Constituents of the Atom

Understanding protons, neutrons, electrons, specific charge and nuclide notation

SPEC 3.2.1.1 • PARTICLES

The Simple Atomic Model

All matter is made up of atoms. The atom consists of a small, dense, positively charged nucleus surrounded by negatively charged electrons.

p+ n

Simple atomic model: Dense nucleus containing protons and neutrons, with electrons orbiting outside

Key Concept

The nucleus contains protons (positive charge) and neutrons (no charge). Electrons orbit the nucleus at relatively large distances.

Subatomic Particle Properties

You must know the charge and mass of each particle in both SI units and relative units.

ParticleCharge (C)Relative ChargeMass (kg)Relative Mass

Proton+1.60 × 10⁻¹⁹+11.67 × 10⁻²⁷1

Neutron001.67 × 10⁻²⁷1

Electron−1.60 × 10⁻¹⁹−19.11 × 10⁻³¹≈ 0 (1/1836)

Exam Tip

The proton and neutron have almost identical masses. The electron is approximately 1836 times lighter than a proton.

Specific Charge

Definition

Specific charge is the ratio of charge to mass of a particle.

Specific Charge = \frac{Q}{m}

SymbolMeaningUnit

Q

Charge of the particleC (coulombs)

m

Mass of the particlekg (kilograms)

Q / m

Specific chargeC kg⁻¹

Important

For ions, you must account for the total charge (number of electrons gained or lost) and the total mass (protons + neutrons).

01Worked Example

Calculate the specific charge of a proton.

Proton charge: +1.60 × 10⁻¹⁹ C | Proton mass: 1.67 × 10⁻²⁷ kg

1

Write the formula:

Specific charge = \frac{Q}{m}

2

Substitute values:

\frac{+ 1.60 \times 1 0 ^{- 19}}{1.67 \times 1 0 ^{- 27}}

3

Calculate:

= 9.58 \times 1 0^{7} C kg^{- 1}

Answer: +9.58 × 10⁷ C kg⁻¹

02Worked Example

Calculate the specific charge of an electron.

Electron charge: −1.60 × 10⁻¹⁹ C | Electron mass: 9.11 × 10⁻³¹ kg

1

Write the formula:

Specific charge = \frac{Q}{m}

2

Substitute values:

\frac{- 1.60 \times 1 0 ^{- 19}}{9.11 \times 1 0 ^{- 31}}

3

Calculate:

= - 1.76 \times 1 0^{11} C kg^{- 1}

Answer: −1.76 × 10¹¹ C kg⁻¹

Key Insight

The electron has a much higher magnitude of specific charge than the proton because it has the same magnitude of charge but much smaller mass.

Nuclide Notation

We use standard notation to describe any nucleus or nuclide:

Notation

_{Z}^{A} X

SymbolNameDefinition

XElement symbolChemical symbol (e.g., C, O, Fe)

ANucleon number (mass number)Total number of protons + neutrons

ZProton number (atomic number)Number of protons in the nucleus

Remember

Number of neutrons = A − Z (nucleon number minus proton number)

Worked Example03

Calculate the specific charge of a helium nucleus (alpha particle), $_{2}^{4} He$ .

1

Find the charge: 2 protons × (+1.60 × 10⁻¹⁹) = +3.20 × 10⁻¹⁹ C

2

Find the mass: 4 nucleons × (1.67 × 10⁻²⁷) = 6.68 × 10⁻²⁷ kg

3

Calculate specific charge:

\frac{3.20 \times 1 0 ^{- 19}}{6.68 \times 1 0 ^{- 27}} = 4.79 \times 1 0^{7} C kg^{- 1}

Answer: +4.79 × 10⁷ C kg⁻¹

04Worked Example

Calculate the specific charge of a lithium-7 ion, Li⁺.

Lithium-7:

_{3}^{7} Li

, Ion charge: +1

1

Find the charge: Li⁺ has lost 1 electron, so charge = +1 × (1.60 × 10⁻¹⁹) = +1.60 × 10⁻¹⁹ C

2

Find the mass: 7 nucleons × (1.67 × 10⁻²⁷) = 1.169 × 10⁻²⁶ kg

3

Specific charge:

\frac{1.60 \times 1 0 ^{- 19}}{1.169 \times 1 0 ^{- 26}} = 1.37 \times 1 0^{7} C kg^{- 1}

Answer: +1.37 × 10⁷ C kg⁻¹

Common Error

For ions, the charge comes from electrons lost or gained, NOT from protons. The mass is still determined by nucleons only (electron mass is negligible).

