Structure of the Atom
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| 日付: | 2026年 11月 9日(月曜日) 16:40 |
1. Historical Development of Atomic Models
Learning Outcomes
- I can describe how atomic models changed as new evidence became available.
- I can explain the contributions of Dalton, Thomson, and Rutherford.
- I can describe how experiments challenged earlier atomic theories.
- I can explain why scientific models evolve over time.
- I can evaluate the role of evidence in developing atomic theory.
What is a scientific model?
A scientific model is a simplified representation used to describe and explain something that cannot always be observed directly.
Atoms are far too small to see using ordinary microscopes. Scientists therefore developed models based on:
- Experimental observations.
- Measurements.
- Patterns in chemical reactions.
- Predictions made by existing theories.
- Results that earlier models could not explain.
An atomic model is not intended to be a perfect miniature picture of an atom. It is a tool for explaining evidence and making testable predictions.
As new evidence became available, scientists revised or replaced earlier atomic models.
The development of atomic theory
The sequence shows an important feature of science: a model may be useful for a time but still require revision when new evidence becomes available.
Early ideas about atoms
The idea that matter might be made from tiny particles is ancient.
Around the fifth century BCE, the Greek thinker Democritus suggested that matter consisted of indivisible particles called atomos, meaning “uncuttable.”
This was a philosophical proposal rather than an experimental scientific theory. Democritus did not have quantitative chemical evidence or equipment capable of testing the idea.
Modern atomic theory began much later, when scientists used measurements from chemical reactions to develop evidence-based explanations.
John Dalton and the solid-sphere model
In the early 1800s, the English scientist John Dalton developed the first modern scientific atomic theory.
Dalton studied gases and patterns in chemical reactions. His model helped explain why elements combine in consistent proportions.
Dalton represented atoms as small, solid spheres. Different elements had different kinds of atoms.
Dalton’s atomic theory
Dalton proposed that:
- All matter is made of atoms.
- Atoms of the same element are identical.
- Atoms of different elements differ in mass and properties.
- Atoms combine in simple whole-number ratios to form compounds.
- Chemical reactions rearrange atoms.
- Atoms are not created or destroyed during ordinary chemical reactions.
For example, water always contains hydrogen and oxygen in a fixed composition. Dalton explained this by proposing that atoms combine in fixed ratios.
Evidence supporting Dalton’s model
Dalton’s theory brought together several chemical laws.
Conservation of mass
In a closed system, the total mass remains constant during a chemical reaction.
Dalton explained this by proposing that reactions rearrange atoms rather than creating or destroying them.
Definite proportions
A pure compound contains the same elements in the same mass proportions.
For example, pure water has a consistent ratio of hydrogen to oxygen. Dalton explained this using a fixed combination of atoms.
Multiple proportions
When the same elements form more than one compound, their masses combine in simple whole-number ratios.
Carbon and oxygen form both carbon monoxide and carbon dioxide. The different compositions can be explained by different numbers of oxygen atoms combining with carbon atoms.
These regular numerical patterns supported the idea that matter consists of discrete particles.
Strengths of Dalton’s model
Dalton’s model:
- Connected atomic ideas to experimental chemistry.
- Explained fixed composition in compounds.
- Explained conservation of atoms during reactions.
- Helped scientists interpret chemical formulae.
- Made chemical reactions easier to represent.
It was a major improvement over unsupported philosophical ideas.
Limitations of Dalton’s model
Later evidence showed that several parts of Dalton’s theory were incomplete.
Atoms are not indivisible. They contain smaller particles, including:
- Electrons.
- Protons.
- Neutrons.
Atoms of the same element are not always identical in mass. Isotopes of an element have the same number of protons but different numbers of neutrons.
Atoms can also be changed in nuclear reactions, although they are conserved in ordinary chemical reactions.
Dalton’s model was not useless because it was incomplete. The idea of atoms as conserved units remains valuable for explaining many chemical reactions.
J. J. Thomson and the electron
During the nineteenth century, scientists investigated electrical discharges through gases in sealed glass tubes.
In 1897, the British physicist J. J. Thomson used cathode-ray experiments to provide evidence for a negatively charged particle smaller than an atom. This particle became known as the electron.
The discovery of the electron directly challenged Dalton’s claim that atoms were indivisible.
The cathode-ray experiment
A cathode-ray tube contained:
- A sealed glass tube.
- Gas at very low pressure.
- Two metal electrodes.
- A high potential difference.
When the voltage was applied, a beam travelled through the tube.
Thomson investigated how this beam responded to electric and magnetic fields.
He observed that:
- The beam was deflected by electric and magnetic fields.
- It behaved as though it contained negatively charged particles.
- The behaviour was similar even when different gases and electrode materials were used.
- The particles had a very large charge-to-mass ratio, indicating that they were much lighter than atoms.
Thomson’s reasoning
Thomson reasoned that:
- The beam contained particles because it could be deflected.
- The particles were negatively charged because of the direction of deflection.
- The particles were present in different kinds of matter.
- Atoms must contain smaller negatively charged components.
This evidence required a new atomic model.
Thomson’s atomic model
Atoms are electrically neutral overall. If they contain negative electrons, they must also contain positive charge.
Thomson proposed that an atom was:
- A sphere of positive charge.
- Embedded with negatively charged electrons.
- Electrically neutral because the charges balanced.
This became known as the plum-pudding model.
The electrons were compared with pieces of fruit distributed through positively charged material.
How Thomson challenged Dalton
Dalton’s model described the atom as a solid, indivisible particle.
Thomson’s evidence showed that:
- Atoms contain smaller particles.
- Atoms have internal structure.
- Negative charge exists within atoms.
- An atom cannot be a completely solid, indivisible sphere.
Thomson retained Dalton’s idea that atoms were units of matter, but changed the proposed internal structure.
Strengths and limitations of Thomson’s model
Thomson’s model successfully included:
- Electrons.
- Positive charge.
- Overall electrical neutrality.
However, it did not include:
- A central nucleus.
- Concentrated positive charge.
- The large amount of empty space within an atom.
Its most serious problem appeared when scientists studied how alpha particles passed through thin metal foil.
The gold-foil experiment
In 1909, Hans Geiger and Ernest Marsden carried out an experiment under the direction of Ernest Rutherford.
They aimed positively charged alpha particles at an extremely thin sheet of gold foil.
A fluorescent screen surrounding the foil detected where the alpha particles travelled.
Gold was useful because it could be hammered into a foil only a few atoms thick.
Predictions from Thomson’s model
According to Thomson’s model:
- Positive charge was spread throughout the atom.
- The atom’s mass was also broadly distributed.
- Alpha particles should pass through with little deflection.
- Large deflections should be extremely unlikely.
Alpha particles are positively charged and relatively massive. A spread-out positive charge was not expected to exert a concentrated force strong enough to turn them sharply.
The observations
The experiment produced three important observations:
- Most alpha particles passed straight through the foil.
- Some were deflected through small angles.
- A very small number were deflected through large angles or returned towards the source.
Most of the results were consistent with particles passing through little material. The rare large deflections were incompatible with Thomson’s diffuse positive charge.
Rutherford’s interpretation
Rutherford used the scattering evidence to infer that:
- Most of the atom is empty space.
- Nearly all its mass is concentrated in a tiny central region.
- Positive charge is concentrated in this central region.
- Electrons occupy the space surrounding the centre.
The small central region became known as the nucleus.
Large deflections occurred when a positively charged alpha particle passed close to the positively charged nucleus. Electrostatic repulsion caused the particle to change direction sharply.
Linking observations to conclusions
| Experimental observation | Rutherford’s conclusion |
|---|---|
| Most alpha particles passed straight through | Most of the atom is empty space |
| Some particles were slightly deflected | Positive charge affects particles passing near the centre |
| Very few particles were strongly deflected | Positive charge and mass are concentrated in a very small region |
| A tiny number returned towards the source | The central region is dense and strongly repels nearby alpha particles |
The conclusions were not direct photographs of atoms. They were inferences that best explained the observed scattering pattern.
Rutherford’s nuclear model
In 1911, Rutherford proposed the nuclear model of the atom.
In this model:
- A tiny, dense nucleus occupies the centre.
- The nucleus has positive charge.
- Most atomic mass is concentrated in the nucleus.
- Electrons move through the space around the nucleus.
- Most of the atom’s volume is empty space.
The nucleus is extremely small compared with the complete atom.
If an atom were enlarged to the size of a large stadium, its nucleus would be much smaller than the stadium itself, while the electrons would occupy the surrounding region.
How the gold-foil experiment challenged Thomson
Thomson’s model placed positive charge throughout the atom. This could explain small deflections but not the rare, very large deflections.
Rutherford’s model explained both major results:
- Most alpha particles encountered empty space and continued straight through.
- A small number passed close to the concentrated positive nucleus and were strongly repelled.
The new model was accepted because it explained the evidence more successfully.
Comparing the three models
| Feature | Dalton | Thomson | Rutherford |
|---|---|---|---|
| General structure | Solid sphere | Positive sphere containing electrons | Tiny nucleus surrounded by electrons |
| Smaller particles included | None | Electrons | Electrons and central positive region |
| Positive charge | Not separately described | Spread throughout the atom | Concentrated in the nucleus |
| Empty space | Not included | Very little or none | Most of the atom |
| Main supporting evidence | Chemical combination and mass ratios | Cathode-ray deflection | Alpha-particle scattering |
| Main limitation | Atom treated as indivisible | No nucleus | Could not fully explain electron behaviour or atomic spectra |
Each model retained useful ideas while changing features that no longer matched the evidence.
