Introduction And Atomic Models

ChemistryAtomic StructureFor JEE aspirants

An atom is the smallest unit of matter that retains an element's chemical identity, and it is made of three fundamental subatomic particles - the negatively charged electron, the positively charged proton, and the electrically neutral neutron. Early atomic models progressed from J.J. Thomson's "plum pudding" (uniform positive charge with embedded electrons) to Rutherford's nuclear model (electrons orbiting a tiny dense positive nucleus), based on the famous $α$ -particle scattering experiment. This chapter builds the foundation for every model that follows, including Bohr's quantum picture.

Key Formulas - Quick Reference

Charge on electron: $e = 1.602 \times 1 0^{- 19}$ C; mass $m_{e} = 9.11 \times 1 0^{- 31}$ kg
Mass of proton: $m_{p} = 1.673 \times 1 0^{- 27}$ kg $\approx$ 1.007 amu
Mass of neutron: $m_{n} = 1.675 \times 1 0^{- 27}$ kg $\approx$ 1.009 amu
Charge/mass ratio (electron): $e / m = 1.76 \times 1 0^{8}$ C/g
Mass number: $A = Z + N$ (protons + neutrons)
Atomic mass unit: $1$ amu $= 1.66 \times 1 0^{- 27}$ kg (1/12th of C-12 mass)
Radius of nucleus: $r = 1.33 \times 1 0^{- 13} A^{1/3}$ cm

1. Fundamental Particles of the Atom

Atoms consist of three fundamental particles differing in mass and electric charge. All three are present in every atom except the ordinary hydrogen atom ( $_{1}^{1}$ H), which has no neutron.

Particle	Symbol	Charge	Mass (kg)	Mass (amu)
Electron	$e$ or $e^{-}$	$- 1.602 \times 1 0^{- 19}$ C	$9.11 \times 1 0^{- 31}$	0.000549
Proton	$p$ or $p^{+}$	$+ 1.602 \times 1 0^{- 19}$ C	$1.673 \times 1 0^{- 27}$	1.007
Neutron	$n$	0 (neutral)	$1.675 \times 1 0^{- 27}$	1.009

Atomic mass unit (amu): Defined as 1/12th of the mass of one atom of

_{6}^{12}

C. 1 amu

= 1.6605 \times 1 0^{- 27}

kg.

2. Discovery of the Electron (Cathode Rays)

When high voltage (~10,000 V) is passed through a discharge tube containing gas at very low pressure (0.01-0.03 mm Hg), rays emerge from the cathode. These are called cathode rays. J.J. Thomson (1897) studied these rays and identified the electron.

Figure 1: Cathode ray discharge tube used by J.J. Thomson to discover the electron.

Key properties of cathode rays

Travel in straight lines, perpendicular to the cathode surface
Cause fluorescence on ZnS and glass; blacken photographic plates
Deflected by both electric and magnetic fields - therefore charged (negative)
Produce X-rays when they strike a metal
Ionize gases through which they pass
Rotate a light paddle wheel - so they have mass and momentum
The $e / m$ ratio is the same for all gases - proving they are a universal subatomic particle

Thomson's $e / m$ measurement: By balancing electric and magnetic deflections, Thomson found

e / m = 1.76 \times 1 0^{8}

C/g for cathode ray particles - independent of the gas or electrode material used.

Millikan's oil-drop experiment (charge on electron)

In 1909, Robert Millikan measured the charge on the electron by suspending charged oil droplets in an electric field. He found that the charge on every droplet was always a whole-number multiple of a smallest value:

e = 1.602 \times 1 0^{- 19}

C (charge on one electron)

Figure 2: Millikan's oil-drop apparatus used to measure the charge on an electron (

e = 1.602 \times 1 0^{- 19}

C).

Combining Millikan's charge with Thomson's $e / m$ ratio gives the electron mass:

m_{e} = \frac{e}{e / m} = \frac{1.602 \times 1 0 ^{- 19}}{1.76 \times 1 0 ^{8}} = 9.11 \times 1 0^{- 31}

3. Discovery of the Proton (Canal Rays)

In 1886, E. Goldstein observed that when cathode rays travel through a perforated cathode, another set of rays travels in the opposite direction. He called these positive rays or canal rays. J.J. Thomson later measured their $e / m$ ratio.

