Particle physics

Class 11 Physics, unit 14, book pages 257–283

Practise 77 MCQs Take the unit quiz Review 54 cards

This unit studies the smallest known building blocks of matter and the rules they obey. It starts with the conservation of charge and nucleons in alpha, beta and gamma processes, then introduces anti-matter, quarks, leptons and the force carriers (photon, gluon, W, Z and the Higgs boson) of the Standard Model. It ends with annihilation, in which a particle and its anti-particle turn into energy or into new particles, and with the accelerators (linear, synchrotron and cyclotron) used to produce and study them.

Learning outcomes (20)

State that nucleon number and charge are conserved in nuclear processes.
Describe the composition, mass and charge of alpha, beta and gamma radiations (both $β^{-}$ (electron) and $β^{+}$ (positron) are included).
Explain that an antiparticle has the same mass but opposite charge to the corresponding particle (give the example that a positron is the anti-particle of an electron).
State that electron-antineutrinos are produced during $β^{-}$ decay and electron-neutrinos are produced during $β^{+}$ decay.
Explain that $α$ -particles have discrete energies but $β$ -particles have a continuous range of energies because anti-neutrinos are emitted in $β^{-}$ decay.
Describe quarks and anti-quarks as fundamental particles (including that there are six flavours (types) of quark: up, down, strange, charm, top and bottom).
Describe protons and neutrons in terms of their quark composition.
State that a hadron may be either a baryon (consisting of three quarks) or a meson (consisting of one quark and one anti-quark).
Describe the changes to quark composition that take place during $β^{-}$ and $β^{+}$ decay.
State that electrons and neutrinos are fundamental particles called leptons.
State that W, Z, gluon and photons are fundamental particles called exchange particles or force carriers.
State that the Higgs boson is a fundamental particle which is responsible for the particle's mass.
Explain that every subatomic particle has a corresponding anti-particle that has the same mass as the given particle but opposite electric or magnetic properties according to the Standard Model of Particle Physics.
Explain that there are various contending theories about what ‘mass’ and ‘force’ are and how they are generated (e.g. that these are generated from quantum fields when they are energized, or from multidimensional ‘strings’ that vibrate in higher dimensions to give rise to particles).
Explain the working principle of particle accelerators and also their uses.
Explain that anti-matter is the counter part of matter (e.g. a positron is the anti-matter counterpart to an electron).
Illustrate that the anti-particles usually have the same mass but opposite charge compared to their matter counterparts.
State that most of the matter in the observable universe is matter.
Describe the asymmetry of matter and anti-matter in the universe as an unresolved mystery.
Describe annihilation reactions (a particle meets its corresponding anti-particle; they undergo annihilation, in which either all the mass is converted to heat and light energy, or some mass is left over in the form of new subatomic particles).

Introduction

From the beginning of time humans have had an urge to discover the mysteries of the universe. In the search for the fundamental building block of the cosmos they have opened many doors to scientific knowledge. This unit looks at the smallest building blocks we know and at the rules they obey.

Particle physics: Also called high-energy physics. The branch of physics that deals with the fundamental constituents of matter and radiation and their interactions. It studies subatomic particles such as quarks, leptons and bosons, and the forces between them.
Standard Model: The model of particle physics that describes these particles and their interactions, excluding gravity. It aims to understand the universe at the smallest scales and highest energies.

Particle physics experiments are often done at large facilities such as the Large Hadron Collider (LHC). There particles are accelerated to high energies and collided, so that physicists can probe the fundamental structure of the universe.

14.1Conservation laws in nuclear reactionsOften examined

Charge and nucleon conservation are fundamental principles in nuclear processes. They are deeply rooted in the laws of physics and in the interactions among subatomic particles. These conservation laws play a crucial role in maintaining the integrity of atomic nuclei and make sure that the fundamental properties of matter are preserved during nuclear reactions and decays.

Conservation of charge: In any closed system the total charge before and after an interaction or decay must remain the same.
Conservation of nucleons: The total number of nucleons (protons plus neutrons) remains constant before and after a nuclear reaction or decay.

14.1.1Conservation of chargeOften examined

The conservation of charge is a fundamental property of particle interactions. In any closed system the total charge before and after an interaction or decay must remain the same. Consider the example of alpha decay of radium:

_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He + Q

Alpha decay of radium-226

Checking charge in the decay

Write down the charge (atomic number) of each nucleus: radium has $88$ , radon has $86$ and the alpha particle (helium nucleus) has $2$ . In units of $e$ these are the charges.
Total charge on the left-hand side:
$Q_{L.H.S} = + 88 e$
Total charge on the right-hand side:
$Q_{R.H.S} = + 86 e + 2 e = + 88 e$
The two totals are equal, so charge is conserved in this decay.
$88 e = 88 e ✓$

Charge is a conserved quantity: it cannot be created or destroyed. It can only be transferred from one particle to another. In nuclear processes the charges of the particles involved (protons, electrons, positrons) are carefully balanced so that the overall charge stays the same.

14.1.2Conservation of nucleonsOften examined

Nucleons, which are protons and neutrons, are the building blocks of atomic nuclei. The conservation of nucleons says that the total number of nucleons remains constant before and after a nuclear reaction or decay. Consider this nuclear reaction:

_{7}^{14} N +_{2}^{4} He \to_{8}^{17} O +_{1}^{1} H + Q

Nitrogen-14 and an alpha particle

Counting nucleons

The nucleon number is the upper number ( $A$ ) of each nucleus. Count it on the left-hand side:
$14 + 4 = 18$
Count it on the right-hand side:
$17 + 1 = 18$
Both sides have $18$ nucleons, so nucleon number is conserved.
The charge (lower number) is conserved in the same reaction as well:
$7 + 2 = 9 = 8 + 1$

The conservation of nucleons is a consequence of the strong nuclear force that binds nucleons together in the nucleus. The strong force is a short-range force that acts between quarks (the constituents of protons and neutrons). It makes sure that nucleons are not spontaneously created or destroyed during nuclear processes.

The conservation of charge and nucleons is upheld in various nuclear reactions, such as alpha decay, beta decay and fusion reactions. These laws have been extensively tested and confirmed by experiments and observations in particle physics and nuclear science. They give a deep insight into the symmetries and interactions within the subatomic world.

14.1Alpha, beta and gamma radiationsOften examined

Alpha particles, beta particles and gamma rays are three types of radiation commonly associated with radioactive decay. They have distinct properties of mass and charge. Here $e = 1.6 \times 1 0^{- 19} C$ is the elementary charge.

1) Alpha particle

Composition: two protons and two neutrons bound together as a single entity.
Mass: about $4$ atomic mass units ( $u$ ), equal to $6.64 \times 1 0^{- 27} kg$ .
Charge: $+ 2 e$ , twice the charge of a single proton.

2) Beta particle

Composition: high-energy electrons ( $β^{-}$ ) or positrons ( $β^{+}$ , beta-plus) emitted during certain types of radioactive decay.
Mass: much smaller than an alpha particle. An electron has a mass of about $9.11 \times 1 0^{- 31} kg$ ; a positron has the same mass.
Charge: $β^{-}$ particles (electrons) carry $- 1 e$ ; $β^{+}$ particles (positrons) carry $+ 1 e$ .

3) Gamma ray

Composition: electromagnetic waves, like X-rays. They do not consist of particles in the traditional sense; they are made of photons.
Mass: none, because photons are massless.
Charge: neutral, they carry no electric charge.

The three radiations compared (Section 14.1).
Property	Alpha particle	Beta particle	Gamma ray
Composition	two protons and two neutrons	electron ( $β^{-}$ ) or positron ( $β^{+}$ )	photons (electromagnetic waves)
Mass	$\approx 4 u = 6.64 \times 1 0^{- 27} kg$	$9.11 \times 1 0^{- 31} kg$	none
Charge	$+ 2 e$	$- 1 e$ or $+ 1 e$	$0$

14.2Particle's decays

Antiparticles are the counterparts of regular particles. They share the same mass as their corresponding particles but have opposite electric charge. This is a consequence of the symmetries in the fundamental laws of physics. Two examples illustrate the idea.

Electron and positron

Electron: a subatomic particle with a negative charge of $- 1 e$ and a mass of about $9.11 \times 1 0^{- 31} kg$ .
Positron: the antiparticle of the electron. It has the same mass ( $9.11 \times 1 0^{- 31} kg$ ) but a positive charge of $+ 1 e$ .

The pairing of an electron and a positron is an example of matter-antimatter annihilation. When an electron and a positron collide, they annihilate each other and their masses are converted into energy in the form of gamma rays.

Proton and antiproton

Proton: found in the nucleus of an atom, with a positive charge of $+ 1 e$ and a mass of about $1.67 \times 1 0^{- 27} kg$ .
Antiproton: the antiparticle of the proton. It has the same mass ( $1.67 \times 1 0^{- 27} kg$ ) but a negative charge of $- 1 e$ . Antiprotons can be produced artificially in high-energy particle accelerators and are used in scientific experiments to study the fundamental properties of matter.

Particle and antiparticle pairs: same mass, opposite charge.
	Mass	Charge
Electron	$9.11 \times 1 0^{- 31} kg$	$- 1 e$
Positron	$9.11 \times 1 0^{- 31} kg$	$+ 1 e$
Proton	$1.67 \times 1 0^{- 27} kg$	$+ 1 e$
Antiproton	$1.67 \times 1 0^{- 27} kg$	$- 1 e$

These examples highlight the principle of charge conservation in particle-antiparticle pairs: the total charge stays conserved while the two members of a pair show opposite signs. The masses of particles and their antiparticles are identical.

14.2.1Alpha decaysOften examined

In alpha decay the atomic number of the parent nucleus reduces by $2$ and the mass number reduces by $4$ units. The decay product is called the daughter nucleus. The daughter nucleus may also remain unstable and undergo further disintegration until it attains stability. Alpha decay of radium is shown below:

_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He + Q

Alpha decay of radium-226

$Q$ is the disintegration energy, which is always positive because the process is spontaneous.

Reading the equation

An alpha particle is a helium nucleus $_{2}^{4} He$ , so it carries away $4$ nucleons and $2$ units of charge.
Mass number: the parent has $226$ , the daughter has $226 - 4 = 222$ .
$226 \to 222 + 4$
Atomic number: the parent has $88$ , the daughter has $88 - 2 = 86$ (radon).
$88 \to 86 + 2$
For any parent $X$ the alpha decay is therefore
$_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He + Q$

Worked example: two alpha decays in a row

Radium-226 (

Z = 88

) undergoes alpha decay, and its daughter nucleus decays by alpha emission again. Write both decay equations and find the mass number and atomic number of the final nucleus.

Two successive alpha decays starting from radium-226. Each decay lowers

A

4

and

Z

2

. The final nucleus is shown in accent.

In alpha decay the parent loses an alpha particle $_{2}^{4} He$ , so the mass number falls by $4$ and the atomic number by $2$ . First decay, mass number:
$226 - 4 = 222$
First decay, atomic number. A nucleus with $Z = 86$ is radon (Rn).
$88 - 2 = 86$
Write the first equation and check that charge ( $88 = 86 + 2$ ) and nucleon number ( $226 = 222 + 4$ ) are conserved.
$_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He + Q$
The radon nucleus is the new parent. Subtract $4$ and $2$ again: $A = 222 - 4 = 218$ and $Z = 86 - 2 = 84$ , which is polonium (Po) in the periodic table.
$_{86}^{222} Rn \to_{84}^{218} Po +_{2}^{4} He + Q$
Check the whole chain. Two alpha particles have taken away $8$ nucleons and $4$ units of charge.
$226 = 218 + 2 (4), 88 = 84 + 2 (2)$

_{84}^{218} Po : A = 218, Z = 84

14.2.2Beta decaysOften examined

Beta-minus ( $β^{-}$ ) decay and beta-plus ( $β^{+}$ ) decay are two types of radioactive decay in which beta particles are emitted. They occur in atomic nuclei that need to reach a more stable nuclear configuration.

Beta-minus decay

In beta-minus decay a neutron ( $n$ ) inside the nucleus is transformed into a proton ( $p$ ), an electron (the beta-minus particle $β^{-}$ ) and an antineutrino ( $\overset{ν}{ˉ}_{e}$ ). The antineutrino is a nearly massless, electrically neutral particle that carries away some of the energy released in the decay. The emitted beta-minus particle is an electron.

n \to p^{+} + β^{-} + \overset{ν}{ˉ}_{e}

Eq. 14.1

The newly formed proton stays in the nucleus and increases the atomic number by $1$ , while the beta-minus particle and the antineutrino leave the nucleus.

Beta-plus decay

In beta-plus decay a proton inside the nucleus is transformed into a neutron, a positron (the beta-plus particle $β^{+}$ ) and a neutrino ( $ν_{e}$ ). The emitted beta-plus particle is a positron, which is the antiparticle of the electron.

p^{+} \to n + β^{+} + ν_{e}

Eq. 14.2

The newly formed neutron stays in the nucleus and reduces the atomic number by $1$ , while the beta-plus particle and the neutrino leave the nucleus.