Rutherford’s contribution
Rutherford’s major contribution was not simply drawing a different atom. He used experimental evidence to locate positive charge and mass within a tiny nucleus.
His work established the basic distinction between:
- The dense central nucleus.
- The much larger region occupied by electrons.
This remains part of modern atomic theory.
Rutherford’s interpretation transformed the scientific understanding of atomic structure.
Later developments
Rutherford’s model was a major advance, but it was not the final atomic model.
Later scientists added further evidence and explanations.
Niels Bohr
In 1913, Bohr proposed that electrons occupy specific energy levels. His model helped explain the line spectra produced by atoms.
The proton
Evidence developed that the nucleus contains positively charged particles called protons. Rutherford is associated with identifying the hydrogen nucleus as a fundamental positive particle.
James Chadwick
In 1932, Chadwick provided evidence for the neutron, an uncharged particle in the nucleus.
The quantum-mechanical model
Modern atomic theory describes electrons using orbitals and probability distributions rather than simple circular paths.
Modern electron-cloud images represent regions where electrons are likely to be found. They are not photographs of electrons following fixed paths.
Why scientific models change
Scientific models change because scientific knowledge is based on evidence rather than permanent authority.
A model may be revised when:
- New technology allows more precise measurements.
- A new experiment produces unexpected results.
- Existing evidence can no longer be explained.
- A new model explains a wider range of observations.
- A new model makes more accurate predictions.
- Several independent investigations support the new explanation.
Changing a model is a strength of science. It shows that scientific explanations respond to evidence.
Models are evaluated, not simply declared true
Scientists evaluate models by asking:
- Does the model explain existing observations?
- Does it make testable predictions?
- Do experiments support those predictions?
- Can the results be repeated?
- Does the model explain evidence better than alternatives?
- Are there observations that the model cannot explain?
No model needs to look exactly like the object it represents. Its value comes from how effectively it explains and predicts.
Unexpected results drive scientific progress
The most important results in the gold-foil experiment were not the most common ones.
Most alpha particles passed straight through, as expected. However, the rare large deflections revealed that Thomson’s model was incomplete.
This shows that:
- Rare results should not automatically be ignored.
- Unexpected evidence may be scientifically important.
- A theory must explain all reliable observations.
- Scientists must be willing to revise prior assumptions.
Before accepting rare results, researchers must also check for errors, contamination or equipment problems.
Evaluating the quality of evidence
Strong experimental evidence should be:
- Measurable.
- Repeatable.
- Consistent.
- Open to examination by other scientists.
- Collected using controlled methods.
- Capable of distinguishing between competing explanations.
A single surprising observation may begin an investigation, but repeated evidence builds confidence.
Thomson’s experiments were repeated using different materials. Alpha-scattering experiments also produced consistent patterns that other scientists could investigate.
Evidence and inference
Evidence and inference are related but different.
Evidence consists of observations and measurements.
Examples include:
- A beam bends towards a positively charged plate.
- Most alpha particles pass through foil.
- A few alpha particles reverse direction.
An inference is a conclusion drawn from the evidence.
Examples include:
- The beam contains negatively charged particles.
- Most of an atom is empty space.
- Positive charge is concentrated in a small nucleus.
Scientists do not observe every part of an atom directly. They construct explanations from how matter and radiation behave.
Science as a collaborative process
Scientific discoveries rarely result from one person working entirely alone.
The development of atomic theory involved:
- Scientists who proposed models.
- Researchers who designed and performed experiments.
- Technicians who built equipment.
- Other scientists who repeated and challenged results.
- Improvements in vacuum tubes, detectors and radiation sources.
The gold-foil experiment is often called Rutherford’s experiment, but Geiger and Marsden performed the scattering work under his direction. Recognizing the wider team gives a more accurate picture of scientific progress.
A worked evidence-analysis example
Consider this statement:
“Most alpha particles passed through gold foil, but a very small number were deflected backwards.”
A strong analysis would explain:
- Thomson’s model predicted only small deflections because positive charge was spread out.
- Most particles passing through suggested that atoms contain a large amount of empty space.
- Backward deflections required a strong, concentrated repulsive force.
- The positive charge and most mass must therefore occupy a tiny nucleus.
- Rutherford’s nuclear model explained the complete pattern better than Thomson’s model.
This links the prediction, observation, inference and revised model.
Real-world connection: indirect evidence
Scientists often study objects that cannot be observed directly by analyzing their effects.
Examples include:
- Medical imaging reconstructing structures from transmitted signals.
- Astronomers detecting planets through changes in a star’s light or motion.
- Seismologists studying Earth’s interior using earthquake waves.
- Particle physicists identifying particles from detector tracks.
Rutherford’s team similarly inferred atomic structure from the paths of scattered particles.
Common misconceptions
- “Scientific models are exact pictures.” They are simplified explanatory tools.
- “Dalton discovered the atom.” Atomic ideas existed earlier, but Dalton developed a modern evidence-based theory.
- “Dalton’s model became completely useless.” It still helps explain conservation and chemical combination.
- “Thomson discovered the nucleus.” Thomson identified the electron; Rutherford developed the nuclear model.
- “Rutherford personally performed every part of the gold-foil experiment.” Geiger and Marsden conducted the scattering experiment under his direction.
- “Most alpha particles bounced backwards.” Only a very small proportion underwent large deflections.
- “Rutherford saw the nucleus directly.” He inferred it from scattering evidence.
- “A revised model proves earlier scientists were careless.” Earlier models were based on the best available evidence.
- “Scientific knowledge changes randomly.” Models change in response to evidence and improved explanations.
Did you know?
Rutherford reportedly compared the backward deflection of alpha particles to firing a projectile at thin material and having it return towards the source.
The result was surprising because Thomson’s model contained no small, concentrated region capable of producing such a strong repulsion.
Key terms
- Atom: The smallest particle of an element that retains its chemical identity.
- Atomic model: A representation used to explain atomic structure and behaviour.
- Scientific evidence: Observations and measurements used to evaluate explanations.
- Inference: A conclusion drawn from evidence.
- Cathode ray: A beam of electrons travelling through a low-pressure tube.
- Electron: A negatively charged subatomic particle.
- Plum-pudding model: Thomson’s model of electrons embedded in positive matter.
- Alpha particle: A positively charged particle containing two protons and two neutrons.
- Scattering: A change in direction caused by an interaction.
- Nucleus: The tiny, dense central region of an atom.
- Nuclear model: Rutherford’s model with concentrated positive charge in a nucleus.
- Repeatability: The ability to obtain consistent results when an investigation is repeated.
- Theory: A well-supported scientific explanation based on evidence.
Key takeaways
- Dalton used chemical evidence to develop the solid-sphere model.
- Thomson’s cathode-ray experiments provided evidence for electrons.
- The electron showed that atoms were divisible and had internal structure.
- The gold-foil results could not be fully explained by Thomson’s model.
- Rutherford inferred that atoms are mostly empty space with a tiny positive nucleus.
- Observations provide evidence, while models provide explanations.
- Reliable evidence should be measurable, repeatable and open to testing.
- Scientific models evolve when new evidence supports a better explanation.
- Revising a model is a normal and necessary part of scientific progress.
2. The Nuclear Atom
Learning Outcomes
- I can identify the main components of an atom.
- I can compare the properties of protons, neutrons, and electrons.
- I can determine the composition of atoms using nuclear notation.
- I can explain the concept of isotopes.
- I can relate atomic structure to atomic number and nucleon number.
Inside the Atom
All ordinary matter is made of atoms. Atoms are extremely small, but they have an internal structure made from even smaller particles.
The three main subatomic particles are:
- Protons
- Neutrons
- Electrons
Protons and neutrons are found together in the nucleus at the centre of the atom. Electrons occupy the region surrounding the nucleus.
Although diagrams often make the nucleus look quite large, the nucleus is actually tiny compared with the overall size of an atom. Most of an atom's volume is the region occupied by its electrons.
The Nucleus
The nucleus is the small, dense central region of an atom.
It contains:
- Protons
- Neutrons
Together, protons and neutrons are sometimes called nucleons because they are particles found in the nucleus.
Almost all of the atom's mass is concentrated in the nucleus.
The nucleus also has an overall positive charge because protons are positively charged while neutrons have no electric charge.
Comparing the Three Main Subatomic Particles
The three main subatomic particles have different charges, masses, and locations.
| Particle | Relative charge | Relative mass | Location |
|---|---|---|---|
| Proton | +1 | 1 | Nucleus |
| Neutron | 0 | 1 | Nucleus |
| Electron | −1 | About 1/1836 | Outside nucleus |
A proton and neutron have approximately the same mass.
An electron has much less mass. For many basic atomic calculations, the mass of an electron is treated as negligible compared with the masses of protons and neutrons.
Protons
A proton has:
- A positive charge
- A relative charge of +1
- A relative mass of approximately 1
- A location inside the nucleus
The number of protons is particularly important because it determines the identity of the element.
For example:
- An atom with 1 proton is hydrogen.
- An atom with 6 protons is carbon.
- An atom with 8 protons is oxygen.
- An atom with 11 protons is sodium.
Changing the number of protons changes the element itself.
Neutrons
A neutron has:
- No electric charge
- A relative charge of 0
- A relative mass of approximately 1
- A location inside the nucleus
Neutrons contribute to the mass of the atom but do not determine which element it is.
