What is the basic structure of matter according to the Edexcel International A Levels Physics curriculum?

In the Edexcel International A Levels Physics curriculum, the basic structure of matter is explained through the atomic model. Matter is composed of atoms, which consist of a nucleus containing protons and neutrons, surrounded by electrons in orbitals. This atomic structure forms the foundation for understanding the properties and behavior of different materials.

How do protons, neutrons, and electrons contribute to the structure of matter?

Protons, neutrons, and electrons are the fundamental particles that make up an atom. Protons, which are positively charged, and neutrons, which are neutral, reside in the nucleus and contribute to the atom's mass. Electrons, which are negatively charged, orbit the nucleus and are involved in chemical bonding and reactions. The arrangement and interaction of these particles determine the physical and chemical properties of matter.

What role do quarks play in the structure of matter?

Quarks are elementary particles that combine to form protons and neutrons, which are the components of atomic nuclei. There are six types of quarks, but protons and neutrons are primarily composed of 'up' and 'down' quarks. Understanding quarks is essential for studying the fundamental forces and interactions that govern the structure of matter at a subatomic level, as covered in the Edexcel International A Levels Physics curriculum.

How does the concept of fields relate to the structure of matter?

Fields, such as electromagnetic and gravitational fields, play a crucial role in the structure of matter by mediating forces between particles. For example, electromagnetic fields are responsible for the attraction and repulsion between charged particles, influencing the structure and stability of atoms and molecules. Gravitational fields affect the large-scale structure of matter in the universe. These concepts are integral to understanding the interactions and behavior of matter in the Edexcel International A Levels Physics syllabus.

What is the significance of the atomic model in understanding the structure of matter?

The atomic model is significant because it provides a framework for understanding the composition and behavior of matter. It explains how atoms combine to form molecules and how the arrangement of electrons influences chemical properties and reactions. The atomic model also helps in understanding the periodic table and the classification of elements. This foundational knowledge is essential for further studies in physics, chemistry, and material science, as emphasized in the Edexcel International A Levels curriculum.

Why is "Reading the nucleon number $A$ as the number of neutrons" a common mistake in Structure of Matter?

Why it happens: A is the larger number in ^A_ZX, so students assume it is the neutron count. How to avoid it: A is protons **plus** neutrons. The number of neutrons is N = A - Z. For ^14_6C, neutrons = 14 - 6 = 8, not 14. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Describing the alpha-scattering observations without linking each to its specific conclusion" a common mistake in Structure of Matter?

Why it happens: Students list what was seen and separately list conclusions, instead of pairing observation with deduction. How to avoid it: Match each observation to its deduction: **most pass straight through → mostly empty space**; **small fraction deflected greatly → concentrated positive charge**; **very few bounce back → tiny, dense, massive nucleus**. Examiners award marks for the linked reasoning, not the description alone. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Substituting an energy in eV or MeV directly into $\tfrac{1}{2}mv^2$ or $kQq/r$" a common mistake in Structure of Matter?

Why it happens: The energy is quoted in MeV, so students forget it is not already in joules. How to avoid it: Convert first: 1\ eV = 1.60 × 10^-19\ J, so 5.0\ MeV = 5.0 × 10^6 × 1.60 × 10^-19 = 8.0 × 10^-13\ J. Only use joules inside energy equations. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Using the charge of the alpha particle as $e$ instead of $2e$ in closest-approach calculations" a common mistake in Structure of Matter?

Why it happens: Students use the elementary charge e automatically, forgetting the alpha is a helium nucleus with two protons. How to avoid it: The alpha particle carries charge q = 2e = 3.20 × 10^-19\ C, and a nucleus of proton number Z carries Q = Ze. Put both correct charges into E_k = 1/4\varepsilon_0Qq/r. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Confusing the sign of the charge / direction of conventional current when accelerating particles" a common mistake in Structure of Matter?

Why it happens: Electrons are negative, so the direction they are pushed is opposite to the electric field and to conventional current; students mix these up. How to avoid it: A field accelerates a **positive** charge along the field direction and a **negative** charge (electron) **against** it. Whatever the sign, the energy gained crossing a p.d. V is |q|V; conventional current flows from + to –, opposite to the electron drift. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Failing to explain *why* high energies are needed by linking it to de Broglie wavelength" a common mistake in Structure of Matter?

Why it happens: Students state that 'more energy means more detail' without invoking = h/p. How to avoid it: Argue the chain explicitly: to resolve a small structure you need a **small wavelength**; by = h/p a small wavelength needs a **large momentum**, which means a **high energy**. That is why accelerators are used to probe the nucleus. Source: Edexcel IAL Physics Unit 4 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Structure of Matter

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Short Notes - Structure of Matter

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Detailed Study Notes

Detailed notes on Further Mechanics, Fields and Particles for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Structure of Matter — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 4)

How we know what an atom is made of. Nuclide notation and isotopes; the alpha-scattering experiment and its evidence for a tiny dense nucleus; estimating nuclear size from closest approach and electron diffraction; and accelerating charged particles — thermionic emission, the electronvolt, and the linear accelerator and cyclotron ( $r = mv/Bq$ ) — with de Broglie ( $\lambda = h/p$ ) explaining why high energies are needed to probe small structures.

At a glance

Nuclide notation $^{A}_{Z}\text{X}$ : $Z$ = proton number (fixes the element), $A$ = nucleon number (protons + neutrons), so neutrons $= A - Z$ .
Isotopes are nuclei of the same element — same $Z$ , different $A$ (different neutron number).
Alpha scattering: most alphas pass straight through → atom is mostly empty space; a few deflect greatly → concentrated positive charge; very few bounce back → a tiny, dense, massive nucleus.
A nucleus ( $\sim 10^{-15}\ \text{m}$ ) is about $10^{5}$ times smaller than an atom ( $\sim 10^{-10}\ \text{m}$ ).
Closest approach: at the turning point all the incoming KE has become electric PE, so $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r}$ — an upper estimate of nuclear radius.
Thermionic emission frees electrons from a hot metal; accelerating a charge through a p.d. gives $eV = \tfrac{1}{2}mv^2$ . $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .
LINAC: alternating p.d. + drift tubes that get longer as particles speed up. Cyclotron: magnetic field bends the path ( $r = mv/Bq$ ), alternating p.d. across dees accelerates; period $T = 2\pi m/Bq$ is speed-independent.
Why high energy? By $\lambda = h/p$ , resolving small structures needs a small wavelength → large momentum → high energy.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Use nuclide notation $^{A}_{Z}\text{X}$ ; state proton, neutron and nucleon numbers; define and identify isotopes.
Describe the alpha-particle scattering experiment and explain how each observation supports the nuclear model of the atom.
Appreciate the relative sizes of atom and nucleus, and estimate nuclear size from the closest approach of an alpha particle and from electron diffraction (de Broglie).
Explain thermionic emission and calculate the speed/energy of a charged particle accelerated through a p.d. using $eV = \tfrac{1}{2}mv^2$ ; use the electronvolt.
Describe the operation of a linear accelerator and a cyclotron, using $r = mv/Bq$ , and explain why particles are given very high energies to probe matter ( $\lambda = h/p$ ).

The nucleus: nuclide notation and isotopes

Z fixes the element, A counts the nucleons, and neutrons = A − Z.

Every atom has a tiny central nucleus made of protons and neutrons (together called nucleons), surrounded by electrons. Two numbers describe any nucleus:

the proton number (atomic number) $Z$ — the number of protons, which fixes the element;
the nucleon number (mass number) $A$ — the total number of protons and neutrons.