Beta decays compared.
	Beta-minus ( $β^{-}$ ) decay	Beta-plus ( $β^{+}$ ) decay
Change inside the nucleus	neutron becomes a proton	proton becomes a neutron
Particles emitted	electron and antineutrino ( $\overset{ν}{ˉ}_{e}$ )	positron and neutrino ( $ν_{e}$ )
Atomic number	increases by $1$	decreases by $1$
Mass (nucleon) number	unchanged	unchanged
Equation	$n \to p^{+} + β^{-} + \overset{ν}{ˉ}_{e}$	$p^{+} \to n + β^{+} + ν_{e}$

What happens to the nucleus

A neutron turning into a proton leaves the number of nucleons unchanged but adds one proton. So the atomic number goes up by $1$ and the mass number stays the same.
$_{Z}^{A} X \to_{Z + 1}^{A} Y + β^{-} + \overset{ν}{ˉ}_{e}$
A proton turning into a neutron again leaves the number of nucleons unchanged, but removes one proton. So the atomic number goes down by $1$ .
$_{Z}^{A} X \to_{Z - 1}^{A} Y + β^{+} + ν_{e}$
Charge balance for the neutron decay: the neutron is neutral; the products carry $+ 1$ , $- 1$ and $0$ .
$0 = (+ 1) + (- 1) + 0$

Both beta-minus and beta-plus decay are governed by the weak nuclear force, one of the fundamental forces in nature responsible for mediating certain types of particle interactions. These processes play a crucial role in the stability and transformation of atomic nuclei, allowing them to reach more balanced configurations and release excess energy.

14.2.3Energies for alpha and beta decayOften examined

Alpha and beta particles are products of radioactive decay, and they carry different amounts of energy because of their different properties and mechanisms of emission.

Energy of alpha particles: discrete

Alpha particles have discrete energies in nuclear processes. These energies are determined by the energy levels and quantum states of the particles within the nucleus and by the binding forces that hold the nucleus together. Inside a nucleus the protons and neutrons are arranged in energy levels or shells, similar to the way electrons are arranged around an atom's nucleus. These energy levels are quantized: they can only take certain discrete values.

Quantization of energy levels: Just as electrons in an atom exist only at certain discrete energy levels, nucleons within a nucleus also occupy quantized energy levels. When nucleons rearrange themselves to form an alpha particle, the energy change corresponds to the difference between the initial and final energy states.
Binding energies: The formation of an alpha particle involves the strong nuclear force. As nucleons combine to form an alpha particle, the binding energies change and the difference is released as energy, which is carried away by the alpha particle.
Conservation of energy: The total energy of the system before and after the alpha emission must be conserved. This leads to the emission of alpha particles with specific, well-defined energies.

Consequently the energies of alpha particles emitted in nuclear processes are quantized and discrete. These specific energies give valuable information about the structure of atomic nuclei, the binding forces involved and the dynamics of nuclear reactions.

Energy of beta particles: continuous

Unlike alpha particles, the energies of beta particles are spread out over a range of values. This continuous energy spectrum is attributed to the probabilistic nature of quantum mechanics together with conservation of energy and momentum. It arises from several factors:

Quantum uncertainty: in quantum mechanics particles are described by wave functions that have inherent uncertainty in certain properties, including energy. This allows a range of possible energies. When a beta particle is emitted, its specific energy is not predetermined.
Neutrinos in beta decay: in $β^{-}$ and $β^{+}$ decay an additional particle is involved, the antineutrino or neutrino. These are extremely light, with negligible mass, and they carry away some of the energy released. The neutrinos can take varying amounts of energy, leaving less for the beta particle, so the beta particles come out with a spread of energies.
Conservation laws: the total energy before and after the decay must be conserved, but there is flexibility in how it is shared between the beta particle and the accompanying neutrino. This gives a continuum of possible energies for the beta particles.
Final state interactions: the emitted beta particles can interact with other particles or fields as they travel through matter. This affects their energy and momentum and contributes to the continuous spectrum seen in experiments.

Why alpha energies are discrete and beta energies are continuous.
	Alpha particles	Beta particles
Energies	discrete (fixed, specific values)	continuous (any value over a range)
Main reason	quantized nuclear energy levels and conservation of energy	the (anti)neutrino carries away a varying share of the energy
Extra particle emitted	none	antineutrino ( $β^{-}$ ) or neutrino ( $β^{+}$ )

14.3Matter and anti-matterOften examined

In 1928 the British physicist Paul Dirac studied the behaviour of electrons at very high speeds (speeds comparable to the speed of light). By combining special relativity and quantum theory he derived an equation for the energy of an electron. For ordinary speeds the relation between energy and momentum of a particle is

E = \frac{p ^{2}}{2 m}

Eq. 14.2 (book numbering)

Solving the relation for relativistic speeds using special relativity and quantum theory, Dirac arrived at

E = \pm m^{2} c^{4} + p^{2} c^{2}

Eq. 14.4

Here $E$ is the energy of the particle, $m$ its mass, $p$ its momentum and $c$ the speed of light. The positive and negative solutions for energy give a striking idea about anti-particles.

Reading the two solutions

Put $p = 0$ (particle at rest) in Eq. 14.4.
$E = \pm m^{2} c^{4} = \pm m c^{2}$
The positive solution $+ m c^{2}$ is the familiar rest energy of the electron. The negative solution $- m c^{2}$ is the surprise: it is not an electron.
The gap between the two solutions at $p = 0$ is
$m c^{2} - (- m c^{2}) = 2 m c^{2}$
This gap, labelled $2 m c^{2}$ , is the narrow waist between the two cones of Fig. 14.1.

Energy of a particle from Dirac's equation

−3

−2

−1

−3

−2

−1

p c / m c^{2}

E / m c^{2}

+ m c^{2}

- m c^{2}

2 m c^{2}

E = + m^{2} c^{4} + p^{2} c^{2}

E = - m^{2} c^{4} + p^{2} c^{2}

Figure 14.1 (slice of the energy cones):

E

against

p c

in units of

m c^{2}

. The upper curve is the positive solution (the electron), the lower curve is the negative solution (the positron). At

p = 0

the two are separated by

2 m c^{2}

If the positive solution of Dirac's equation is for the electron (the particle), then the negative solution for energy must be for a particle which has the mass and momentum of the electron but the opposite charge and magnetic moment. Dirac said this energy solution is for the positron, the anti-particle of the electron. The energy cones for the electron and positron are shown in Fig. 14.1.

Soon after Dirac predicted anti-particles, Anderson discovered the positron. In 1932, while studying the tracks of cosmic ray particles in a cloud chamber at the California Institute of Technology, he found a particle with the mass of an electron but the opposite (positive) charge. This was the first experimental discovery of an anti-particle. Many more anti-particles were found afterwards. Electron, proton and neutron, with their anti-particles, are listed in Table 14.1.

Table 14.1: Particles and anti-particles.
Property	Electron	Positron	Proton	Anti-proton	Neutron	Anti-neutron
Mass (kg)	$9.1 \times 1 0^{- 31}$	$9.1 \times 1 0^{- 31}$	$1.67 \times 1 0^{- 27}$	$1.67 \times 1 0^{- 27}$	$1.67 \times 1 0^{- 27}$	$1.67 \times 1 0^{- 27}$
Mass ( $MeV / c^{2}$ )	$0.51$	$0.51$	$938.28$	$938.28$	$939.57$	$939.57$
Charge	$- 1 e$	$+ 1 e$	$+ 1 e$	$- 1 e$	$0$	$0$
Spin	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
Family	fermion	fermion	fermion	fermion	fermion	fermion

Boson: A particle with integer spin ( $0, 1, 2, \dots$ ).
Fermion: A particle with half-integer spin ( $\frac{1}{2}, \frac{3}{2}, \dots$ ).

Particles are divided into these two groups according to their spins. Electron, positron, proton, anti-proton, neutron and anti-neutron are all fermions.

Example 14.1: Momentum of a particle of known energy

The energy of a particle system is

2 MeV

moving with relativistic speed, and the particles approach each other. If their masses are

0.511 MeV / c^{2}

, find their momentum. Also justify why momentum has two values, positive and negative.

Given

$E = 2 MeV = (2 \times 1 0^{6}) (1.6 \times 1 0^{- 19}) J = 3.2 \times 1 0^{- 13} J$ energy
$m = 0.511 MeV / c^{2} = \frac{( 0.511 \times 1 0 ^{6} ) ( 1.6 \times 1 0 ^{- 19} )}{9 \times 1 0 ^{16}} kg \approx 9.11 \times 1 0^{- 31} kg$ mass
$c = 3 \times 1 0^{8} m s^{- 1}$ speed of light

To find

momentum $p$

Use the energy-momentum relation of Eq. 14.4.
$E = \pm m^{2} c^{4} + p^{2} c^{2}$
Square both sides to remove the root and the $\pm$ .
$E^{2} = m^{2} c^{4} + p^{2} c^{2}$
Rearrange for $p^{2}$ .
$p^{2} = \frac{E ^{2} - m ^{2} c ^{4}}{c ^{2}}$
Put in the values. First the rest energy: $m c^{2} = (9.11 \times 1 0^{- 31}) (3 \times 1 0^{8})^{2} = 8.2 \times 1 0^{- 14} J$ .
$p = \frac{( 3.2 \times 1 0 ^{- 13} ) ^{2} - ( 9.11 \times 1 0 ^{- 31} ) ^{2} ( 3 \times 1 0 ^{8} ) ^{4}}{( 3 \times 1 0 ^{8} ) ^{2}}$
Evaluate. The numerator is $1.024 \times 1 0^{- 25} - 6.7 \times 1 0^{- 27} = 9.57 \times 1 0^{- 26} J^{2}$ ; divide by $9 \times 1 0^{16}$ and take the root.
$p = 1.06 \times 1 0^{- 42} = \pm 1.03 \times 1 0^{- 21} kg m s^{- 1}$
Cross-check in energy units: $p c = E^{2} - (m c^{2})^{2} = 4 - 0.261 = 1.93 MeV$ , so $p = 1.93 MeV / c$ , which is the same $1.03 \times 1 0^{- 21} kg m s^{- 1}$ .

Why two values? The momentum appears only as $p^{2}$ in the energy relation, so both $+ p$ and $- p$ satisfy it. The two values are the momenta of the two particles moving towards each other, equal in size and opposite in direction. The book links the positive and negative values to the particle and the anti-particle.

p = \pm 1.03 \times 1 0^{- 21} kg m s^{- 1}

Assignment 14.1: Momentum of a positron

Find the momentum of a positron moving with relativistic speed. The energy of the positron is

0.47 MeV

Given

$K = 0.47 MeV$ kinetic energy of the positron
$m c^{2} = 0.511 MeV$ rest energy

To find

momentum $p$

Total energy is kinetic plus rest energy.
$E = K + m c^{2} = 0.47 + 0.511 = 0.981 MeV$
Use $E^{2} = (p c)^{2} + (m c^{2})^{2}$ and solve for $p c$ .
$p c = E^{2} - (m c^{2})^{2} = (0.981)^{2} - (0.511)^{2} MeV$
Evaluate.
$p c = 0.962 - 0.261 = 0.837 MeV$
Convert to SI: $1 MeV = 1.6 \times 1 0^{- 13} J$ and divide by $c$ .
$p = \frac{0.837 \times 1.6 \times 1 0 ^{- 13}}{3 \times 1 0 ^{8}} = 4.47 \times 1 0^{- 22} kg m s^{- 1}$

p \approx 0.84 MeV / c \approx 4.5 \times 1 0^{- 22} kg m s^{- 1}

Worked example: the two energies of an electron at a given momentum

An electron has momentum

p

such that

p c = 1.00 MeV

. Use the energy relation

E = \pm m^{2} c^{4} + p^{2} c^{2}

to find the two energies, the kinetic energy of the positive-energy solution, and compare the gap between the two solutions with the gap at

p = 0

. Take

m c^{2} = 0.511 MeV

Given

$m c^{2} = 0.511 MeV$ rest energy of the electron
$p c = 1.00 MeV$

To find

$E$ (both signs)
kinetic energy of the positive solution
the gap between the two solutions

Square the rest energy.
$(m c^{2})^{2} = 0.51 1^{2} = 0.261 Me V^{2}$
Put it into the relation with $(p c)^{2} = 1.00$ .
$E = \pm 0.261 + 1.00 = \pm 1.261$
Take the root.
$E = \pm 1.12 MeV$
The kinetic energy of the positive solution is the total energy minus the rest energy.
$K = E - m c^{2} = 1.12 - 0.511 = 0.61 MeV$
The gap between the two solutions is $2 E$ . At $p = 0$ it was $2 m c^{2} = 1.022 MeV$ , so at this momentum the gap is wider.
$2 E = 2 (1.12) = 2.25 MeV$

The two solutions at pc = 1.00 MeV

−3

−2

−1

−3

−2

−1

p c (MeV)

E (MeV)

(1.00, 1.12)

(1.00, - 1.12)

E = + m^{2} c^{4} + p^{2} c^{2}

E = - m^{2} c^{4} + p^{2} c^{2}

The upper curve is the positive solution and the lower curve the negative solution. At

p c = 1.00 MeV

they sit at

\pm 1.12 MeV

, further apart than the gap

2 m c^{2}

p = 0

E = \pm 1.12 MeV, K = 0.61 MeV

Physicists at the Relativistic Heavy Ion Collider (RHIC) in New York created the anti-nuclei of helium-4. Anti-helium-4 is the heaviest anti-particle produced. Scientists at CERN (the European research laboratory for particle physics) artificially created an atom of anti-matter, which has opened the doors of further research in the field of anti-matter. Hydrogen along with its anti-hydrogen is shown in Fig. 14.2.

Figure 14.2: Hydrogen and anti-hydrogen. Anti-hydrogen is built from the anti-particles: an antiproton (made of anti-quarks) and a positron.