Atoms of the same element can contain different numbers of neutrons. These different forms are called isotopes.
Electrons
An electron has:
- A negative charge
- A relative charge of −1
- A very small mass
- A location outside the nucleus
Electrons are involved in many important processes, particularly:
- Chemical bonding
- Ion formation
- Electricity
- Chemical reactions
In a neutral atom, the number of electrons equals the number of protons.
For example, a neutral atom containing 8 protons also contains 8 electrons.
The positive and negative charges balance, giving an overall charge of zero.
Atomic Number
Every element has an atomic number.
The atomic number tells us the number of protons in the nucleus.
Atomic number = number of protons
The symbol Z is often used for atomic number.
For example, carbon has an atomic number of 6.
Therefore, every carbon atom has:
6 protons
A neutral carbon atom also has:
6 electrons
The atomic number is what defines an element.
Nucleon Number
The nucleon number, also commonly called the mass number, is the total number of protons and neutrons in the nucleus.
Nucleon number = number of protons + number of neutrons
The symbol A is commonly used for nucleon number.
For example, suppose an atom contains:
- 6 protons
- 8 neutrons
Its nucleon number is:
6 + 8 = 14
Therefore:
Nucleon number = 14
Notice that electrons are not included in the nucleon number.
Finding the Number of Neutrons
If you know the atomic number and nucleon number, you can determine the number of neutrons.
Use:
Number of neutrons = nucleon number − atomic number
For example, an atom has:
- Atomic number = 11
- Nucleon number = 23
Therefore:
Number of neutrons = 23 − 11 = 12
The atom contains:
- 11 protons
- 12 neutrons
- 11 electrons if it is neutral
Because it has 11 protons, the element is sodium.
Nuclear Notation
Scientists use nuclear notation to show important information about an atom.
A common format is:
ᴬ_Z X
where:
- X = chemical symbol
- A = nucleon number
- Z = atomic number
For example, carbon-12 can be represented as:
¹²₆C
This tells us:
- Nucleon number = 12
- Atomic number = 6
- Protons = 6
- Neutrons = 12 − 6 = 6
If the atom is neutral:
- Electrons = 6
This notation allows us to determine the composition of an atom very quickly.
Worked Example: Sodium-23
Consider:
²³₁₁Na
The lower number is the atomic number.
Therefore:
Protons = 11
The upper number is the nucleon number.
Therefore:
Protons + neutrons = 23
Calculate the neutrons:
Neutrons = 23 − 11 = 12
Because no ionic charge is shown, we treat the atom as neutral:
Electrons = 11
So sodium-23 contains:
- 11 protons
- 12 neutrons
- 11 electrons
Worked Example: Chlorine-37
Consider:
³⁷₁₇Cl
Atomic number = 17
Therefore:
Protons = 17
Nucleon number = 37
Therefore:
Neutrons = 37 − 17 = 20
For a neutral atom:
Electrons = 17
So chlorine-37 contains:
- 17 protons
- 20 neutrons
- 17 electrons
What Are Isotopes?
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons.
Because they have the same number of protons, they have the same atomic number and are the same element.
Because they have different numbers of neutrons, they have different nucleon numbers.
For example, carbon has several isotopes.
Carbon-12:
- 6 protons
- 6 neutrons
- Nucleon number = 12
Carbon-13:
- 6 protons
- 7 neutrons
- Nucleon number = 13
Carbon-14:
- 6 protons
- 8 neutrons
- Nucleon number = 14
All three are carbon because all three contain 6 protons.
Comparing Isotopes
Consider these two atoms:
³⁵₁₇Cl
and
³⁷₁₇Cl
Both have atomic number 17.
Therefore, both contain:
17 protons
This means both are chlorine.
For chlorine-35:
Neutrons = 35 − 17 = 18
For chlorine-37:
Neutrons = 37 − 17 = 20
The two atoms are isotopes because they have:
- The same number of protons
- Different numbers of neutrons
What Changes an Element's Identity?
It is important to distinguish between changes in protons, neutrons, and electrons.
Changing the Number of Protons
Changing the number of protons changes the element.
For example:
6 protons → carbon
7 protons → nitrogen
Changing the Number of Neutrons
Changing the number of neutrons changes the isotope, but not the element.
For example:
Carbon-12 → 6 protons and 6 neutrons
Carbon-14 → 6 protons and 8 neutrons
Both are still carbon.
Changing the Number of Electrons
Changing the number of electrons changes the electrical charge of the atom.
The atom becomes an ion, but it remains the same element.
Neutral Atoms and Ions
In a neutral atom:
Number of protons = number of electrons
If an atom loses electrons, it has more protons than electrons and becomes positively charged.
If an atom gains electrons, it has more electrons than protons and becomes negatively charged.
For example, sodium contains 11 protons.
Neutral sodium:
- 11 protons
- 11 electrons
- Net charge = 0
A sodium ion, Na⁺:
- 11 protons
- 10 electrons
- Net charge = +1
The number of protons has not changed, so it is still sodium.
Atomic Number and the Periodic Table
Elements on the periodic table are arranged according to atomic number.
As you move from one element to the next, the atomic number increases.
For example:
- Hydrogen: atomic number 1
- Helium: atomic number 2
- Lithium: atomic number 3
- Beryllium: atomic number 4
- Boron: atomic number 5
- Carbon: atomic number 6
This means that the number of protons increases by one as you move through these elements.
The periodic table is therefore fundamentally organized around the structure of the atomic nucleus.
Nucleon Number vs Relative Atomic Mass
The nucleon number of a particular atom is always a whole number because it counts individual protons and neutrons.
For example:
Carbon-12 has a nucleon number of 12.
However, the periodic table often shows carbon's relative atomic mass as approximately 12.01.
These are not the same thing.
The decimal value on the periodic table reflects the relative abundances and masses of the naturally occurring isotopes of an element.
Therefore:
Nucleon number → describes one particular isotope
Relative atomic mass → represents an average based on naturally occurring isotopes
This is why many relative atomic masses on the periodic table are not whole numbers.
Why Isotopes Matter
Different isotopes of an element have very similar chemical behaviour because they have the same number of protons and, for neutral atoms, the same number and arrangement of electrons.
However, their nuclear properties can be different.
Some isotopes are stable, while others are radioactive.
Radioactive isotopes have many applications, including:
- Medical diagnosis
- Medical treatment
- Scientific research
- Archaeological dating
- Industrial monitoring
Carbon-14, for example, is radioactive and can be used in dating once-living materials.
A Quick Method for Atomic Composition Problems
When given nuclear notation, work through the information systematically.
Find the Protons
Protons = atomic number
Find the Neutrons
Neutrons = nucleon number − atomic number
Find the Electrons
For a neutral atom:
Electrons = protons
If the particle is an ion, adjust the electron number according to its charge.
Identify the Element
The number of protons determines the element.
Worked Example
An atom has:
- Atomic number = 15
- Nucleon number = 31
Determine its composition.
First:
Protons = 15
Next:
Neutrons = 31 − 15 = 16
If the atom is neutral:
Electrons = 15
Atomic number 15 corresponds to phosphorus.
Therefore, the atom contains:
- 15 protons
- 16 neutrons
- 15 electrons
It is phosphorus-31.
Common Mistakes
Confusing Atomic Number and Nucleon Number
Remember:
Atomic number = protons
Nucleon number = protons + neutrons
Including Electrons in the Nucleon Number
Electrons are not nucleons.
Only protons and neutrons are counted.
Assuming Isotopes Have Different Numbers of Protons
If the proton number changes, the element changes.
Isotopes have the same number of protons but different numbers of neutrons.
Assuming All Atoms Have Equal Numbers of Protons and Electrons
This is true only for neutral atoms.
Ions have unequal numbers of protons and electrons.
Using Relative Atomic Mass as the Nucleon Number
The decimal number on the periodic table is usually a relative atomic mass, not the nucleon number of one specific atom.
Thinking Electrons Contribute Most of the Atom's Mass
Almost all atomic mass is concentrated in the nucleus because electrons have very little mass compared with protons and neutrons.
Check Your Understanding
1. Name the three main subatomic particles.
2. Which particles are found in the nucleus?
3. Compare the charges of protons, neutrons, and electrons.
4. Which particle determines the identity of an element?
5. What does atomic number tell us?
6. What does nucleon number tell us?
7. An atom has atomic number 12 and nucleon number 24. How many protons, neutrons, and electrons does a neutral atom contain?
8. Determine the numbers of protons, neutrons, and electrons in ²⁷₁₃Al.
9. Determine the numbers of protons and neutrons in ⁴⁰₂₀Ca.
10. Define an isotope.
11. Carbon-12 and carbon-14 are isotopes. What is the same about their nuclei, and what is different?
12. Why does changing the number of neutrons not change the identity of an element?
13. What happens to an atom if its number of electrons changes?
14. Explain why nucleon numbers are whole numbers while relative atomic masses are often decimals.
15. An atom contains 17 protons and 18 neutrons. What are its atomic number and nucleon number?
Key Terms
- Atom – the smallest particle of an element that retains the properties of that element.
- Nucleus – the small, dense central region of an atom containing protons and neutrons.
- Proton – a positively charged subatomic particle found in the nucleus.
- Neutron – an electrically neutral subatomic particle found in the nucleus.
- Electron – a negatively charged subatomic particle found outside the nucleus.
- Nucleon – a proton or neutron found in the nucleus.
