick as characters.
Create an animation/simulation of the atomic models using any digital tool.
Draw a bar graph of the number of electrons in each energy level for any three elements.
Explore the PhET simulations: phet.colorado.edu → “Rutherford Scattering” and
“Isotopes and Atomic Mass”
— End of Chapter 8 · Journey Inside the Atom — This may not be the end … the atom is still being discovered.