14.3.1Asymmetry of matter and anti-matter

Particles and anti-particles mirror each other almost perfectly, so for every particle in the universe there ought to be a particle of anti-matter. Our Earth, the Sun and the dust between the galaxies are made of matter. The whole universe looks to be made of matter, so when we look around we do not see any anti-matter. Why? This asymmetry in the universe is essential for the existence of stars and even life itself, so asymmetry is a good thing in itself.

C-symmetry (charge conjugation): A particle and its anti-particle are charge conjugates of each other (a positive and a negative pair). Whenever a pair of particle and anti-particle is made in a reaction, the law of conservation of charge requires this conjugation.
Parity (P): The particle and anti-particle are like ‘mirrors’ to each other, as in the inversion of spatial coordinates. The state of being equal with reference to some coordinate is called parity (the space inversion).
CP-symmetry: Charge conjugation and parity taken together.
CP-violation: A phenomenon where the same decay process has a different probability for a particle than for an anti-particle.

Figure 14.3: A simple depiction of C-symmetry (swap positive and negative) and P-symmetry (mirror the space).

If matter and anti-matter are perfectly balanced, why is anti-matter so rare? What happened to it? The answer lies in the early universe.

The story in steps

The most accepted theory of the creation of the universe is the big bang theory. About 13.7 billion years ago an explosion produced matter and anti-matter in perfectly equal amounts, with CP-symmetry.
With perfect symmetry, all the matter would have annihilated with the anti-matter, leaving an empty universe.
The real universe has an excess of matter, the part that survived the annihilation. This is due to CP-violation.
In particle physics CP-violation is the violation of the combined conservation laws, charge conjugation and parity, by the weak force. It seems to be the main cause of the asymmetry.
In the early universe equal amounts of matter and anti-matter were produced, but after the first second, when nuclei formed, matter started dominating.

Figure 14.4: The timeline of the universe. Early on there were equal amounts of matter and anti-matter; the universe later became matter dominated.

In the nucleus two forces are prominent: the strong nuclear force, which binds the nucleus, and the weak force, responsible for nuclear decays. Both violate CP. The evidence for CP-violation of the strong force in the early universe is less than for the weak force.

14.4QuarksOften examined

In the early 20th century protons and neutrons were considered elementary particles. However, in 1964 Murray Gell-Mann and George Zweig independently proposed the quark theory, which described protons and neutrons as being composed of quarks. The existence of quarks was first confirmed in a deep inelastic scattering experiment at the Stanford Linear Accelerator Center (SLAC) in 1968.

Elementary particles: Particles which are not composed of other particles.

Elementary particles are broadly of two groups: the matter particles (fermions, with half-integer spin) and the force carriers (bosons, with integer spin). Like the periodic table of elements, elementary particles are arranged in a table known as the Standard Model of elementary particles (Fig. 14.5).

Figure 14.5: The standard model of particles. Matter particles (quarks and leptons, in three generations I, II, III) on the left; force carriers (gauge bosons) and the scalar Higgs boson on the right.

The first group of matter particles in the table is the quarks, which have a size of roughly 10^{-18}\,\mathrm{m}. Quarks are elementary particles and the basic building blocks of mesons and baryons, the main constituents of all matter. Each quark has a corresponding anti-quark with the same properties but opposite charge. Quarks combine to form composite particles called hadrons. Quarks come in six flavours (a flavour is a property of the particle): up, down, charm, strange, top and bottom. All the quarks have spin \tfrac12.

Figure 14.6: The quarks (spin \tfrac12 for all). Charges are in units of the electron charge e.
Quark (anti-quark)	Generation	Charge	Mass	Notes
Up u (\bar u)	first	+\tfrac23e	1.7 to 3.1\,\mathrm{MeV}/c^2 (\approx2.2)	lightest quark, so the most stable
Down d (\bar d)	first	-\tfrac13e	4.1 to 5.7\,\mathrm{MeV}/c^2 (\approx4.7)	heavier than up, lighter than the rest; highly stable
Charm c (\bar c)	second	+\tfrac23e	\approx1280\,\mathrm{MeV}/c^2 (1.28\,\mathrm{GeV}/c^2)	hundreds of times more massive than up and down
Strange s (\bar s)	second	-\tfrac13e	\approx96\,\mathrm{MeV}/c^2	very long life, about 100\,000 times longer than similar particles
Top t (\bar t)	third	+\tfrac23e	\approx173.1\,\mathrm{GeV}/c^2	heaviest quark; not stable, decays before forming hadronic bound states
Bottom b (\bar b)	third	-\tfrac13e	\approx4.18\,\mathrm{GeV}/c^2	decays mostly into a charm quark, rarely into an up quark

Colour charge

Another property of quarks and gluons is the colour charge, related to the particle's strong interaction in the theory of quantum chromodynamics (QCD). Quantum chromodynamics is the model that governs the behaviour of quarks. This colour property is totally different from the everyday meaning of colour. Quarks are found in three colour charges, red, blue and green, along with three anti-colour charges, anti-red, anti-blue and anti-green.

14.4.1HadronsOften examined

Hadron: A subatomic particle made up of quarks and anti-quarks held together by the strong force with the help of gluons.

Hadrons are the heaviest particles. Quarks have fractional charges, but hadrons, which are made up of quarks, have integer charges. Examples are kaons, protons, neutrons, anti-protons and anti-neutrons. Hadrons experience the strong nuclear force. Their properties are:

They carry no net colour charge, although the quarks that make them up carry colour charge.
Hadrons are unstable and can decay. The only hadron that is stable even in the free state is the proton, while the neutron is stable only within the nucleus.

Depending on the quarks taking part, hadrons are divided into two groups: baryons and mesons.

1. Baryons

Baryon: A hadron made up of three quarks. Baryons are approximately equal in mass to a proton or greater. They are fermions, with half-integer spin because of the odd number of quarks. The proton and the neutron are well-known baryons.
Baryon number: A quantum number equal to the number of baryons minus the number of anti-baryons in a system of subatomic particles. Baryons have baryon number +1 and anti-baryons have baryon number -1.

Table 14.2: Some baryons with their properties.
Particle	Rest mass (\mathrm{MeV}/c^2)	Spin	Electric charge	Quark composition	Baryon number
Proton (p)	938.3	\tfrac12	+1	uud	+1
Neutron (n)	939.6	\tfrac12	0	ddu	+1
Omega (\Omega)	1672	\tfrac32	-1	sss	+1
Delta (\Delta)	1232	\tfrac32	+2	uuu	+1

2. Mesons

Meson: A hadron made up of a quark and an anti-quark pair. Mesons are lighter than baryons. They are bosons, with integral spin (-1,0,+1), because the number of quarks in them is even. Their baryon number is 0. Commonly known mesons are the pion, kaon and rho.

Figure 14.7: Left, a baryon (three quarks); right, a meson (a quark and an anti-quark).

Table 14.3: Some mesons with their properties.
Particle	Rest mass (\mathrm{GeV}/c^2)	Spin	Electric charge	Quark composition
Pion (\pi^{+})	0.140	0	+1	u\bar d
Kaon (K^{-})	0.494	0	-1	s\bar u
Kaon (K^{0})	0.498	0	0	d\bar s
Rho (\rho^{+})	0.770	1	+1	u\bar d

Baryons and mesons compared.
	Baryon	Meson
Quark content	three quarks	a quark and an anti-quark
Spin	half-integer (fermion)	integer (boson)
Baryon number	+1	0
Examples	proton, neutron, omega, delta	pion, kaon, rho

Worked example: quarks inside a meson and a baryon

Using the quark charges u=+\tfrac23e, d=-\tfrac13e and the rule that an anti-quark has the opposite charge, find the charge of (a) a particle d\bar u, (b) a particle uuu, (c) a particle \bar u\bar u\bar d. Say whether each is a meson or a baryon (or anti-baryon), and name the ones that appear in Tables 14.2 and 14.3.

A meson has a quark and an anti-quark; a baryon has three quarks. Anti-quarks have opposite charges: \bar u=-\tfrac23e and \bar d=+\tfrac13e.
(a) d\bar u is a quark plus an anti-quark, so a meson.
-\tfrac13-\tfrac23=-1\quad\Rightarrow\ \text{the pion }\pi^{-}\ (\text{the anti-particle of }\pi^{+}=u\bar d)
(b) uuu has three quarks, so a baryon.
+\tfrac23+\tfrac23+\tfrac23=+2\quad\Rightarrow\ \text{the delta }\Delta\ \text{of Table 14.2}
(c) \bar u\bar u\bar d has three anti-quarks, so an anti-baryon.
-\tfrac23-\tfrac23+\tfrac13=-1\quad\Rightarrow\ \text{the anti-proton}\ (p=uud)
Check the rule for hadrons: every total is a whole number, as it must be, because the thirds always cancel in a valid hadron.

Quark pictures of a meson (a quark and an anti-quark) and a baryon (three quarks). The charges of the quarks add to the integer charge of the hadron (totals in accent).

\text{(a) }-1\ (\text{meson});\quad \text{(b) }+2\ (\text{baryon});\quad \text{(c) }-1\ (\text{anti-baryon})

14.4.2Composition of proton and neutronOften examined

Protons and neutrons are made up of quarks. A proton is made up of two up quarks and one down quark. An up quark has a charge of +\tfrac23e and a down quark has a charge of -\tfrac13e. Adding the charges of the three quarks gives +1e, the charge of the proton. The anti-proton consists of the anti-quarks of the same type.

Proton

Quark content of the proton:
p=u\,u\,d
Add the charges of the quarks.
q_p=\left(+\dfrac23\right)+\left(+\dfrac23\right)+\left(-\dfrac13\right)=+1
Add the spins as well (two quarks up, one down, in the figure): \tfrac12+\tfrac12-\tfrac12=\tfrac12, which is the spin of the proton.

In terms of colour charge a proton consists of a blue-up, a red-up and a green-down quark.

Neutron

Quark content of the neutron:
n=d\,d\,u
Add the charges of the quarks.
q_n=\left(-\dfrac13\right)+\left(-\dfrac13\right)+\left(+\dfrac23\right)=0
The spins add up as -\tfrac12+\tfrac12+\tfrac12=\tfrac12, the spin of the neutron.

A neutron is made up of two down quarks and one up quark. Both down quarks have charge -\tfrac13e and the up quark has +\tfrac23e, so the total is zero. The anti-neutron consists of the anti-quarks of the same type. In terms of colour charge a neutron consists of a blue-down, a red-down and a green-up quark.

Figure 14.8: Composition of proton and neutron. Spins: \tfrac12+\tfrac12-\tfrac12=\tfrac12 for the proton and -\tfrac12+\tfrac12+\tfrac12=\tfrac12 for the neutron. Charges: +\tfrac23+\tfrac23-\tfrac13=1 and +\tfrac23-\tfrac13-\tfrac13=0.

Figure 14.9: Composition of the anti-proton and the anti-neutron (lower row), compared with the proton and the neutron.

Quark composition of the nucleons and their anti-particles.
Particle	Quarks	Charge
Proton	u\,u\,d	+1
Neutron	d\,d\,u	0
Anti-proton	\bar u\,\bar u\,\bar d	-1
Anti-neutron	\bar d\,\bar d\,\bar u	0

14.4.3Change in quarks during beta decaysOften examined

Whenever a proton changes into a neutron or a neutron changes into a proton, the process is called beta decay (Section 14.2.2). We now study how the quarks change their type during these decays.

1) Beta-positive decay

A proton is converted into a neutron, a positron and a neutrino. This decay is represented by Eq. 14.2:

p^{+}\;\to\;n+e^{+}+\nu_e

Only the proton and the neutron are made of quarks. A proton consists of two up quarks and one down quark, and a neutron of two down quarks and one up quark. So in this reaction one of the up quarks turns into a down quark, with the production of a weak boson W^{+}. The W^{+} is very short-lived and soon decays into a positron and a neutrino through the weak interaction.

The decaying proton has the quarks u\,u\,d. One u changes into a d and a W^+ boson is emitted.
u\;\to\;d+W^{+}
The short-lived W^+ decays by the weak interaction.
W^{+}\;\to\;e^{+}+\nu_e
The quarks that are left, u\,d\,d, form a neutron. Charge check on the first step: +\tfrac23=-\tfrac13+1.
(u\,u\,d)\;\to\;(u\,d\,d)+e^{+}+\nu_e

Figure 14.10: Beta-positive decay. An up quark turns into a down quark (shown in accent) with the emission of a W^+ boson, which decays into e^+ and \nu_e.

2) Beta-negative decay

A neutron is converted into a proton, an electron and an anti-neutrino. This decay is given by Eq. 14.1:

n\;\to\;p^{+}+e^{-}+\bar\nu_e

Again only the neutron and the proton are made of quarks. A neutron consists of two down quarks and one up quark, while a proton consists of two up quarks and one down quark. In this reaction one of the down quarks turns into an up quark, with the production of a weak boson W^{-}, which is very short-lived and soon decays into an electron and an anti-neutrino through the weak interaction.

The decaying neutron has the quarks u\,d\,d. One d changes into a u and a W^- boson is emitted.
d\;\to\;u+W^{-}
The W^- decays by the weak interaction.
W^{-}\;\to\;e^{-}+\bar\nu_e
The quarks that are left, u\,d\,u, form a proton. Charge check on the first step: -\tfrac13=+\tfrac23-1.
(u\,d\,d)\;\to\;(u\,u\,d)+e^{-}+\bar\nu_e

Figure 14.11: Beta-negative decay. A down quark turns into an up quark (shown in accent) with the emission of a W^- boson, which decays into e^- and \bar\nu_e.