- Atomic number – the number of protons in an atom's nucleus.
- Nucleon number – the total number of protons and neutrons in a nucleus.
- Nuclear notation – a notation showing an element's chemical symbol, atomic number, and nucleon number.
- Isotope – an atom of an element with the same number of protons but a different number of neutrons.
- Ion – an atom or group of atoms with an overall electric charge.
- Relative atomic mass – the weighted average mass of an element's naturally occurring isotopes relative to a standard.
Key Takeaways
- Atoms contain protons, neutrons, and electrons.
- Protons and neutrons are located in the nucleus.
- Electrons occupy the region surrounding the nucleus.
- Protons have charge +1, neutrons have charge 0, and electrons have charge −1.
- Protons and neutrons have much greater masses than electrons.
- The atomic number equals the number of protons.
- The number of protons determines the identity of the element.
- The nucleon number equals the number of protons plus neutrons.
- The number of neutrons can be calculated using: neutrons = nucleon number − atomic number.
- In a neutral atom, the number of electrons equals the number of protons.
- Isotopes have the same number of protons but different numbers of neutrons.
- Different isotopes therefore have the same atomic number but different nucleon numbers.
- Changing the number of electrons produces an ion; changing the number of neutrons produces a different isotope; changing the number of protons produces a different element.
3. High-Energy Scattering Experiments
Learning Outcomes
- I can describe how scattering experiments provide information about atomic structure.
- I can explain the results of Rutherford's gold foil experiment.
- I can describe how high-energy electrons are used to probe nuclei.
- I can explain how scattering angles reveal structural information.
- I can evaluate evidence obtained from scattering experiments.
How Can We Study Something Too Small to See?
Atoms and atomic nuclei are far too small to examine directly with an ordinary microscope. Instead, scientists can investigate their structure by firing particles at them and studying what happens.
This technique is called scattering.
The basic idea is simple:
- Fire particles toward a target.
- Detect where the particles go after interacting with the target.
- Measure how much their paths change.
- Use the pattern of scattering to infer the structure of the target.
This is similar to trying to determine the shape of an invisible object by throwing small balls toward it and observing how the balls bounce away.
Scattering experiments have provided some of the most important evidence for our modern understanding of atomic and nuclear structure.
Rutherford's Gold Foil Experiment
One of the most famous scattering experiments was carried out in the early twentieth century by Hans Geiger and Ernest Marsden, working under the direction of Ernest Rutherford.
The experiment tested the accepted model of the atom at the time.
According to the plum pudding model, the atom consisted of positive charge spread throughout the atom, with negatively charged electrons embedded within it.
If this model were correct, a fast-moving positively charged particle passing through an atom should experience only relatively small deflections.
The scattering experiment showed something very different.
The Experimental Setup
The experiment used alpha particles.
Alpha particles are:
- Positively charged
- Relatively massive compared with electrons
- Fast-moving when emitted by radioactive materials
A radioactive source produced a narrow beam of alpha particles.
The beam was directed toward a very thin sheet of gold foil.
A detector surrounding the foil allowed the researchers to determine where the alpha particles travelled after passing through or interacting with the gold atoms.
Gold was useful because it can be made into extremely thin sheets only a few hundred atoms thick.
What Scientists Expected
Under the plum pudding model, the positive charge of an atom was thought to be spread throughout its volume.
Because this positive charge was diffuse, scientists expected the alpha particles to pass through the foil with only small changes in direction.
The expected result was therefore:
Almost all alpha particles should travel approximately straight through the foil.
Large-angle deflections were not expected.
Particles travelling backward toward the source seemed especially unlikely.
What They Actually Observed
The experiment produced three important observations.
Most Alpha Particles Passed Straight Through
The overwhelming majority of alpha particles passed through the gold foil with little or no change in direction.
Some Alpha Particles Were Deflected
A smaller number changed direction as they passed through the foil.
Some experienced noticeable deflections.
A Very Small Number Were Deflected Through Large Angles
A tiny fraction of the alpha particles were scattered through very large angles.
Some even travelled approximately back toward the source.
This final observation was especially surprising.
What Did the Results Mean?
Scientists needed a model of the atom that could explain all three observations.
Most Particles Passed Straight Through
This suggested that most of an atom's volume contains relatively little matter capable of strongly deflecting an alpha particle.
Therefore:
Most of the atom is empty space.
Some Particles Were Deflected
Because alpha particles are positively charged, their deflection suggested that they were being repelled by another concentration of positive charge.
Therefore:
The positive charge of the atom must be concentrated rather than spread uniformly throughout the atom.
A Few Particles Were Strongly Deflected
Large-angle scattering required a very strong interaction concentrated within a very small region.
This suggested that most of the atom's mass and positive charge were concentrated in a tiny central region.
This region became known as the nucleus.
The Nuclear Model of the Atom
The evidence led to a new model of atomic structure.
According to the nuclear model:
- The atom contains a very small central nucleus.
- The nucleus contains positive charge.
- Most of the atom's mass is concentrated in the nucleus.
- The nucleus occupies only a tiny fraction of the atom's total volume.
- Most of the atom is empty space.
- Electrons occupy the region surrounding the nucleus.
This model replaced the plum pudding model.
Evidence and Conclusions
A useful way to understand Rutherford scattering is to connect each observation directly to a conclusion.
| Observation | Evidence about atomic structure |
|---|---|
| Most alpha particles passed straight through | Most of the atom is empty space |
| Some particles were deflected | Positive charge is concentrated within the atom |
| Very few particles were strongly deflected | The concentrated positive region is extremely small |
| Some particles scattered backward | The nucleus is dense and capable of producing very strong repulsive interactions |
The strength of the experiment was not simply that particles scattered. It was the pattern and frequency of different scattering angles that provided information about the atom.
Why Positive Alpha Particles Are Deflected
Alpha particles have a positive electric charge.
The atomic nucleus also has a positive charge.
Like charges repel.
An alpha particle passing far from a nucleus experiences relatively little repulsion and therefore changes direction only slightly.
An alpha particle passing closer to the nucleus experiences a much stronger repulsive force and is deflected through a larger angle.
A particle approaching the nucleus very closely can experience a very large deflection.
Scattering Angle
The scattering angle describes how much the direction of a particle changes during an interaction.
A small scattering angle means the particle's path changed only slightly.
A large scattering angle means the particle experienced a much greater change in direction.
The distribution of scattering angles provides information about the structure of the target.
In Rutherford scattering:
Small deflection → particle passed relatively far from a nucleus
Large deflection → particle passed relatively close to a nucleus
Very large or backward deflection → particle experienced a particularly strong interaction with a nucleus
The exact relationship is more complicated than this simple description, but the central principle remains: the pattern of scattering reveals information about structures that cannot be directly seen.
Why Large-Angle Scattering Was So Important
The rare large-angle deflections were particularly significant.
A model with positive charge spread throughout the atom could not easily explain how a heavy, fast-moving alpha particle could suddenly change direction by a large amount.
A small, concentrated positive nucleus could.
The rarity of these events was also important.
If the nucleus occupied a large fraction of the atom, large deflections would have occurred much more frequently.
Instead, they were rare.
This provided evidence that the nucleus occupies a very small fraction of the atom's volume.
From Alpha Particles to High-Energy Electrons
The same basic idea can be taken much further.
Modern scattering experiments can use high-energy electrons as probes.
Electrons are accelerated to very high speeds and directed toward a target.
Detectors then measure:
- How many electrons pass through
- How many are scattered
- Their scattering angles
- Their energies after scattering
Scientists compare the observed scattering pattern with predictions from different models of nuclear structure.
Why Use High-Energy Electrons?
Electrons are useful probes because they are extremely small and interact electrically with charged particles in the nucleus.
Increasing the energy of the electrons also improves their ability to investigate very small structures.
In general:
Higher-energy particles can probe smaller distance scales.
This is related to the wave behaviour of matter.
Particles such as electrons have an associated wavelength. As their momentum increases, their wavelength becomes shorter.
Shorter wavelengths allow scientists to investigate smaller structures, much as shorter-wavelength electromagnetic radiation can reveal finer details than longer-wavelength radiation.
Electron Wavelength and Resolution
The wavelength associated with a particle is described by the de Broglie relationship:
λ = h / p
where:
- λ = wavelength
- h = Planck's constant
- p = momentum
You do not always need to calculate this relationship to understand scattering experiments.
The important idea is:
Greater momentum → shorter wavelength → ability to probe smaller structures
This is one reason particle accelerators are designed to produce extremely high-energy beams.
How Electron Scattering Reveals Nuclear Structure
Imagine firing high-energy electrons toward a nucleus.
If the nucleus behaved exactly like a tiny point charge, scientists would expect a particular scattering pattern.
If the experimental results differ from that prediction, the difference provides evidence that the nucleus has internal size and structure.
By carefully analysing:
- scattering angles
- numbers of scattered particles
- particle energies
- scattering probabilities
scientists can estimate properties such as:
- nuclear size
- charge distribution
- internal nuclear structure
Scattering therefore acts as an indirect form of imaging.
A Useful Analogy: Exploring an Invisible Object
Imagine that you are standing in a dark room containing an invisible object.
You throw hundreds of small balls toward it.
Suppose:
- Most continue straight.
- Some bounce slightly sideways.
- A few bounce sharply backward.
Even though you cannot see the object, the paths of the balls give you information about:
- Its location
- Its size
- Its shape
- How solid or concentrated it is
Particle scattering uses the same basic reasoning, but at atomic and subatomic scales.
