OxfordAQA International GCSE Physics: Atomic structure, isotopes and types of ionising radiation (9203)

This guide is written for the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. It covers sections 3.7.1 a to f (atomic structure) and 3.7.2 a to g (ionizing radiation from the nucleus). The qualification is untiered. Half-life, contamination and irradiation, fission and fusion are separate parts of the specification and are not taught here.

Use the course hub to see the other topics and the printable checklist to tick off each outcome. To test yourself, use the revision notes, the practice questions and the 10-minute diagnostics. General technique is on the exam preparation page.

What this unit covers

Section	What you must be able to do
3.7.1 a	State the size of an atom and describe the nuclear model, including where the mass is
3.7.1 b	Explain what alpha particle scattering by thin metal foil shows
3.7.1 c	Give the relative mass and relative charge of a proton, neutron and electron
3.7.1 d, e	Explain why atoms are neutral; describe how ions form
3.7.1 f	Define isotope, atomic number and mass number, and use the notation
3.7.2 a to c	Describe unstable nuclei and random decay; know that the four emissions are unaffected by chemical or physical processes
3.7.2 d	Give sources of background radiation
3.7.2 e	Describe the nature of alpha, beta and gamma radiation
3.7.2 f	Complete nuclear equations for single alpha and beta decay
3.7.2 g	Compare ionising power, penetration and range in air

The structure of an atom (3.7.1 a)

An atom has a radius of about 10⁻¹⁰ m. The simple model has a small, central, positively charged nucleus made of protons and neutrons, with electrons around it. The nucleus is far smaller than the atom as a whole, and holds almost all of the mass. A nuclear radius of about 10⁻¹⁵ m is a good working value.

Worked example 1. Scale the nucleus up to a pea of radius 1 cm. How big is the atom?

• Ratio of radii = 10⁻¹⁰ / 10⁻¹⁵ = 10⁵.
• Atom radius = 1 cm × 10⁵ = 10⁵ cm = 1 km.

So an atom is mostly empty space, with a tiny dense core at the centre and electrons at various distances from it.

Evidence from alpha particle scattering (3.7.1 b)

In the experiment, a narrow beam of alpha particles is aimed at very thin metal foil, with a detector that can be moved round the foil. The apparatus is in a vacuum, because alpha particles travel only a few centimetres in air.

Observation	Conclusion about the atom
Most alpha particles go straight on	Most of an atom is empty space
A small number are deflected through small angles	They were repelled by something positive, so the nucleus is positively charged
Very few bounce back through large angles	A tiny nucleus holds the positive charge and most of the mass

Alpha particles are positive, so they are only repelled strongly when they pass very close to a concentrated positive charge. Only a heavy nucleus can turn a fast alpha particle back. Before this experiment, the positive charge was thought to be spread through the whole atom. Spread-out charge could not produce large deflections, so that model had to be replaced by the nuclear model.

Protons, neutrons and electrons (3.7.1 c, d, e)

• Proton: charge +1, mass 1.
• Neutron: charge 0, mass 1.
• Electron: charge –1, mass very small.

The electron’s mass is roughly one two-thousandth of a proton’s, so it contributes almost nothing to the total.

A neutral atom contains as many electrons as protons. The charges cancel, so the atom has no overall charge. The electrons are arranged at various distances from the nucleus. Atoms can lose or gain outer electrons and become charged particles called ions.

• Lose electrons: protons outnumber electrons, so a positive ion forms.
• Gain electrons: electrons outnumber protons, so a negative ion forms.
• Charge of the ion = number of protons – number of electrons.

Only electrons are lost or gained. The nucleus is not changed when an ion forms.

Atomic number, mass number and isotopes (3.7.1 f)

• Atomic number (also called proton number): how many protons the nucleus contains.
• Mass number = number of protons + number of neutrons.
• Number of neutrons = mass number – atomic number.

An atom is written with the mass number as a superscript before the symbol and the atomic number as a subscript, for example ²³₁₁Na. The atoms of one element always have the same number of protons. Isotopes of an element have identical atomic numbers and different mass numbers, because their neutron counts differ.

Worked example 2. Find the particles in ²³₁₁Na, in the ion ²³₁₁Na⁺ and in ³⁵₁₇Cl⁻.

Species	Protons	Neutrons	Electrons	Overall charge
²³₁₁Na atom	11	23 – 11 = 12	11	0
²³₁₁Na⁺ ion	11	12	10	11 – 10 = +1
³⁵₁₇Cl⁻ ion	17	35 – 17 = 18	18	17 – 18 = –1

Worked example 3. Carbon-12, carbon-13 and carbon-14 all have atomic number 6.

• Protons: 6 in each. Electrons in a neutral atom: 6 in each.
• Neutrons: 12 – 6 = 6, 13 – 6 = 7 and 14 – 6 = 8.

They are isotopes of carbon. They have the same number of electrons, so their chemical behaviour is the same. Their nuclei differ, and some isotopes are unstable.

Unstable nuclei and random decay (3.7.2 a, b, c)

Some atomic nuclei are unstable. An unstable nucleus emits particles or radiation and changes to the nucleus of a different element, which is more stable. This is radioactive decay. Changes in the nucleus are what release the energy.