Quark-level picture of the two beta decays.
	Beta-plus decay	Beta-minus decay
Nucleon change	p\to n	n\to p
Quark change	up \to down	down \to up
Weak boson	W^+	W^-
Leptons emitted	e^+ and \nu_e	e^- and \bar\nu_e

14.5LeptonsOften examined

In the Standard Model the second group of matter particles is the leptons. Leptons are also elementary particles; they are not composed of other particles. Leptons can have a unit positive or a unit negative electric charge or may be neutral. The smallest lepton is the electron, which is the most stable among all leptons. The other leptons, the muon and the tau, exist only in high-energy collisions and last for a very short period.

Lepton: An elementary (fundamental) particle of spin \tfrac12 that does not take part in strong interactions. The electron, muon, tau and their three neutrinos are leptons.

Leptons have spin \tfrac12 and do not take part in strong interactions; the strong nuclear force is blind for leptons. Leptons cannot be found within the nucleus or nucleons; if they are produced in a nucleus, they escape with large kinetic energies.

Figure 14.12 and Table 14.4: Leptons with their properties (the tau mass is 1.7768\,\mathrm{GeV}/c^2 in Fig. 14.12).
Particle	Symbol	Rest mass	Spin	Electric charge	Lepton number
Electron	e^-	0.511\,\mathrm{MeV}/c^2	\tfrac12	-1	+1
Muon	\mu	105.66\,\mathrm{MeV}/c^2	\tfrac12	-1	+1
Tau	\tau	1776.8\,\mathrm{MeV}/c^2	\tfrac12	-1	+1
Electron neutrino	\nu_e	<1.0\,\mathrm{eV}/c^2	\tfrac12	0	+1
Muon neutrino	\nu_\mu	<0.17\,\mathrm{MeV}/c^2	\tfrac12	0	+1
Tau neutrino	\nu_\tau	<18.2\,\mathrm{MeV}/c^2	\tfrac12	0	+1

Lepton number: A conserved quantum number. For leptons its value is +1 and for anti-leptons it is -1. For any reaction the lepton number is the number of leptons minus the number of anti-leptons.

14.5.1Conservation laws of lepton and baryonOften examined

Besides the conservation laws of mass, energy, momentum and charge, we need two more: the lepton-number conservation and the baryon-number conservation. Every particle interaction must be according to these laws. An interaction which does not satisfy any of these laws cannot take place in the real world.

Law of conservation of lepton number: The lepton number of a reaction on both sides should essentially be the same.
Law of conservation of baryon number: The baryon number of a reaction on both sides should essentially be the same.

Values to use when checking a reaction.
Quantity	Particle	Anti-particle
Lepton number	+1 (electron, muon, tau, neutrinos)	-1
Baryon number	+1 (proton, neutron, other baryons)	-1
Mesons, photons, bosons	0	0

Example 14.2: Conservation laws in beta decay

Show that the beta-positive decay and the beta-negative decay satisfy the laws of conservation of charge, lepton number and baryon number.

Given

\text{Beta-negative} = n\to p^{+}+e^{-}+\bar\nu_e
\text{Beta-positive} = p^{+}\to n+e^{+}+\nu_e

To find

charge, lepton number and baryon number are the same on both sides

Values used for each particle.
Particle	Charge	Lepton number	Baryon number
n	0	0	+1
p^{+}	+1	0	+1
e^{-}	-1	+1	0
e^{+}	+1	-1	0
\nu_e	0	+1	0
\bar\nu_e	0	-1	0

Conservation of charge

Beta-negative decay: charge before =0; charge after =(+1)+(-1)+(0).
0=(+1)+(-1)+(0)\;\Rightarrow\;0=0\quad\text{satisfied}
Beta-positive decay: charge before =+1; charge after =(0)+(+1)+(0).
+1=(0)+(+1)+(0)\;\Rightarrow\;+1=+1\quad\text{satisfied}

Conservation of lepton number

Beta-negative decay: lepton number before =0; after =(0)+(+1)+(-1), because e^- is a lepton (+1) and \bar\nu_e is an anti-lepton (-1).
0=(0)+(+1)+(-1)\;\Rightarrow\;0=0\quad\text{satisfied}
Beta-positive decay: lepton number before =0; after =(0)+(-1)+(+1), because e^+ is an anti-lepton (-1) and \nu_e is a lepton (+1).
0=(0)+(-1)+(+1)\;\Rightarrow\;0=0\quad\text{satisfied}

Conservation of baryon number

Beta-negative decay: baryon number before =+1 (neutron); after =(+1)+(0)+(0), since only the proton is a baryon.
+1=(+1)+(0)+(0)\;\Rightarrow\;+1=+1\quad\text{satisfied}
Beta-positive decay: baryon number before =+1 (proton); after =(+1)+(0)+(0), since only the neutron is a baryon.
+1=(+1)+(0)+(0)\;\Rightarrow\;+1=+1\quad\text{satisfied}

As all the basic conservation laws are satisfied, the beta-positive and beta-negative decays are real (they can happen).

\text{Charge, lepton number and baryon number are conserved in both decays}

Assignment 14.2: Checking particle reactions

Show that the particle reactions given below satisfy the laws of conservation of energy, charge, lepton number and baryon number. (i) \pi^{-}+p\to\pi^{0}+n+\pi^{-}+\pi^{+} (ii) \mu^{+}\to e^{+}+\nu_e+\bar\nu_\mu

The pions are mesons, so they carry no lepton number and no baryon number (Table 14.3). The \mu^+ is the anti-particle of the muon, so its lepton number is -1.

(i) \pi^{-}+p\to\pi^{0}+n+\pi^{-}+\pi^{+}

Totals for reaction (i): all three are equal on the two sides.
	Charge	Lepton number	Baryon number
Left: \pi^-+p	-1+1=0	0+0=0	0+1=+1
Right: \pi^0+n+\pi^-+\pi^+	0+0-1+1=0	0	0+1+0+0=+1

Energy: the total energy (rest mass energy plus kinetic energy) is conserved. The right-hand side has more rest mass: about 0.140\times3+0.940=1.36\,\mathrm{GeV} against 0.140+0.938=1.08\,\mathrm{GeV} on the left. The extra \approx0.28\,\mathrm{GeV} is supplied by the kinetic energy of the incoming pion, so energy is conserved when the pion arrives with at least that much kinetic energy.

(ii) \mu^{+}\to e^{+}+\nu_e+\bar\nu_\mu

Totals for reaction (ii).
	Charge	Lepton number	Baryon number
Left: \mu^+	+1	-1	0
Right: e^++\nu_e+\bar\nu_\mu	+1+0+0=+1	-1+1-1=-1	0

Energy: the muon is much heavier (105.66\,\mathrm{MeV}/c^2) than the positron (0.511\,\mathrm{MeV}/c^2) and the neutrinos are almost massless, so about 105\,\mathrm{MeV} of rest energy is released as kinetic energy of the products. The total energy is conserved and the decay is allowed.

\text{All four quantities are conserved in both reactions}

Worked example: why the antineutrino must be there

A student writes the decay of a neutron as n\to p^{+}+e^{-}. Test it against the conservation of charge, baryon number and lepton number. What must be added to make the decay allowed?

Values used.
Particle	Charge	Baryon number	Lepton number
n	0	+1	0
p^{+}	+1	+1	0
e^{-}	-1	0	+1
\bar\nu_e	0	0	-1

Charge: left =0; right =(+1)+(-1)=0.
0=0\quad\text{satisfied}
Baryon number: left =+1; right =(+1)+(0)=+1.
+1=+1\quad\text{satisfied}
Lepton number: left =0; right =(0)+(+1)=+1.
0\neq+1\quad\text{violated}
The electron has lepton number +1, so something with lepton number -1 and no charge or baryon number must come out as well. The anti-neutrino \bar\nu_e has exactly these values.
n\to p^{+}+e^{-}+\bar\nu_e
Recheck lepton number with the antineutrino included.
0=(0)+(+1)+(-1)=0\quad\text{satisfied}

n\to p^{+}+e^{-}+\bar\nu_e

Feynman diagram of two approaching protons repelling each other by exchanging a photon (the carrier of the electromagnetic force).

14.6Force carriersOften examined

In the Standard Model there is a group of particles which do not make up matter but are responsible for binding matter together. The interactions between matter particles are mainly due to four fundamental forces: the strong nuclear force, the weak nuclear force, the electromagnetic force and the gravitational force. Each force acts through the exchange of particles called force carriers or mediators. The force carriers are shown in Fig. 14.13 and described below.

Figure 14.13: Force carriers (interactions / bosons) of the Standard Model.
Force carrier	Force it carries	Mass	Charge	Spin
Gluon (g)	strong nuclear force	0	0	1
Photon (\gamma)	electromagnetic force	0	0	1
Z boson	weak nuclear force	\approx91.19\,\mathrm{GeV}/c^2	0	1
W boson (W^+, W^-)	weak nuclear force	\approx80.360\,\mathrm{GeV}/c^2	\pm1	1
Higgs boson (H)	gives mass (a scalar boson, not a force carrier)	\approx124.97\,\mathrm{GeV}/c^2	0	0

14.6.1The photonOften examined

In interactions involving the electromagnetic force, the exchange particle is the photon. A photon is the smallest quantum of electromagnetic radiation and the basic unit of light. Photons travel through vacuum with a speed of 3\times10^{8}\,\mathrm{m\,s^{-1}}. According to Max Planck, the energy of a photon is the product of its oscillation frequency and Planck's constant:

E=hf

Eq. 14.5

Photons have zero rest mass energy; they exist only as moving particles.
They are neutral and stable particles.
They have a spin of 1, so they belong to the boson family.
They are elementary particles which have energy, momentum and frequency of oscillation.

According to quantum field theory, photons are exchanged between charged particles when they exert the electromagnetic force on each other. Hence they are the mediators of the electromagnetic force. During electromagnetic interactions every photon is emitted by one charged particle and absorbed by another, so the photon carries the electromagnetic force.

Figure 14.14: Photons as mediators of the electromagnetic force. Two approaching electrons repel each other by exchanging a photon.

14.6.2The gluon

Gluon: A particle that is exchanged between quarks to bind them together to form particles like protons and neutrons.

Gluons are massless particles and travel at the speed of light.
They are vector bosons, which means they have a spin of 1.
They are the exchange particles for the colour force (the strong nuclear force).
Like quarks, gluons also have the property of colour, and they can exchange colour to conserve colour charge.
They have no electric charge.

During strong interactions one quark or anti-quark emits a gluon and another quark or anti-quark absorbs the gluon; hence gluons are called the quanta of the strong force. The exchange of a gluon between two quarks is shown in Fig. 14.15.

Figure 14.15: Exchange of a gluon between two quarks.

14.6.3The Z-boson

The Z boson was discovered in 1983 at CERN. It is electrically neutral. Its mass is 91.19\,\mathrm{GeV}/c^2, which is nearly a hundred times that of the proton, making it one of the heaviest elementary particles. Bosons of this kind are very short-lived, with a lifetime of the order of 10^{-25}\,\mathrm{s}.

The Z boson is the carrier of the weak nuclear force and is a partner to the W^+ and W^- bosons, which mediate radioactive decay processes.
In weak interactions the W and Z bosons interact with each other as well as with quarks and leptons.

14.6.4The W-boson

W bosons are found in two different charge compositions, W^{+} and W^{-}, so they can have both positive and negative electric charge. The W and Z bosons are gauge bosons that mediate the weak nuclear force. They have masses of 80.360\,\mathrm{GeV}/c^2 and 91.2\,\mathrm{GeV}/c^2 respectively and spin 1.

They are responsible for many nuclear processes, such as the reactions that power stars (and the beta decays met in Section 14.4.3). They are very short-lived particles, with a lifetime of the order of 3\times10^{-25}\,\mathrm{s}. According to the book they do not have their anti-particles.

The weak-force carriers compared.
	W^{\pm} bosons	Z boson
Force mediated	weak nuclear force	weak nuclear force
Charge	+1 or -1	0
Mass	80.360\,\mathrm{GeV}/c^2	91.19\,\mathrm{GeV}/c^2
Spin	1	1
Lifetime	\sim3\times10^{-25}\,\mathrm{s}	\sim10^{-25}\,\mathrm{s}

14.6.5The Higgs bosonOften examined

Up to the year 1964 the mass of a particle was considered to be an intrinsic property. In 1964 Peter Higgs proposed a particle which is responsible for the mass of matter, which he called the Higgs particle. In July 2012 this particle was experimentally confirmed at CERN. The Higgs particle has a mass of 124.97\,\mathrm{GeV}/c^2, which makes it about 130 times more massive than a proton.

Peter Higgs gave the idea of a new type of field, called the Higgs field, that fills the universe and gives mass to all elementary particles. It is a scalar field and the Higgs boson is a wave in that field. It has zero spin, zero electric charge and no strong force interactions.

The four particles above (photon, gluon, Z and W) are all gauge bosons, that is, forms of force carrier, while the Higgs boson is an elementary particle which is produced due to the quantum excitation of the Higgs field. How strongly a particle interacts with the Higgs field decides its mass:

Quarks interact strongly with the Higgs field, gaining relatively large mass.
Electrons (leptons) interact only slightly with the Higgs field, making them extremely light particles.
Photons have no mass, and hence do not interact with the Higgs field.