Scattering Patterns
Scientists do not usually draw conclusions from a single scattered particle.
Instead, they examine large numbers of scattering events.
The result is a scattering pattern or distribution.
For example, scientists might measure how many particles are detected at:
- 10°
- 20°
- 30°
- 60°
- 90°
- 120°
- 150°
A successful scientific model should predict a scattering distribution that agrees with the experimental observations.
Models Make Predictions
Scattering experiments are powerful because scientific models make testable predictions.
The general process is:
Model → prediction → experiment → evidence → evaluation
For example, the plum pudding model predicted that large-angle alpha-particle scattering should be extremely unlikely.
The experimental observations showed significant deviations from what that model could explain.
Scientists therefore needed to revise the model.
The nuclear model provided a much better explanation of the evidence.
This is an important example of how scientific knowledge develops.
Evaluating Evidence from Scattering Experiments
Good scientific evidence should be evaluated carefully.
Scientists consider questions such as:
- Were enough particles detected?
- Were the measurements repeatable?
- How accurately were scattering angles measured?
- Could background radiation affect the results?
- Could the particles interact with something other than the intended target?
- Does the proposed model explain the entire scattering pattern?
- Do independent experiments produce similar results?
A model becomes more convincing when it successfully explains many observations and makes predictions that are confirmed by further experiments.
Evidence Does Not Give a Direct Photograph
A scattering experiment does not normally produce a direct photograph of a nucleus.
Instead, scientists make inferences from measurements.
For example:
Observation: Very few alpha particles scatter backward.
Inference: A very small region of the atom can produce extremely strong repulsion.
Model: The atom contains a tiny, dense, positively charged nucleus.
It is important to distinguish between the observation and the interpretation of that observation.
Scattering and Scientific Models
Scientific models are not simply guesses.
A useful model must:
- Explain existing observations
- Agree with experimental evidence
- Make testable predictions
- Be modified or rejected if new evidence contradicts it
Rutherford's work is a classic example.
The gold foil results did not fit the existing atomic model, so scientists developed a model that better explained the evidence.
Later experiments refined the nuclear model even further.
From the Atom to the Nucleus
Rutherford scattering showed that atoms contain nuclei.
Later, higher-energy scattering experiments allowed scientists to investigate the nuclei themselves.
As experimental energies increased, scientists were able to probe increasingly small scales.
This eventually contributed to evidence that protons and neutrons are themselves not fundamental particles but have an internal structure involving quarks.
Why Higher Energy Reveals Smaller Structures
This is one of the most important ideas in modern experimental physics.
To investigate something extremely small, scientists need a probe capable of resolving structures at that scale.
Higher-energy particles generally have shorter associated wavelengths.
Therefore:
Low energy → longer wavelength → less detail
High energy → shorter wavelength → greater ability to investigate small structures
Modern particle accelerators use this principle to investigate matter at scales far smaller than atomic nuclei.
Comparing Rutherford and Modern Scattering Experiments
| Feature | Rutherford scattering | Modern high-energy scattering |
|---|---|---|
| Probe | Alpha particles | Often electrons or other high-energy particles |
| Target | Thin gold foil | Nuclei, protons, or other particles |
| Measurement | Scattering direction | Angle, energy, momentum and other properties |
| Main purpose | Investigate atomic structure | Investigate nuclear and subnuclear structure |
| Key idea | Scattering reveals hidden structure | Scattering reveals hidden structure |
Although the technology has changed enormously, the underlying reasoning is remarkably similar.
Common Mistakes
Saying All Alpha Particles Were Deflected
Most alpha particles passed through the gold foil with little or no deflection.
Only a small fraction were significantly deflected.
Saying Rutherford Discovered That Atoms Are Solid
The experiment provided evidence for almost the opposite conclusion.
Most of the atom is empty space.
Saying the Nucleus Is Large
The nucleus contains most of the atom's mass but occupies only a tiny fraction of its volume.
Dense does not mean large.
Confusing Electrons with Alpha Particles
Rutherford's gold foil experiment used alpha particles, not electrons.
High-energy electron scattering refers to later experiments used to probe smaller structures.
Assuming Large Scattering Angles Are the Most Common
Large-angle scattering is relatively rare.
Its rarity is itself important evidence about the small size of the nucleus.
Treating the Atomic Model as a Direct Observation
Scientists do not directly see the nucleus in Rutherford scattering.
They infer its existence and properties from the scattering evidence.
Ignoring the Importance of Scattering Angle
The amount of deflection provides information about the strength and nature of the interaction.
Scattering angle is therefore a key experimental measurement.
Check Your Understanding
1. What is a scattering experiment?
2. Why are scattering experiments useful for studying atoms and nuclei?
3. What type of particle was used in Rutherford's gold foil experiment?
4. What happened to most of the alpha particles?
5. What happened to a small number of alpha particles?
6. What was particularly surprising about a very small fraction of the particles?
7. What did the large number of particles passing straight through suggest about atomic structure?
8. What did large-angle scattering suggest about the distribution of positive charge?
9. Explain why large-angle scattering events were relatively rare.
10. Why are alpha particles repelled by an atomic nucleus?
11. What information can scientists obtain from scattering angles?
12. Why are high-energy electrons useful for investigating nuclei?
13. Explain why higher-energy particles can investigate smaller structures.
14. Why do scientists examine large numbers of scattering events rather than relying on one particle?
15. Explain how Rutherford's experiment demonstrates the relationship between experimental evidence and scientific models.
Key Terms
- Scattering – the change in direction of a particle caused by an interaction with another particle or structure.
- Scattering experiment – an experiment in which particles are directed toward a target and their resulting paths are measured.
- Alpha particle – a positively charged particle consisting of two protons and two neutrons.
- Gold foil experiment – the experiment in which alpha-particle scattering provided evidence for the nuclear atom.
- Scattering angle – the angle through which a particle's direction changes during an interaction.
- Nucleus – the tiny, dense, positively charged central region of an atom.
- Nuclear model – a model of the atom in which most mass and positive charge are concentrated in a small nucleus.
- High-energy electron – an electron accelerated to high energy for use as a probe of small structures.
- de Broglie wavelength – the wavelength associated with a moving particle.
- Resolution – the ability to distinguish small or closely spaced structures.
- Inference – a conclusion reached from observations and evidence.
- Scattering pattern – the distribution of particles detected after a scattering experiment.
Key Takeaways
- Scattering experiments allow scientists to investigate structures that are too small to observe directly.
- Rutherford's experiment fired alpha particles at thin gold foil.
- Most alpha particles passed straight through, showing that most of an atom is empty space.
- Some alpha particles were deflected, showing that positive charge is concentrated within the atom.
- Rare large-angle deflections provided evidence for a tiny, dense, positively charged nucleus.
- Scattering angle provides information about the interaction between a probe particle and the target.
- High-energy electrons can be used to investigate nuclear structure.
- Higher-energy particles have shorter associated wavelengths and can probe smaller distance scales.
- Scientists analyse large numbers of scattering events to identify reliable patterns.
- Scattering experiments provide indirect evidence: scientists infer structure from measured particle behaviour.
- Scientific models can be tested by comparing their predicted scattering patterns with experimental results.
- When experimental evidence disagrees with a model, the model may need to be modified or replaced.
4. Nuclear Radius and Nuclear Density
Learning Outcomes
- I can define nuclear radius.
- I can use the nuclear radius relationship to estimate nuclear sizes.
- I can explain why nuclear radius increases with nucleon number.
- I can calculate nuclear density using radius and mass information.
- I can explain why nuclear density is approximately constant for all nuclei.
The Size of the Nucleus
The nucleus is an extremely small region at the centre of an atom. It contains the atom's protons and neutrons and therefore contains almost all of the atom's mass.
A typical atom has a radius of about:
10⁻¹⁰ m
A typical nucleus has a radius of about:
10⁻¹⁵ m
This means that the radius of an atom is roughly 100,000 times greater than the radius of its nucleus.
The nucleus therefore occupies only a tiny fraction of the atom's volume.
What Is Nuclear Radius?
The nuclear radius is a measure of the size of an atomic nucleus.
Nuclei do not have perfectly sharp boundaries, so nuclear radius should be understood as an approximate measure of the extent of the nucleus.
Nuclear dimensions are often measured in femtometres (fm).
1 fm = 10⁻¹⁵ m
For example:
5 fm = 5 × 10⁻¹⁵ m
The femtometre is particularly useful because most nuclear radii are only a few femtometres.
Measuring Nuclear Size
Nuclei are far too small to measure directly with an ordinary microscope.
Instead, scientists use experiments such as high-energy electron scattering.
High-energy electrons are directed toward nuclei. Scientists measure how the electrons scatter and use the scattering pattern to estimate the size and charge distribution of the nucleus.
This is one reason scattering experiments are so important in nuclear physics: they allow scientists to investigate structures far smaller than can be seen directly.
Nuclear Radius and Nucleon Number
Experiments show that nuclear radius depends on the number of nucleons in the nucleus.
The relationship is approximately:
R = R₀A^(1/3)
where:
- R = nuclear radius
- R₀ = nuclear radius constant
- A = nucleon number
- A^(1/3) means the cube root of A
A commonly used approximate value is:
R₀ ≈ 1.2 × 10⁻¹⁵ m
or:
R₀ ≈ 1.2 fm
Therefore, a convenient version of the relationship is:
R ≈ 1.2A^(1/3) fm
Understanding the Cube Root
The expression A^(1/3) means the cube root of A.