Worked Example05

For the nuclide $_{26}^{56} Fe$ , state the number of protons, neutrons and electrons in a neutral atom.

1

Protons = Z = 26

2

Neutrons = A − Z = 56 − 26 = 30

3

Electrons = Z = 26 (neutral atom has equal protons and electrons)

Answer: 26 protons, 30 neutrons, 26 electrons

Isotopes

Definition

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

Isotopes have:

Same proton number (Z) - same element, same chemical properties

Different nucleon number (A) - different mass

Different number of neutrons - different physical properties

Example: Carbon Isotopes

IsotopeProtonsNeutronsNucleons

_{6}^{12} C

6612

_{6}^{13} C

6713

_{6}^{14} C

6814

Worked Example06

Two nuclei are represented as $_{17}^{35} Cl$ and $_{17}^{37} Cl$ . Explain why these are isotopes.

1

Both have Z = 17, so both have 17 protons. This means they are both chlorine atoms.

2

Cl-35 has 35 − 17 = 18 neutrons

3

Cl-37 has 37 − 17 = 20 neutrons

4

Same protons, different neutrons = isotopes

Answer: They are isotopes because they have the same number of protons (17) but different numbers of neutrons (18 and 20).

Worked Example07

Calculate the specific charge of a doubly-ionised oxygen atom, O²⁺ (oxygen-16).

1

O²⁺ has lost 2 electrons, so charge = +2 × (1.60 × 10⁻¹⁹) = +3.20 × 10⁻¹⁹ C

2

Oxygen-16 has 16 nucleons: mass = 16 × (1.67 × 10⁻²⁷) = 2.672 × 10⁻²⁶ kg

3

Specific charge:

\frac{3.20 \times 1 0 ^{- 19}}{2.672 \times 1 0 ^{- 26}} = 1.20 \times 1 0^{7} C kg^{- 1}

Answer: +1.20 × 10⁷ C kg⁻¹

Comparing Specific Charges

Understanding the relative specific charges is important for exam questions.

Particle/IonSpecific Charge (C kg⁻¹)Notes

Electron−1.76 × 10¹¹Highest magnitude (smallest mass)

Proton+9.58 × 10⁷~1840× smaller than electron

Alpha particle (He²⁺)+4.79 × 10⁷Half the proton value

Why Does This Matter?

Particles with higher specific charge are deflected more by electric and magnetic fields. This is how we separate isotopes in mass spectrometers.

Practice

Question 1: Calculate the specific charge of a carbon-12 nucleus,

_{6}^{12} C

.

1

Charge: 6 protons × (1.60 × 10⁻¹⁹) = 9.60 × 10⁻¹⁹ C

2

Mass: 12 nucleons × (1.67 × 10⁻²⁷) = 2.004 × 10⁻²⁶ kg

3

Specific charge: (9.60 × 10⁻¹⁹) ÷ (2.004 × 10⁻²⁶) = 4.79 × 10⁷ C kg⁻¹

Answer: +4.79 × 10⁷ C kg⁻¹

Practice

Question 2: A nucleus has 20 protons and 20 neutrons. Write its nuclide notation and identify the element.

1

Proton number Z = 20, so element is calcium (Ca)

2

Nucleon number A = 20 + 20 = 40

3

Nuclide notation:

_{20}^{40} Ca

Answer: $_{20}^{40} Ca$ (Calcium-40)

Practice

Question 3: An Fe³⁺ ion is formed from iron-56. Calculate its specific charge.

1

Fe³⁺ has lost 3 electrons: charge = +3 × (1.60 × 10⁻¹⁹) = +4.80 × 10⁻¹⁹ C

2

Mass: 56 nucleons × (1.67 × 10⁻²⁷) = 9.352 × 10⁻²⁶ kg

3

Specific charge: (4.80 × 10⁻¹⁹) ÷ (9.352 × 10⁻²⁶) = 5.13 × 10⁶ C kg⁻¹

Answer: +5.13 × 10⁶ C kg⁻¹

Summary

Key Concepts

ConceptKey Points

Atomic StructureNucleus (protons + neutrons) with orbiting electrons

Specific ChargeQ/m in C kg⁻¹; electron has highest magnitude

Nuclide Notation

_{Z}^{A} X

where A = nucleons, Z = protons

IsotopesSame Z, different A (same protons, different neutrons)

Key Exam Skills

Calculate specific charge for any particle, nucleus or ion

Convert between nuclide notation and particle numbers

Identify isotopes from nuclide notation

Use both SI and relative units correctly