Apply The Nuclear Atom — Apply Structure 1.2—The nuclear atom in problem-solving and explanations - IB Chemistry

SECTION 1

Historical Context & Motivation

For centuries, people imagined the atom as the smallest possible unit of matter — something solid and indivisible, just as the Greek philosopher Democritus proposed around 400 BCE. This picture began to crumble at the turn of the twentieth century when physicists discovered that atoms actually contain even smaller particles arranged in a specific structure. Understanding that structure is the key to explaining why elements behave the way they do, how ions form, and why isotopes of the same element can have different masses.

1897 • Discovery of the Electron

J.J. Thomson used cathode ray tubes to show that atoms contain negatively charged particles called electrons. This proved that atoms are not indivisible after all.

1911 • Rutherford's Gold Foil Experiment

Ernest Rutherford fired alpha particles at gold foil and found that most passed straight through, but a few bounced back sharply. He concluded that the atom has a tiny, dense, positively charged nucleus at its centre.

1913 • Bohr's Atomic Model

Niels Bohr proposed that electrons orbit the nucleus in fixed energy levels, explaining the line spectra of hydrogen and laying the groundwork for modern atomic notation.

1932 • Discovery of the Neutron

James Chadwick identified a neutral particle — the neutron — inside the nucleus. This completed the picture of the three subatomic particles and explained why atoms of the same element could have different masses.

With protons, neutrons, and electrons identified, scientists needed a compact way to communicate atomic composition. The questions this lesson addresses are practical: given an element's symbol, how do you determine the number of each subatomic particle? How do you write isotope notation? And how do you use these numbers to solve real IB Chemistry problems?

SECTION 2

Core Principles & Definitions

The nuclear atom model rests on several foundational ideas that you will apply over and over in IB Chemistry. Each of these principles connects directly to problem-solving, so understanding them clearly is essential before tackling calculations.

Atomic Number (Z)

The number of protons in the nucleus. Z defines the element — change Z and you change the element entirely. In a neutral atom, Z also equals the number of electrons.

Mass Number (A)

The total count of protons plus neutrons (collectively called nucleons) in the nucleus. A is always a whole number and is written as a superscript in isotope notation.

Isotopes

Atoms of the same element (same Z) that differ in their number of neutrons (different A). Isotopes have identical chemical properties but different physical properties such as mass and nuclear stability.

Ions

When a neutral atom gains or loses electrons, it becomes an ion. The number of protons stays the same, but the charge changes. Cations are positive (lost electrons); anions are negative (gained electrons).

Relative Atomic Mass (Aᵣ)

The weighted average of the masses of all naturally occurring isotopes of an element, measured relative to ¹/₁₂ the mass of a carbon-12 atom. This is the non-integer number you see on the periodic table.

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KEY TAKEAWAY

Think of an atom like a house address. The atomic number (Z) is the street number — it uniquely identifies which element you are looking at. The mass number (A) is like the total number of rooms inside the house — same address, but houses on the same street can have different numbers of rooms. Those are your isotopes.

SECTION 3

Visualizing the Nuclear Atom

The diagram below shows the structure of a lithium-7 atom in isotope notation. Pay attention to how the atomic number and mass number are positioned relative to the element symbol, and notice where each subatomic particle is located within the atom.

The isotope notation for lithium-7 shows the mass number (A = 7) as a superscript and the atomic number (Z = 3) as a subscript to the left of the element symbol. The nucleus contains 3 protons (pink) and 4 neutrons (violet), while 3 electrons (cyan) occupy energy levels around the nucleus.

Notice in the diagram that the number of neutrons is not written directly in the isotope notation. You must calculate it by subtracting: neutrons = A − Z. For lithium-7, that gives 7 − 3 = 4 neutrons. This simple subtraction is one of the most frequently tested skills in IB Chemistry. Also notice that the atom is electrically neutral because the number of protons (3+) equals the number of electrons (3−), so the overall charge is zero.

SECTION 4

Mathematical Framework

Although nuclear atom problems do not require advanced mathematics, there are a few essential relationships that you should be able to apply quickly and confidently. These equations connect atomic number, mass number, and charge to the counts of subatomic particles.

NUMBER OF NEUTRONS

N = A − Z

Where N = number of neutrons, A = mass number (protons + neutrons), and Z = atomic number (protons).

ELECTRONS IN AN ION

electrons = Z − charge

For a neutral atom the charge is 0, so electrons = Z. For a cation like Fe³⁺, electrons = 26 − 3 = 23. For an anion like Cl⁻, electrons = 17 − (−1) = 18.

RELATIVE ATOMIC MASS (WEIGHTED AVERAGE)

Aᵣ = Σ (fractional abundance × isotope mass)

Sum each isotope's mass multiplied by its fraction (or percentage ÷ 100) of natural occurrence. For example, if chlorine is 75.77 % ³⁵Cl and 24.23 % ³⁷Cl, then Aᵣ = (0.7577 × 35) + (0.2423 × 37) = 35.48.