Key properties of positive (canal) rays

Positively charged (deflected by fields opposite to cathode rays)
Charge is a whole-number multiple of $e$
Unlike cathode rays, properties depend on the gas in the discharge tube
Deflection is smaller than that of cathode rays (heavier particles)
The lightest particle in positive rays comes from hydrogen and is called the proton

Proton: A subatomic particle with charge

+ e

and mass

\approx 1.007

amu (

1.673 \times 1 0^{- 27}

kg) - approximately 1836 times heavier than an electron.

4. Discovery of the Neutron

Rutherford predicted (1920) the existence of a neutral particle in the nucleus to explain the discrepancy between atomic number and atomic mass. In 1932, James Chadwick confirmed it by bombarding beryllium with $α$ -particles:

_{4}^{9} Be +_{2}^{4} He ⟶_{6}^{12} C +_{0}^{1} n

The emitted radiation was highly penetrating and undeflected by electric or magnetic fields, showing it was neutral. Its mass was very close to that of a proton. This particle was named the neutron.

Solved Example 1

The neutron is attracted towards
(A) Positively charged particles
(B) Negatively charged particles
(C) Not attracted by any charge
(D) None of these

Solution:

Neutron is an uncharged (neutral) particle. It is not attracted by any charge.
Answer: (C).

5. Important Atomic Terms

Term	Definition	Example
Nuclide	A specific atomic species	$_{6}^{12}$ C, $_{17}^{35}$ Cl
Nucleons	Protons + neutrons (particles in nucleus)	-
Atomic number (Z)	Number of protons in the nucleus	Z(Na) = 11
Mass number (A)	Total nucleons: $A = Z + N$	A(Na) = 23
Isotopes	Same Z, different A	$^{1}$ H, $^{2}$ H, $^{3}$ H
Isobars	Same A, different Z	$_{18}^{40}$ Ar, $_{19}^{40}$ K, $_{20}^{40}$ Ca
Isotones	Same N (neutrons), different Z	$_{6}^{14}$ C, $_{7}^{15}$ N, $_{8}^{16}$ O
Isoelectronic	Same number of electrons	N $_{2}$ , CO, CN $^{-}$ , NO $^{+}$ (14 e $^{-}$ each)

Figure 6: Comparison of isotopes, isobars and isotones showing how protons (red), neutrons (grey) and mass number relate.

Solved Example 2

The ion that is isoelectronic with CO is
(A) CN

^{-}

(B) O

_{2}^{+}

_{2}^{-}

(D) N

_{2}^{+}

Solution:

CO has $6 + 8 = 14$ electrons. Count each option:

CN $^{-}$ : $6 + 7 + 1 = 14$ electrons $✓$
O $_{2}^{+}$ : $8 + 8 - 1 = 15$ electrons
O $_{2}^{-}$ : $8 + 8 + 1 = 17$ electrons
N $_{2}^{+}$ : $7 + 7 - 1 = 13$ electrons

Answer: (A) CN $^{-}$ .

Solved Example 3

Find the number of neutrons in a neutral atom having atomic mass 23 and 11 electrons.

Solution:

For a neutral atom, number of electrons = number of protons = $Z = 11$ .
Number of neutrons $= A - Z = 23 - 11 = 12$ .

Solved Example 4

How many protons, electrons and neutrons are present in 0.18 g of

_{15}^{30}

Solution:

Moles of P $= \frac{0.18}{30} = 0.006$ mol
Atoms $= 0.006 \times 6.022 \times 1 0^{23} = 3.6 \times 1 0^{21}$ atoms
Per atom: $Z = 15$ protons, 15 electrons, $A - Z = 15$ neutrons.
Total: Protons $= 15 \times 3.6 \times 1 0^{21} = 5.4 \times 1 0^{22}$ ; Electrons $= 5.4 \times 1 0^{22}$ ; Neutrons $= 5.4 \times 1 0^{22}$ .

6. Thomson's Atomic Model (Plum Pudding, 1904)

J.J. Thomson proposed that an atom is a sphere of uniformly distributed positive charge in which electrons are embedded like plums in a pudding (or seeds in a watermelon). The atom as a whole is electrically neutral because the total positive charge equals the negative charge of the electrons.

Figure 3: Thomson's plum pudding model - electrons embedded in a sphere of uniformly distributed positive charge.

Limitation: Thomson's model could not explain the results of Rutherford's

α

-particle scattering experiment - specifically why some

α

-particles bounced back sharply.

