This MCQ module is based on: Subatomic Particles Atomic Models
TOPIC 5 OF 28
Subatomic Particles Atomic Models
🎓 Class 11
Chemistry
CBSE
Theory
Ch 2 – Structure of Atom
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Subatomic Particles Atomic Models
Upload images, PDFs, or Word documents to include their content in assessment generation.
Discovery of Subatomic Particles and Early Atomic Models
Introduction: Looking Inside the Atom
In Class 9 you learnt that every sample of matter is built out of atoms. Dalton's 1808 theory pictured atoms as hard, indivisible spheres — a useful picture for stoichiometry but one that left deep puzzles unanswered. Why do different elements give different colours when heated (sodium yellow, copper green)? Why does a gas inside a cathode-ray tube glow? Why do metals conduct electricity while sulphur does not?
Between 1895 and 1935 a remarkable chain of experiments cracked the atom open. Electrons, protons and neutrons were discovered one by one, and an entire subatomic zoo was catalogued. This first part of Chapter 2 retraces that journey, from the cathode-ray tube to Rutherford's gold foil, and sets up the quantum ideas you will meet in Parts 2 and 3.
2.1 Discovery of Subatomic Particles
2.1.1 Discovery of the Electron — The Cathode Ray Tube
A cathode ray tube is a sealed glass tube containing a gas at very low pressure with two metal electrodes. William Crookes (1870s) showed that when a very high potential difference (~10 000 V) is applied across the electrodes, an invisible beam travels from the cathode to the anode, causing the glass behind the anode to glow. These rays were named cathode rays.
Careful experiments established four properties of cathode rays:
- They travel in straight lines (objects placed in their path cast a shadow).
- They are deflected towards the positive plate when an electric field is applied — so the rays carry negative charge.
- They are deflected by a magnetic field in a direction consistent with negatively charged particles.
- Their properties are independent of the gas in the tube and of the metal of the cathode — so these particles are a universal component of all matter.
In 1897 J. J. Thomson balanced electric and magnetic deflections to measure the charge-to-mass ratio of the particles:
e/me = 1.758820 × 1011 C kg−1
He named these particles electrons. In 1909 R. A. Millikan's oil-drop experiment measured the charge on a single electron by balancing the gravitational pull on a tiny charged oil droplet against an upward electric field:
The charge always came out a whole-number multiple of a smallest value — so charge itself is quantised. Millikan's value:
e = −1.602176 × 10−19 C
Combining Thomson's e/m with Millikan's e gave the mass of the electron:
me = e ÷ (e/me) = 9.1094 × 10−31 kg ≈ 1⁄1837 × mproton
2.1.2 Discovery of the Proton and the Neutron
As soon as the electron was recognised, a balancing positive component was expected. Eugen Goldstein (1886) had in fact already seen it: using a perforated cathode in a discharge tube, he observed rays streaming in the opposite direction — from the region behind the cathode. These were later christened canal rays or anode rays. Their properties:
- They carry positive charge.
- Their charge-to-mass ratio depends on the gas in the tube (unlike cathode rays).
- Their e/m is largest when the gas is hydrogen — the positive particle from hydrogen is the lightest positive particle known and is called the proton.
Proton: charge = +1.602 × 10−19 C; mass = 1.672 × 10−27 kg
But protons and electrons alone could not explain the mass of most atoms — a helium nucleus, for example, weighs four units but carries only two unit charges. In 1932 James Chadwick bombarded beryllium with α-particles and detected a new, electrically neutral particle of mass slightly greater than the proton:
94Be + 42He → 126C + 10n (neutron)
The neutron carries no charge and has mass 1.675 × 10−27 kg.
| Particle | Symbol | Charge (C) | Relative charge | Mass (kg) | Relative mass (amu) |
|---|---|---|---|---|---|
| Electron | e− | −1.602 × 10−19 | −1 | 9.109 × 10−31 | 0.00055 |
| Proton | p | +1.602 × 10−19 | +1 | 1.672 × 10−27 | 1.00728 |
| Neutron | n | 0 | 0 | 1.675 × 10−27 | 1.00867 |
Worked Numericals on Fundamental Particles
Example 2.1 — Charge on one mole of electrons
Given: Charge on one electron \(e = 1.602 \times 10^{-19}\) C; Avogadro number \(N_A = 6.022 \times 10^{23}\) mol−1.
Solution: One mole contains \(N_A\) electrons, so the total charge is
\[ Q = N_A \cdot e = (6.022 \times 10^{23})(1.602 \times 10^{-19}) \]
\[ Q = 9.647 \times 10^{4}\ \text{C} = 96\,470\ \text{C} \]
This value is the Faraday constant \(F\) — the charge on one mole of electrons.