We write a nucleus in nuclide notation as $^{A}_{Z}\text{X}$ , where $\text{X}$ is the element's symbol. The number of neutrons is then

$N = A - Z.$

For example $^{14}_{6}\text{C}$ has $Z = 6$ protons, $A = 14$ nucleons and therefore $14 - 6 = 8$ neutrons. In a neutral atom the number of electrons equals the number of protons, $Z$ .

Isotopes are nuclei of the same element — they have the same $Z$ — but a different number of neutrons, and so a different $A$ . Chlorine, for instance, occurs as $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ : both have $17$ protons (both are chlorine), but one has $18$ neutrons and the other $20$ . Because chemistry is governed by the electrons (and hence by $Z$ ), isotopes behave identically chemically but differ in mass.

In nuclide notation the upper number is the nucleon number A and the lower number is the proton number Z. The number of neutrons is A − Z. Isotopes of an element share the same Z but have different A.

$^{A}_{Z}\text{X}$ : $Z$ = protons (fixes element), $A$ = nucleons (protons + neutrons).
Neutrons $= A - Z$ ; neutral atom has $Z$ electrons.
Isotopes: same $Z$ , different $A$ (different neutron number).
Isotopes are chemically identical but differ in mass.

See the full worked example for structure of matter →

Alpha scattering — evidence for the nuclear atom

Match each observation to its deduction: empty space, concentrated positive charge, tiny dense nucleus.

Before 1911 the "plum-pudding" model imagined positive charge spread thinly through the whole atom. Geiger and Marsden, directed by Rutherford, tested this by firing alpha particles (positive, fast, from a radioactive source) at a very thin gold foil in a vacuum, and recording where they went with a movable detector.

Three observations, three conclusions. This is the classic exam question — you must link each observation to the deduction it forces:

Observation	Conclusion
Most alphas pass almost straight through	The atom is mostly empty space
A small fraction are deflected through large angles	There is a concentration of positive charge in a very small region (like charges repel)
A very few ( $\sim 1$ in $8000$ ) bounce back ( $> 90°$ )	The positive charge sits in a tiny, extremely dense, massive nucleus carrying almost all the atom's mass

The plum-pudding model could not explain the large-angle deflections: a diffuse positive charge could never repel a fast alpha strongly enough to turn it right around. Only a small, dense, positively charged nucleus does. This is the birth of the nuclear model of the atom.

Alpha particles fired at a nucleus: most pass straight through the empty space, a few passing close are deflected greatly, and the rare one heading almost straight at the tiny dense nucleus is repelled right back.

Why the experimental details matter: the thin foil means an alpha usually meets only a single layer of atoms, so a large deflection comes from one close encounter with a nucleus rather than a build-up of many small ones. The vacuum stops the alphas being absorbed or scattered by air before they reach the foil and the detector.

Most pass through → atom is mostly empty space.
Few deflected greatly → small, concentrated positive charge.
Very few bounce back → tiny, dense, massive nucleus.
Thin foil → single scattering event; vacuum → alphas not absorbed by air.

See the full worked example for structure of matter →

Estimating nuclear size — closest approach and electron diffraction

Turn KE into electric PE for an upper estimate, or diffract electrons whose wavelength matches the nucleus.

Alpha scattering shows the nucleus is tiny; two methods let us estimate how tiny.

1 — Closest approach (energy method). Fire an alpha head-on at a nucleus. As the positive alpha nears the positive nucleus, electrostatic repulsion slows it down: its kinetic energy is converted into electric potential energy. At the closest point the alpha is momentarily at rest, so all its initial kinetic energy has become electric PE. Setting them equal:

$E_k = \frac{1}{4\pi\varepsilon_0}\,\frac{Qq}{r},$

where $q = 2e$ (the alpha), $Q = Ze$ (the nucleus), and $r$ is the closest distance. Rearranging gives $r$ . Because the alpha never actually touches the nuclear surface, this $r$ is an upper estimate of the nuclear radius (typically a few $\times 10^{-14}\ \text{m}$ ).

2 — Electron diffraction. High-energy electrons fired at nuclei have a de Broglie wavelength (Unit 2) $\lambda = h/p$ comparable to nuclear dimensions, so they diffract around the nucleus and form a diffraction pattern. The angle of the first minimum gives the nuclear diameter. This gives a more direct value than closest approach.

For scale, a nucleus is about $1 \times 10^{-15}\ \text{m}$ across while an atom is about $1 \times 10^{-10}\ \text{m}$ — a ratio of about $10^{5}$ . If the nucleus were a $1\ \text{cm}$ marble, the atom would be about $1\ \text{km}$ wide: the atom is overwhelmingly empty space.

In the closest-approach method the alpha is momentarily at rest at distance r, where its initial kinetic energy has all become electric potential energy. Solving E_k = kQq/r for r gives an upper estimate of the nuclear radius.

Closest approach: $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r}$ with $q = 2e$ , $Q = Ze$ → upper estimate of nuclear radius.
Electron diffraction: $\lambda = h/p \sim$ nuclear size → diffraction pattern gives the radius.
Nucleus $\sim 10^{-15}\ \text{m}$ ; atom $\sim 10^{-10}\ \text{m}$ ; ratio $\sim 10^{5}$ .
Convert any MeV energy to joules before using it.

See the full worked example for structure of matter →

Freeing and accelerating charges — thermionic emission and the electronvolt

Heat a metal to release electrons; a p.d. gives them energy eV = ½mv²; 1 eV = 1.60×10⁻¹⁹ J.

To probe matter we need fast charged particles. First we make some, then we accelerate them.

Thermionic emission. Heating a metal (e.g. a filament) gives some of its conduction electrons enough kinetic energy to escape from the surface. This is thermionic emission, and it is how the electron gun in an accelerator (or old cathode-ray tube) supplies electrons.

Accelerating through a potential difference. A charged particle of charge $q$ released from rest and accelerated through a p.d. $V$ has work done on it by the electric field equal to $qV$ . All of this becomes kinetic energy:

$qV = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{\frac{2qV}{m}}.$

For an electron $q = e$ , so $eV = \tfrac{1}{2}mv^2$ . Note that a positive charge is accelerated along the field, a negative charge (electron) against it — but either way the energy gained is $|q|V$ .

The electronvolt. Because $qV$ is an energy, a convenient unit for tiny energies is the electronvolt: the energy gained by an electron accelerated through one volt.

$1\ \text{eV} = e \times 1\ \text{V} = 1.60 \times 10^{-19}\ \text{J}, \qquad 1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}.$

Particle energies are usually quoted in eV or MeV, but you must convert to joules before putting an energy into $\tfrac{1}{2}mv^2$ or $kQq/r$ .

Thermionic emission: heating a metal releases electrons from its surface.
Accelerated through a p.d.: $qV = \tfrac{1}{2}mv^2$ , so $v = \sqrt{2qV/m}$ .
$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ ; $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ .
Positive charge accelerates along the field, electron against it; energy gained is $|q|V$ either way.

See the full worked example for structure of matter →

Accelerators — the linear accelerator and the cyclotron

LINAC: longer and longer drift tubes with an alternating p.d. Cyclotron: a magnetic field spirals the path while an alternating p.d. accelerates.

A single p.d. can only give so much energy; accelerators apply an accelerating p.d. many times over.

Linear accelerator (LINAC). Charged particles travel in a straight line through a series of hollow drift tubes connected alternately to the two terminals of an alternating (a.c.) supply. The particle is accelerated across each gap between tubes (gaining $eV$ ), but inside a tube there is no field, so it coasts at constant speed while the supply reverses. Timed correctly, the field is always in the accelerating direction each time the particle reaches a gap. Because the particle speeds up at every gap but the supply has a fixed period, it must spend the same time in each tube — so each successive drift tube is longer than the last.