14.6.6Theories about generation of force and mass

In classical physics we study only the macroscopic concepts of mass and force. In quantum physics, however, these two have very different meanings. There are four basic forces of nature: the strong nuclear force, the weak nuclear force, the electromagnetic force and the gravitational force. Two main theories about the generation of force are discussed here. The generation of mass, as explained in Section 14.6.5, involves the Higgs field. For force there are two main theories: the quantum field theory and the string theory.

Quantum field theory (QFT)

Quantum field theory is a set of physical principles that combine the elements of quantum mechanics with the theory of relativity to explain the behaviour of subatomic particles and their interactions due to a variety of force fields. Three different quantum field theories address three of the four fundamental forces in nature. Matter interacts through these forces: electromagnetism explains how atoms hold together, the strong nuclear force explains the stability of nuclei, and the weak nuclear force explains why some atoms undergo radioactive decay.

QFT describes particles as excited states of quantum fields, which are more fundamental than the particles themselves.
The interaction between particles is basically an interaction between the fields. Each interaction can be represented by a Feynman diagram.
According to QFT, mass is not the strength of attraction of two bodies or the quantity of matter; rather, it is the tendency of an object to resist changes in its speed and position.
At the smallest level, everything is made up of some entities or fluid-like substance called a quantum field. This field sometimes behaves like particles, sometimes like waves, and they can even interact with each other.

String theory

String theory suggests that the whole universe is made up of tiny vibrating strings. These strings are smaller than the smallest particles. They can twist, vibrate and fold, and by twisting, folding and vibrating they create matter, energy and forces. Phenomena such as electromagnetism and gravity arise from these strings. The theory replaces point-like particles by one-dimensional strings. According to the theory, strings have characteristics like charge and mass, and these characteristics are controlled by their vibrations. The composition of matter according to string theory is shown in Fig. 14.16.

Figure 14.16: Composition of matter according to string theory. Matter is made of molecules, molecules of atoms, atoms of nucleons, and nucleons of tiny vibrating strings.

The two theories compared.
	Quantum field theory	String theory
Basic entity	quantum fields (particles are their excitations)	tiny one-dimensional vibrating strings
Particles are	excited states of the fields	point-like particles replaced by strings
Mass	resistance to a change of speed and position	a characteristic controlled by the string vibration
Interactions shown by	Feynman diagrams	twisting, folding and vibrating strings

14.7Particle annihilationOften examined

Einstein's famous mass-energy equation indicates that mass can be converted into energy and vice versa. Soon after the big bang, matter and anti-matter were formed. What will happen if matter comes close to its anti-matter? They annihilate each other and convert their mass into energy in accordance with Einstein's mass-energy equation. A particle and its anti-particle cannot exist in the same place, because when they come close to each other they annihilate. After the annihilation the product is energy in the form of gamma-ray photons.

If an electron and a positron come in close proximity, they annihilate each other and form two gamma-ray photons:

e^{-}+e^{+}\;\to\;\gamma+\gamma

Electron-positron annihilation

The production of two photons is necessary in this reaction to conserve the momentum of the system. The annihilation of an electron and a positron is shown in Fig. 14.17.

Figure 14.17: Electron-positron annihilation. The two photons leave in opposite directions.

In the reaction above the product is energy, but this energy can further be converted into matter particles. In the Large Hadron Collider at CERN and the Tevatron at Fermilab, the energy from such annihilations has been converted into matter particles which were previously undiscovered.

Worked example: energy and photons from annihilation at rest

An electron and a positron, almost at rest, annihilate into two photons. (a) Find the energy, frequency and wavelength of each photon. (b) A proton and an anti-proton at rest annihilate into two photons: find the energy of each photon. Use m_ec^2=0.511\,\mathrm{MeV}, m_pc^2=938.3\,\mathrm{MeV}, 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}, h=6.63\times10^{-34}\,\mathrm{J\,s} and c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}.

(a) Electron and positron

The particles are at rest, so the total energy before is their rest energy. Two photons share it equally, because the total momentum is zero and the two photons must leave in opposite directions with equal momentum.
E_{\gamma}=\dfrac{2m_ec^{2}}{2}=0.511\,\mathrm{MeV}
Convert to joules.
E=0.511\times10^{6}\times1.6\times10^{-19}=8.18\times10^{-14}\,\mathrm{J}
Frequency from E=hf.
f=\dfrac{E}{h}=\dfrac{8.18\times10^{-14}}{6.63\times10^{-34}}=1.23\times10^{20}\,\mathrm{Hz}
Wavelength from c=f\lambda.
\lambda=\dfrac{c}{f}=\dfrac{3.00\times10^{8}}{1.23\times10^{20}}=2.43\times10^{-12}\,\mathrm{m}

(b) Proton and anti-proton

Same reasoning: the rest energy of the pair is 2m_pc^{2}, shared equally by the two photons.
E_{\gamma}=\dfrac{2(938.3)}{2}=938.3\,\mathrm{MeV}
The total energy released is 1876.6\,\mathrm{MeV}=1.88\,\mathrm{GeV}, which is about 1836 times the 1.022\,\mathrm{MeV} of the electron pair, because a proton is that much heavier than an electron.
\dfrac{2(938.3)}{2(0.511)}=1836

\text{(a) }0.511\,\mathrm{MeV},\ 1.23\times10^{20}\,\mathrm{Hz},\ 2.43\times10^{-12}\,\mathrm{m};\quad \text{(b) }938.3\,\mathrm{MeV}\ \text{each}

14.7.1Annihilation of matter into matter

Annihilation not only gives us energy by totally destroying the particles, but it can also produce some new particles. If quarks come close to their own flavoured anti-quarks, they annihilate each other and produce energy and a new pair of quark and anti-quark. The annihilation of an up quark and an anti-up quark is shown in Fig. 14.18. At first they annihilate each other by producing energy, and then through gluon interactions they finally produce a top quark and an anti-top quark.

Up and anti-up to top and anti-top

An up quark meets an anti-up quark and they annihilate, producing energy.
u+\bar u\;\to\;\text{energy}
The energy appears as a gluon, which then turns into a new quark and anti-quark pair.
\text{gluon}\;\to\;t+\bar t
So a pair of one flavour has become a pair of another flavour: matter has changed into new matter.
u+\bar u\;\to\;t+\bar t

Figure 14.18: Up and anti-up quark annihilation. The energy is carried by a gluon and produces a top quark and an anti-top quark.

14.8Particle acceleratorsOften examined

In high-energy physics (particle physics) we need to accelerate particles to very high speeds to study their behaviour at high energies. We collide these high-energy particles to produce new particles. To reach such high energies we need projectiles that can be accelerated to higher velocities. The instruments used to accelerate particles up to a desired speed and energy are called accelerators.

Particle accelerator: An instrument used to accelerate particles up to a desired speed and energy, for research and technology.

There are two basic types of particle accelerators: linear accelerators and circular accelerators, which include synchrotrons and cyclotrons.

Linear accelerators (LINACS)

We use linear accelerators to get a steady, intense beam of particles. The SLAC National Accelerator Laboratory has the longest LINAC in the world, which is 2 miles long and can accelerate a particle up to 50 billion eV.

Working of a linear accelerator

A linear accelerator consists of a set of cylindrical accelerating tubes (drift tubes) of increasing length, connected alternately in a vacuum chamber.
The odd-numbered cylinders are positive and the even-numbered cylinders are negative, and the potential on the tubes reverses periodically according to the oscillating frequency. An oscillator alternates the polarity of the tubes.
An ion source is used to produce the ions which are to be accelerated.
The accelerated beam of charged particles leaves the drift tubes and moves towards the target.

Figure 14.19: Linear accelerator. The drift tubes (marked + and − at one instant) are joined alternately to a high-frequency AC supply; the beam gains energy in the gaps between the tubes.

Synchrotron accelerators

Synchrotron accelerators, also called the ‘heavy lifters’, are circular accelerators. They are the highest-energy accelerators in the world. They have a closed pathway that moves particles in a ring.

When a particle enters the accelerator it moves repeatedly in the circular path, which is enclosed in a vacuum pipe.
Radiofrequency cavities at intervals around the ring increase the speeds of the particles.
Several magnets create electromagnetic fields to bend the particles' path.
The maximum energy obtained by a synchrotron accelerator is 13.6\,\mathrm{TeV}, produced during the proton collision in the Large Hadron Collider (LHC) at CERN.
Researchers send different beams to collide at the required energy at fixed points, where particle detectors (like the Compact Muon Solenoid, CMS) are positioned to detect the particles produced by these collisions.

Synchrotron accelerators are used to study the basic building blocks of the universe. Accelerators like the LHC can accelerate heavy ions and nuclei like lead. At RHIC in the USA, heavy ions like uranium and gold are used to create quark-gluon plasma.

Figure 14.20 (a): Synchrotron accelerator. Bending electromagnets keep the beam on the ring, the accelerating cavity raises its energy each turn, and a deflecting electromagnet extracts it.

Cyclotron accelerators

Cyclotron accelerators accelerate particles in a spiral pattern, starting from the centre of the spiral. They use one large electromagnet to bend the path of the particles. To move particles in increasingly larger circles they use metal electrodes. Cyclotrons can accelerate particles to 520 million eV.

A cyclotron consists of two circular (D-shaped) metal boxes called dees, which are separated from each other and connected to a radiofrequency oscillator of frequency 10^{7}\,\mathrm{Hz}. An alternating potential of 10\,\mathrm{kV} is applied between the dees. A cyclotron can draw particles from different parts of the accelerator at different energies. The source of ions is placed in the middle, from which the ions move outward in a spiral path in the applied electric and magnetic fields.

Figure 14.20 (b): Cyclotron accelerator. The beam crosses the electric field region (gap) between the dees twice per turn and spirals outward in the magnetic field.

The three accelerators of Section 14.8 compared.
	Linear accelerator	Synchrotron	Cyclotron
Path of the particles	straight line through drift tubes	closed ring in a vacuum pipe	spiral from the centre outward
Acceleration by	alternating potential on the drift tubes	radiofrequency cavities around the ring	alternating potential (10\,\mathrm{kV}, 10^{7}\,\mathrm{Hz}) between two dees
Bending	none (straight)	several magnets	one large electromagnet
Energy / size quoted	50 billion eV; SLAC, 2 miles long	13.6\,\mathrm{TeV} at the LHC	520 million eV
Known for	steady, intense beam	highest energies, the ‘heavy lifters’	spiral path, particles drawn off at different energies

14.8.1Uses of particle accelerators

Particle accelerators are used not only for scientific research but also in various other fields of daily life, including the medical field. Some uses of particle accelerators are:

Linear accelerators are used to produce medical isotopes and to create beams of radiation for cancer treatment. The electron linear accelerator (LINAC) for cancer therapy is the most common type of particle accelerator. It serves as an injector for higher-energy accelerators. LINACs generate X-rays and high-energy electrons for radiation therapy.
Synchrotron accelerators are used to produce a high-energy environment to study the building blocks of the universe. The Higgs boson was discovered using these accelerators. In the medical field they are used for studying cell biology, crystallography, high-resolution imaging and cancer radiation therapy. In environmental sciences they are used in toxicology, clean combustion and atmospheric research. They are also used in agriculture, mineral exploration, engineering and forensics.
Cyclotron accelerators are used to produce large numbers of certain types of particles, such as muons and neutrinos. They are also used to produce medical isotopes. They are the best source of high-energy beams for particle physics experiments. They are used in particle therapy to treat cancer.

Example 14.3: Pair production by a gamma-ray photon

A 3.0\,\mathrm{MeV} photon interacts with a lead nucleus and creates an electron-positron pair. The pair moves perpendicular to the initial direction of travel of the photon. (a) Find the kinetic energies of the electron and positron, assuming the nucleus is at rest after the collision. (b) Find the kinetic energy of the lead nucleus to ensure momentum conservation. Would this amount of kinetic energy greatly affect the results in part (a)? The mass of the lead atom is 193007\,\mathrm{MeV}.

Pair production near a lead nucleus: the photon (along x) creates e^+ and e^- that move along +y and -y.