Example 2.2 — Charge on 1 g of electrons
Step 1. Number of electrons in 1 g: \( n = \dfrac{1\ \text{g}}{9.109 \times 10^{-28}\ \text{g}} = 1.098 \times 10^{27}\) electrons.
Step 2. Total charge:
\[ Q = n \cdot e = (1.098 \times 10^{27})(1.602 \times 10^{-19}) = 1.759 \times 10^{8}\ \text{C} \]
i.e. about 1.76 × 108 coulombs — a colossal charge, which is why bulk matter is almost perfectly neutral.
Example 2.3 — Total mass and charge of an atom with 24 p, 25 n, 24 e
Mass = 24(1.672 × 10−27) + 25(1.675 × 10−27) + 24(9.109 × 10−31)
\[ m = (4.013 + 4.188 + 0.0000219)\times 10^{-26}\ \text{kg} \approx 8.202 \times 10^{-26}\ \text{kg} \]
Charge: 24 protons give +24e and 24 electrons give −24e, so net charge = 0 (neutral atom).
Notice how the electron mass contributes only 0.03 % of the total — for most purposes the mass of an atom ≈ mass of its nucleus.
Example 2.4 — Protons, neutrons and electrons in common species
Identify p, n, e for: (a) 3517Cl−, (b) 5626Fe3+, (c) 168O2−.
Rule: p = Z; n = A − Z; e = Z − (charge) [charge with sign].
(a) Cl−: p = 17, n = 35 − 17 = 18, e = 17 − (−1) = 18.
(b) Fe3+: p = 26, n = 56 − 26 = 30, e = 26 − 3 = 23.
(c) O2−: p = 8, n = 16 − 8 = 8, e = 8 − (−2) = 10.
2.2 Atomic Models
2.2.1 Thomson's Model — the Plum-Pudding Atom (1898)
With electrons in hand, Thomson proposed the first atomic model in which sub-structure was explicit. The atom is pictured as a sphere (radius ≈ 10−10 m) of uniformly distributed positive matter in which the tiny negatively charged electrons are embedded — like plums in a pudding, or watermelon seeds in the pink flesh. The total positive charge balances the total negative charge, so the atom is electrically neutral.
Drawback: The model predicts that α-particles fired at a thin metal foil should pass through with only tiny deflections, because the positive charge and mass are spread smoothly over the whole atom. Rutherford's experiment showed that this prediction fails dramatically.
2.2.2 Rutherford's α-Particle Scattering Experiment (1911)
Ernest Rutherford, with Geiger and Marsden, directed a beam of high-energy α-particles (He2+ nuclei) at a very thin (≈100 nm) gold foil and recorded the deflections on a surrounding zinc-sulphide screen.
Observations:
- The great majority of α-particles passed straight through the foil without deflection.
- A small fraction were deflected through small angles.
- A very few (about 1 in 20 000) were deflected through angles greater than 90° — some even bounced straight back.
Rutherford reportedly said this was like firing a 15-inch naval shell at a sheet of tissue paper and having it bounce back at you — utterly unexpected on Thomson's model.
Rutherford's Nuclear Model of the Atom
To explain the observations Rutherford proposed that:
- Almost all the mass and all the positive charge of the atom is concentrated in a tiny central nucleus (radius ≈ 10−15 m — about 105 times smaller than the atom).
- Electrons revolve around the nucleus in circular orbits, held by electrostatic attraction (like planets around the sun).
- Most of the atom is empty space — hence most α-particles pass straight through.
Limitations of Rutherford's Model L4 Analyse
- Instability. Classical electromagnetism says any accelerating charge radiates energy. An electron moving in a circle is continuously accelerating (centripetal acceleration), so it must lose energy, spiral inward and crash into the nucleus in ≈10−8 s. But atoms are stable for billions of years.
- No explanation of line spectra. A spiralling electron would radiate a continuous range of frequencies; real atoms emit only specific, sharp wavelengths (line spectrum of hydrogen, sodium D-lines, etc.).
- No explanation of why electrons do not fall in, or how they are arranged.
2.2.3 Atomic Number, Mass Number, Isotopes and Isobars
Rutherford's model fixed the grammar of atomic description:
- Atomic number (Z) = number of protons in the nucleus = number of electrons in the neutral atom. Z identifies the element.
- Mass number (A) = number of protons + number of neutrons = Z + n.