Cyclotron. Two hollow D-shaped electrodes (dees) sit in a uniform magnetic field perpendicular to the plane of the dees. The magnetic force $Bqv$ is always perpendicular to the velocity, so it acts as the centripetal force and bends the particle into a circular arc inside each dee:

$Bqv = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{Bq}.$

Each time the particle crosses the gap between the dees, an alternating p.d. accelerates it (gaining $eV$ ). It then follows a larger-radius arc (because $r \propto v$ ), so the path spirals outwards. Remarkably, the time for one full circle,

$T = \frac{2\pi r}{v} = \frac{2\pi}{v}\cdot\frac{mv}{Bq} = \frac{2\pi m}{Bq},$

is independent of the speed ( $v$ cancels), so a single fixed frequency $f = Bq/2\pi m$ keeps the p.d. in step as the particle speeds up.

In a cyclotron the magnetic field bends the particle into semicircles inside the dees (r = mv/Bq) while the alternating p.d. across the gap accelerates it on every crossing. The radius grows with speed, so the path spirals outwards; the period T = 2πm/Bq is independent of speed.

Why go to such trouble — high energy to see small things. To resolve structure on a length scale $d$ , a probing particle needs a de Broglie wavelength $\lambda \lesssim d$ . By $\lambda = h/p$ , a small wavelength demands a large momentum, and a large momentum means a high kinetic energy. A nucleus is only $\sim 10^{-15}\ \text{m}$ across, so accelerators must give particles very high energies before they can probe it. Smaller wavelength → more detail → higher momentum → higher energy.

LINAC: alternating p.d.; accelerate across gaps; drift tubes get longer as the particle speeds up (fixed supply period).
Cyclotron: magnetic field gives centripetal force ( $r = mv/Bq$ ); alternating p.d. across dees accelerates each crossing; path spirals out.
Cyclotron period $T = 2\pi m/Bq$ is independent of speed → fixed-frequency supply works.
High energy is needed because $\lambda = h/p$ : small $\lambda$ (fine detail) needs large $p$ , hence high energy.

See the full worked example for structure of matter →

Quick recap

$^{A}_{Z}\text{X}$ : $Z$ = protons (element), $A$ = nucleons; neutrons $= A - Z$ ; isotopes share $Z$ but differ in $A$ .
Alpha scattering: most through → empty space; few deflected → concentrated positive charge; very few back → tiny dense massive nucleus.
Nuclear size: closest approach $E_k = kQq/r$ (upper estimate) or electron diffraction ( $\lambda = h/p$ ); nucleus $\sim 10^{-15}\ \text{m}$ vs atom $\sim 10^{-10}\ \text{m}$ .
Thermionic emission frees electrons; a p.d. gives $qV = \tfrac{1}{2}mv^2$ ; $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ — convert to joules before use.
LINAC: alternating p.d. + lengthening drift tubes. Cyclotron: $r = mv/Bq$ , alternating p.d. across dees, spiral path, $T = 2\pi m/Bq$ speed-independent.
High energies are needed because $\lambda = h/p$ : to resolve small structures you need a small wavelength, hence large momentum and high energy.
Show full working, convert MeV → J, use $q = 2e$ for an alpha, and quote answers to 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$^{A}_{Z}\text{X}$ : $A$ = nucleon number, $Z$ = proton number, neutrons $= A - Z$ ; isotopes have same $Z$ , different $A$ .
Alpha scattering: most through → empty space; few deflected → small concentrated positive charge; very few back → tiny dense massive nucleus.
Closest approach: $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r}$ (alpha $q = 2e$ , nucleus $Q = Ze$ ) — upper estimate of nuclear radius.
Accelerating a charge: $qV = \tfrac{1}{2}mv^2$ ; $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .
Cyclotron: $r = \dfrac{mv}{Bq}$ , period $T = \dfrac{2\pi m}{Bq}$ (independent of speed).
de Broglie: $\lambda = \dfrac{h}{p}$ — small wavelength (fine detail) needs large momentum, hence high energy.
Nucleus $\sim 10^{-15}\ \text{m}$ ; atom $\sim 10^{-10}\ \text{m}$ ; ratio $\sim 10^{5}$ ; atom is mostly empty space.

How it’s examined

Structure of Matter sits in Unit 4 (WPH14, Further Mechanics, Fields and Particles), a structured IA2 paper with short-answer, calculation and extended-response items. Typical formats: read a nuclide / count protons, neutrons and electrons (1–2 marks, AO1); define isotopes / the electronvolt / thermionic emission (1–2 marks, AO1); the alpha-scattering question where you must pair each observation with its conclusion (up to 6 marks — the linking is where marks are won or lost); closest-approach and accelerating-p.d. calculations using $E_k = kQq/r$ and $qV = \tfrac{1}{2}mv^2$ (2–4 marks, M and A marks, energies converted to joules); and an extended 'explain the LINAC / cyclotron' or 'why are high energies needed' question drawing on $r = mv/Bq$ and $\lambda = h/p$ (5–6 marks, levels-marked). ALWAYS show the equation, substitution with units, and a 3 s.f. answer; for the alpha-scattering answer, always link observation → deduction rather than merely describing.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 4 (WPH14) examiner reports; Edexcel IAL Physics Unit 4 sample assessment materials and mark schemes. Last reviewed 2026-07-05.

Take this whole topic with you

Step-by-step worked examples — Structure of Matter

Step-by-step solutions to past-paper-style questions on structure of matter, written exactly the way a tutor would explain them at the board.

Question type:

1Reading nuclide notation (2 marks)
Getting startedIdentify & classify• nuclide, AO1, Unit 4
Question
A nucleus is written $^{14}_{6}\text{C}$ . State the number of protons, neutrons and nucleons in this nucleus. (2 marks)
Step-by-step solution
1. Step 1
  In $^{A}_{Z}\text{X}$ the lower number is the proton number $Z$ and the upper number is the nucleon (mass) number $A$ . Here $Z = 6$ and $A = 14$ .
2. Step 2
  Number of protons $= Z = 6$ . Number of nucleons $= A = 14$ . Number of neutrons $= A - Z = 14 - 6 = 8$ .
  $N = A - Z = 14 - 6 = 8$
Answer
Protons $= 6$ , neutrons $= 8$ , nucleons (protons + neutrons) $= 14$ .
Examiner tip
Mark scheme: (1) protons $= 6$ and nucleons $= 14$ read correctly from $Z$ and $A$ ; (2) neutrons $= A - Z = 8$ . The commonest slip is quoting $A = 14$ as the neutron number — neutrons are $A - Z$ , not $A$ .
2Identifying isotopes (2 marks)
Getting startedIdentify & classify• isotopes, AO1, Unit 4
Question
Three nuclei are $^{35}_{17}\text{Cl}$ , $^{37}_{17}\text{Cl}$ and $^{37}_{18}\text{Ar}$ . State, with a reason, which two are isotopes of the same element. (2 marks)
Step-by-step solution
1. Step 1
  Isotopes are nuclei of the same element — they have the same proton number $Z$ — but different nucleon numbers $A$ (different numbers of neutrons).
2. Step 2
  $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ both have $Z = 17$ , so they are the same element (chlorine), but $A = 35$ and $A = 37$ differ. $^{37}_{18}\text{Ar}$ has $Z = 18$ , a different element, even though its $A$ matches.
Answer
$^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ are isotopes: same proton number ( $Z = 17$ ) but different nucleon numbers (35 and 37).
Examiner tip
Mark scheme: (1) correct pair identified; (2) reason — same $Z$ (same element), different $A$ / different neutron number. Choosing the two with equal $A$ ( $^{37}\text{Cl}$ and $^{37}\text{Ar}$ ) is the trap: isotopes share $Z$ , not $A$ .
3Speed of an electron accelerated through a p.d. (3 marks)
Getting startedDirect calculation• thermionic emission, electronvolt, AO2, Unit 4
Question
An electron is released from rest at a hot filament and accelerated through a potential difference of $2500\ \text{V}$ . Calculate its final speed. (electron mass $= 9.11 \times 10^{-31}\ \text{kg}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ ) (3 marks)
Step-by-step solution
1. Step 1
  The work done by the field on the charge equals the kinetic energy gained. For a charge $e$ moving through a p.d. $V$ : $eV = \tfrac{1}{2}mv^2$ .
  $eV = \tfrac{1}{2}mv^2$
2. Step 2
  Rearrange for the speed $v$ .
  $v = \sqrt{\dfrac{2eV}{m}}$
3. Step 3
  Substitute and evaluate to 3 significant figures.
  $v = \sqrt{\dfrac{2 \times (1.60 \times 10^{-19}) \times 2500}{9.11 \times 10^{-31}}} = 2.96 \times 10^{7}\ \text{m s}^{-1}$
Answer
$v = 2.96 \times 10^{7}\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) $eV = \tfrac{1}{2}mv^2$ ; (2) correct rearrangement; (3) $2.96 \times 10^{7}\ \text{m s}^{-1}$ . Forgetting the factor of 2, or using $e = 2500$ instead of $1.60 \times 10^{-19}$ , are the usual errors.
4Converting an energy in electronvolts to joules (3 marks)
Building confidenceDirect calculation• electronvolt, energy, AO2, Unit 4
Question
An alpha particle has kinetic energy $5.0\ \text{MeV}$ . (a) Convert this energy to joules. (b) Given the alpha particle has mass $6.64 \times 10^{-27}\ \text{kg}$ , calculate its speed. ( $e = 1.60 \times 10^{-19}\ \text{C}$ ) (3 marks)
Step-by-step solution
1. Step 1
  (a) One electronvolt is the energy gained by an electron moving through $1\ \text{V}$ : $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . A megaelectronvolt is $10^{6}\ \text{eV}$ .
  $E = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-13}\ \text{J}$
2. Step 2
  (b) Use $E_k = \tfrac{1}{2}mv^2$ and rearrange for the speed.
  $v = \sqrt{\dfrac{2E_k}{m}}$
3. Step 3
  Substitute the energy in joules (not eV) and evaluate.
  $v = \sqrt{\dfrac{2 \times 8.0 \times 10^{-13}}{6.64 \times 10^{-27}}} = 1.55 \times 10^{7}\ \text{m s}^{-1}$
Answer
(a) $E = 8.0 \times 10^{-13}\ \text{J}$ . (b) $v = 1.55 \times 10^{7}\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (a) multiply by $1.60 \times 10^{-19}$ and by $10^{6}$ → $8.0 \times 10^{-13}\ \text{J}$ ; (b) $E_k = \tfrac{1}{2}mv^2$ used with the energy in joules. Substituting $5.0 \times 10^{6}$ straight into $\tfrac{1}{2}mv^2$ (energy left in eV) is the classic fatal slip.
5Radius of a proton's path in a cyclotron (3 marks)
Building confidenceMulti-step problem• cyclotron, magnetic field, AO2, Unit 4
Question
A proton (mass $1.67 \times 10^{-27}\ \text{kg}$ , charge $1.60 \times 10^{-19}\ \text{C}$ ) moves at $1.5 \times 10^{7}\ \text{m s}^{-1}$ inside a cyclotron whose magnetic field has magnitude $0.50\ \text{T}$ . Calculate the radius of its circular path. (3 marks)
Step-by-step solution
1. Step 1
  The magnetic force $Bqv$ provides the centripetal force $\dfrac{mv^2}{r}$ . Equating and rearranging gives the radius of the path.
  $Bqv = \dfrac{mv^2}{r} \implies r = \dfrac{mv}{Bq}$
2. Step 2
  Substitute the proton values.
  $r = \dfrac{(1.67 \times 10^{-27})(1.5 \times 10^{7})}{(0.50)(1.60 \times 10^{-19})}$
3. Step 3
  Evaluate to 3 significant figures.
  $r = 0.313\ \text{m}$
Answer
$r = 0.313\ \text{m}$ (about $31\ \text{cm}$ ).
Examiner tip
Mark scheme: (1) $Bqv = mv^2/r$ leading to $r = mv/Bq$ ; (2) correct substitution; (3) $0.313\ \text{m}$ . Note one power of $v$ cancels — a common error is leaving $v^2$ in and getting the wrong power. Faster particles travel on larger-radius arcs, which is why the cyclotron path spirals outwards.
6Why the cyclotron frequency is independent of speed (4 marks)
Building confidenceShow that / prove• cyclotron, period, AO2, Unit 4
Question
In a cyclotron a proton follows a circular path of radius $r = mv/Bq$ . (a) Show that the time for one complete revolution is $T = \dfrac{2\pi m}{Bq}$ . (b) Hence calculate the frequency of the alternating p.d. needed for a proton in a field of $0.80\ \text{T}$ . ( $m_p = 1.67 \times 10^{-27}\ \text{kg}$ , $q = 1.60 \times 10^{-19}\ \text{C}$ ) (4 marks)
Step-by-step solution
1. Step 1
  (a) The period is the circumference divided by the speed: $T = \dfrac{2\pi r}{v}$ .
  $T = \dfrac{2\pi r}{v}$
2. Step 2
  Substitute $r = \dfrac{mv}{Bq}$ . The speed $v$ cancels, so the period does not depend on the speed (at non-relativistic energies).
  $T = \dfrac{2\pi}{v} \times \dfrac{mv}{Bq} = \dfrac{2\pi m}{Bq}$
3. Step 3
  (b) The required frequency is $f = \dfrac{1}{T} = \dfrac{Bq}{2\pi m}$ .
  $f = \dfrac{Bq}{2\pi m}$
4. Step 4
  Substitute and evaluate.
  $f = \dfrac{(0.80)(1.60 \times 10^{-19})}{2\pi (1.67 \times 10^{-27})} = 1.22 \times 10^{7}\ \text{Hz}$
Answer
(a) $T = \dfrac{2\pi m}{Bq}$ , independent of $v$ . (b) $f = 1.22 \times 10^{7}\ \text{Hz}$ ( $\approx 12.2\ \text{MHz}$ ).
Examiner tip
Mark scheme: (a) $T = 2\pi r/v$ and substitute $r = mv/Bq$ so $v$ cancels (2 marks); (b) $f = 1/T = Bq/2\pi m$ and $1.22 \times 10^{7}\ \text{Hz}$ (2 marks). The key physical point — that $T$ is speed-independent — is why a fixed-frequency alternating p.d. keeps a proton in step as it spirals out.
7Closest approach of an alpha particle to a nucleus (4 marks)
StretchMulti-step problem• closest approach, Rutherford, AO2, Unit 4
Question
A head-on alpha particle of kinetic energy $5.0\ \text{MeV}$ approaches a gold nucleus ( $Z = 79$ ). Estimate the closest distance of approach. (charge on alpha $= 2e$ , $e = 1.60 \times 10^{-19}\ \text{C}$ , $\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}$ ) (4 marks)
Step-by-step solution
1. Step 1
  At the closest approach the alpha momentarily stops: all its initial kinetic energy has become electric potential energy in the field of the nucleus. Set $E_k$ equal to the electrostatic PE.
  $E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r}$
2. Step 2
  Convert the kinetic energy to joules.
  $E_k = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-13}\ \text{J}$
3. Step 3
  Rearrange for the closest distance $r$ .
  $r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{E_k}$
4. Step 4
  Substitute ( $2e = 3.20 \times 10^{-19}\ \text{C}$ , $Ze = 79 \times 1.60 \times 10^{-19}\ \text{C}$ ) and evaluate.
  $r = \dfrac{(8.99 \times 10^{9})(3.20 \times 10^{-19})(79 \times 1.60 \times 10^{-19})}{8.0 \times 10^{-13}} = 4.55 \times 10^{-14}\ \text{m}$
Answer
$r \approx 4.55 \times 10^{-14}\ \text{m}$ (about $4.5 \times 10^{-14}\ \text{m}$ , a few $\times 10^{-14}\ \text{m}$ ).
Examiner tip
Mark scheme: (1) equate initial KE to electric PE $= kQq/r$ ; (2) KE in joules; (3)/(4) rearrange and $r \approx 4.5 \times 10^{-14}\ \text{m}$ . This value is an upper estimate of the nuclear radius — the alpha never quite reaches the surface. Leaving the energy in MeV, or using $e$ instead of $2e$ for the alpha, are the common errors.
8de Broglie wavelength and probing the nucleus (5 marks)
StretchMulti-step problem• de Broglie, electron diffraction, AO2, Unit 4
Question
To resolve detail on the scale of a nucleus, a probing particle must have a de Broglie wavelength of about $1.0 \times 10^{-15}\ \text{m}$ . (a) Calculate the momentum a particle needs for this wavelength. (b) An electron of mass $9.11 \times 10^{-31}\ \text{kg}$ is accelerated through $4000\ \text{V}$ ; calculate its de Broglie wavelength and comment on whether it can resolve nuclear-scale detail. ( $h = 6.63 \times 10^{-34}\ \text{J s}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ ) (5 marks)
Step-by-step solution
1. Step 1
  (a) The de Broglie relation is $\lambda = \dfrac{h}{p}$ , so the required momentum is $p = \dfrac{h}{\lambda}$ .
  $p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{1.0 \times 10^{-15}} = 6.63 \times 10^{-19}\ \text{kg m s}^{-1}$
2. Step 2
  (b) First find the electron's speed from $eV = \tfrac{1}{2}mv^2$ .
  $v = \sqrt{\dfrac{2eV}{m}} = \sqrt{\dfrac{2(1.60 \times 10^{-19})(4000)}{9.11 \times 10^{-31}}} = 3.75 \times 10^{7}\ \text{m s}^{-1}$
3. Step 3
  Its momentum is $p = mv$ .
  $p = (9.11 \times 10^{-31})(3.75 \times 10^{7}) = 3.42 \times 10^{-23}\ \text{kg m s}^{-1}$
4. Step 4
  Its de Broglie wavelength is $\lambda = h/p$ .
  $\lambda = \dfrac{6.63 \times 10^{-34}}{3.42 \times 10^{-23}} = 1.94 \times 10^{-11}\ \text{m}$
5. Step 5
  This wavelength ( $\sim 10^{-11}\ \text{m}$ ) is about $10^{4}$ times larger than a nucleus ( $\sim 10^{-15}\ \text{m}$ ), so a $4\ \text{keV}$ electron cannot resolve nuclear detail. To reach $\lambda \sim 10^{-15}\ \text{m}$ the electron needs a much higher momentum, hence far higher energy — which is why particle accelerators are used.
Answer
(a) $p = 6.63 \times 10^{-19}\ \text{kg m s}^{-1}$ . (b) $\lambda = 1.94 \times 10^{-11}\ \text{m}$ — far too long to resolve nuclear detail, so much higher energy (larger momentum, smaller $\lambda$ ) is required.
Examiner tip
Mark scheme: (a) $p = h/\lambda = 6.63 \times 10^{-19}\ \text{kg m s}^{-1}$ ; (b) $v = 3.75 \times 10^{7}\ \text{m s}^{-1}$ , $p = mv$ , $\lambda = 1.94 \times 10^{-11}\ \text{m}$ ; comment that this $\gg$ nuclear size so higher energy is needed. The link 'smaller wavelength → more detail → higher momentum → higher energy' is the examined idea, not just the numbers.
9Interpreting the alpha-scattering results (6 marks)
StretchShow that / prove• Adapted from Edexcel IAL Physics Unit 4• Rutherford, nuclear model, AO1, Unit 4
Question
In the alpha-scattering experiment, alpha particles are fired at a thin gold foil. Most pass almost straight through, a small fraction are deflected through large angles, and a very few ( $\sim 1$ in $8000$ ) bounce back through more than $90°$ . State the conclusion drawn from each of these three observations. (6 marks)
Step-by-step solution
1. Step 1
  Observation 1 — most pass almost straight through with little or no deflection.
2. Step 2
  Conclusion 1: the atom is mostly empty space; the alpha particles travel through the gaps in the foil without meeting anything to deflect them.
3. Step 3
  Observation 2 — a small fraction are deflected through large angles.
4. Step 4
  Conclusion 2: there is a concentration of positive charge (the alpha is positive and is repelled), and this charge occupies a very small region — only alphas passing close to it are deflected strongly.
5. Step 5
  Observation 3 — a very few bounce straight back (deflected through more than $90°$ ).
6. Step 6
  Conclusion 3: the positive charge is carried by a tiny, extremely dense, massive core — the nucleus — able to repel a fast alpha completely; this also shows almost all of the atom's mass is concentrated there.
Answer
Most pass straight through → atom is mostly empty space. Small fraction deflected greatly → small, concentrated positive charge. Very few bounce back → the positive charge sits in a tiny, dense, massive nucleus.
Examiner tip
Top-band answer: each mark pairs an observation with its specific deduction (empty space; concentrated positive charge; tiny dense massive nucleus). Merely describing the observations, or listing conclusions without linking them, is capped — the examiner wants observation → deduction each time.

