PARTICLESAND RADIATION

AQA A-level Physics · 3.2 · Lessons 1–3
⬇ Student Booklet ⬇ Teacher Booklet

Key Terms — Particles and Radiation

All Terms

01
Proton
DefinitionA positively charged particle in the nucleus. Charge +1.6×10−19 C, mass 1.673×10−27 kg.
02
Neutron
DefinitionAn uncharged particle in the nucleus. Charge 0, mass 1.675×10−27 kg. It is very slightly heavier than a proton.
03
Electron
DefinitionA negatively charged particle that orbits the nucleus. Charge −1.6×10−19 C, mass 9.11×10−31 kg — about 1/1800 of the mass of a nucleon.
04
Nucleus
DefinitionThe tiny, dense centre of the atom containing the protons and neutrons. It holds almost all of the mass of the atom but occupies almost none of its volume.
05
Nucleon
DefinitionA particle found in the nucleus — a proton or a neutron.
06
Nucleon number (A)
DefinitionThe total number of protons and neutrons in the nucleus. Also called the mass number.
07
Proton number (Z)
DefinitionThe number of protons in the nucleus. Also called the atomic number; it identifies the element, and equals the number of electrons in a neutral atom.
08
Neutron number (N)
DefinitionThe number of neutrons in the nucleus: N = A − Z.
09
Nuclide notation
DefinitionThe way a nucleus is written: AZX, with the nucleon number A above the proton number Z, in front of the element symbol X.
10
Elementary charge (e)
DefinitionThe smallest unit of charge carried by a free particle, e = 1.6×10−19 C. All charges on ions are whole-number multiples of e.
11
Specific charge
DefinitionThe charge-to-mass ratio of a particle: charge ÷ mass, measured in C kg−1. The electron has the largest specific charge; the neutron's is zero.
12
Isotope
DefinitionAtoms of the same element (same proton number) with different numbers of neutrons, and so different nucleon numbers. Isotopes are chemically identical.
13
Ion
DefinitionAn atom that has gained or lost electrons, giving it a net negative or positive charge — always a whole-number multiple of 1.6×10−19 C.
14
Relative atomic mass
DefinitionThe mean mass of an atom of an element relative to carbon-12. Chlorine's value of 35.5 arises from a mixture of 3517Cl (about 75%) and 3717Cl (about 25%).
15
Photon
DefinitionA discrete packet (quantum) of electromagnetic energy. Its energy is E = hf = hcλ.
16
Electronvolt (eV)
DefinitionThe energy transferred when an electron moves through a potential difference of 1 V: 1 eV = 1.6×10−19 J. 1 MeV = 106 eV.
17
Antiparticle
DefinitionA particle with the same mass and rest energy as its corresponding particle but the opposite charge, baryon number, lepton number and strangeness.
18
Positron
DefinitionThe antiparticle of the electron (e⁺): same mass and rest energy (0.51 MeV), charge +1.6×10−19 C.
19
Rest energy
DefinitionThe energy locked up in a particle's rest mass, E = mc2. Normally quoted in MeV.
20
Annihilation
DefinitionWhen a particle meets its antiparticle, their total rest energy (plus any kinetic energy) is converted into two gamma photons, which travel in opposite directions so that momentum is conserved.
21
Pair production
DefinitionThe conversion of a single high-energy photon into a particle–antiparticle pair, usually close to a nucleus, which recoils to conserve momentum.
22
Minimum photon energy (pair production)
DefinitionThe photon must supply at least the total rest energy of the pair: Emin = 2mc2 (1.02 MeV for an electron–positron pair).
23
Quark
DefinitionA fundamental particle that makes up hadrons. At A-level you need up (u), charge +⅔e; down (d), charge −⅓e; and strange (s), charge −⅓e. All three have baryon number +⅓.
24
Antiquark
DefinitionThe antiparticle of a quark, written with a bar (u̅, d̅, s̅). It has the opposite sign of charge, baryon number and strangeness.
25
Hadron
DefinitionA particle that feels the strong interaction and is made of quarks. Every hadron is either a baryon or a meson.
26
Baryon
DefinitionA hadron made of three quarks (qqq), such as the proton (uud) or the neutron (udd). Baryon number +1.
27
Antibaryon
DefinitionA hadron made of three antiquarks (q̅q̅q̅), such as the antiproton u̅u̅d̅. Baryon number −1.
28
Meson
DefinitionA hadron made of one quark and one antiquark (qq̅), such as a pion or a kaon. Baryon number 0, and all mesons are unstable.
29
Pion (π)
DefinitionA meson made only of up and down quarks: π⁺ = ud̅, π⁻ = u̅d, π⁰ = uu̅ or dd̅. Strangeness 0.
30
Kaon (K)
DefinitionA meson containing a strange quark or antiquark: K⁺ = us̅, K⁻ = u̅s, K⁰ = ds̅. Strangeness ±1.
31
Baryon number
DefinitionA quantum number carried by hadrons: +⅓ for each quark and −⅓ for each antiquark. It is +1 for a baryon, −1 for an antibaryon and 0 for a meson.
32
Strangeness
DefinitionA quantum number carried by particles containing strange quarks: −1 for each s quark and +1 for each s̅ antiquark.

Lesson 1 · The Nuclear Atom

Do Now

Q1
What are the three constituents of an atom and their charges?
Model AnswerProton (+1.6×10−19 C), neutron (0), electron (−1.6×10−19 C).
Q2
What does the proton (atomic) number tell you?
Model AnswerThe number of protons in the nucleus (equals the number of electrons in a neutral atom).
Q3
What is an isotope?
Model AnswerAtoms of the same element (same proton number) with different numbers of neutrons.