- Notation: AZX — e.g. 2311Na means sodium, Z = 11, A = 23, so n = 12.
| Class | Same | Different | Examples |
|---|---|---|---|
| Isotopes | Z (protons) | A (neutrons) | 11H (protium), 21H (deuterium), 31H (tritium); 126C, 146C |
| Isobars | A | Z | 146C and 147N; 4018Ar, 4019K, 4020Ca |
| Isotones | n (neutrons) | Z, A | 146C, 157N, 168O (each has 8 neutrons) |
| Isoelectronic | electrons | Z | Na+, Mg2+, F−, O2−, Ne — all have 10 e |
Remember: Chemical properties are set by electron configuration (and therefore by Z), so all isotopes of the same element behave almost identically chemically. Differences show up in mass-dependent physical properties (rate of diffusion, density, NMR frequency).
Activity 2.1 — Why a Solid-Sphere Atom Cannot Be Right L3 Apply
Objective: Use a simple thought experiment to see why α-scattering falsifies Thomson's model.
Materials: Marbles, a thin cardboard sheet, a table.
- Drop marbles vertically onto a flat cardboard sheet lying on a table. Note how many bounce straight back.
- Now imagine you replace the cardboard with a cardboard containing a few tiny dense steel bolts glued on its underside. Predict how often a marble would now rebound sharply.
- Relate this to Rutherford's observations.
Predict first: Before reading on, write down what fraction of α-particles you expect to bounce back if the atom were a uniform pudding vs. if the atom contained a tiny dense nucleus.
For a uniform Thomson pudding, the electric field on any α-particle is small and smooth — almost no particle would come back. Rutherford's rare bounce-backs (1 in 20 000) can only be produced by very rare, head-on encounters with a tiny, very dense positive charge. The rarity of large deflections is itself evidence that the nucleus is small — the smaller the target, the less often you hit it.
Atomic Model Timeline — Click to Explore
Step through how the picture of the atom evolved between 1808 and 1913.
Competency-Based Questions L3 Apply
A laboratory has a cathode-ray tube and a vacuum pump. Students perform Thomson's experiment and independently also study canal rays. They then record data on α-scattering from a gold foil using an old Rutherford-style apparatus.
Q1. Which of the following experimental facts about cathode rays is the single strongest evidence that electrons are a universal component of matter?
C. A universal particle must have the same properties no matter what source it comes from. Since e/m is independent of both gas and electrode, the cathode-ray particle must be present in every element.
Q2. An ion X2+ contains 18 electrons and a mass number of 40. State its number of protons and neutrons.
Electrons in ion = 18; since charge = +2, protons = 18 + 2 = 20 (calcium). Neutrons = A − Z = 40 − 20 = 20. So the ion is Ca2+.
Q3. Fill in the blank: In Millikan's oil-drop experiment, the charges on individual oil drops were always ______ of 1.602 × 10−19 C.
Whole-number multiples (integral multiples). This proved that electric charge is quantised.
Q4. True or False: Two isobars can belong to the same element. Justify.
False. Isobars are defined by having the same mass number but different atomic numbers, so by definition they belong to different elements. Same-element, same-A atoms would be identical.
Q5. A gold foil 100 nm thick contains roughly 300 layers of atoms. Explain qualitatively why almost all α-particles pass straight through, yet a very small number bounce back — referring to nuclear size.
The nucleus occupies only about (10−15/10−10)2 = 10−10 of the cross-section of an atom. So in 300 atomic layers, the chance of any α-particle scoring a near head-on hit on a nucleus is extremely small — but when it happens the Coulomb repulsion from the concentrated positive charge is huge, and the α-particle is violently deflected or reversed.
Assertion–Reason Questions
Options: A. Both A and R are true and R is the correct explanation of A. B. Both A and R are true but R is NOT the correct explanation. C. A is true, R is false. D. A is false, R is true.
Assertion (A): The charge-to-mass ratio of cathode rays is independent of the gas used in the tube.
Reason (R): Electrons are fundamental particles present in the atoms of every element.
A. Both statements are true and R is exactly why A holds: the electron is a universal particle, so its e/m cannot depend on the gas.
Assertion (A): Rutherford's model cannot explain the stability of atoms.
Reason (R): According to classical electrodynamics, an accelerating electron radiates energy continuously.
A. Both true; the radiation predicted by R would make the orbiting electron spiral into the nucleus, giving the instability mentioned in A.
Assertion (A): 4018Ar and 4020Ca are isotopes.
Reason (R): Both have the same mass number.
D. A is false — they are isobars, not isotopes, because they have different Z. R is true in isolation.
AI Tutor
Chemistry Class 11 Part I – NCERT (2025-26)
Ready