Unstable nuclei emit alpha particles, beta particles or neutrons, together with electromagnetic radiation as gamma waves. Substances that do this are radioactive. Three points to know:

• Decay is a random process. There is no way to say when any one nucleus will decay, or which nucleus in a sample will go next.
• Neither chemical nor physical processes affect this behaviour. Heating a sample, compressing it or combining it chemically with other elements does not change how its nuclei decay.
• The radiations are called ionising because they can remove electrons from atoms and make ions.

Background radiation (3.7.2 d)

Background radiation is always present, wherever you are. It has several sources:

• radioactive substances in the environment, such as radon gas given off by some rocks and soils, and materials in buildings;
• radiation from space, often called cosmic rays;
• devices such as X-ray machines in hospitals.

A detector with no source nearby still records counts. Worked example 4. A detector records 84 counts in 10 minutes with no source present. A source is then placed beside it and 1200 counts are recorded in 2 minutes.

• Background count rate = 84 / 10 = 8.4 counts per minute.
• Count rate with the source = 1200 / 2 = 600 counts per minute.
• Corrected count rate = 600 – 8.4 = 591.6, which is 592 counts per minute (3 significant figures).

Because decay is random, repeating the measurement gives slightly different counts each time.

The nature of the radiations (3.7.2 e)

• Alpha particle: two protons and two neutrons, which is a helium nucleus. Charge +2, written ⁴₂He or α.
• Beta particle: a fast-moving electron that leaves the nucleus when one of its neutrons changes into a proton. Written ⁰₋₁e or β.
• Gamma radiation: electromagnetic radiation from the nucleus. No mass and no charge.
• Neutron emission: some unstable nuclei emit a neutron, so the mass number falls by one and the atomic number does not change.

A beta particle is made in the nucleus. It is a different electron from those arranged around the atom.

Nuclear equations (3.7.2 f)

A nuclear equation represents a decay. You are required to balance equations for single alpha and beta decay by completing the atomic number and mass number. Identifying the new element is not required. The rule is that the total mass number and the total atomic number are the same on both sides.

Emission	Mass number of nucleus	Atomic number of nucleus
Alpha	falls by 4	falls by 2
Beta	unchanged	rises by 1
Gamma	unchanged	unchanged

Worked example 5 (alpha). ²²⁶₈₈Ra → ᴬ_Z X + ⁴₂He. Find A and Z.

• Mass number: 226 = A + 4, so A = 222.
• Atomic number: 88 = Z + 2, so Z = 86.
• ²²²₈₆X + ⁴₂He

Worked example 6 (beta). ¹⁴₆C → ᴬ_Z X + ⁰₋₁e. Find A and Z.

• Mass number: 14 = A + 0, so A = 14.
• Atomic number: 6 = Z + (–1), so Z = 7.
• ¹⁴₇X + ⁰₋₁e

The atomic number rises by one in beta decay because a neutron becomes a proton. The beta particle carries away a charge of –1, so the nucleus gains +1.

Worked example 7. ²⁴¹₉₅Am emits an alpha particle, and ⁹⁰₃₈Sr emits a beta particle. Find the numbers of the new nuclei.

• Alpha: A = 241 – 4 = 237, Z = 95 – 2 = 93.
• Beta: A = 90, Z = 38 + 1 = 39.

Ionising power, penetration and range in air (3.7.2 g)

Ionising power is how readily the radiation knocks electrons off atoms it passes. Penetration is how far it passes through a material. Range in air is how far it travels in air before it can no longer be detected above background.

	Alpha	Beta	Gamma
Relative ionising power	strongest	moderate	weakest
Stopped by	a sheet of paper, or skin	a few millimetres of aluminium	reduced by thick lead or concrete, never completely stopped
Range in air	a few centimetres	tens of centimetres to a few metres	very long; intensity keeps falling with distance

The three properties are linked. A strongly ionising radiation loses energy at every ionisation, so it is soon used up and stops quickly. A weakly ionising radiation loses energy slowly, so it travels much further. That is why the most ionising radiation (alpha) has the shortest range and the least ionising (gamma) the longest. The ionisation also explains why the radiations are hazardous: ionisation in living cells can damage them.

Worked example 8. A source emits a mixture. The background count rate is 30 counts per minute.

Absorber between source and detector	Count rate (per minute)	Corrected (minus 30)
None	1480	1450
Paper	1474	1444
3 mm aluminium	505	475
Thick lead block	52	22

• Paper makes almost no difference (a fall of 6 counts, which is within the random variation), so there is no alpha.
• Aluminium removes 1444 – 475 = 969 counts per minute, so beta is present.
• What gets through aluminium is gamma, and lead cuts it almost to background.

Common errors

• Writing that a beta particle is “an electron from the electron shells”. It comes from the nucleus.
• Saying the nucleus is “mostly empty space”. The atom is mostly empty space.
• Calculating neutrons as the atomic number. It is mass number minus atomic number.
• Using the atom’s electrons for the nuclear charge. The nuclear charge is the number of protons.
• Putting the atomic number on top in the notation. The mass number goes on top.
• Saying that gamma is “stopped by lead”. Thick lead reduces it, but does not stop it completely.
• Explaining the alpha scattering result with “alpha particles hit electrons”. Electrons are far too light to deflect them backwards.
• Saying heating a source makes it decay faster. Physical and chemical processes do not affect decay.

Next, use the revision notes to condense this, then try the practice questions.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.7.1 a to f and 3.7.2 a to g.