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IB Exam Tip

On IB papers, the periodic table gives relative atomic mass (Aᵣ), not mass number (A). Aᵣ is a decimal because it is a weighted average. If a question asks for the mass number of a specific isotope, it will always be a whole number.

SECTION 5

Isotopes, Ions, and Particle Counts

A common IB Chemistry task is to compare different species — atoms, isotopes, and ions — in terms of their subatomic particle counts. The table below shows several examples, illustrating how changing the mass number or charge affects the particle breakdown while the number of protons remains the defining feature of the element.

Species	Z (protons)	A (mass number)	Neutrons (A − Z)	Electrons	Charge
¹²C	6	12	6	6	0
¹⁴C	6	14	8	6	0
²³Na	11	23	12	11	0
²³Na⁺	11	23	12	10	+1
³⁵Cl⁻	17	35	18	18	−1
⁵⁶Fe³⁺	26	56	30	23	+3

Subatomic particle counts for selected atoms and ions

The three isotopes of hydrogen — protium, deuterium, and tritium — all share Z = 1 (one proton), but their mass numbers differ because they contain 0, 1, and 2 neutrons respectively. This is the simplest illustration of how isotopes work.

The hydrogen isotopes provide the clearest demonstration of the isotope concept because the nucleus is small enough to visualize easily. Notice that the chemical identity of each atom remains hydrogen because Z never changes. The only difference is in the number of neutrons, which affects the atom's mass but not its chemistry. In nature, protium makes up 99.98 % of all hydrogen atoms, deuterium about 0.02 %, and tritium is radioactive and extremely rare.

SECTION 6

Worked Example

Let's work through a multi-part problem that combines isotope notation, particle counting, and relative atomic mass — exactly the kind of question that appears on IB Chemistry Paper 1 and Paper 2.

Determining Subatomic Particles and Relative Atomic Mass of Copper

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Step 1 — Read the Problem

Copper has two naturally occurring isotopes: ⁶³Cu (69.17 %) and ⁶⁵Cu (30.83 %). Copper's atomic number is 29. (a) Determine the number of protons, neutrons, and electrons in a neutral atom of ⁶³Cu. (b) Determine the number of electrons in a Cu²⁺ ion. (c) Calculate the relative atomic mass of copper.
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Step 2 — Part (a): Particle Count for ⁶³Cu

Protons = Z = 29. Neutrons = A − Z = 63 − 29 = 34. For a neutral atom, electrons = protons = 29.

29 protons, 34 neutrons, 29 electrons
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Step 3 — Part (b): Electrons in Cu²⁺

A Cu²⁺ ion has lost 2 electrons compared to the neutral atom. Electrons = Z − charge = 29 − 2 = 27.

27 electrons
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Step 4 — Part (c): Relative Atomic Mass

Aᵣ = (fractional abundance of ⁶³Cu × 63) + (fractional abundance of ⁶⁵Cu × 65). Convert percentages to fractions: 69.17 % = 0.6917 and 30.83 % = 0.3083. Aᵣ = (0.6917 × 63) + (0.3083 × 65) = 43.577 + 20.040 = 63.62.

Aᵣ = 63.62
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Step 5 — Verify

The relative atomic mass of 63.62 is between 63 and 65 but closer to 63, which makes sense because the lighter isotope (⁶³Cu) is more abundant (69.17 %). This value matches the value on the IB periodic table (63.55, with the small difference due to rounding of abundances in this problem).

SECTION 7

Strengths and Limitations of the Nuclear Atom Model

The nuclear atom model — with its distinct protons, neutrons, and electrons — is powerful for many IB Chemistry tasks, but it does have boundaries. Knowing where the model works well and where it falls short helps you choose the right approach for different types of questions.

Strengths	Limitations
Accurately predicts the number of subatomic particles in any atom or ion	Does not explain electron arrangement in energy levels (shells/subshells)
Explains the existence of isotopes through different neutron counts	Cannot explain why certain isotopes are radioactive while others are stable
Enables calculation of relative atomic mass from isotopic abundances	Does not account for mass defect or binding energy (nuclear physics)
Explains ion formation by electron gain/loss	Cannot predict which ions an element will form or explain variable oxidation states

Strengths vs. limitations of the nuclear atom model at the IB Chemistry level

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KEY TAKEAWAY

Think of the nuclear atom model as a blueprint of a building. It tells you how many rooms (protons, neutrons, electrons) the building has and where the main structure (nucleus vs. outer space) is, but it does not tell you the colour of the walls or how the furniture is arranged. For questions about electron configurations, bonding, and reactivity, you will need more detailed models introduced later in the IB course.

SECTION 8

Connection to Advanced Theory

Structure 1.2 gives you the foundation: what atoms are made of and how to count their parts. Later topics in IB Chemistry build directly on this foundation, requiring you to go beyond simple particle counts. The table below previews how the nuclear atom connects to more advanced models you will encounter.