Part 1 of 3 · Constituents of the Atom

An atom has a tiny, dense nucleus of protons and neutrons, surrounded by orbiting electrons.

Charge and mass of the constituents
  • Proton: charge +1.6×10−19 C, mass 1.673×10−27 kg.
  • Neutron: charge 0, mass 1.675×10−27 kg.
  • Electron: charge −1.6×10−19 C, mass 9.11×10−31 kg.
XAZA = nucleon numberZ = proton numberelement symbol
Fig 1.1 — Nuclide notation for an element X.

A nuclide is written AZX, where A is the nucleon number and Z the proton number; the neutron number is N = A − Z.

Questions — Constituents of the Atom

Q1
State the number of protons, neutrons and electrons in a neutral atom of 168O, 2311Na and 19779Au. (3 marks)
Model AnswerO: 8 p, 8 n, 8 e. Na: 11 p, 12 n, 11 e. Au: 79 p, 118 n, 79 e.
Q2
Name the part of the atom that has (2 marks)(a) almost all the mass, (1 mark)(b) most of the volume. (1 mark)
Model Answer(a) The nucleus.(b) The space occupied by the orbiting electrons.
Q3
One isotope of nitrogen is 157N. State the number of protons and neutrons it contains. (2 marks)
Model Answer7 protons and 8 neutrons.
Q4
Determine the charge, in C, of a 23992U nucleus. (2 marks)
Model AnswerQ = 92 × 1.6×10−19 = 1.5×10−17 C.
Q5
A carbon nucleus 126C has a charge of 9.6×10−19 C. Show how this value arises. (1 mark)
Model Answer6 protons × 1.6×10−19 C = 9.6×10−19 C.
Q6
Calculate the total mass of a neutral atom of 42He. (3 marks)
Model Answer2 protons + 2 neutrons + 2 electrons = 2(1.673×10−27) + 2(1.675×10−27) + 2(9.11×10−31) = 3.346×10−27 + 3.350×10−27 + 1.82×10−30 = 6.70×10−27 kg.
Q7
Calculate the charge, in C, of the nucleus of an aluminium atom 2713Al. (2 marks)
Model AnswerQ = 13 × 1.6×10−19 = 2.1×10−18 C.
Q8
An atom of iron is represented by 5626Fe. (5 marks)(a) Calculate the mass of its nucleus. (2 marks)(b) Show that the electrons account for less than 0.1% of the mass of the whole atom. (3 marks)
Model Answer(a) m = 26(1.673×10−27) + 30(1.675×10−27) = 4.35×10−26 + 5.03×10−26 = 9.4×10−26 kg.(b) Mass of 26 electrons = 26 × 9.11×10−31 = 2.37×10−29 kg. As a fraction of the atom: 2.37×10−29 / 9.4×10−26 = 2.5×10−4, i.e. 0.025% — less than 0.1%.
Q9
A neutral atom of 2311Na loses one electron. (4 marks)(a) Calculate the charge of the resulting ion. (1 mark)(b) Calculate the mass of the resulting ion. (3 marks)
Model Answer(a) +1.6×10−19 C.(b) 11 protons + 12 neutrons + 10 electrons = 11(1.673×10−27) + 12(1.675×10−27) + 10(9.11×10−31) = 1.840×10−26 + 2.010×10−26 + 9.1×10−30 = 3.85×10−26 kg.
Q10
Estimate the mass of a 23892U nucleus, taking the mass of a nucleon to be 1.67×10−27 kg, and state one assumption you have made. (3 marks)
Model Answerm ≈ 238 × 1.67×10−27 = 4.0×10−25 kg. Assumption: protons and neutrons have the same mass (and the mass lost as binding energy is ignored).

Part 2 of 3 · Specific Charge

Specific charge is the charge-to-mass ratio of a particle:

specific charge = chargemass   (unit C kg−1)

The electron has the largest specific charge (small mass); the neutron's is zero (no charge).

Find the specific charge of a proton (charge 1.6×10−19 C, mass 1.673×10−27 kg).

Worked Example (VESSU)
V
Q = 1.6×10−19 C, m = 1.673×10−27 kg
E
specific charge = Qm
S
= 1.6×10−191.673×10−27
S
= 9.6×107
U
C kg−1
Answer: specific charge of a proton = 9.6×107 C kg−1

Questions — Specific Charge

Q11
Define specific charge and state its unit. (2 marks)
Model AnswerThe charge per unit mass of a particle (charge ÷ mass); unit C kg−1.
Q12
Calculate the specific charge of (e = 1.6×10−19 C, mp = 1.673×10−27 kg, me = 9.11×10−31 kg): (3 marks)(a) a proton, (2 marks)(b) an electron. (1 mark)
Model Answer(a) 9.6×107 C kg−1.(b) 1.76×1011 C kg−1.
Q13
State, with a reason, which of the proton, neutron and electron has (2 marks)(a) the largest specific charge, (1 mark)(b) the smallest specific charge. (1 mark)
Model Answer(a) Electron — the same magnitude of charge as the proton but a much smaller mass.(b) Neutron — zero charge, so zero specific charge.
Q14
A 42He nucleus (alpha particle) has mass 6.64×10−27 kg. Calculate its specific charge. (3 marks)
Model AnswerQ = 2×1.6×10−19 = 3.2×10−19 C; specific charge = 3.2×10−19/6.64×10−27 = 4.8×107 C kg−1.

Part 3 of 3 · Isotopes and Ions

Isotopes have the same proton number but different neutron numbers; they are chemically identical.

An ion forms when an atom gains or loses electrons, giving it a net negative or positive charge (a multiple of 1.6×10−19 C).

Chlorine

Chlorine's relative atomic mass of 35.5 arises from a mixture of 3517Cl (about 75%) and 3717Cl (about 25%).