Given

E_{\text{photon}} = 3\,\mathrm{MeV}
m_{\text{lead}}c^2 = 193007\,\mathrm{MeV} rest energy of lead atom
m_ec^2 = 0.511\,\mathrm{MeV} rest energy of electron and positron

To find

(a) K_{e^-} and K_{e^+}
(b) K_{\text{nucleus}}

(a) Kinetic energies of the pair

Conservation of energy, ignoring the energy of the nucleus: the photon energy becomes the total energy of the two particles.
E_{\text{photon}}=E_{e^-}+E_{e^+}\quad\Rightarrow\quad3\,\mathrm{MeV}=E_{e^-}+E_{e^+}\quad(1)
Conservation of the y-component of momentum (ignoring the nucleus): the photon has none along y, so the electron and positron have equal and opposite y-momenta.
0=-p_{e^-}c+p_{e^+}c\quad\text{or}\quad p_{e^-}c=p_{e^+}c
Equal momentum and equal mass mean equal energy, E^2=(pc)^2+(mc^2)^2. Put E_{e^-}=E_{e^+}=E in Eq. (1).
3\,\mathrm{MeV}=2E
Solve for the energy of each particle.
E=\dfrac{3}{2}\,\mathrm{MeV}=1.5\,\mathrm{MeV}
The kinetic energy is the total energy minus the rest energy 0.511\,\mathrm{MeV}.
K=E-mc^2=(1.5-0.511)\,\mathrm{MeV}=0.989\,\mathrm{MeV}

(b) Kinetic energy of the nucleus

The x-component of momentum must also be conserved. The pair moves along y, so the whole photon momentum goes to the nucleus. For a photon E=pc.
p_{\text{photon}}c=p_{\text{nucleus}}c\quad\Rightarrow\quad p_{\text{nucleus}}c=3\,\mathrm{MeV}
Total energy of the nucleus from E^2=(pc)^2+(mc^2)^2.
E_{\text{nucleus}}=\sqrt{(pc)^2+(mc^2)^2}=\sqrt{(3\,\mathrm{MeV})^2+(193007\,\mathrm{MeV})^2}
Take 193007 outside the root.
E_{\text{nucleus}}=193007\,\mathrm{MeV}\sqrt{1+\dfrac{9}{(193007)^{2}}}
Use the binomial expansion \sqrt{1+x}\approx1+\tfrac12x for a very small x.
E_{\text{nucleus}}\approx193007\,\mathrm{MeV}\left(1+\dfrac12\cdot\dfrac{9}{(193007)^2}\right)
The kinetic energy is the part above the rest energy.
K_{\text{nucleus}}=\dfrac{9}{2\times193007}\,\mathrm{MeV}=2.33\times10^{-5}\,\mathrm{MeV}=23.3\,\mathrm{eV}

Does this affect part (a)? No. The nucleus takes only 2.33\times10^{-5}\,\mathrm{MeV} out of 3\,\mathrm{MeV}, a fraction of about 8\times10^{-6}. So the nucleus can preserve momentum conservation while taking a very small portion of the total energy, and we can ignore it when dividing the photon energy between the electron and the positron.

K_{e^-}=K_{e^+}=0.989\,\mathrm{MeV},\qquad K_{\text{nucleus}}=2.33\times10^{-5}\,\mathrm{MeV}\approx23.3\,\mathrm{eV}

Assignment 14.3: Energy of the gamma ray

A gamma-ray photon produces an electron-positron pair. If the rest mass energy of the electron is 0.51\,\mathrm{MeV} and the total kinetic energy of the pair is 0.78\,\mathrm{MeV}, find the energy of the gamma-ray photon in MeV.

Given

m_ec^2 = 0.51\,\mathrm{MeV} rest energy of electron (and of positron)
K_{\text{total}} = 0.78\,\mathrm{MeV} kinetic energy of the pair

To find

E_\gamma

The photon energy becomes the rest energy of both particles plus their kinetic energy.
E_\gamma=m_ec^2+m_ec^2+K_{\text{total}}
Put in the values.
E_\gamma=0.51+0.51+0.78\,\mathrm{MeV}
Add.
E_\gamma=1.80\,\mathrm{MeV}

E_\gamma=1.80\,\mathrm{MeV}

Formula sheet

Alpha decay

{}^{A}_{Z}X\to{}^{A-4}_{Z-2}Y+{}^{4}_{2}\mathrm{He}+Q

X: parent nucleus, Y: daughter nucleus, Q: disintegration energy (always positive)

Units: Q in \mathrm{J} or \mathrm{MeV}

Beta-minus decay (Eq. 14.1)

n\to p^{+}+\beta^{-}+\bar\nu_e

n: neutron, p^+: proton, \beta^-: electron, \bar\nu_e: electron anti-neutrino

Beta-plus decay (Eq. 14.2)

p^{+}\to n+\beta^{+}+\nu_e

\beta^+: positron, \nu_e: electron neutrino

Energy of a particle (Dirac, Eq. 14.4)

E=\pm\sqrt{m^{2}c^{4}+p^{2}c^{2}}

E: energy, m: mass, p: momentum, c: speed of light. The positive sign is the particle, the negative sign the anti-particle

Units: \mathrm{J} (or \mathrm{MeV})

Momentum from the energy

p^{2}=\dfrac{E^{2}-m^{2}c^{4}}{c^{2}}

obtained by squaring Eq. 14.4; p has two values, \pm

Unit: \mathrm{kg\,m\,s^{-1}}

Energy of a photon (Eq. 14.5)

E=hf

h: Planck's constant, f: frequency of the photon

Unit: \mathrm{J}

Electron-positron annihilation

e^{-}+e^{+}\to\gamma+\gamma

two photons are needed to conserve momentum

Pair production: photon energy

E_\gamma=2m_ec^{2}+K_{\text{total}}

m_ec^2=0.511\,\mathrm{MeV}: rest energy of the electron (and positron), K_{\text{total}}: total kinetic energy of the pair

Unit: \mathrm{MeV}

Kinetic energy of a particle

K=E-mc^{2}

E: total energy, mc^2: rest energy

Unit: \mathrm{MeV}

Quark composition of the proton

p=uud,\quad q_p=\tfrac23+\tfrac23-\tfrac13=+1

up quark +\tfrac23e, down quark -\tfrac13e

Unit: e

Quark composition of the neutron

n=udd,\quad q_n=\tfrac23-\tfrac13-\tfrac13=0

up quark +\tfrac23e, down quark -\tfrac13e

Unit: e

Quark change in beta-plus decay

u\to d+W^{+},\quad W^{+}\to e^{+}+\nu_e

an up quark turns into a down quark

Quark change in beta-minus decay

d\to u+W^{-},\quad W^{-}\to e^{-}+\bar\nu_e

a down quark turns into an up quark

Conservation laws in a reaction

\sum Q,\ \sum L,\ \sum B\ \text{(left)}=\sum Q,\ \sum L,\ \sum B\ \text{(right)}

Q: charge, L: lepton number (+1 lepton, -1 anti-lepton), B: baryon number (+1 baryon, -1 anti-baryon)

Key terms

Particle physics: Also called high-energy physics: the branch of physics that deals with the fundamental constituents of matter and radiation and their interactions.
Standard Model: The model of particle physics that describes the elementary particles (quarks, leptons, bosons) and their interactions, excluding gravity.
Conservation of charge: The total charge before and after any interaction or decay in a closed system stays the same.
Conservation of nucleons: The total number of protons and neutrons stays the same before and after a nuclear reaction or decay.
Alpha particle: Two protons and two neutrons bound together (a helium nucleus): mass about 4\,\mathrm{u} (6.64\times10^{-27}\,\mathrm{kg}), charge +2e.
Beta particle: A high-energy electron (\beta^-, charge -1e) or positron (\beta^+, charge +1e) of mass 9.11\times10^{-31}\,\mathrm{kg} emitted in radioactive decay.
Gamma ray: An electromagnetic wave (photons): no mass and no charge.
Anti-particle: A particle with the same mass as its particle but opposite charge (and magnetic moment); e.g. the positron is the anti-particle of the electron.
Alpha decay: Decay in which the atomic number of the parent falls by 2 and the mass number by 4; the disintegration energy Q is positive.
Beta-minus decay: A neutron changes into a proton, an electron and an anti-neutrino (n\to p^++\beta^-+\bar\nu_e); the atomic number rises by 1.
Beta-plus decay: A proton changes into a neutron, a positron and a neutrino (p^+\to n+\beta^++\nu_e); the atomic number falls by 1.
Neutrino and anti-neutrino: Nearly massless, electrically neutral leptons emitted in beta decay; they carry away a variable share of the energy.
Boson: A particle with integer spin.
Fermion: A particle with half-integer spin.
Dirac's equation: E=\pm\sqrt{m^2c^4+p^2c^2}: its negative solution led to the prediction of the anti-particle.
CP-symmetry: Charge conjugation (C: swap particle and anti-particle) and parity (P: space inversion) taken together.
CP-violation: A phenomenon where the same decay process has a different probability for a particle than for an anti-particle; the main cause of the matter-anti-matter asymmetry.
Quark: An elementary particle of spin \tfrac12, of six flavours (up, down, charm, strange, top, bottom), with charge \pm\tfrac13e or \pm\tfrac23e.
Elementary particle: A particle that is not composed of other particles.
Colour charge: The property of quarks and gluons (red, blue, green and the anti-colours) related to the strong interaction in quantum chromodynamics.
Hadron: A particle made of quarks and anti-quarks held by the strong force: a baryon or a meson.
Baryon: A hadron of three quarks, with half-integer spin and baryon number +1; e.g. proton, neutron.
Meson: A hadron of a quark and an anti-quark, with integer spin and baryon number 0; e.g. pion, kaon, rho.
Baryon number: Number of baryons minus number of anti-baryons: +1 for a baryon and -1 for an anti-baryon; conserved.
Lepton: An elementary spin-\tfrac12 particle that does not feel the strong force: electron, muon, tau and their neutrinos.
Lepton number: +1 for leptons and -1 for anti-leptons; conserved in every reaction.
Force carrier (mediator): A boson exchanged between particles to produce a force: photon (electromagnetic), gluon (strong), W and Z (weak).
Photon: The quantum of electromagnetic radiation: massless, neutral, spin 1, energy E=hf; carrier of the electromagnetic force.
Gluon: A massless, electrically neutral spin-1 particle exchanged between quarks; carrier of the strong (colour) force.
Z and W bosons: Heavy gauge bosons (spin 1) that mediate the weak nuclear force; W has charge \pm1, Z is neutral.
Higgs boson: The particle (a wave in the Higgs field) associated with the field that gives mass to elementary particles; spin 0, charge 0, mass 124.97\,\mathrm{GeV}/c^2.
Quantum field theory: Combines quantum mechanics and relativity: particles are excited states of quantum fields and interactions are interactions between fields.
String theory: Proposes that the universe is made of tiny vibrating one-dimensional strings that create matter, energy and forces.
Feynman diagram: A diagram that represents a particle interaction, with the exchanged force carrier drawn between the particles.
Annihilation: A particle and its anti-particle meeting and converting their mass into energy (gamma photons) or into new particles.
Particle accelerator: An instrument that accelerates particles to very high speeds and energies: linear accelerator (linac), synchrotron or cyclotron.
Linear accelerator (LINAC): Accelerator with drift tubes of increasing length in a vacuum chamber, with alternating polarity, giving a steady, intense beam.
Synchrotron: Circular accelerator with a closed ring, radiofrequency cavities and bending magnets; the highest-energy accelerator (LHC).
Cyclotron: Circular accelerator with two dees, one large electromagnet and an RF oscillator; particles move in an outward spiral.
Quark-gluon plasma: A state of baryonic matter at very high energy in which quarks and gluons are free of the strong-force confinement, like the universe just after the big bang.

Multiple choice questionsp. 281

The book prints no answer key for these multiple choice questions. Each key below is worked out from the unit's own teaching, and the reason is given with it.

Which fundamental conservation principle states that the total number of nucleons (protons and neutrons) remains constant in a closed system?

Acharge
Benergy
CnucleonAnswer
Dmomentum

Why

The principle of nucleon conservation says that the total number of nucleons (protons plus neutrons) is the same before and after a nuclear reaction (Section 14.1.2). The book's example is {}_{7}\mathrm{N}^{14}+{}_{2}\mathrm{He}^{4}\rightarrow{}_{8}\mathrm{O}^{17}+{}_{1}\mathrm{H}^{1}, where 14+4=17+1=18.

Charge (A) is conserved too, but it counts electric charge, not nucleons. Energy (B) and momentum (D) are conserved in every process, but neither is a count of nucleons.

According to the principle of charge conservation, in any physical process, the total electric charge before and after the process must:

Adecrease
Bincrease
Cbecome zero
Dbecome neutralAnswer

Why

The principle of charge conservation says that the net electric charge of a closed system is the same before and after any process (Section 14.1.1). Options A and B describe a change in the total charge, which the principle forbids. Option C is also wrong, because conservation does not force the charge to zero: the alpha decay {}_{88}\mathrm{Ra}^{226}\rightarrow{}_{86}\mathrm{Rn}^{222}+{}_{2}\mathrm{He}^{4} has charge +88 before and +86+2=+88 after.

No option reads 'remains the same', which is what the principle requires. Option D is the nearest: a process that starts with zero total charge, as in Example 14.2, ends with zero total charge, so the total charge is not changed by it.

Note. The book prints no key, and the wording 'remains the same' is missing from the options. D is the nearest option.

In a nuclear reaction, if a nucleus emits an alpha particle (helium nucleus), what happens to the nucleon number (A) of the original nucleus?

Adecreases by 1
Bdecreases by 2Answer
Cremains same
Dincreases by 1

Why

An alpha particle is {}_{2}\mathrm{He}^{4}. When the nucleus emits it, the atomic number falls by 2 and the mass number falls by 4, as in {}_{88}\mathrm{Ra}^{226}\rightarrow{}_{86}\mathrm{Rn}^{222}+{}_{2}\mathrm{He}^{4} (Section 14.2.1): 226-222=4 and 88-86=2.

No option gives the change in the nucleon number, which is a decrease by 4. Option B is the only one that matches the change in the atomic number (88-86=2), so the question seems to be about the atomic number. Options A and D give the wrong direction and size, and option C is wrong because the number does change.

Note. The change in the nucleon number A in alpha decay is 4, which no option gives. B (decrease by 2) is the change in the atomic number Z, so the question was probably meant to ask about Z. Key B.

During beta-minus (β⁻) decay, a neutron is transformed into a proton, emitting an electron (beta particle) in the process. What happens to the total charge during this decay?