Structure 1.2 (This Lesson)	Where It Leads
Counting electrons in atoms and ions	Electron configurations (Structure 1.3) — arranging electrons in energy levels and subshells
Understanding isotopes differ in neutron count	Mass spectrometry (Structure 1.2 data) — measuring isotopic masses and abundances experimentally
Calculating relative atomic mass from abundances	Mole calculations (Structure 1.4) — converting between mass, moles, and number of particles
Ion formation by gaining or losing electrons	Ionic bonding (Structure 2.1) — electrostatic attraction between cations and anions in lattices

How Structure 1.2 concepts connect to later IB Chemistry topics

Mastering the nuclear atom model now pays dividends throughout the course. Every time you write an electron configuration, balance a nuclear equation, or calculate molar mass, you are using the same Z and A values introduced in this lesson. Build a strong habit of identifying protons, neutrons, and electrons before attempting any calculation, and you will find subsequent topics much more manageable.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two atoms have the same atomic number but different mass numbers. Explain how this is possible and state the term used to describe these atoms.

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These atoms are called isotopes. They have the same number of protons (same Z, so they are the same element), but they differ in the number of neutrons, which changes their mass number (A = protons + neutrons). Because chemical behaviour depends on electron number (which equals proton number in neutral atoms), isotopes of the same element undergo identical chemical reactions. However, their different masses give them different physical properties such as rate of diffusion and density.

Question 2: BASIC CALCULATION

Determine the number of protons, neutrons, and electrons in a neutral atom of ⁸⁰Se (selenium, Z = 34).

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Protons = Z = 34. Neutrons = A − Z = 80 − 34 = 46. Since the atom is neutral, electrons = protons = 34. The selenium-80 atom therefore contains 34 protons, 46 neutrons, and 34 electrons. The protons and neutrons are located in the nucleus, while the electrons occupy energy levels surrounding the nucleus.

Question 3: INTERMEDIATE

An ion of an element has 18 electrons, 20 protons, and 20 neutrons. Write the full isotope notation for this ion, including its charge.

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The element with Z = 20 is calcium (Ca). The mass number A = protons + neutrons = 20 + 20 = 40. The charge equals protons − electrons = 20 − 18 = +2. Therefore the full isotope notation is ⁴⁰₂₀Ca²⁺. This is a calcium-40 cation with a 2+ charge, meaning the neutral calcium atom has lost two electrons to form this ion. This is consistent with calcium's position in Group 2 of the periodic table, where elements typically form 2+ ions.

Question 4: APPLIED

Silicon has three naturally occurring isotopes: ²⁸Si (92.23 %, mass 27.977), ²⁹Si (4.67 %, mass 28.976), and ³⁰Si (3.10 %, mass 29.974). Calculate the relative atomic mass of silicon and compare your answer to the value on the periodic table (28.09).

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Aᵣ = (0.9223 × 27.977) + (0.0467 × 28.976) + (0.0310 × 29.974). Calculating each term: 0.9223 × 27.977 = 25.803, 0.0467 × 28.976 = 1.354, 0.0310 × 29.974 = 0.929. Summing: Aᵣ = 25.803 + 1.354 + 0.929 = 28.086 ≈ 28.09. This is very close to the periodic table value. The slight difference arises because the isotopic masses and abundances used here are rounded values. The result is much closer to 28 than to 30 because the lightest isotope (²⁸Si) is overwhelmingly the most abundant at 92.23 %.

Question 5: CRITICAL THINKING

Element X has two isotopes. The relative atomic mass of X is 10.81. One isotope has a mass number of 11 and an abundance of 80.1 %. The other isotope has an abundance of 19.9 %. Determine the mass number of the second isotope, and hence identify element X.

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Let the mass number of the second isotope be m. Using the weighted average equation: 10.81 = (0.801 × 11) + (0.199 × m). First, 0.801 × 11 = 8.811. So 10.81 = 8.811 + 0.199m. Rearranging: 0.199m = 10.81 − 8.811 = 1.999. Therefore m = 1.999 ÷ 0.199 = 10.05 ≈ 10. The second isotope has a mass number of 10. Since the relative atomic mass of 10.81 matches boron on the periodic table (Z = 5), element X is boron. Its two stable isotopes are ¹⁰B and ¹¹B. This problem demonstrates how you can work the relative atomic mass equation in reverse to find an unknown isotope's mass number.

SUMMARY

Lesson Summary

Summary

The nuclear atom model describes the atom as a tiny, dense nucleus containing positively charged protons and neutral neutrons, surrounded by negatively charged electrons in energy levels. The atomic number (Z) defines the element and equals the proton count, while the mass number (A) equals protons plus neutrons. Isotopes are atoms with the same Z but different A, and ions form when atoms gain or lose electrons, changing the electron count but not Z or A.

To solve problems, use the key relationships: neutrons = A − Z and electrons = Z − charge. The relative atomic mass (Aᵣ) is the weighted average of isotopic masses. Being fluent in these calculations provides the foundation for electron configurations, mole calculations, ionic bonding, and every subsequent topic in IB Chemistry.