Questions — Isotopes and Ions

Q15
State what is meant by an isotope. (2 marks)
Model AnswerAtoms of the same element (same proton number) that have different numbers of neutrons (different nucleon numbers).
Q16
Explain why isotopes of an element behave identically in chemical reactions. (2 marks)
Model AnswerThey have the same number of protons and therefore the same number and arrangement of electrons, which determine chemical behaviour.
Q17
A neutral chlorine atom 3717Cl gains one electron to become an ion. (3 marks)(a) State the charge of the ion in C. (1 mark)(b) State the number of protons, neutrons and electrons in the ion. (2 marks)
Model Answer(a) −1.6×10−19 C.(b) 17 protons, 20 neutrons, 18 electrons.
Q18
An oxygen atom loses two electrons. State the charge of the resulting ion in C, and explain whether it is positive or negative. (3 marks)
Model AnswerCharge = +2×1.6×10−19 = +3.2×10−19 C; positive, because it has lost negative electrons leaving more protons than electrons.

Exam Practice — Lesson 1

Charge of electron = −1.6×10−19 C  ·  proton rest mass = 1.673×10−27 kg  ·  neutron rest mass = 1.675×10−27 kg. Take the mass of a nucleon as 1.67×10−27 kg.
Q1
(Total: 5 marks)(a) State the number of electrons, protons and neutrons in 4020Ca, 5626Fe and 6329Cu. (3 marks)(b) Which of these three nuclei has the largest specific charge? Justify your answer. (2 marks)
Mark Scheme(a) Ca: 20 p, 20 n, 20 e. Fe: 26 p, 30 n, 26 e. Cu: 29 p, 34 n, 29 e.(b) Calcium-40. Specific charge depends on the proton-to-nucleon ratio: Ca 20/40 = 0.50, Fe 26/56 = 0.46, Cu 29/63 = 0.46 — Ca has the highest proportion of protons per unit mass.
Q2
Name the part of an atom that (Total: 3 marks)(a) has no charge, (1 mark)(b) has the largest specific charge, (1 mark)(c) when removed, leaves a different isotope of the same element. (1 mark)
Mark Scheme(a) The neutron.(b) The electron.(c) The neutron — removing it changes the nucleon number but not the proton number.
Q3
One isotope of nitrogen may be represented 157N. (Total: 6 marks)(a)(i) State the number of each type of particle in its nucleus. (1 mark)(a)(ii) Determine the specific charge, in C kg−1, of its nucleus. (3 marks)(b)(i) What is the charge, in C, of an atom of nitrogen from which a single electron has been removed? (1 mark)(b)(ii) What name is used to describe an atom from which an electron has been removed? (1 mark)
Mark Scheme(a)(i) 7 protons and 8 neutrons.(a)(ii) Q = 7 × 1.6×10−19 = 1.12×10−18 C; m = 15 × 1.67×10−27 = 2.51×10−26 kg; specific charge = 1.12×10−18 / 2.51×10−26 = 4.5×107 C kg−1.(b)(i) +1.6×10−19 C.(b)(ii) A (positive) ion.
Q4
(Total: 4 marks)(a) Determine the charge, in C, of a 23992U nucleus. (2 marks)(b) A positive ion with a uranium nucleus has a charge of +4.8×10−19 C. Determine how many electrons are in this ion. (2 marks)
Mark Scheme(a) Q = 92 × 1.6×10−19 = 1.5×10−17 C.(b) +4.8×10−19 / 1.6×10−19 = +3, so the ion has 3 fewer electrons than protons: 92 − 3 = 89 electrons.
Q5
A radioactive isotope of carbon is represented by 166C. (Total: 6 marks)(a) Using the same notation, give the isotope of carbon that has two fewer neutrons. (1 mark)(b) Calculate the charge on the ion formed when two electrons are removed from an atom of carbon-16. (2 marks)(c) Calculate the specific charge of the nucleus of an atom of carbon-16. (3 marks)
Mark Scheme(a) 146C.(b) +2 × 1.6×10−19 = +3.2×10−19 C.(c) Q = 6 × 1.6×10−19 = 9.6×10−19 C; m = 16 × 1.67×10−27 = 2.67×10−26 kg; specific charge = 3.6×107 C kg−1.
Q6
(Total: 5 marks)(a) Calculate the mass of an ion with a specific charge of 1.20×107 C kg−1 and a negative charge of 3.2×10−19 C. (2 marks)(b) The ion has eight protons in its nucleus. Calculate its number of electrons and its number of neutrons. (3 marks)
Mark Scheme(a) m = Q / specific charge = 3.2×10−19 / 1.20×107 = 2.7×10−26 kg.(b) A charge of −3.2×10−19 C means 2 extra electrons, so 8 + 2 = 10 electrons. A = 2.67×10−26 / 1.67×10−27 = 16, so neutrons = 16 − 8 = 8.
Q7
An ion of plutonium 23994Pu has an overall charge of +1.6×10−19 C. (Total: 5 marks)(a)(i) For this ion, state the number of protons. (1 mark)(a)(ii) State the number of neutrons. (1 mark)(a)(iii) State the number of electrons. (1 mark)(b) Plutonium has several isotopes. Explain the meaning of the word isotopes. (2 marks)
Mark Scheme(a)(i) 94.(a)(ii) 239 − 94 = 145.(a)(iii) A charge of +1 means one electron fewer than protons: 93.(b) Atoms of the same element with the same proton number but different numbers of neutrons, and therefore different nucleon numbers.
Q8
The nucleus of a particular atom has a nucleon number of 14 and a proton number of 6. (Total: 10 marks)(a)(i) State what is meant by nucleon number and by proton number. (1 mark)(a)(ii) Calculate the number of neutrons in the nucleus of this atom. (1 mark)(a)(iii) Calculate the specific charge of the nucleus. (3 marks)(b)(i) The specific charge of the nucleus of another isotope of the element is 4.8×107 C kg−1. State what is meant by an isotope. (2 marks)(b)(ii) Calculate the number of neutrons in this isotope. (3 marks)
Mark Scheme(a)(i) Nucleon number: the total number of protons and neutrons in the nucleus. Proton number: the number of protons in the nucleus.(a)(ii) 14 − 6 = 8 neutrons.(a)(iii) Q = 6 × 1.6×10−19 = 9.6×10−19 C; m = 14 × 1.67×10−27 = 2.34×10−26 kg; specific charge = 4.1×107 C kg−1.(b)(i) Nuclei of the same element (same proton number) with different numbers of neutrons.(b)(ii) Same element, so Q = 9.6×10−19 C. m = 9.6×10−19 / 4.8×107 = 2.0×10−26 kg; A = 2.0×10−26 / 1.67×10−27 = 12; neutrons = 12 − 6 = 6.
Q9
(Total: 8 marks)(a) State what is meant by the specific charge of a nucleus and give an appropriate unit for this quantity. (2 marks)(b)(i) Nucleus X has the same nucleon number as nucleus Y. The specific charge of X is 1.25 times greater than that of Y. Explain, in terms of protons and neutrons, why the specific charge of X is greater than that of Y. (2 marks)(b)(ii) Nucleus X is 105B. Deduce the number of protons and the number of neutrons in nucleus Y. (4 marks)
Mark Scheme(a) The charge of the nucleus divided by its mass (charge per unit mass); unit C kg−1.(b)(i) Equal nucleon numbers means the two nuclei have essentially the same mass, so the one with the greater specific charge must carry the greater charge. X therefore has more protons (and correspondingly fewer neutrons) than Y.(b)(ii) Masses are equal, so the charge ratio equals the specific charge ratio: ZX = 1.25 ZY. ZY = 5 / 1.25 = 4 protons. Y has the same nucleon number (10), so neutrons = 10 − 4 = 6.
Q10
(Total: 10 marks)(a) Explain what is meant by an isotope. (2 marks)(b) Two isotopes of uranium are compared. The first has 92 protons and 143 neutrons; the specific charge of the second is 3.7×107 C kg−1.(b)(i) Write the unit for specific charge. (1 mark)(b)(ii) State the number of protons in the second isotope. (1 mark)(b)(iii) Calculate the specific charge of the first isotope. (3 marks)(b)(iv) Calculate the number of neutrons in the second isotope. (3 marks)
Mark Scheme(a) Atoms of the same element that have the same number of protons but different numbers of neutrons.(b)(i) C kg−1.(b)(ii) 92 — both are uranium.(b)(iii) Q = 92 × 1.6×10−19 = 1.47×10−17 C; m = 235 × 1.67×10−27 = 3.92×10−25 kg; specific charge = 3.8×107 C kg−1.(b)(iv) m = 1.47×10−17 / 3.7×107 = 3.98×10−25 kg; A = 3.98×10−25 / 1.67×10−27 = 238; neutrons = 238 − 92 = 146.