Aincreases
Bdecreases
Cremains sameAnswer
Dbecomes neutral

Why

In \beta^{-} decay the neutron (charge 0) becomes a proton (charge +1) and the electron (charge -1) is emitted with an antineutrino (charge 0): n\rightarrow p^{+}+e^{-}+\bar\nu_e (eq 14.1). The total charge is 0 before and (+1)+(-1)+0=0 after, as in the check in Example 14.2.

The total charge therefore remains the same. Options A and B describe a change, which conservation forbids, and option D is wrong because the total charge was already zero before the decay; it did not become neutral.

Positrons in our atmosphere can be produced during the natural process of:

Aspinning of Earth
BlightningAnswer
CEarth's magnetic field
Dwinds

Why

Positrons form in the atmosphere in high-energy natural events, and lightning is the standard example. Lightning produces very energetic electrons, which create electron-positron pairs. The book does not explain this, so the key is worked out here.

The spinning of the Earth, its magnetic field and the winds do not produce positrons. The magnetic field only bends charged particles that are already present, and the winds move air without making new particles.

Note. The book prints no key and does not discuss this process. The key is worked out from general physics; lightning is the only option that is a high-energy natural event.

Which type of carrier particle has not been found yet?

AW-boson
BZ-boson
Cpion
DgravitonAnswer

Why

The Z boson was discovered in 1983 at CERN (Section 14.6.3) and the W boson is a known force carrier of the weak force (Section 14.6.4), so both have been found. The pion is a meson, made of a quark and an antiquark, and it is not a force carrier.

The graviton is the proposed carrier of gravity. The book notes that the Standard Model excludes gravity (the unit introduction), and no graviton has been observed, so it is the carrier not yet found.

What type of hadron is always constructed partially of an anti-quark?

Abaryon
Blepton
CmesonAnswer
Dphoton

Why

A meson is made of one quark and one anti-quark (Section 14.4.1, Table 14.3). It is the only hadron that always contains an anti-quark.

A baryon is made of three quarks and no anti-quarks (proton uud, neutron ddu). A lepton contains no quarks at all, and a photon is a force carrier, not a hadron.

Analysis of which particles resulted into the search of Higgs boson?

AW and Z bosonAnswer
Bup and down quark
Cmesons and baryons
Dneutrinos and photons

Why

The Higgs boson gives mass to the elementary particles through the Higgs field (Section 14.6.5). It was searched for through the particles that carry the weak force, and the analysis of the W and Z bosons and their decays gave the signal that confirmed it at CERN in 2012.

Up and down quarks make up ordinary matter, and mesons, baryons, neutrinos and photons are not the particles the search was built around. Option A is the best answer.

Note. The book does not name the particles analysed in the Higgs search, so the key is worked out from the unit's account of the Higgs field and the weak carriers.

Which device is used for proton decay?

ACyclotron
BSynchrotron
Csuper-kamiokandeAnswer
DLarge Hadron Collider

Why

Proton decay, if it occurs, is so rare that it is studied in very large underground detectors. Super-Kamiokande is such a detector, built to look for proton decay and for neutrinos.

The cyclotron, the synchrotron and the Large Hadron Collider are particle accelerators (Section 14.8). They speed up and collide particles, but they do not watch for the decay of protons.

Note. The book does not mention Super-Kamiokande. The key is worked out from general physics.

Lepton number for 'delta' particle is:

A-1
B0Answer
C+1/2
D+1

Why

The delta (\Delta) is a baryon (Table 14.2). Leptons have lepton number +1 and anti-leptons -1, but baryons have lepton number 0; every baryon in Table 14.2 has lepton number 0.

Option A (-1) is the lepton number of an anti-lepton, option C (+\frac{1}{2}) is a spin value, and option D (+1) is the lepton number of an electron, not of a baryon.

Which particle has a baryon number of +1?

Amuon
Btau
Ckaon
DomegaAnswer

Why

Baryons have baryon number +1 (Section 14.4.1, Table 14.2). The omega \Omega has quarks sss, so it is a baryon with baryon number +1.

The muon and tau are leptons, which have baryon number 0. The kaon is a meson, made of a quark and an antiquark, so its baryon number is 0.

The particle which has the colour property is:

Apion
Bkaon
CgluonAnswer
Dproton

Why

Colour charge belongs to quarks and to gluons: the gluon carries colour between quarks and is the quantum of the strong force (Sections 14.4 and 14.6.2).

The pion, the kaon and the proton are hadrons, which carry no net colour charge (Section 14.4.1), even though the quarks inside them carry colour.

Short questionsp. 281–282

13 marks

Why does beta-minus (\beta^{-}) decay lead to an increase in the atomic number of the nucleus, while beta-plus (\beta^{+}) decay results in a decrease in the atomic number?

Solution

In \beta^{-} decay a neutron inside the nucleus changes into a proton, and an electron (the beta-minus particle) and an antineutrino are emitted: n\rightarrow p^{+}+e^{-}+\bar\nu_e (eq 14.1). The new proton stays in the nucleus, so the number of protons, which is the atomic number Z, rises by 1.

In \beta^{+} decay a proton changes into a neutron, and a positron and a neutrino are emitted: p^{+}\rightarrow n+e^{+}+\nu_e (eq 14.2). The new neutron stays in the nucleus, so Z falls by 1.

Charge is conserved in both cases. The emitted electron carries -1 away in \beta^{-} decay, so the nucleus must gain +1; the emitted positron carries +1 away in \beta^{+} decay, so the nucleus loses one unit of charge. The nucleon number A is unchanged in both.

23 marks

Explain how the mass-energy equivalence principle, as expressed by Einstein's famous equation E=mc^{2}, is relevant to understanding the energy release in nuclear processes such as alpha decay.

Solution

Einstein's relation E=mc^{2} says that a mass m is equivalent to an energy mc^{2}, so mass can be converted into energy. In alpha decay the parent nucleus splits into a daughter nucleus and an alpha particle, and the total rest mass of the products is less than the mass of the parent.

Q=\left(m_{\text{parent}}-m_{\text{daughter}}-m_{\alpha}\right)c^{2}

The lost mass \Delta m appears as energy Q=\Delta m\,c^{2}, which becomes the kinetic energy of the alpha particle and the recoiling daughter nucleus. Because alpha decay happens spontaneously, Q is always positive (Section 14.2.1). With 1\,\mathrm{u}=931.5\,\mathrm{MeV}/c^{2}, a mass difference of 0.00459\,\mathrm{u} gives about 4.27\,\mathrm{MeV} (Numerical problem 1).

33 marks

How do the principles of nucleon conservation and charge conservation in nuclear reactions reflect deeper symmetries and conservation laws present in the fundamental interactions of subatomic particles?

Solution

Nucleon number and charge are conserved in every nuclear reaction and decay (Section 14.1). Nucleon number is kept because the strong force, which binds the nucleons, does not create or destroy them. Charge is conserved in all processes, and the book uses both laws to check every reaction it describes.

The book connects these laws with symmetries of nature. Charge conjugation (C) pairs every particle with its antiparticle of opposite charge, and parity (P) relates a particle to its mirror image. Together they form CP symmetry (Section 14.3.1). The weak force, which governs beta decay, violates CP, which is why the laws of nature are not perfectly symmetric.

Lepton number and baryon number are conserved in particle reactions in the same way (Section 14.5.1). Each law is a statement about what the interactions allow, so a reaction is possible only if all the conserved quantities balance on both sides.

43 marks

Can anti-matter burn? Explain.

Solution

Burning is a chemical reaction in which a fuel combines with oxygen and releases energy. Anti-matter does not burn in this sense, because it does not need oxygen and its energy is not released by a chemical reaction.

Anti-matter does release energy when it meets matter. The particle and its antiparticle annihilate, and their mass is converted to energy, as in e^{-}+e^{+}\rightarrow\gamma+\gamma (Section 14.7). The energy is far larger than any burning: the book states that 1 kg of anti-matter meeting 1 kg of matter would release about 43 megatons of TNT (information box, page 278).

So anti-matter does not burn, but it releases a very large amount of energy when it comes into contact with matter.

53 marks

List at least three applications of anti-matter.

Solution

Medicine (cancer treatment): anti-protons are used to treat some types of cancer (Section 14.3, anti-matter and medical technology).
Medical imaging: positrons are used in the positron emission tomography (PET) scan to create images of internal organs such as the heart and brain, and to detect cancers.
Research: anti-protons are used in scientific experiments to study the fundamental properties of matter, and anti-hydrogen atoms are artificially created at CERN to research anti-matter.

The PET scan works through annihilation: a positron meets an electron and the pair annihilates into two photons that travel 180^{\circ} apart. A scanner detects these photons and uses them to build an image of the internal body part.

63 marks

Why are gluons short range exchange particles? Explain.

Solution

Gluons are the exchange particles of the strong nuclear force. A gluon is exchanged between quarks and binds them together to form protons and neutrons (Section 14.6.2, Figure 14.15). Each gluon also carries colour charge, so the strong force acts only between coloured particles that are close together inside a hadron.

The strong force is a short-range force (Section 14.1.2), and its effect is confined to distances about the size of a nucleon. Gluons are exchanged only between quarks inside a nucleon or nucleus, so their interaction does not reach far beyond the hadron that contains the quarks. This limited range is why gluons are called short-range exchange particles.

73 marks

Why do gluons interact only with particles in the first two rows of the standard model?

Solution

A gluon carries colour charge, and so it interacts only with particles that carry colour. In the standard model (Figure 14.5) the first two rows of the table contain the six quarks, up, charm, top, down, strange and bottom, which are the coloured matter particles.

Leptons (electron, muon, tau and their neutrinos) have no colour charge, so gluons do not act on them. The photon, the Z boson and the W bosons carry no colour either, so the gluon does not interact with them. The gluon acts only on quarks, which are the particles in the first two rows.

Note. 'Rows' refers to the layout of the standard model table in Figure 14.5.

83 marks

What are the Feynman diagrams? What information about particles these diagrams give?

Solution

A Feynman diagram is a picture of a particle interaction. Time runs up the page and space runs across it. Straight lines show matter particles (fermions), and wavy lines show the force carrier (boson) that is exchanged between them (Section 14.6, Point to Ponder). Each interaction can be represented by one such diagram.

The diagram shows which particles take part in the interaction, which force carrier is exchanged, and the order of events in time. In Figure 14.14 two electrons approach, exchange a photon and move apart, which represents their electromagnetic repulsion. The points where lines meet are the vertices, where the force carrier is emitted or absorbed.

Figure 14.14: Feynman diagram of two electrons repelling by exchanging a photon (γ). Time runs upward.

93 marks

What are the differences between quarks and leptons?

Solution

Property	Quarks	Leptons
Electric charge	Fractional: +\frac{2}{3}e or -\frac{1}{3}e	Whole numbers: 0 or -1e (electron, muon, tau)
Strong nuclear force	Feel it, because they carry colour charge	Do not feel it
Where they are found	Only inside hadrons such as protons and neutrons	Free particles, found outside the nucleus
Spin	\frac{1}{2}	\frac{1}{2}

Quarks combine to form hadrons, while leptons are elementary particles that exist on their own (Sections 14.4 and 14.5). Both groups are matter particles with spin \frac{1}{2}, and each has six types arranged in three generations (Figure 14.5).

103 marks

What advantages do circular particle accelerators have over the linear particle accelerators?

Solution

A linear accelerator (LINAC) gives a steady, intense beam, and the SLAC LINAC reaches about 50 billion eV. However, a particle passes each accelerating gap only once, so the machine must be made longer as the energy required grows.

A circular accelerator sends the particles round a closed path many times, so each pass through the radio-frequency cavities adds more energy. The same magnets and cavities are reused, so the machine stays compact while its energy rises. The cyclotron reaches 520\,\mathrm{MeV}, and the synchrotron at the LHC reaches 13.6\,\mathrm{TeV}, the highest energy in the world (Section 14.8).

Circular machines can also collide beams at fixed points where detectors are placed, so one machine serves several experiments at high energy (Section 14.8).

113 marks

What is the difference between quantum field theory and the string theory?

Solution

Quantum field theory (QFT) describes particles as excited states of quantum fields. The interaction between particles is an interaction between fields, and each interaction can be drawn as a Feynman diagram. QFT addresses three of the four fundamental forces (Section 14.6.6).

String theory replaces point-like particles by tiny one-dimensional vibrating strings. The charge and mass of each particle are set by the way its string vibrates, and phenomena such as electromagnetism and gravity arise from the strings.

The main difference is the basic object. QFT is built on fields, whose excitations behave like particles, while string theory is built on vibrating strings. The book presents string theory as the theory that also includes gravity, which QFT does not describe.

123 marks

How do we determine if a particle reaction or decay occurs?

Solution

A reaction or decay can occur only if it obeys every conservation law. We check charge, nucleon number, lepton number and baryon number on both sides of the reaction (Section 14.5.1), and check that the energy balances.

Write the reaction and give every particle its charge, lepton number and baryon number from the tables.
Add the values on each side of the reaction and compare them.
If every quantity is the same on both sides, and the energy released is positive, the reaction is allowed. If any quantity fails to balance, the reaction cannot happen.

Example 14.2 applies this to beta-minus and beta-plus decay. Charge, lepton number and baryon number are the same on both sides in each case, so both decays are real.

Comprehensive questionsp. 282–283

Explain how the principles of nucleon conservation and charge conservation play a fundamental role in understanding nuclear processes. Provide examples to illustrate the application of these conservation laws in nuclear reactions.