For example:
8^(1/3) = 2
because:
2 × 2 × 2 = 8
Similarly:
27^(1/3) = 3
because:
3 × 3 × 3 = 27
And:
64^(1/3) = 4
because:
4 × 4 × 4 = 64
Therefore, nuclear radius does increase as nucleon number increases, but it increases much more slowly than A itself.
Worked Example: Nuclear Radius
Estimate the radius of a nucleus with nucleon number:
A = 64
Use:
R = R₀A^(1/3)
Since:
R₀ = 1.2 fm
and:
64^(1/3) = 4
then:
R = 1.2 × 4
Therefore:
R = 4.8 fm
In metres:
R = 4.8 × 10⁻¹⁵ m
So the estimated nuclear radius is 4.8 fm.
Worked Example: A Larger Nucleus
Consider a nucleus with:
A = 125
Since:
125^(1/3) = 5
use:
R = 1.2 × 5
Therefore:
R = 6.0 fm
or:
R = 6.0 × 10⁻¹⁵ m
The nucleus with more nucleons is larger.
Why Does Nuclear Radius Increase?
A larger nucleon number means the nucleus contains more:
- Protons
- Neutrons
Adding nucleons increases the size of the nucleus.
However, the radius does not increase directly in proportion to nucleon number.
For example, multiplying A by 8 only doubles the radius because:
8^(1/3) = 2
Therefore:
Nucleon number × 8 → radius × 2
This relationship tells us something important about the internal structure of nuclei.
Nuclear Volume
To understand nuclear density, we first need to consider nuclear volume.
A nucleus can be approximated as a sphere.
The volume of a sphere is:
V = (4/3)πR³
where:
- V = volume
- R = radius
Because nuclear radius follows:
R = R₀A^(1/3)
we can cube both sides conceptually:
R³ = R₀³A
Therefore:
Nuclear volume is proportional to A.
In simpler terms:
More nucleons → proportionally larger nuclear volume
This relationship is extremely important when explaining nuclear density.
Worked Example: Nuclear Volume
Suppose a nucleus has radius:
R = 4.0 × 10⁻¹⁵ m
Use:
V = (4/3)πR³
First cube the radius:
R³ = (4.0 × 10⁻¹⁵)³
R³ = 64 × 10⁻⁴⁵
which is:
R³ = 6.4 × 10⁻⁴⁴ m³
Now calculate the volume:
V = (4/3)π(6.4 × 10⁻⁴⁴)
This gives approximately:
V = 2.7 × 10⁻⁴³ m³
This is an extraordinarily small volume.
Nuclear Mass
The mass of a nucleus depends mainly on the number of nucleons it contains.
A proton and neutron each have a mass of approximately:
1.67 × 10⁻²⁷ kg
For an estimate, we can therefore write:
Nuclear mass ≈ A × 1.67 × 10⁻²⁷ kg
where A is the nucleon number.
For example, a nucleus containing 100 nucleons has an approximate mass of:
m ≈ 100 × 1.67 × 10⁻²⁷
Therefore:
m ≈ 1.67 × 10⁻²⁵ kg
This is an approximation because actual nuclear masses are affected slightly by nuclear binding energy.
Nuclear Density
Density describes how much mass is contained within a particular volume.
The density equation is:
Density = mass / volume
Using symbols:
ρ = m / V
where:
- ρ = density in kg/m³
- m = mass in kg
- V = volume in m³
Because nuclei contain considerable mass in extremely small volumes, nuclear density is enormous.
Worked Example: Calculating Nuclear Density
Suppose a nucleus has:
Mass = 1.0 × 10⁻²⁵ kg
and:
Radius = 5.0 × 10⁻¹⁵ m
First calculate its volume.
V = (4/3)πR³
Substitute the radius:
V = (4/3)π(5.0 × 10⁻¹⁵)³
Cube the radius:
(5.0 × 10⁻¹⁵)³ = 1.25 × 10⁻⁴³ m³
Therefore:
V ≈ 5.24 × 10⁻⁴³ m³
Now calculate density:
ρ = m / V
ρ = (1.0 × 10⁻²⁵) / (5.24 × 10⁻⁴³)
Therefore:
ρ ≈ 1.9 × 10¹⁷ kg/m³
This is an extraordinarily high density.
Typical nuclear densities are on the order of:
10¹⁷ kg/m³
How Dense Is Nuclear Matter?
Nuclear density is difficult to imagine because it is so much greater than densities encountered in everyday life.
For comparison:
- Water has a density of about 1.0 × 10³ kg/m³.
- Many metals have densities around 10³ to 10⁴ kg/m³.
- Nuclear matter has a density around 10¹⁷ kg/m³.
This means nuclear matter is roughly 100 trillion times denser than water.
The enormous density results from packing protons and neutrons into an extremely small volume.
Why Is Nuclear Density Approximately Constant?
One of the most interesting properties of nuclei is that their densities are approximately the same, regardless of their nucleon number.
We can understand why from the relationships for mass and volume.
Nuclear mass is approximately proportional to nucleon number:
m ∝ A
Nuclear radius follows:
R ∝ A^(1/3)
Because volume depends on radius cubed:
V ∝ R³
Substitute the radius relationship:
V ∝ [A^(1/3)]³
Therefore:
V ∝ A
So both mass and volume increase approximately in proportion to A.
Density is:
ρ = m / V
Since both the numerator and denominator increase approximately in the same proportion, their ratio stays approximately constant.
Therefore:
Nuclear density is approximately independent of nucleon number.
A Simple Comparison
Imagine one nucleus has 8 times as many nucleons as another.
Its mass is approximately:
8 times greater
Its radius is:
8^(1/3) = 2 times greater
If radius doubles, volume increases by:
2³ = 8
So:
- Mass increases × 8
- Volume increases × 8
Density therefore remains approximately the same.
This is a powerful consequence of the relationship:
R ∝ A^(1/3)
What Constant Nuclear Density Tells Us
The approximately constant density of nuclei suggests that nucleons are packed together in a broadly similar way in different nuclei.
A larger nucleus is not simply a much more tightly packed version of a smaller nucleus.
Instead, adding nucleons generally increases the volume available to contain them.
This behaviour is sometimes compared with adding more closely packed objects to a cluster: the cluster grows while the average packing density remains similar.
Radius Does Not Increase Linearly with Nucleon Number
A common mistake is to assume:
Twice as many nucleons → twice the radius
This is incorrect.
Because:
R ∝ A^(1/3)
doubling A changes the radius by a factor of:
2^(1/3) ≈ 1.26
So if the nucleon number doubles, the nuclear radius increases by only about 26%.
The volume, however, approximately doubles.
Worked Example: Comparing Nuclear Radii
Nucleus A has nucleon number:
A = 27
Nucleus B has nucleon number:
A = 216
Calculate their approximate radii.
For Nucleus A:
27^(1/3) = 3
Therefore:
R = 1.2 × 3 = 3.6 fm
For Nucleus B:
216^(1/3) = 6
Therefore:
R = 1.2 × 6 = 7.2 fm
Nucleus B contains:
216 / 27 = 8 times as many nucleons
but its radius is only:
7.2 / 3.6 = 2 times larger
Its volume, however, is approximately 8 times larger.
Estimating Nuclear Density from the Radius Relationship
We can combine the nuclear radius relationship with the density equation.
Use:
R = R₀A^(1/3)
Nuclear volume is:
V = (4/3)πR³
Substituting the radius relationship gives:
V = (4/3)πR₀³A
Approximate nuclear mass is:
m = Amₙ
where mₙ represents the approximate mass of one nucleon.
Density is:
ρ = m / V
Therefore:
ρ = Amₙ / [(4/3)πR₀³A]
The A appears in both the numerator and denominator, so it cancels.
This leaves:
ρ = mₙ / [(4/3)πR₀³]
The important point is that A is no longer present.
This mathematical result explains why nuclear density is approximately constant for nuclei of different sizes.
Nuclear Density and Neutron Stars
Nuclear density is not normally encountered in everyday matter because atoms contain enormous amounts of empty space compared with the size of their nuclei.
However, extremely dense astronomical objects called neutron stars contain matter compressed to densities comparable to those found in atomic nuclei.
Neutron stars form from the collapsed cores of certain massive stars.
They provide an extraordinary example of matter existing under conditions approaching nuclear density on an astronomical scale.
Evidence and Limitations
The relationship:
R = R₀A^(1/3)
is an empirical approximation based on experimental evidence.
Real nuclei are more complicated than perfectly uniform spheres.
For example:
- Some nuclei are not perfectly spherical.
- Nuclear charge and matter do not have perfectly sharp boundaries.
- Proton and neutron distributions can differ slightly.
- Very light or unusual nuclei may show departures from the simple model.
Nevertheless, the relationship is extremely useful for estimating nuclear dimensions and explaining the approximately constant density of nuclear matter.
Common Mistakes
Confusing Atomic Radius with Nuclear Radius
Atomic radii are typically around:
10⁻¹⁰ m
Nuclear radii are typically around:
10⁻¹⁵ m
These differ by roughly five orders of magnitude.
Forgetting the Cube Root
The relationship is:
R = R₀A^(1/3)
not:
R = R₀A
Assuming Twice the Nucleon Number Means Twice the Radius
Because radius depends on the cube root of A, doubling A increases the radius by only about 26%.