7. Rutherford's Atomic Model (Nuclear Model)

Ernest Rutherford (1911) bombarded a thin gold foil with $α$ -particles from a radioactive source and studied their scattering pattern using a ZnS-coated screen.

Figure 4: Rutherford's

α

-particle scattering by thin gold foil. Only rare particles hitting the tiny dense nucleus bounce back.

Observations from $α$ -scattering

Most $α$ -particles passed straight through - so the atom is mostly empty space
Some $α$ -particles were deflected through small angles - so there is a concentrated positive region
A very few (1 in 20,000) bounced back through angles greater than $9 0^{\circ}$ - implying a tiny, dense positive core

Conclusions - Rutherford's nuclear model

The atom has a tiny, dense, positively charged nucleus at its centre containing nearly all of the mass.
The nucleus is about $1 0^{- 15}$ m in radius (atom $\sim 1 0^{- 10}$ m) - so nucleus:atom $\approx 1 : 1 0^{5}$ by radius.
Electrons revolve around the nucleus in circular orbits like planets around the sun.
The centripetal force needed for the orbit is supplied by electrostatic attraction between electron and nucleus.

Figure 5: Rutherford's nuclear model of the atom - dense central nucleus with electrons revolving around it.

Solved Example 5

α

-particle travelling towards a gold nucleus returns back from a distance of

1 0^{- 10}

m. Calculate the velocity of the

α

-particle. (Given: 1 amu

= 1.66 \times 1 0^{- 27}

kg, atomic mass of He = 4, Z(Au) = 79,

1/ (4 π ε_{0}) = 9 \times 1 0^{9}

N m

^{2}

^{2}

Solution:

At the closest approach the $α$ -particle stops instantaneously - all KE has been converted to electrostatic PE:

$\frac{1}{2} m v^{2} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( 79 e )}{r}$

Mass $m = 4 \times 1.66 \times 1 0^{- 27} = 6.64 \times 1 0^{- 27}$ kg

$\frac{1}{2} (6.64 \times 1 0^{- 27}) v^{2} = 9 \times 1 0^{9} \times \frac{158 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1 0 ^{- 10}}$

Solving: $v \approx 6.3 \times 1 0^{6}$ m/s.

Defects of Rutherford's model

Stability problem: According to classical electrodynamics, an accelerating charged particle must radiate energy continuously. An electron in a circular orbit is constantly accelerating (changing direction), so it should spiral into the nucleus in less than $1 0^{- 8}$ seconds - yet atoms are stable.
Discrete spectrum problem: As the electron spiralled in, it would emit a continuous spectrum. But atomic spectra are found to consist of discrete lines, not a continuum.
Electron positions unspecified: The model does not tell us anything about the exact positions or energies of electrons.

Niels Bohr addressed these defects (1913) by proposing quantised electron orbits - the subject of the next concept.

Solved Example 6

Prove that the density of the nucleus is constant (independent of the element).

Solution:

Radius of nucleus: $r = 1.33 \times 1 0^{- 15} A^{1/3}$ m (where $A$ is mass number).
Mass of nucleus: $m = A \times 1.66 \times 1 0^{- 27}$ kg.

Density $ρ = \frac{mass}{volume} = \frac{A \times 1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.33 \times 1 0 ^{- 15} A ^{1/3} ) ^{3}}$

$= \frac{A \times 1.66 \times 1 0 ^{- 27} \times 3}{4 π \times ( 1.33 ) ^{3} \times 1 0 ^{- 45} \times A} = \frac{3 \times 1.66 \times 1 0 ^{- 27}}{4 π \times ( 1.33 ) ^{3} \times 1 0 ^{- 45}}$

The $A$ cancels, giving $ρ \approx 1.68 \times 1 0^{17}$ kg/m $^{3}$ - a constant independent of the element.

Common Mistakes to Avoid

Watch out

Isotopes vs isobars vs isotones - keep them straight. Isotopes: same $Z$ . Isobars: same $A$ . Isotones: same $N$ (neutrons). Isoelectronic: same electron count.
Do not confuse mass number with atomic mass. $A$ is an integer (nucleon count); atomic mass is the weighted average of isotopic masses (usually not an integer).
Cathode rays vs canal rays. Cathode ray properties are independent of gas (all electrons). Canal ray properties depend on the gas (each gas produces its own positive ion).
The neutron is attracted to nothing. It has zero net charge; it does not respond to electric fields.
Rutherford's model does not prove electron orbits are quantised. It only asserts electrons revolve; the quantisation comes from Bohr.
In an ion, electrons $\neq =$ Z. Electrons = Z - (charge). E.g. Na $^{+}$ has 10 electrons, not 11.