Model Answers — Structure of Matter

High-scoring sample answers for structure of matter on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Explain what is meant by two nuclei being isotopes of the same element.
Model answer
Isotopes are nuclei of the same element, so they have the same number of protons (the same proton number $Z$ ).

They differ in their number of neutrons, and therefore have different nucleon (mass) numbers $A$ . For example $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ both have $17$ protons but $18$ and $20$ neutrons respectively.
Why this scores
Why this scores 2/2. (1) same proton number / same number of protons (same element); (2) different number of neutrons / different nucleon number. Saying only 'different mass' is too vague — you must attribute the difference to neutrons and the sameness to protons.
Question 2
2 marks
Q (2 marks). A neutral atom of an element is represented by $^{40}_{19}\text{K}$ . State the number of protons, neutrons and electrons in this neutral atom.
Model answer
The lower number is the proton number, $Z = 19$ , and the upper number is the nucleon number, $A = 40$ .

Protons $= Z = 19$ .

Neutrons $= A - Z = 40 - 19 = 21$ .

Electrons $= 19$ (a neutral atom has equal numbers of protons and electrons).
Why this scores
Why this scores 2/2. (1) protons $= 19$ and neutrons $= 21$ (using $A - Z$ ); (2) electrons $= 19$ from neutrality. A frequent slip is giving neutrons $= 40$ — that is the nucleon number, not the neutron number.
Question 3
3 marks
Q (3 marks). (a) Define the electronvolt. (b) State what is meant by thermionic emission.
Model answer
(a) One electronvolt is the energy transferred to (or gained by) an electron when it is accelerated through a potential difference of one volt. In joules, $1\ \text{eV} = e \times 1\ \text{V} = 1.60 \times 10^{-19}\ \text{J}$ .