Lesson 2 · Particles and Antiparticles

Do Now

Q1
Last lesson: State what is meant by the specific charge of a nucleus and give its unit.
Model AnswerThe charge of the nucleus divided by its mass (charge per unit mass); unit C kg−1.
Q2
Calculation: An atom of 2713Al loses three electrons. Calculate the specific charge of the ion produced. (mass of a nucleon = 1.67×10−27 kg, e = 1.6×10−19 C)
Model AnswerThe ion has lost 3 electrons, so its charge is +3 × 1.6×10−19 = 4.8×10−19 C — not the full nuclear charge of 13e, because 10 electrons remain to cancel most of it. Mass ≈ 27 × 1.67×10−27 = 4.51×10−26 kg. Specific charge = 4.8×10−19 / 4.51×10−26 = 1.1×107 C kg−1.
Q3
Recall: List the electromagnetic spectrum in order of increasing frequency.
Model AnswerRadio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
Q4
Stretch: An electron and a positron each have a rest mass of 9.11×10−31 kg. Using E = mc2 (c = 3.00×108 m s−1), calculate the total energy released when they annihilate, in J and in MeV.
Model AnswerTotal mass = 2 × 9.11×10−31 = 1.822×10−30 kg. E = 1.822×10−30 × (3.00×108)2 = 1.64×10−13 J. In eV: 1.64×10−13 / 1.6×10−19 = 1.02×106 eV = 1.02 MeV, shared equally between two photons of 0.51 MeV each.

Part 1 of 3 · Photons and the Electronvolt

Electromagnetic radiation is emitted in packets (photons) of energy:

E = h f = h cλ

The electronvolt (eV) is the energy transferred when an electron moves through a p.d. of 1 V: 1 eV = 1.6×10−19 J.

Calculate the energy of a photon of frequency 5.0×1014 Hz (h = 6.63×10−34 J s).

Worked Example (VESSU)
V
f = 5.0×1014 Hz, h = 6.63×10−34 J s
E
E = h f
S
E = 6.63×10−34 × 5.0×1014
S
E = 3.3×10−19
U
J
Answer: energy of the photon E = 3.3×10−19 J

Questions — Photons and the Electronvolt

Q1
Calculate the energy of a photon of light of frequency 6.0×1014 Hz. (2 marks)
Model AnswerE = hf = 6.63×10−34×6.0×1014 = 4.0×10−19 J.
Q2
Convert a photon energy of 4.0×10−19 J into electronvolts. (1 mark)
Model Answer4.0×10−19/1.6×10−19 = 2.5 eV.
Q3
Calculate the wavelength of a photon of energy 8.2×10−14 J (c = 3.0×108 m s−1). (3 marks)
Model Answerλ = hc/E = (6.63×10−34×3.0×108)/8.2×10−14 = 2.4×10−12 m.
Q4
State what is meant by the electronvolt and give its value in joules. (2 marks)
Model AnswerThe energy transferred when an electron moves through a potential difference of 1 V; 1 eV = 1.6×10−19 J.
Q5
An X-ray photon has energy 12 keV. Calculate its frequency. (2 marks)
Model AnswerE = 12×103×1.6×10−19 = 1.92×10−15 J; f = E/h = 2.9×1018 Hz.

