Nuclear atom, nuclides and amount
Key idea: Rutherford scattering turned observations into a nuclear model. Nuclide notation then records proton and nucleon numbers, while the mole links microscopic particles to measurable amounts.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 1 of 3
Check your understandingBy the end of this lesson, you should be able to
- Infer nuclear size and concentrated positive charge from Rutherford-scattering observations.
- Use proton number, nucleon number, neutron number, isotope and nuclide notation correctly.
- Use Avogadro’s constant to convert between amount of substance and particle number.
Learn the idea
Big question: How did a rare scattering event overturn the old picture of the atom?
Link each observation to one inference
Most alpha particles passed through thin foil almost undeflected, showing that most atomic volume is empty. Some were deflected, showing repulsion from positive charge.
A very small fraction turned through large angles. Such a strong interaction in a rare encounter requires positive charge and most atomic mass to be concentrated in a tiny nucleus.
Check your understanding: Why does the rarity of large-angle scattering matter?
It shows that the strongly repelling region occupies only a tiny fraction of the atom's volume.
Read nuclides and count particles
In ᴬ_ZX, Z is proton number and identifies the element; A is nucleon number, so neutron number is A − Z. Isotopes share Z but have different neutron numbers and A values.
One mole contains 6.02 × 10²³ specified entities. Say whether you are counting atoms, molecules, ions or nuclei before using N = nNA.
Check your understanding: Can two nuclides with different Z be isotopes?
No. Different Z means different elements.
Key ideas
- A is written upper left and Z lower left of the symbol.
- Neutron number = A − Z.
- Large-angle scattering is rare because the nucleus occupies a tiny fraction of atomic volume.
Relationships to know
A = proton number + neutron numberN = nNₐNₐ = 6.02 × 10²³ mol⁻¹
Follow the reasoning
Worked example
Use nuclide notation and the mole together
Question: A 0.250 g sample of carbon-14, ¹⁴₆C, is treated as pure. State its proton, neutron and electron numbers per neutral atom and estimate the number of nuclei. Use molar mass 14.0 g mol⁻¹ and NA = 6.02 × 10²³ mol⁻¹.
Step 1: Read the nuclide
Why: Z fixes protons; neutral atoms have matching electron number.
Working: Protons = 6, electrons = 6, neutrons = 14 − 6 = 8.
Step 2: Find amount of substance
Why: Mass must be divided by molar mass before using Avogadro's constant.
Working: n = 0.250/14.0 = 1.786 × 10⁻² mol.
Step 3: Count the stated entity
Why: Each carbon atom contains one carbon nucleus.
Working: N = nNA = 1.786 × 10⁻²(6.02 × 10²³) = 1.08 × 10²² nuclei.
Answer: Each neutral atom has 6 protons, 8 neutrons and 6 electrons; the sample contains about 1.08 × 10²² nuclei.
Check: The mass is roughly 1/56 of a mole, so a particle count near 10²² is sensible.
Now try it with support
Practise with support
For ²³⁵₉₂U, state the proton and neutron numbers. How many nuclei are in 0.020 mol?
Hints
- Neutrons = A − Z.
- Use N = nNₐ.
View the guided answer
There are 92 protons and 235 − 92 = 143 neutrons. N = 0.020(6.02 × 10²³) = 1.20 × 10²² nuclei.
Your turn
Practise independently
Use Rutherford-scattering observations to distinguish what establishes nuclear size, positive charge and mostly empty atomic space.
Check your answer
Most alpha particles being undeflected shows that most of the atom is empty space. The very small fraction scattered through large angles shows that positive charge and most mass are concentrated in a tiny nucleus; the rarity of those events establishes its small size.
Common mistakes and exam guidance
Watch out for
- Calling isotopes nuclides with different proton numbers.
- Using molar amount without stating what particle is being counted.
In an exam
- Link each Rutherford observation to one inference rather than listing conclusions without evidence.
- Check that A and Z balance in any later nuclear equation.
Put the ideas together
Exam-style practice [7 marks]
In a scattering experiment, 99.0% of alpha particles pass through within 1°, 0.9% are moderately deflected and 0.1% exceed 90°. Explain what each result shows and why the observations are inconsistent with positive charge spread throughout the atom.
Plan before you answer
- Match evidence and inference line by line.
- Use the alpha particle's positive charge.
- Explain both strength and rarity of large deflections.
View the marking points and model answer
Marking points
- Most pass through → mostly empty space.
- Deflection due to electrostatic repulsion.
- Repelling region is positive because alpha is positive.
- Large angles require a strong force.
- Strong force requires concentrated charge/close approach.
- Rarity shows the region is tiny.
- Concludes positive charge and most mass are concentrated in a small nucleus, unlike a spread-out model.
Model answer
The 99.0% almost undeflected result shows that most atomic volume is empty. Deflections arise because positive alpha particles are repelled by positive charge. A turn beyond 90° requires a very strong force during a close approach, so charge and most mass must be concentrated. Since only 0.1% encounter this strong region, it must be tiny. Spread-out positive charge could not provide the intense, localised repulsion needed for rare backward scattering.
Finish from memory
Three-question recap
What does Z represent?
Check
Proton number.
How do isotopes differ?
Check
They have the same proton number but different neutron numbers.
What is Avogadro's constant?
Check
6.02 × 10²³ specified entities per mole.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027