(b) Thermionic emission is the release of electrons from the surface of a heated metal: heating gives some conduction electrons enough kinetic energy to escape from the metal surface.
Why this scores
Why this scores 3/3. (a) 2 marks — energy gained by an electron through a p.d. of $1\ \text{V}$ (1), equals $1.60 \times 10^{-19}\ \text{J}$ (1). (b) 1 mark — electrons freed from a heated metal. Defining the eV as 'a small unit of energy' with no reference to charge $\times$ p.d. earns nothing.
Question 4
4 marks
Q (4 marks). The radius of a typical atom is about $1 \times 10^{-10}\ \text{m}$ and the radius of its nucleus is about $1 \times 10^{-15}\ \text{m}$ . (a) Calculate the ratio of the atomic radius to the nuclear radius. (b) State what this tells you about the structure of the atom, and identify which particles occupy the space outside the nucleus.
Model answer
(a) Ratio: $\frac{r_{atom}}{r_{nucleus}} = \frac{1 \times 10^{-10}}{1 \times 10^{-15}} = 1 \times 10^{5}$ The atom is about 100 000 times wider than its nucleus.

(b) Because the nucleus is so tiny compared with the whole atom, the atom is mostly empty space. Almost all of the atom's mass and all of its positive charge are concentrated in the central nucleus (protons and neutrons), while the vastly larger surrounding volume is occupied by the electrons.
Why this scores
Why this scores 4/4. (a) 2 marks — ratio $= 10^{5}$ (correct powers of ten). (b) 2 marks — atom is mostly empty space / nucleus holds the mass and positive charge; electrons occupy the outer space. A common error is a ratio of $10^{15}/10^{10}$ inverted or mis-subtracted powers.
Question 5
5 marks
Q (5 marks). In Rutherford's alpha-scattering experiment, most alpha particles passed straight through the gold foil but a very small number were deflected through angles greater than $90°$ . (a) State the conclusion drawn from each observation. (b) Explain why a thin foil and a vacuum were used.
Model answer
(a)

Most pass straight through: the atom is mostly empty space, so the alphas travel through without being deflected.

A very few bounce back ( $> 90°$ ): there is a tiny, dense, positively charged nucleus — only an alpha meeting this concentrated positive charge head-on is repelled strongly enough to reverse. This also shows nearly all the atom's mass is concentrated there.

(b) A thin foil ensures the alphas usually pass through only a single layer of atoms, so any large deflection is due to one close encounter with a nucleus (not a build-up of many small deflections). The vacuum stops the alpha particles from being absorbed or scattered by air molecules before they reach the foil and detector.
Why this scores
Why this scores 5/5. (a) 3 marks — empty space (1); small dense positive nucleus (1); links each conclusion to its observation (1). (b) 2 marks — thin foil → single scattering event (1); vacuum → alphas not absorbed/scattered by air (1). The discriminator in (a) is pairing each observation with its specific deduction, not just listing conclusions.
Question 6
5 marks
Q (5 marks). A linear accelerator (LINAC) accelerates protons using a series of cylindrical drift tubes connected to an alternating potential difference. (a) Explain how the protons are accelerated. (b) Explain why the drift tubes are made progressively longer along the accelerator.
Model answer
(a) The drift tubes are connected alternately to the two terminals of an alternating (a.c.) supply. A proton is accelerated across the gap between two tubes by the electric field there, gaining kinetic energy ( $eV$ per gap). While the proton is inside a drift tube there is no field, so it moves at constant speed (the tube shields it). The supply reverses while the proton is shielded inside the tube, so that when it reaches the next gap the field is again in the direction that accelerates it. Repeating this over many gaps raises the proton to high energy.

(b) Each time the proton crosses a gap it gains energy and so moves faster. Because the alternating supply has a fixed period, the proton must spend the same time in each tube for the field to be correctly reversed at every gap. Travelling faster in the same time means it covers a greater distance, so each successive drift tube must be longer than the last.
Why this scores
Why this scores 5/5. (a) 3 marks — alternating p.d.; accelerated across gaps (gains $eV$ ); shielded/no field inside tubes while the supply reverses. (b) 2 marks — proton speeds up each gap; fixed period means equal time per tube, so higher speed → longer tube. Saying the field accelerates the proton inside the tube is the level-capping error — the tube shields it; acceleration happens only in the gaps.
Question 7
5 marks
Q (5 marks). An electron gun uses thermionic emission to release electrons, which are then accelerated from rest through a potential difference of $3.0 \times 10^{3}\ \text{V}$ . (a) Calculate the kinetic energy gained, in joules. (b) Calculate the final speed of an electron. ( $m_e = 9.11 \times 10^{-31}\ \text{kg}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ )
Model answer
(a) The work done by the field equals the kinetic energy gained: $E_k = eV$ . $E_k = (1.60 \times 10^{-19})(3.0 \times 10^{3}) = 4.8 \times 10^{-16}\ \text{J}$ (Equivalently this is $3.0 \times 10^{3}\ \text{eV} = 3.0\ \text{keV}$ .)

(b) Using $E_k = \tfrac{1}{2}m v^2$ , rearrange for the speed: $v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 4.8 \times 10^{-16}}{9.11 \times 10^{-31}}} = 3.25 \times 10^{7}\ \text{m s}^{-1}$
Why this scores
Why this scores 5/5. (a) 2 marks — $E_k = eV$ (1), $= 4.8 \times 10^{-16}\ \text{J}$ (1). (b) 3 marks — $E_k = \tfrac{1}{2}mv^2$ (1), rearranged to $v = \sqrt{2E_k/m}$ (1), $3.25 \times 10^{7}\ \text{m s}^{-1}$ (1). Using the energy in eV inside $\tfrac{1}{2}mv^2$ , or dropping the factor of 2, are the marked errors.
Question 8
6 marks
Q (6 marks). A cyclotron uses a uniform magnetic field and an alternating potential difference across two hollow 'dees'. (a) Explain the roles of the magnetic field and of the alternating p.d. (b) Explain why the frequency of the alternating p.d. can be kept constant even though the proton speeds up. (This question is marked using levels.)
Model answer
(a) The magnetic field is perpendicular to the plane of the dees. It exerts a force $Bqv$ on the moving proton that is always perpendicular to its velocity, providing the centripetal force that makes the proton follow a circular arc inside each dee ( $Bqv = mv^2/r$ ). Inside a hollow dee there is no electric field, so the proton's speed is unchanged there — only its direction changes.

The alternating p.d. is applied across the gap between the two dees. Each time the proton crosses the gap it is accelerated by the electric field and gains kinetic energy ( $eV$ ). Because $r = mv/Bq$ , a faster proton follows a larger radius, so the path spirals outwards as the proton gains energy.

(b) The time for one full circle is $T = \frac{2\pi r}{v} = \frac{2\pi}{v}\cdot\frac{mv}{Bq} = \frac{2\pi m}{Bq}$ The speed $v$ cancels, so the period (and hence the time to complete each half-circle inside a dee) is the same regardless of how fast the proton is moving (at non-relativistic speeds). The proton therefore always arrives back at the gap in step with the supply, so a single fixed frequency $f = Bq/2\pi m$ keeps accelerating it on every crossing — the p.d. does not need to change as the proton spirals out and speeds up.
Why this scores
Why this reaches the top band (5–6/6). (a) magnetic field provides centripetal force / circular path (no speed change in dees); alternating p.d. accelerates across the gap; radius grows so path spirals out. (b) derive $T = 2\pi m/Bq$ and state $v$ cancels → period independent of speed → fixed frequency works. Continuous linked reasoning that names the centripetal role of the field and shows $v$ cancelling is what distinguishes the top level from a bare description.
Question 9
6 marks
Q (6 marks). Explain, with reference to the de Broglie relation, why particles must be accelerated to very high energies in order to investigate the structure of the nucleus. Include an estimate of the wavelength required.
Model answer
To 'see' or resolve detail on a given scale, the wavelength of the probing radiation or particle must be comparable to or smaller than the size of that detail (just as a microscope cannot resolve features smaller than the wavelength of the light it uses). A nucleus is only about $1 \times 10^{-15}\ \text{m}$ across, so a probing particle must have a de Broglie wavelength of roughly $\lambda \sim 1 \times 10^{-15}\ \text{m}.$

By the de Broglie relation $\lambda = \dfrac{h}{p}$ , a small wavelength requires a large momentum $p$ . Since $h$ is fixed, halving the wavelength doubles the required momentum. A momentum this large corresponds to a very high kinetic energy, which is why the particles (e.g. electrons or protons) must be accelerated in a LINAC or cyclotron to very high energies before they can resolve nuclear-scale structure.