Part 2 of 3 · Antiparticles and Annihilation

Every particle has a corresponding antiparticle with the same mass and rest energy but opposite charge, baryon number, lepton number and strangeness.

When a particle meets its antiparticle they annihilate, converting their total rest energy into two gamma photons.

particleantiparticlemomentum +pmomentum −pγγannihilationtotal momentum before = (+p) + (−p) = 0
Fig 2.1 — Annihilation of a particle–antiparticle pair at rest.

The pair approach with equal and opposite momenta, so the total momentum beforehand is zero. The two photons must therefore travel in exactly opposite directions — which is why two photons are produced and never one.

The energy locked up in a particle's mass is its rest energy, given by Einstein's mass–energy equation:

E = m c2
Using E = mc2
  • m is the rest mass in kg and c = 3.00×108 m s−1, which gives E in joules.
  • Rest energies are normally quoted in MeV: divide by 1.6×10−19 to reach eV, then by 106 to reach MeV.
  • A particle and its antiparticle have the same rest mass, and therefore the same rest energy.
  • In annihilation the total rest energy becomes photon energy; in pair production photon energy becomes rest energy.
  • In annihilation of a particle–antiparticle pair at rest, each of the two photons carries an energy of m c2.

Calculate the rest energy of an electron (mass 9.11×10−31 kg) in J and in MeV.

Worked Example (VESSU)
V
m = 9.11×10−31 kg, c = 3.00×108 m s−1
E
E = m c2
S
E = 9.11×10−31 × (3.00×108)2
S
E = 8.2×10−14 J; 8.2×10−14 / 1.6×10−19 = 5.1×105 eV
U
0.51 MeV
Answer: rest energy of an electron = 8.2×10−14 J = 0.51 MeV

Questions — Antiparticles and Annihilation

Q6
Show that the rest energy of an electron (mass 9.11×10−31 kg) is about 8.2×10−14 J (c = 3.0×108 m s−1). (2 marks)
Model AnswerE = mc2 = 9.11×10−31×(3.0×108)2 = 8.2×10−14 J.
Q7
Show that this rest energy is equivalent to 0.51 MeV. (2 marks)
Model Answer8.2×10−14/1.6×10−19 = 5.1×105 eV = 0.51 MeV.
Q8
State the rest energy of a positron and give a reason. (1 mark)
Model Answer0.51 MeV — an antiparticle has the same mass (and rest energy) as its particle.
Q9
An electron and a positron, both at rest, annihilate to produce two identical gamma photons. (3 marks)(a) State why two photons are produced rather than one. (1 mark)(b) Calculate the energy of each photon in MeV. (2 marks)
Model Answer(a) To conserve momentum (the initial momentum is zero, so the photons travel in opposite directions).(b) Total rest energy = 2×0.51 = 1.02 MeV; each photon = 0.51 MeV.
Q10
Calculate the frequency of each gamma photon produced when an electron and a positron, both at rest, annihilate (each photon energy 8.2×10−14 J). (2 marks)
Model Answerf = E/h = 8.2×10−14/6.63×10−34 = 1.2×1020 Hz.

Part 3 of 3 · Pair Production

Pair production is the reverse of annihilation: a high-energy photon creates a particle-antiparticle pair, usually near a nucleus (which recoils to conserve momentum).

γ photonnucleuse⁻ (electron)e⁺ (positron)pair production
Fig 2.2 — Pair production near a nucleus.

The photon must have at least the total rest energy of the pair: Emin = 2 m c2.

Questions — Pair Production

Q11
Explain the process of pair production. (3 marks)
Model AnswerA high-energy photon, passing near a nucleus, is converted into a particle and its antiparticle; the nucleus recoils to conserve momentum; charge, energy and momentum are conserved.
Q12
Explain why pair production cannot occur if the photon frequency is below a certain value. (2 marks)
Model AnswerThe photon energy hf must be at least the total rest energy (2mc2) of the pair; below the corresponding frequency there is insufficient energy to create the masses.
Q13
Calculate the minimum photon energy, in MeV, needed to create an electron-positron pair (electron rest energy 0.51 MeV). (2 marks)
Model AnswerEmin = 2×0.51 = 1.02 MeV.
Q14
Calculate the minimum frequency of a photon that can produce an electron-positron pair. (3 marks)
Model AnswerE = 1.02×106×1.6×10−19 = 1.63×10−13 J; f = E/h = 1.63×10−13/6.63×10−34 = 2.5×1020 Hz.