Forgetting to Cube the Radius When Finding Volume
For a spherical nucleus:
V = (4/3)πR³
The radius must be cubed.
Forgetting to Cube the Power of Ten
For example:
(10⁻¹⁵)³ = 10⁻⁴⁵
not 10⁻¹⁵.
Confusing Nuclear Mass with Nucleon Number
Nucleon number A is a count and has no unit.
Mass is measured in kilograms or atomic mass units.
Assuming Larger Nuclei Are Much Denser
Larger nuclei contain more nucleons, but they also have proportionally greater volume.
Their average nuclear densities therefore remain approximately constant.
Check Your Understanding
1. What is meant by nuclear radius?
2. What is a femtometre?
3. State the approximate relationship between nuclear radius and nucleon number.
4. What does A represent in the nuclear radius relationship?
5. Estimate the radius of a nucleus with A = 8 using R₀ = 1.2 fm.
6. Estimate the radius of a nucleus with A = 27.
7. A nucleus has 8 times as many nucleons as another nucleus. How many times larger is its radius?
8. Why does nuclear radius increase as nucleon number increases?
9. State the equation for the volume of a spherical nucleus.
10. State the equation for density.
11. Why is nuclear matter extremely dense?
12. Explain why nuclear volume is approximately proportional to nucleon number.
13. Explain why nuclear mass is approximately proportional to nucleon number.
14. Use these relationships to explain why nuclear density is approximately constant.
15. Why should the simple nuclear radius relationship be regarded as an approximation rather than an exact description of every nucleus?
Key Terms
- Nuclear radius – an approximate measure of the size of an atomic nucleus.
- Femtometre (fm) – a unit of length equal to 10⁻¹⁵ m.
- Nucleon – a proton or neutron within a nucleus.
- Nucleon number (A) – the total number of protons and neutrons in a nucleus.
- Nuclear radius constant (R₀) – the constant used in the approximate nuclear radius relationship.
- Nuclear volume – the volume occupied by a nucleus, often estimated by treating it as a sphere.
- Nuclear density – the mass of a nucleus divided by its volume.
- Density – mass per unit volume.
- Nuclear matter – matter consisting of closely packed nucleons.
- Empirical relationship – a relationship established from experimental observations.
Key Takeaways
- Atomic nuclei have radii on the order of 10⁻¹⁵ m.
- Nuclear dimensions are often expressed in femtometres (fm).
- Nuclear radius can be estimated using R = R₀A^(1/3).
- A commonly used value is R₀ ≈ 1.2 fm.
- Nuclear radius increases as nucleon number increases, but it increases according to the cube root of A.
- Nuclear volume is approximately proportional to nucleon number.
- Nuclear mass is also approximately proportional to nucleon number.
- Nuclear density can be calculated using ρ = m / V.
- Typical nuclear densities are on the order of 10¹⁷ kg/m³.
- Because nuclear mass and nuclear volume both increase approximately in proportion to A, nuclear density remains approximately constant.
- The approximately constant density suggests that nucleons are packed with broadly similar average density in nuclei of different sizes.
5. Distance of Closest Approach
Learning Outcomes
- I can explain the concept of distance of closest approach.
- I can apply conservation of energy to nuclear interactions.
- I can calculate the distance of closest approach for alpha particles approaching nuclei.
- I can relate closest-approach calculations to nuclear size estimates.
- I can explain how closest-approach experiments support the nuclear model of the atom.
What Is the Distance of Closest Approach?
In scattering experiments, a positively charged particle such as an alpha particle can be fired toward a positively charged atomic nucleus.
Because both the alpha particle and nucleus are positively charged, they repel each other electrically.
As the alpha particle approaches the nucleus:
- The electric repulsion becomes stronger.
- The alpha particle slows down.
- Its kinetic energy decreases.
- Its electric potential energy increases.
If the alpha particle is directed straight toward the centre of the nucleus, there is a point at which it momentarily stops before being repelled backward.
The distance between the centre of the alpha particle and the centre of the nucleus at this point is called the distance of closest approach.
Why Doesn't the Alpha Particle Reach the Nucleus?
An alpha particle has a charge of:
+2e
A nucleus containing Z protons has a charge of:
+Ze
Since both charges are positive, there is an electrostatic repulsive force between them.
As the distance decreases, this force becomes stronger.
For a head-on approach, the alpha particle loses speed until its kinetic energy has been converted into electric potential energy.
At the closest point:
Speed = 0
and therefore:
Kinetic energy = 0
The alpha particle then accelerates away from the nucleus.
Conservation of Energy
The calculation of closest approach relies on the principle of conservation of energy.
Energy cannot be created or destroyed. It can only be transferred or transformed.
Initially, far from the nucleus, the alpha particle has kinetic energy.
As it approaches the nucleus:
Kinetic energy → electric potential energy
At the distance of closest approach, for the idealized head-on case:
Initial kinetic energy = maximum electric potential energy
This provides the equation needed to calculate the closest distance.
Electric Potential Energy
The electric potential energy of two point charges separated by distance r is:
Electric potential energy = kQ₁Q₂ / r
where:
- k = Coulomb constant
- Q₁ = charge of the alpha particle
- Q₂ = charge of the nucleus
- r = separation between their centres
Use:
k = 8.99 × 10⁹ N·m²/C²
The elementary charge is:
e = 1.60 × 10⁻¹⁹ C
Therefore, an alpha particle has charge:
Qα = +2e = +3.20 × 10⁻¹⁹ C
A nucleus with atomic number Z has charge:
Qnucleus = +Ze
Closest-Approach Equation
At the distance of closest approach:
Initial kinetic energy = electric potential energy
Therefore:
KE = kQ₁Q₂ / r
Rearranging for r:
r = kQ₁Q₂ / KE
For an alpha particle approaching a nucleus:
Q₁ = 2e
and:
Q₂ = Ze
So we can also write:
r = 2kZe² / KE
This gives the closest centre-to-centre separation for an idealized head-on approach.
Worked Example: Alpha Particle Approaching Gold
An alpha particle approaches a gold nucleus directly.
Gold has atomic number:
Z = 79
Suppose the alpha particle has kinetic energy:
KE = 8.0 × 10⁻¹³ J
Calculate the distance of closest approach.
The alpha particle charge is:
Q₁ = 2e = 3.20 × 10⁻¹⁹ C
The gold nucleus has charge:
Q₂ = 79e
So:
Q₂ = 79 × 1.60 × 10⁻¹⁹
Q₂ = 1.264 × 10⁻¹⁷ C
Now use:
r = kQ₁Q₂ / KE
Substitute:
r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(1.264 × 10⁻¹⁷) / (8.0 × 10⁻¹³)
First multiply the charges:
(3.20 × 10⁻¹⁹)(1.264 × 10⁻¹⁷) = 4.0448 × 10⁻³⁶
Then multiply by k:
(8.99 × 10⁹)(4.0448 × 10⁻³⁶) ≈ 3.64 × 10⁻²⁶
Now divide by the kinetic energy:
r = (3.64 × 10⁻²⁶) / (8.0 × 10⁻¹³)
Therefore:
r ≈ 4.55 × 10⁻¹⁴ m
So the distance of closest approach is approximately:
4.6 × 10⁻¹⁴ m
or:
46 fm
Interpreting the Result
The calculated value does not necessarily represent the radius of the gold nucleus.
Instead, it tells us that the alpha particle could not approach any closer than about 46 fm under those conditions.
The actual radius of a gold nucleus is considerably smaller.
Therefore, the result gives an upper limit on nuclear size:
Nuclear radius < distance of closest approach
This was historically important because scattering experiments showed that the positive charge and mass of the atom must be concentrated within a region much smaller than the atom itself.
Why Faster Alpha Particles Get Closer
From:
r = kQ₁Q₂ / KE
we can see that:
r is inversely proportional to kinetic energy.
Therefore:
Greater kinetic energy → smaller closest approach
A faster alpha particle can overcome more of the electrical repulsion and travel closer to the nucleus before stopping.
For example, if the kinetic energy doubles:
KE × 2 → r ÷ 2
If the kinetic energy triples:
KE × 3 → r ÷ 3
This makes high-energy particles particularly useful for investigating very small structures.
Comparing Two Alpha Particles
Suppose Alpha Particle A has kinetic energy:
4.0 × 10⁻¹³ J
Alpha Particle B has:
8.0 × 10⁻¹³ J
Both approach the same nucleus directly.
Particle B has twice the kinetic energy.
Therefore:
rB = 1/2 rA
Particle B can approach twice as close to the nucleus.
Effect of Nuclear Charge
The closest-approach equation also shows:
r ∝ Z
for alpha particles with the same kinetic energy.
A nucleus with more protons has a larger positive charge and therefore produces stronger electrostatic repulsion.
For the same incoming alpha-particle energy:
Larger Z → larger distance of closest approach
For example, an alpha particle approaching a nucleus with 80 protons will be stopped farther away than the same alpha particle approaching a nucleus with 20 protons.
From Kinetic Energy to Closest Approach
Sometimes a problem gives the alpha particle's mass and speed instead of its kinetic energy.
In this case, first calculate:
KE = ½mv²
Then use:
r = kQ₁Q₂ / KE
For an alpha particle, the mass is approximately:
mα = 6.64 × 10⁻²⁷ kg
Worked Example Using Speed
An alpha particle travels directly toward a nucleus at:
1.0 × 10⁷ m/s
Suppose the nucleus has:
Z = 50
First calculate the alpha particle's kinetic energy.