Quantitatively, $p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{1 \times 10^{-15}} \approx 6.6 \times 10^{-19}\ \text{kg m s}^{-1}$ — an enormous momentum for a subatomic particle, confirming that only high-energy accelerators can probe the nucleus.
Why this scores
Why this scores 6/6. Resolution needs $\lambda \le$ feature size (1); nucleus $\sim 10^{-15}\ \text{m}$ so $\lambda \sim 10^{-15}\ \text{m}$ (1); quote/apply $\lambda = h/p$ (1); smaller $\lambda$ → larger $p$ (1) → higher energy (1); supporting calculation $p \approx 6.6 \times 10^{-19}\ \text{kg m s}^{-1}$ (1). The chain 'small wavelength → large momentum → high energy' is the examined argument.
Question 10
6 marks
Q (6 marks). An alpha particle of kinetic energy $6.0\ \text{MeV}$ makes a head-on approach to a gold nucleus ( $Z = 79$ ). (a) Explain the energy transfer that takes place as the alpha approaches and then recedes. (b) Estimate the closest distance of approach. ( $e = 1.60 \times 10^{-19}\ \text{C}$ , $\dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}$ )
Model answer
(a) As the positive alpha particle approaches the positive nucleus, the electrostatic repulsion does negative work on it: the alpha slows down, so its kinetic energy is transferred to electric potential energy. At the closest approach the alpha is momentarily at rest, and all its initial kinetic energy has become electric potential energy. As it then recedes, the repulsion pushes it away, converting the potential energy back into kinetic energy, so it leaves with the same speed it arrived with (energy is conserved).

(b) At the closest approach, initial KE $=$ electric PE: $E_k = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r}$ Convert the energy: $E_k = 6.0 \times 10^{6} \times 1.60 \times 10^{-19} = 9.6 \times 10^{-13}\ \text{J}$ . Rearranging for $r$ : $r = \frac{(8.99 \times 10^{9})(2 \times 1.60 \times 10^{-19})(79 \times 1.60 \times 10^{-19})}{9.6 \times 10^{-13}} = 3.79 \times 10^{-14}\ \text{m}$ So the closest approach is about $3.8 \times 10^{-14}\ \text{m}$ , which is an upper estimate of the gold nuclear radius.
Why this scores
Why this scores 6/6. (a) 3 marks — KE → electric PE as it approaches (1); momentarily at rest / all KE becomes PE at closest approach (1); PE → KE on receding, leaves at same speed / energy conserved (1). (b) 3 marks — equate KE to $kQq/r$ with alpha charge $2e$ (1); convert MeV to J (1); $r \approx 3.8 \times 10^{-14}\ \text{m}$ (1). Using $e$ rather than $2e$ for the alpha, or leaving the energy in MeV, are the usual fatal errors.

Key Formulae — Structure of Matter

The formulae you need to memorise for structure of matter on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Nuclide (isotope) notation
$^{A}_{Z}\text{X}, \quad N = A - Z$
$A$
nucleon (mass) number = protons + neutrons
$Z$
proton (atomic) number = number of protons
$N$
number of neutrons $= A - Z$
$\text{X}$
chemical symbol of the element
When to use
To read or write a nucleus. $Z$ fixes the element; isotopes share $Z$ but differ in $A$ (hence in $N = A - Z$ ).
Example
$^{14}_{6}\text{C}$ has $6$ protons, $14$ nucleons and $14 - 6 = 8$ neutrons.
Energy gained accelerating a charge through a p.d.
$eV = \tfrac{1}{2}mv^2$
$e$
charge (magnitude) of the particle / $\text{C}$
$V$
accelerating potential difference / $\text{V}$
$m$
mass of the particle / $\text{kg}$
$v$
final speed / $\text{m s}^{-1}$
When to use
When a charged particle starting from rest is accelerated through a p.d. (thermionic gun, LINAC gap, cyclotron dee gap). Work done $= eV$ becomes kinetic energy. Rearrange to $v = \sqrt{2eV/m}$ .
Example
An electron through $2500\ \text{V}$ : $v = \sqrt{2(1.60 \times 10^{-19})(2500)/(9.11 \times 10^{-31})} = 2.96 \times 10^{7}\ \text{m s}^{-1}$ .
The electronvolt
$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$
$1\ \text{eV}$
energy gained by an electron (charge $e$ ) accelerated through $1\ \text{V}$
$1\ \text{MeV}$
$10^{6}\ \text{eV} = 1.60 \times 10^{-13}\ \text{J}$
When to use
To convert between particle energies quoted in eV/MeV and joules. Always convert to joules before using energies in $\tfrac{1}{2}mv^2$ or $kQq/r$ .
Example
$5.0\ \text{MeV} = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-13}\ \text{J}$ .
Radius of a charged particle in a magnetic field
$r = \dfrac{mv}{Bq}$
$r$
radius of the circular path / $\text{m}$
$m$
mass of the particle / $\text{kg}$
$v$
speed of the particle / $\text{m s}^{-1}$
$B$
magnetic flux density / $\text{T}$
$q$
charge on the particle / $\text{C}$
When to use
For the circular motion inside a cyclotron (or any $Bqv = mv^2/r$ situation). Faster particles curve on larger radii, so the path spirals outwards. The period $T = 2\pi m/Bq$ is independent of $v$ .
Example
A proton at $1.5 \times 10^{7}\ \text{m s}^{-1}$ in $0.50\ \text{T}$ : $r = (1.67 \times 10^{-27})(1.5 \times 10^{7})/[(0.50)(1.60 \times 10^{-19})] = 0.313\ \text{m}$ .
Closest approach (energy conservation)
$E_k = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r}$
$E_k$
initial kinetic energy of the incoming charge / $\text{J}$
$Q$
charge of the target nucleus ( $= Ze$ ) / $\text{C}$
$q$
charge of the incoming particle (alpha $= 2e$ ) / $\text{C}$
$r$
closest distance of approach / $\text{m}$
When to use
For a head-on approach: at the closest point the incoming particle stops, so all its initial KE has become electric PE. Gives an upper estimate of nuclear radius.
Example
A $5.0\ \text{MeV}$ alpha to gold ( $Z = 79$ ): $r = (8.99 \times 10^{9})(2e)(79e)/(8.0 \times 10^{-13}) = 4.55 \times 10^{-14}\ \text{m}$ .
de Broglie wavelength
$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$
$\lambda$
de Broglie wavelength / $\text{m}$
$h$
Planck constant $= 6.63 \times 10^{-34}\ \text{J s}$
$p$
momentum of the particle / $\text{kg m s}^{-1}$
When to use
To find the wavelength associated with a moving particle and to explain electron diffraction of the nucleus. Smaller $\lambda$ resolves finer detail, so probing small structures needs large momentum → high energy.
Example
A $4\ \text{keV}$ electron ( $p = 3.42 \times 10^{-23}\ \text{kg m s}^{-1}$ ) has $\lambda = 6.63 \times 10^{-34}/3.42 \times 10^{-23} = 1.94 \times 10^{-11}\ \text{m}$ .