Exam Practice — Lesson 2

h = 6.63×10−34 J s  ·  c = 3.00×108 m s−1  ·  rest energy of an electron = 0.51 MeV  ·  1 eV = 1.6×10−19 J.
Q1
(Total: 3 marks)(a) State the name of the antiparticle of a positron. (1 mark)(b) Describe what happens when a positron and its antiparticle meet. (2 marks)
Mark Scheme(a) The electron.(b) They annihilate. Their total rest energy is converted into two gamma photons, which travel in opposite directions so that momentum is conserved.
Q2
Under certain conditions a photon may be converted into an electron and a positron. (Total: 9 marks)(a) State the name of this process. (1 mark)(b) For the conversion to take place the photon has to have an energy equal to or greater than a certain minimum energy.(b)(i) Explain why there is a minimum energy. (2 marks)(b)(ii) Show that this minimum energy is about 1 MeV. (1 mark)(b)(iii) Explain what happens to the excess energy when the photon energy is greater than the minimum energy. (1 mark)(b)(iv) A photon has an energy of 1.0 MeV. Calculate the frequency associated with this photon energy, and state an appropriate unit in your answer. (4 marks)
Mark Scheme(a) Pair production.(b)(i) The photon energy has to create the rest mass of both particles, so it must be at least the total rest energy of the electron and the positron. Below this there is not enough energy to create the two masses.(b)(ii) Emin = 2 × 0.51 = 1.02 MeV ≈ 1 MeV.(b)(iii) It is carried away as kinetic energy of the electron and the positron (with a small amount as recoil kinetic energy of the nucleus).(b)(iv) E = 1.0×106 × 1.6×10−19 = 1.6×10−13 J. f = E/h = 1.6×10−13 / 6.63×10−34 = 2.4×1020. Unit: Hz.
Q3
(Total: 5 marks)(a) Pair production can occur when a photon interacts with matter. Explain the process of pair production. (2 marks)(b) Explain why pair production cannot take place if the frequency of the photon is below a certain value. (3 marks)
Mark Scheme(a) A high-energy photon passing close to a nucleus is converted into a particle and its corresponding antiparticle, for example an electron and a positron. The nucleus recoils so that momentum is conserved; charge and energy are also conserved.(b) The photon energy is hf, and this must be at least the total rest energy of the pair, 2mc2. This sets a minimum frequency fmin = 2mc2/h. Below that frequency the photon simply does not carry enough energy to create the two rest masses, so no pair can be produced.

Lesson 3 · Quarks and Hadrons

Do Now

Q1
Last lesson: State the minimum energy a photon must have to produce a particle–antiparticle pair.
Model AnswerTwice the rest energy of one of the particles, E = 2mc².
Q2
Calculation: A gamma photon has a frequency of 3.0×1020 Hz. Calculate its energy in MeV, and state whether it could create an electron–positron pair. (h = 6.63×10−34 J s, 1 eV = 1.6×10−19 J)
Model AnswerE = hf = 6.63×10−34 × 3.0×1020 = 1.99×10−13 J. In eV: 1.99×10−13 / 1.6×10−19 = 1.24×106 eV = 1.24 MeV. This exceeds the 1.02 MeV needed, so yes — a pair could be created.
Q3
Recall: Name the three particles found in an atom and state where each is located.
Model AnswerProtons and neutrons in the nucleus; electrons orbiting outside it.
Q4
Stretch: In a high-energy collision a proton and an antiproton are created together by pair production. The rest energy of a proton is 938 MeV. Calculate the minimum energy, in J, needed to create the pair. Then, given that a proton is uud, suggest the quark composition of the antiproton.
Model AnswerBoth particles have the same rest energy, so Emin = 2 × 938 = 1876 MeV = 1.876×109 × 1.6×10−19 = 3.0×10−10 J. The antiproton is built from the corresponding antiquarks, so it is u̅u̅d̅ — giving a charge of −⅔ − ⅔ + ⅓ = −1e, as expected for the antiparticle of the proton.

Part 1 of 3 · Baryons

The three quarks at A-level are up (u), down (d) and strange (s). Antiquarks have the opposite sign of charge, baryon number and strangeness.

A baryon is three quarks (qqq). The two you must know are the proton and the neutron:

uudproton (uud)uddneutron (udd)
Fig 3.1 — Quark composition of the proton and neutron.

Quarks carry fractional charges, but no free particle has ever been observed with a fractional charge. Every baryon and every meson must therefore have a whole-number charge — 0, ±1e or ±2e. This is a useful check: if a proposed combination of quarks does not come to an integer, it cannot be a real hadron.

The properties of the quarks are given to you in the exam:

QuarkChargeBaryon No.StrangenessCharmnessBottomnessTopness
d−⅓+⅓0000
u+⅔+⅓0000
s−⅓+⅓−1000
c+⅔+⅓0+100
b−⅓+⅓00−10
t+⅔+⅓000+1
Properties of the quarks. Only up, down and strange, and the first three property columns, are tested at A-level. The shaded entries are shown for completeness — if a question involves a charm, bottom or top quark, AQA will supply the data you need.

Questions — Baryons

Q1
State the charge, in units of e, of each combination below, and therefore state whether it is a valid combination of quarks. (12 marks)(a) uuu, (2 marks)(b) uud, (2 marks)(c) udd, (2 marks)(d) ddd, (2 marks)(e) ud, (2 marks)(f) uud̅. (2 marks)
Model Answer(a) +⅔ + ⅔ + ⅔ = +2. Integer, so valid (this is the Δ⁺⁺).(b) +⅔ + ⅔ − ⅓ = +1. Integer, so valid (a proton).(c) +⅔ − ⅓ − ⅓ = 0. Integer, so valid (a neutron).(d) −⅓ − ⅓ − ⅓ = −1. Integer, so valid (the Δ⁻).(e) +⅔ − ⅓ = +⅓. Fractional, so not valid — two quarks can never form a hadron.(f) +⅔ + ⅔ − ⅔ = +⅔. Fractional, so not valid.
Q2
Which combination has the correct charge to be (2 marks)(a) a proton, (1 mark)(b) a neutron? (1 mark)
Model Answer(a) uud (charge +1).(b) udd (charge 0).
Q3
The Δ⁻ particle has a charge of −1e. State its quark combination. (1 mark)
Model Answerddd.
Q4
The Δ⁺⁺ particle has a charge of +2e. State its quark combination. (1 mark)
Model Answeruuu.
Q5
By referring to the charges on quarks, explain why the neutron is uncharged. (2 marks)
Model AnswerA neutron is udd: +⅔ − ⅓ − ⅓ = 0, so the quark charges cancel exactly.
Q6
State the baryon number of a proton, and show how it follows from its quark structure. (2 marks)
Model Answer+1. Each of the three quarks has baryon number +⅓, and 3 × ⅓ = 1.