Frequently Asked Questions

What are the three fundamental subatomic particles that make up an atom?

An atom consists of three fundamental subatomic particles: the electron (negative charge, mass 9.11 × 10^-31 kg), the proton (positive charge, mass 1.673 × 10^-27 kg), and the neutron (no charge, mass 1.675 × 10^-27 kg). Protons and neutrons sit in the nucleus, while electrons move around it.

How did J.J. Thomson discover the electron?

In 1897, J.J. Thomson passed high-voltage electricity through a discharge tube at low pressure and observed cathode rays. He measured the charge-to-mass ratio ( $e / m = 1.76 \times 1 0^{8}$ C/g) and found it was the same for every gas, proving the rays were a universal, subatomic, negatively charged particle. He called it the electron.

What did Millikan's oil drop experiment prove?

Millikan (1909) suspended charged oil droplets in an electric field and measured how the field balanced gravity. He found that the charge on every droplet was always a whole-number multiple of $1.602 \times 1 0^{- 19}$ C. This confirmed that electric charge is quantised and gave the precise charge on a single electron.

What is the difference between isotopes, isobars and isotones?

Isotopes have the same atomic number (Z) but different mass numbers (A) - e.g. $^{1}$ H, $^{2}$ H, $^{3}$ H. Isobars have the same A but different Z - e.g. $_{18}^{40}$ Ar and $_{20}^{40}$ Ca. Isotones have the same number of neutrons but different Z - e.g. $_{6}^{14}$ C and $_{7}^{15}$ N (both have 8 neutrons).

How does Rutherford's model differ from Thomson's plum pudding model?

Thomson's model imagined the atom as a sphere of uniformly-spread positive charge with electrons embedded inside like plums in a pudding. Rutherford's $α$ -scattering experiment showed that positive charge is concentrated in a tiny, dense central nucleus (about $1 0^{- 15}$ m), with electrons orbiting far away. The atom is mostly empty space - very different from Thomson's uniform sphere.

Why did Rutherford's atomic model fail?

According to classical electrodynamics, an electron accelerating in a circular orbit should continuously radiate energy and spiral into the nucleus in about $1 0^{- 8}$ s - but atoms are stable. Rutherford's model also predicts a continuous spectrum, whereas atoms show discrete line spectra. Both defects were addressed by Bohr's quantised orbits.

What is the atomic mass unit (amu)?

1 amu is defined as one-twelfth of the mass of one atom of carbon-12. Numerically, $1$ amu $= 1.6605 \times 1 0^{- 27}$ kg. It is a convenient unit because a proton or neutron weighs almost exactly 1 amu, and the mass number A of any nuclide roughly equals its mass in amu.

How do you find the number of neutrons in an atom?

Number of neutrons $N = A - Z$ , where $A$ is the mass number and $Z$ is the atomic number. Example: for $_{11}^{23}$ Na, $N = 23 - 11 = 12$ neutrons.

Previous year questions on Introduction And Atomic Models

2 questions from past papers, each with a step-by-step solution.

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1. Fundamental Particles of the Atom

2. Discovery of the Electron (Cathode Rays)

Key properties of cathode rays

Millikan's oil-drop experiment (charge on electron)

3. Discovery of the Proton (Canal Rays)

Key properties of positive (canal) rays

4. Discovery of the Neutron

5. Important Atomic Terms

6. Thomson's Atomic Model (Plum Pudding, 1904)

7. Rutherford's Atomic Model (Nuclear Model)

Observations from α-scattering

Conclusions - Rutherford's nuclear model

Defects of Rutherford's model

Common Mistakes to Avoid

Frequently Asked Questions

What are the three fundamental subatomic particles that make up an atom?

How did J.J. Thomson discover the electron?

What did Millikan's oil drop experiment prove?

What is the difference between isotopes, isobars and isotones?

How does Rutherford's model differ from Thomson's plum pudding model?

Why did Rutherford's atomic model fail?

What is the atomic mass unit (amu)?

How do you find the number of neutrons in an atom?

Previous year questions on Introduction And Atomic Models

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Observations from $α$ -scattering