Key Definitions and Keywords — Structure of Matter

Definitions to memorise and the exact keywords mark schemes credit for structure of matter answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Nucleon (mass) number, A
Examiner keyword
The total number of nucleons — protons plus neutrons — in a nucleus. It is the upper number in the notation $^{A}_{Z}\text{X}$ .
Proton (atomic) number, Z
Examiner keyword
The number of protons in a nucleus. It fixes which element the atom is and is the lower number in $^{A}_{Z}\text{X}$ . In a neutral atom it also equals the number of electrons.
Nucleon
Examiner keyword
A particle found in the nucleus — a proton or a neutron. The nucleon number $A$ counts them together.
Isotope
Examiner keyword
One of two or more nuclei of the same element (same proton number $Z$ ) that have different numbers of neutrons and hence different nucleon numbers $A$ .
Nuclide
Examiner keyword
A particular type of nucleus specified by its proton number and nucleon number, written $^{A}_{Z}\text{X}$ .
Nuclear model of the atom
Examiner keyword
The model, established by alpha-scattering, in which almost all the mass and all the positive charge of an atom are concentrated in a tiny central nucleus, with electrons occupying the much larger surrounding space; the atom is mostly empty space.
Alpha-particle scattering experiment
Examiner keyword
The Geiger–Marsden experiment in which alpha particles are fired at a thin metal foil: most pass straight through, a few deflect greatly, and a very few bounce back — providing evidence for the nuclear model of the atom.
Closest distance of approach
Examiner keyword
The least separation reached by a charged particle heading straight for a nucleus, found by equating its initial kinetic energy to the electric potential energy $\frac{1}{4\pi\varepsilon_0}\frac{Qq}{r}$ . It gives an upper estimate of the nuclear radius.
Thermionic emission
Examiner keyword
The release of electrons from the surface of a hot metal, caused by heating giving some electrons enough kinetic energy to escape. It is the electron source in an electron gun.
Electronvolt (eV)
Examiner keyword
The energy transferred to an electron when it is accelerated through a potential difference of one volt: $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .
Linear accelerator (LINAC)
Examiner keyword
An accelerator in which charged particles are accelerated in a straight line across the gaps between a series of drift tubes connected to an alternating p.d.; the tubes are made progressively longer as the particles speed up.
Cyclotron
Examiner keyword
An accelerator in which a uniform magnetic field bends charged particles into a spiral inside two hollow dees, while an alternating p.d. across the gap accelerates them on each crossing. The period $T = 2\pi m/Bq$ is independent of speed.
de Broglie wavelength
Examiner keyword
The wavelength $\lambda = h/p$ associated with a moving particle of momentum $p$ . Smaller wavelengths (larger momenta) are needed to resolve smaller structures, e.g. by electron diffraction.
Electron diffraction (of the nucleus)
Examiner keyword
A method of estimating nuclear size by firing high-energy electrons at nuclei; their de Broglie wavelength is comparable to nuclear dimensions, so a diffraction pattern forms from which the nuclear radius can be found.
Magnetic centripetal force
The force $Bqv$ on a charged particle moving perpendicular to a magnetic field; being always perpendicular to the velocity, it provides the centripetal force for circular motion, giving $r = mv/Bq$ .

Common Mistakes and Misconceptions — Structure of Matter

The traps other students keep falling into on structure of matter questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Reading the nucleon number $A$ as the number of neutrons
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
$A$ is the larger number in $^{A}_{Z}\text{X}$ , so students assume it is the neutron count.
How to avoid it
$A$ is protons plus neutrons. The number of neutrons is $N = A - Z$ . For $^{14}_{6}\text{C}$ , neutrons $= 14 - 6 = 8$ , not $14$ .
✕Describing the alpha-scattering observations without linking each to its specific conclusion
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students list what was seen and separately list conclusions, instead of pairing observation with deduction.
How to avoid it
Match each observation to its deduction: most pass straight through → mostly empty space; small fraction deflected greatly → concentrated positive charge; very few bounce back → tiny, dense, massive nucleus. Examiners award marks for the linked reasoning, not the description alone.
✕Substituting an energy in eV or MeV directly into $\tfrac{1}{2}mv^2$ or $kQq/r$
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
The energy is quoted in MeV, so students forget it is not already in joules.
How to avoid it
Convert first: $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ , so $5.0\ \text{MeV} = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-13}\ \text{J}$ . Only use joules inside energy equations.
✕Using the charge of the alpha particle as $e$ instead of $2e$ in closest-approach calculations
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students use the elementary charge $e$ automatically, forgetting the alpha is a helium nucleus with two protons.
How to avoid it
The alpha particle carries charge $q = 2e = 3.20 \times 10^{-19}\ \text{C}$ , and a nucleus of proton number $Z$ carries $Q = Ze$ . Put both correct charges into $E_k = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{r}$ .
✕Confusing the sign of the charge / direction of conventional current when accelerating particles
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Electrons are negative, so the direction they are pushed is opposite to the electric field and to conventional current; students mix these up.
How to avoid it
A field accelerates a positive charge along the field direction and a negative charge (electron) against it. Whatever the sign, the energy gained crossing a p.d. $V$ is $|q|V$ ; conventional current flows from + to –, opposite to the electron drift.
✕Failing to explain why high energies are needed by linking it to de Broglie wavelength
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students state that 'more energy means more detail' without invoking $\lambda = h/p$ .
How to avoid it
Argue the chain explicitly: to resolve a small structure you need a small wavelength; by $\lambda = h/p$ a small wavelength needs a large momentum, which means a high energy. That is why accelerators are used to probe the nucleus.

Structure of Matter — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Structure of Matter

Short Notes - Structure of Matter

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Reading nuclide notation (2 marks)

2Identifying isotopes (2 marks)

3Speed of an electron accelerated through a p.d. (3 marks)

4Converting an energy in electronvolts to joules (3 marks)

5Radius of a proton's path in a cyclotron (3 marks)

6Why the cyclotron frequency is independent of speed (4 marks)

7Closest approach of an alpha particle to a nucleus (4 marks)

8de Broglie wavelength and probing the nucleus (5 marks)

9Interpreting the alpha-scattering results (6 marks)

Nuclide (isotope) notation

Energy gained accelerating a charge through a p.d.

The electronvolt

Radius of a charged particle in a magnetic field

Closest approach (energy conservation)

de Broglie wavelength

Nucleon (mass) number, A

Proton (atomic) number, Z

Nucleon

Isotope

Nuclide

Nuclear model of the atom

Alpha-particle scattering experiment

Closest distance of approach

Thermionic emission

Electronvolt (eV)

Linear accelerator (LINAC)

Cyclotron

de Broglie wavelength

Electron diffraction (of the nucleus)

Magnetic centripetal force

✕Reading the nucleon number AAA as the number of neutrons

✕Describing the alpha-scattering observations without linking each to its specific conclusion

✕Substituting an energy in eV or MeV directly into 12mv2\tfrac{1}{2}mv^221​mv2 or kQq/rkQq/rkQq/r

✕Using the charge of the alpha particle as eee instead of 2e2e2e in closest-approach calculations

✕Confusing the sign of the charge / direction of conventional current when accelerating particles

✕Failing to explain why high energies are needed by linking it to de Broglie wavelength

Structure of Matter — frequently asked questions

✕Reading the nucleon number $A$ as the number of neutrons

✕Substituting an energy in eV or MeV directly into $\tfrac{1}{2}mv^2$ or $kQq/r$

✕Using the charge of the alpha particle as $e$ instead of $2e$ in closest-approach calculations