Part 2 of 3 · Classification of Hadrons

All hadrons feel the strong interaction and are made of quarks. They come in exactly two kinds:

hadronsbaryonsqqq(p, n)mesonsq q(π, K)three quarks, or a quark and an antiquark
Fig 3.2 — The two classes of hadron.
The two classes of hadron
  • Baryons: three quarks (e.g. proton, neutron). The proton is the only stable baryon, and every other baryon eventually decays into a proton.
  • Mesons: one quark and one antiquark (e.g. pions π, kaons K). All mesons are unstable.

A meson is always one quark and one antiquark (qq̅). The two families you need are the pions and the kaons.

udπ⁺ pionusK⁺ kaona meson is one quark + one antiquarkthe bar marks the antiquark; a kaon contains a strange quark
Fig 3.3 — Quark structure of the pion and the kaon.
Quark structure of the mesons you need
  • Pions: π⁺ = ud̅   π⁻ = u̅d   π⁰ = uu̅ or dd̅.
  • Kaons: K⁺ = us̅   K⁻ = u̅s   K⁰ = ds̅ (or d̅s).
  • Every kaon contains a strange quark or antiquark, so kaons have strangeness ±1; pions have strangeness 0.
  • Every meson has baryon number 0 (quark +⅓ and antiquark −⅓).

Questions — Classification of Hadrons

Q7
State what is meant by a hadron. (2 marks)
Model AnswerA particle that feels the strong interaction and is made of quarks.
Q8
State the quark structure of each, and give an example: (4 marks)(a) a baryon, (2 marks)(b) a meson. (2 marks)
Model Answer(a) Three quarks — e.g. proton or neutron.(b) A quark and an antiquark — e.g. a pion or kaon.
Q9
Name the only stable baryon, and state what happens to all the others. (2 marks)
Model AnswerThe proton. Every other baryon is unstable and eventually decays into a proton.
Q10
Classify each of the following as a baryon or a meson: proton, pion, neutron, kaon. (2 marks)
Model AnswerBaryons: proton, neutron. Mesons: pion, kaon.
Q11
A meson consists of a quark and an antiquark. Show that a u quark and a d antiquark give a charge of +1e. (2 marks)
Model Answeru = +⅔e, d̅ = +⅓e; total = +⅔ + ⅓ = +1e.
Q12
State the quark structure of (3 marks)(a) π⁺, (1 mark)(b) π⁻, (1 mark)(c) π⁰. (1 mark)
Model Answer(a) ud̅.(b) u̅d.(c) uu̅ or dd̅.
Q13
State the quark structure of (a) K⁺ and (b) K⁻, and state the strangeness of each. (4 marks)
Model Answer(a) us̅, strangeness +1 (the s̅ antiquark has strangeness +1). (b) u̅s, strangeness −1.
Q14
Show that the baryon number of any meson is zero. (2 marks)
Model AnswerA quark has baryon number +⅓ and an antiquark −⅓, so +⅓ − ⅓ = 0.
Q15
Explain why a K⁰ meson has strangeness +1 but a π⁰ meson has strangeness 0. (2 marks)
Model AnswerK⁰ = ds̅ and contains a strange antiquark (strangeness +1); π⁰ contains only up and down quarks, which have strangeness 0.

Part 3 of 3 · Antibaryons and Antiparticles of Hadrons

Every hadron has an antiparticle, built from the corresponding antiquarks. An antibaryon is three antiquarks (q̅q̅q̅).

The antibaryons you need
  • Antiproton p̅ = u̅u̅d̅ — charge −1e, baryon number −1.
  • Antineutron n̅ = u̅d̅d̅ — charge 0, baryon number −1.
  • An antiparticle has the same mass and rest energy as its particle.
  • It has the opposite charge, baryon number and strangeness.

Note that the antineutron is not the same as the neutron, even though both are uncharged: their baryon numbers are −1 and +1, so they are genuinely different particles.

Mesons can be each other's antiparticles

Because a meson already contains an antiquark, the antiparticle of a meson is another meson. Replacing every quark by its antiquark turns π⁺ (ud̅) into π⁻ (u̅d), so the two charged pions are antiparticles of each other. The same is true of K⁺ and K⁻. The π⁰ is its own antiparticle.

Questions — Antibaryons and Antiparticles of Hadrons

Q16
State the quark structure of an antibaryon. (1 mark)
Model AnswerThree antiquarks, q̅q̅q̅.
Q17
State the quark structure of (2 marks)(a) an antiproton, (1 mark)(b) an antineutron. (1 mark)
Model Answer(a) u̅u̅d̅.(b) u̅d̅d̅.
Q18
Show that the antiproton has a charge of −1e. (2 marks)
Model Answeru̅ = −⅔e and d̅ = +⅓e, so −⅔ − ⅔ + ⅓ = −1e.
Q19
Give one property of an antiparticle that is the same as its corresponding particle, and one that is different. (2 marks)
Model AnswerSame: mass (and rest energy). Different: charge (also baryon number and strangeness).
Q20
Explain why the antineutron is a different particle from the neutron, even though both are uncharged. (2 marks)
Model AnswerThe neutron has baryon number +1 and the antineutron −1; they are built from quarks and antiquarks respectively.
Q21
Name the antiparticle of (2 marks)(a) the π⁺, (1 mark)(b) the K⁺. (1 mark)
Model Answer(a) The π⁻.(b) The K⁻.