Use:
KE = ½mv²
Substitute:
KE = ½(6.64 × 10⁻²⁷)(1.0 × 10⁷)²
Square the speed:
(1.0 × 10⁷)² = 1.0 × 10¹⁴
Therefore:
KE = ½(6.64 × 10⁻²⁷)(1.0 × 10¹⁴)
KE = 3.32 × 10⁻¹³ J
The nucleus has charge:
Q = 50e
Q = 50(1.60 × 10⁻¹⁹)
Q = 8.00 × 10⁻¹⁸ C
Now use:
r = kQ₁Q₂ / KE
r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(8.00 × 10⁻¹⁸) / (3.32 × 10⁻¹³)
This gives approximately:
r = 6.9 × 10⁻¹⁴ m
or:
r ≈ 69 fm
Energy in Electronvolts
Nuclear energies are often given in electronvolts (eV) rather than joules.
One electronvolt is:
1 eV = 1.60 × 10⁻¹⁹ J
Common nuclear units include:
1 keV = 10³ eV
1 MeV = 10⁶ eV
Therefore:
1 MeV = 1.60 × 10⁻¹³ J
An alpha particle with energy 5.0 MeV therefore has:
KE = 5.0 × 1.60 × 10⁻¹³ J
KE = 8.0 × 10⁻¹³ J
This is the kinetic energy used in our earlier gold-nucleus example.
Worked Example Using MeV
An alpha particle with kinetic energy 6.0 MeV approaches a nucleus with atomic number 60.
First convert the energy to joules:
KE = 6.0 × 1.60 × 10⁻¹³
KE = 9.60 × 10⁻¹³ J
The alpha particle has charge:
Q₁ = 3.20 × 10⁻¹⁹ C
The nucleus has charge:
Q₂ = 60 × 1.60 × 10⁻¹⁹
Q₂ = 9.60 × 10⁻¹⁸ C
Now:
r = kQ₁Q₂ / KE
r = (8.99 × 10⁹)(3.20 × 10⁻¹⁹)(9.60 × 10⁻¹⁸) / (9.60 × 10⁻¹³)
Therefore:
r ≈ 2.88 × 10⁻¹⁴ m
or:
r ≈ 28.8 fm
Closest Approach and Nuclear Size
Closest-approach experiments can help establish limits on nuclear size.
Suppose an alpha particle turns around at:
r = 3.0 × 10⁻¹⁴ m
If the interaction can be explained entirely by electrostatic repulsion at this distance, then the nucleus must fit inside that separation.
Therefore:
Nuclear radius must be less than about 3.0 × 10⁻¹⁴ m.
Experiments using particles with greater kinetic energies can probe even smaller distances.
This provides progressively better information about nuclear dimensions.
Connection to Rutherford Scattering
Rutherford scattering provided powerful evidence for the nuclear model of the atom.
When alpha particles were fired at thin metal foil:
- Most passed through almost undeflected.
- Some were deflected through small angles.
- A very small number were deflected through large angles.
- A tiny fraction were scattered backward.
The large-angle scattering could only be explained if the atom's positive charge and most of its mass were concentrated in a very small region.
That region is the nucleus.
Closest Approach and the Nuclear Model
The distance-of-closest-approach calculation strengthens the nuclear interpretation.
If positive charge were spread throughout the entire atom, as suggested by the older plum pudding model, an alpha particle would not experience the intense concentrated repulsion required to reverse its motion over such a small distance.
Instead, the results indicate that:
- Positive charge is concentrated.
- The concentration occupies a tiny region.
- The region is much smaller than the atom.
- Strong electrostatic forces occur when alpha particles approach this region.
These observations support the existence of a small, positively charged nucleus.
Head-On and Non-Head-On Approaches
The simple closest-approach calculation assumes a head-on collision.
The alpha particle travels directly toward the centre of the nucleus.
In this special case, the alpha particle can momentarily stop before reversing direction.
Most particles in a scattering experiment do not approach the nucleus exactly head-on.
A particle passing to one side of the nucleus follows a curved path and is deflected through some angle rather than simply reversing direction.
The closer its path takes it to the nucleus, the greater the deflection generally becomes.
Assumptions in the Simple Model
The basic closest-approach calculation makes several simplifying assumptions.
It assumes:
- The nucleus and alpha particle can be treated approximately as point charges until very small separations.
- The interaction is dominated by electrostatic repulsion.
- The nucleus remains approximately stationary.
- The approach is directly head-on.
- Energy is conserved.
- Other interactions can be neglected at the calculated separation.
These assumptions make the model useful, but they also establish its limitations.
When the Simple Model Breaks Down
At sufficiently small distances, the simple electrostatic model becomes incomplete.
When particles approach nuclear dimensions, the strong nuclear interaction can become important.
The nucleus also has a finite size rather than being a perfect point charge.
Therefore, closest-approach calculations based only on Coulomb repulsion work best when the particles remain sufficiently far apart that electrostatic interaction dominates.
Deviations from the simple model can themselves provide information about nuclear structure.
Connecting Closest Approach to Nuclear Radius
It is useful to distinguish these two quantities:
Nuclear radius
An estimate of the physical size of the nucleus.
Distance of closest approach
The minimum centre-to-centre separation reached by an incoming charged particle under specified conditions.
They are not automatically equal.
If:
rclosest > Rnucleus
then the alpha particle turns around before reaching the nuclear surface.
Increasing the alpha particle's energy reduces the distance of closest approach and allows scientists to probe smaller distances.
Common Mistakes
Assuming Closest Approach Equals Nuclear Radius
The closest approach usually gives an upper limit on nuclear size, not necessarily the actual nuclear radius.
Forgetting That Both Charges Are Positive
An alpha particle and a normal atomic nucleus are both positively charged.
Their interaction is therefore repulsive.
Forgetting the Alpha Particle Has Charge +2e
Use:
Qα = +2e
not +e.
Using Atomic Mass Instead of Atomic Number
The nuclear electric charge depends on the number of protons.
Therefore:
Qnucleus = +Ze
where Z is the atomic number, not the nucleon number.
Forgetting to Convert MeV to Joules
If using SI values for k and charge, energy should be converted to joules.
Remember:
1 MeV = 1.60 × 10⁻¹³ J
Assuming the Alpha Particle Has Zero Energy at Closest Approach
Its kinetic energy is momentarily zero in the idealized head-on model.
Its energy has not disappeared. It has been converted into electric potential energy.
Applying the Simple Equation to Every Scattering Path
The straightforward energy calculation applies to a head-on approach.
Non-head-on scattering requires consideration of the particle's trajectory and angular momentum.
Check Your Understanding
1. Define the distance of closest approach.
2. Why does an alpha particle slow down as it approaches a positively charged nucleus?
3. What happens to the alpha particle's kinetic energy during its approach?
4. What happens to its electric potential energy?
5. What is the alpha particle's kinetic energy at the closest point in an ideal head-on approach?
6. State the energy relationship used to calculate the distance of closest approach.
7. What is the charge of an alpha particle in terms of e?
8. What is the charge of a nucleus with atomic number Z?
9. Why can a higher-energy alpha particle approach a nucleus more closely?
10. If an alpha particle's kinetic energy doubles, what happens to its distance of closest approach to the same nucleus?
11. Why does a nucleus with a larger atomic number produce a larger closest-approach distance for alpha particles of equal energy?
12. Convert 4.0 MeV into joules.
13. Explain why a calculated closest approach can provide an upper limit for nuclear radius.
14. Why is the simple calculation most appropriate for a head-on approach?
15. Explain how closest-approach and scattering experiments provide evidence for a small, positively charged nucleus.
Key Terms
- Distance of closest approach – the minimum separation between an incoming charged particle and a nucleus during an interaction.
- Alpha particle – a particle containing two protons and two neutrons with charge +2e.
- Electrostatic repulsion – the repulsive force between charges of the same sign.
- Kinetic energy – energy associated with motion.
- Electric potential energy – energy associated with the positions of electric charges.
- Conservation of energy – the principle that total energy remains constant in an isolated system.
- Electronvolt (eV) – a unit of energy commonly used in atomic and nuclear physics.
- Mega-electronvolt (MeV) – one million electronvolts.
- Head-on approach – a trajectory directed toward the centre of the target nucleus.
- Nuclear radius – an approximate measure of the size of a nucleus.
- Scattering – a change in the direction of a particle caused by an interaction.
- Coulomb interaction – the electrical attraction or repulsion between charged particles.
Key Takeaways
- The distance of closest approach is the minimum separation reached by an incoming charged particle and a nucleus.
- An alpha particle and an atomic nucleus are both positively charged, so they repel.
- As an alpha particle approaches a nucleus, kinetic energy is converted into electric potential energy.
- For an ideal head-on approach, the alpha particle momentarily stops at its closest point.
- Conservation of energy gives: KE = kQ₁Q₂ / r.
- The closest distance can therefore be calculated using: r = kQ₁Q₂ / KE.
- An alpha particle has charge +2e, while a nucleus has charge +Ze.
- Increasing the alpha particle's kinetic energy decreases its distance of closest approach.
- A larger nuclear charge produces stronger electrostatic repulsion.
- Closest-approach calculations can provide an upper limit on nuclear size.
- Distance of closest approach and nuclear radius are related concepts, but they are not necessarily equal.
- Closest-approach and scattering experiments support the model of the atom containing a tiny, concentrated, positively charged nucleus.