Exam Practice — Lesson 3

Quark properties are given on the data sheet. h = 6.63×10−34 J s · c = 3.00×108 m s−1 · me = 9.11×10−31 kg · 1 eV = 1.6×10−19 J.
Q1
What is the quark structure of an antiproton? (Total: 1 mark)A u̅d̅B d̅d̅s̅C d̅d̅u̅D u̅u̅d̅
Mark SchemeD. A proton is uud, so the antiproton is u̅u̅d̅ (charge −⅔ − ⅔ + ⅓ = −1).
Q2
(Total: 4 marks)(a) Name three types (or flavours) of quark. (2 marks)(b) By referring to the charges on quarks, explain why the neutron is uncharged. (2 marks)
Mark Scheme(a) Up, down and strange.(b) A neutron is udd. The charges are +⅔e, −⅓e and −⅓e, which sum to zero.
Q3
Which of the following is not true? (Total: 1 mark)A Each meson consists of a single quark and a single antiquark.B Each baryon consists of three quarks.C The magnitude of the charge on every quark is ⅓.D A particle consisting of a single quark has not been observed.
Mark SchemeC. The down and strange quarks carry ⅓e, but the up quark carries ⅔e, so the magnitudes are not all ⅓.
Q4
Which line correctly classifies the particle, its category and its quark combination? (Total: 1 mark)A neutron  ·  baryon  ·  u̅dB neutron  ·  meson  ·  uddC proton  ·  baryon  ·  uudD positive pion  ·  meson  ·  u̅d
Mark SchemeC. A proton is a baryon of structure uud. A is wrong (u̅d is not a baryon), B is wrong (a neutron is a baryon), D is wrong (π⁺ is ud̅, not u̅d).
Q5
What are the numbers of hadrons, baryons and mesons in an atom of 73Li?   (hadrons · baryons · mesons) (Total: 1 mark)A 7  ·  3  ·  3B 7  ·  4  ·  4C 7  ·  7  ·  0D 10  ·  7  ·  0
Mark SchemeC. The nucleus holds 3 protons and 4 neutrons: 7 nucleons, all of which are baryons and therefore all hadrons. There are no mesons. The 3 electrons are leptons, not hadrons.
Q6
(Total: 2 marks)State the differences in quark structure between a meson and a baryon. (2 marks)
Mark SchemeA baryon is three quarks; a meson is one quark and one antiquark.
Q7
(Total: 2 marks)(a) Give the name of a particle that is a hadron. (1 mark)(b) Pions are mesons. Give a possible quark structure for a pion. (1 mark)
Mark Scheme(a) Any of: proton, neutron, pion, kaon (or an antiparticle of these).(b) ud̅ (or u̅d, uu̅, dd̅).
Q8
The table below gives information about some fundamental particles. (Total: 11 marks)(a) Complete the table. Row 1: quark structure uud, strangeness 0. Row 2: Sigma⁺, quark structure uus, charge +1. Row 3: quark structure ud̅, strangeness 0, baryon number 0. (7 marks)(b)(i) Each particle in the table has an antiparticle. Give one example of a baryon and its corresponding antiparticle. (1 mark)(b)(ii) State the quark structure of an antibaryon. (1 mark)(b)(iii) Give one property of an antiparticle that is the same as its corresponding particle, and one that is different. (2 marks)
Mark Scheme(a) Row 1 — particle: proton; charge +1; baryon number +1. Row 2 — strangeness −1 (one s quark); baryon number +1. Row 3 — particle: π⁺; charge +1.(b)(i) Proton and antiproton (or neutron and antineutron).(b)(ii) Three antiquarks, q̅q̅q̅.(b)(iii) Same: mass or rest energy. Different: charge (or baryon number, or strangeness).
Q9
Cosmic rays are high-energy particles from space. They collide with air molecules in the Earth’s atmosphere to produce pions and kaons. (Total: 9 marks)(a) Pions and kaons are mesons. Identify the quark–antiquark composition of a meson: qqq, qq̅q̅, qq̅ or qq. (1 mark)(b) A positron with a kinetic energy of 2.0 keV collides with an electron at rest, creating two photons of equal energy. Show that the energy of each photon is 8.2×10−14 J. (3 marks)(c) Calculate the wavelength of a photon of energy 8.2×10−14 J. (2 marks)(d) Show that the speed of the positron before the collision was about 2.7×107 m s−1. (3 marks)
Mark Scheme(a) qq̅ — one quark and one antiquark.(b) Total energy = 2 rest energies + the positron's kinetic energy = 2(9.11×10−31 × (3.00×108)2) + 2.0×103 × 1.6×10−19 = 1.640×10−13 + 3.2×10−16 = 1.643×10−13 J. Each photon takes half: 8.2×10−14 J.(c) λ = hc/E = (6.63×10−34 × 3.00×108) / 8.2×10−14 = 2.4×10−12 m.(d) Ek = 2.0×103 × 1.6×10−19 = 3.2×10−16 J. v = √(2Ek/m) = √(2 × 3.2×10−16 / 9.11×10−31) = 2.7×107 m s−1.
Q10
The positive kaon, K⁺, has a strangeness of +1. (Total: 3 marks)(a) What is the quark structure of the K⁺? (1 mark)(b) What is the baryon number of the K⁺? (1 mark)(c) What is the antiparticle of the K⁺? (1 mark)
Mark Scheme(a) us̅ — the s̅ gives strangeness +1 and, with u, a charge of +⅔ + ⅓ = +1.(b) 0 — the quark (+⅓) and antiquark (−⅓) cancel.(c) The K⁻, of structure u̅s.
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