Solution

Conservation of charge

In any closed system the total electric charge before and after an interaction or decay is the same. Charge is transferred between particles but is never created or destroyed (Section 14.1.1).

{}_{88}\mathrm{Ra}^{226}\longrightarrow{}_{86}\mathrm{Rn}^{222}+{}_{2}\mathrm{He}^{4}+Q

Add the charges on the left: the radium nucleus has +88.
Q_{\text{left}}=+88
Add the charges on the right: the radon nucleus has +86 and the alpha particle has +2.
Q_{\text{right}}=(+86)+(+2)=+88
The two sides agree, so the charge is conserved.
+88=+88

Conservation of nucleons

The total number of nucleons (protons plus neutrons) is the same before and after a nuclear reaction. The strong force binds the nucleons, and it does not create or destroy them (Section 14.1.2).

{}_{7}\mathrm{N}^{14}+{}_{2}\mathrm{He}^{4}\rightarrow{}_{8}\mathrm{O}^{17}+{}_{1}\mathrm{H}^{1}+Q

Count the nucleons on the left.
14+4=18
Count the nucleons on the right.
17+1=18
Check the charge as well: the left side has 7+2=9 and the right side has 8+1=9.
9=9

Role in nuclear processes

The two laws decide which reactions can occur and what the products are. In the alpha decay of radium-226 the mass number falls by 4 and the atomic number by 2, so the daughter is radon-222, which is what the balance requires. In the nitrogen reaction both the nucleon number and the charge match, so the reaction is allowed and produces an oxygen-17 nucleus and a proton.

The same laws apply to beta decay. In \beta^{-} decay a neutron becomes a proton and an electron is emitted, so the charge balances while the nucleon number does not change (Section 14.2.2). Together, the two laws keep every nucleon and every unit of charge accounted for in each nuclear process.

Describe the process of beta-minus (\beta^{-}) decay and beta-plus (\beta^{+}) decay. How do these decay processes violate the conservation of nucleon number within a nucleus while still respecting the conservation of charge? How is the energy released during these processes related to the mass difference between the initial and final particles?

Solution

Beta-minus decay

In \beta^{-} decay a neutron inside the nucleus becomes a proton, and an electron and an antineutrino are emitted: n\rightarrow p^{+}+\beta^{-}+\bar\nu_e (eq 14.1). In quark terms, a down quark changes into an up quark and emits a W^{-} boson, which decays into the electron and the antineutrino (Figure 14.11).

Beta-plus decay

In \beta^{+} decay a proton becomes a neutron, and a positron and a neutrino are emitted: p^{+}\rightarrow n+\beta^{+}+\nu_e (eq 14.2). An up quark changes into a down quark and emits a W^{+} boson, which decays into the positron and the neutrino (Figure 14.10).

Nucleon number and charge

The question says these decays violate the conservation of nucleon number. They do not. In both decays the nucleon number A is unchanged: one nucleon changes its type, neutron to proton or proton to neutron, and the nucleus keeps the same A. Only the atomic number Z changes, by +1 in \beta^{-} decay and by -1 in \beta^{+} decay.

Charge is conserved in both decays. For \beta^{-} decay, 0=(+1)+(-1)+0, and for \beta^{+} decay, +1=0+(+1)+0. The emitted electron or positron carries away the charge that the nucleus gains or loses.

Energy released and the mass difference

Derivation Energy released in beta-minus decay

Aim. Relate the energy released in a beta-minus decay to the mass difference between the initial and final particles.

Assumptions.

The parent nucleus is at rest before the decay.
The antineutrino has negligible mass, so the energy is shared between the electron and the antineutrino (Section 14.2.3).

The energy released equals the mass that disappears, multiplied by c^{2}.
Q=\left(m_{i}-m_{f}\right)c^{2}
The final mass includes the daughter nucleus, the electron and the antineutrino.
m_{f}=m_{\text{daughter}}+m_{e}+m_{\bar\nu}
Using atomic masses, the electron that is created is accounted for by the extra electron on the daughter atom, so the electron mass cancels.
Q=\left(M_{\text{parent}}-M_{\text{daughter}}\right)c^{2}
With the antineutrino mass negligible, the energy is shared between the electron and the antineutrino, so the largest kinetic energy of the electron equals Q.
K_{\max}\approx Q

ResultQ=\left(M_{\text{parent}}-M_{\text{daughter}}\right)c^{2}\quad\text{(atomic masses)}

The energy released depends only on the mass difference, so a decay can occur only if the parent is heavier than the daughter plus the emitted particles.

Note. The question's premise that beta decay violates conservation of nucleon number is not correct: A is unchanged, and it is the atomic number Z that changes. The answer corrects this.

Discuss the nature of gamma radiation in the context of nuclear decay. How is gamma radiation different from alpha and beta particles? Explain how the emission of gamma rays is related to the de-excitation of an excited nucleus and the transition to a lower energy state.

Solution

Nature of gamma radiation

Gamma rays are electromagnetic waves, similar to X-rays. They are not particles in the traditional sense, but they are emitted as photons, each with energy E=hf (eq 14.5). Gamma rays have no mass and no charge, and they travel at the speed of light, c (Section 14.2.3 and the radiation description on page 259).

Difference from alpha and beta particles

Radiation	Composition	Charge	Mass
Alpha	Two protons and two neutrons (a helium nucleus)	+2e	About 4\,\mathrm{u}
Beta (\beta^{-} or \beta^{+})	An electron or a positron	-e or +e	Much smaller than an alpha particle
Gamma	A photon (an electromagnetic wave)	0	0

Alpha and beta radiation are particles emitted from the nucleus, and each changes A or Z when it leaves. Gamma rays carry energy away from the nucleus but change neither A nor Z.

Gamma emission and de-excitation

After an alpha or beta decay the daughter nucleus is often left in an excited state, with more energy than its ground state. It returns to a lower energy state and emits the difference as a gamma photon. Since A and Z do not change, the nucleus stays the same, only in a lower energy state.

Frequency and wavelength of the emitted gamma ray

Write the energy of the photon as the energy lost by the nucleus.
E_{\gamma}=E_{\text{excited}}-E_{\text{lower}}
Since the photon energy is E=hf, its frequency is found from the energy difference.
f=\dfrac{E_{\gamma}}{h}
Its wavelength then follows from c=f\lambda.
\lambda=\dfrac{c}{f}

Discuss the fundamental differences between alpha, beta, and gamma radiation in terms of their origins, properties, interactions with matter, and potential biological hazards. How do these types of radiation impact various applications, such as medical imaging and nuclear energy production?

Solution

Origins and properties

Property	Alpha	Beta	Gamma
Origin	Rearrangement of nucleons in the nucleus (Section 14.2.3)	A neutron or proton changes type in the nucleus (Section 14.2.2)	De-excitation of an excited nucleus (Question 3)
Charge	+2e	-e or +e	0
Mass	About 4\,\mathrm{u}	Very small compared with alpha	0
Nature	Helium nucleus	Electron or positron	Photon (electromagnetic wave)

Interactions with matter

Because alpha particles are heavy and doubly charged, they ionise the matter they pass through strongly and travel only a short distance. Beta particles are lighter and less strongly charged, so they penetrate further. Gamma rays have no charge and pass through matter much more readily, so they need dense shielding. These statements go beyond the book, which covers the origins and properties, and they are given qualitatively.

Biological hazards

Alpha and beta sources are most dangerous when they enter the body, where they deposit their energy in the tissue next to them. Gamma rays penetrate the body and can cause damage from outside as well.

Applications

Medical imaging: the PET scan uses positrons, which annihilate with electrons to form two gamma photons travelling 180^{\circ} apart. A scanner detects these photons to form images of the heart, brain and cancers (Section 14.3, anti-matter and medical technology).
Nuclear energy production: the book does not discuss nuclear reactors. The radiation properties above explain why shielding is needed, but the book itself does not describe nuclear energy production, so this part is not answered from its content.

Note. The book does not cover nuclear energy production. The answer gives the radiation properties that matter for it, and says that the book does not describe reactors.

What is matter and anti-matter? How did Dirac propose the existence of anti-matter?

Solution

Matter and anti-matter

Matter is made of particles such as electrons, protons and neutrons. Every particle has an antiparticle with the same mass and spin but the opposite charge and magnetic moment. Anti-matter is made of these antiparticles. The positron is the antiparticle of the electron, and the anti-proton is the antiparticle of the proton (Table 14.1).

Table 14.1 (extract): particles and their antiparticles
Particle	Charge	Mass (MeV/c^{2})	Spin
Electron	-1e	0.51	\frac{1}{2}
Positron	+1e	0.51	\frac{1}{2}
Proton	+1e	938.28	\frac{1}{2}
Anti-proton	-1e	938.28	\frac{1}{2}

Dirac's energy solution

In 1928 Paul Dirac combined special relativity and quantum theory to find the energy of an electron in terms of its mass m and momentum p. The result is a relation with two solutions, one positive and one negative.

Derivation Dirac's energy solution

Aim. Find the two energy solutions of the relativistic relation and say what each means.

Assumptions.

The particle is free, so it has no potential energy.
The relativistic relation between energy, mass and momentum holds.

Start from the relativistic energy-momentum relation and square the energy.
E^{2}=m^{2}c^{4}+p^{2}c^{2}
Take the square root. Both signs satisfy the equation, so there are two solutions.
E=\pm\sqrt{m^{2}c^{4}+p^{2}c^{2}}
The positive solution is the energy of the particle itself. The negative solution has the same mass and momentum but negative energy.
E_{+}=+\sqrt{m^{2}c^{4}+p^{2}c^{2}},\qquad E_{-}=-\sqrt{m^{2}c^{4}+p^{2}c^{2}}

ResultE_{+}\ \text{is the particle (electron); }E_{-}\ \text{is a particle of opposite charge (positron)}

Dirac proposed that the negative-energy solution describes an antiparticle with the same mass and momentum but opposite charge. Anderson found the positron in 1932 in cosmic-ray tracks in a cloud chamber (Figure 14.1).

Example 14.1 uses the same relation, rearranged as p^{2}=\dfrac{E^{2}-m^{2}c^{4}}{c^{2}}. For a system with energy 2\,\mathrm{MeV} and electron mass, it gives p=\pm1.03\times10^{-21}\,\mathrm{kg\,m\,s^{-1}}, the two signs corresponding to the particle and the anti-particle.

Why is there asymmetry between matter and anti-matter in the known universe?

Solution

Matter and anti-matter are mirror images

Particles and antiparticles mirror each other almost perfectly. In every reaction a particle and its antiparticle are created as a pair, which is why charge conservation requires charge conjugation (C symmetry). Parity (P symmetry) relates a particle to its mirror image. Together they form CP symmetry (Section 14.3.1, Figure 14.3).

Why matter should be rare

The big bang is thought to have produced matter and anti-matter in equal amounts. With perfect CP symmetry they would have annihilated each other, leaving an empty universe. The universe we see is made mostly of matter, so the symmetry must have been slightly broken in the early universe.

CP violation

CP violation means that a decay has a different probability for a particle than for its antiparticle (Section 14.3.1). The weak force violates CP, and this is the main cause of the asymmetry. Most matter and anti-matter annihilated in the early universe, but the small surplus of matter survived. Matter began to dominate after the first second, when nuclei formed (Figure 14.4), and it now forms the stars, planets and dust that we see.

The strong force shows much less CP violation than the weak force, so it contributes less to the asymmetry.

72025 Annual I, Set D, Q.4 OR

Explain the quark family of particles.

Solution

Quarks

Quarks are elementary particles and the building blocks of hadrons. There are six flavours, grouped into three generations. Each quark has spin \frac{1}{2}, and each has an antiquark with the same mass and opposite charge (Figure 14.6).

Quark	Symbol	Charge	Mass (book)	Generation
Up	u	+\frac{2}{3}e	1.7 to 3.1\,\mathrm{MeV}/c^{2}	I
Down	d	-\frac{1}{3}e	4.1 to 5.7\,\mathrm{MeV}/c^{2}	I
Charm	c	+\frac{2}{3}e	1280\,\mathrm{MeV}/c^{2}	II
Strange	s	-\frac{1}{3}e	96\,\mathrm{MeV}/c^{2}	II
Top	t	+\frac{2}{3}e	173.1\,\mathrm{GeV}/c^{2}	III
Bottom	b	-\frac{1}{3}e	4.18\,\mathrm{GeV}/c^{2}	III

Colour charge

Quarks carry colour charge, which can be red, blue or green, and antiquarks carry the anti-colours. Colour is the property on which the strong force acts, and it has nothing to do with the everyday meaning of colour (Section 14.4.1).

Generations

The up and down quarks are the lightest and most stable, and they make up protons and neutrons. The heavier quarks belong to the second and third generations. The charm and strange quarks are the second generation. The top quark is the heaviest and decays before it can form bound hadrons, and the bottom quark decays mostly into a charm quark.

What are hadrons, baryons, mesons and leptons?

Solution

Hadrons

Hadrons are particles made of quarks and antiquarks, held together by the strong force with the help of gluons. They are the heaviest particles of matter, they carry no net colour charge, and their charges are whole numbers. Hadrons divide into baryons and mesons (Section 14.4.1).

Baryons

A baryon is made of three quarks. Its spin is half-integer, so it is a fermion. Baryons have baryon number +1 and anti-baryons have -1. The proton and the neutron are the best known baryons.

Table 14.2: some baryons with their properties
Particle	Quarks	Charge	Spin	Baryon number
Proton p	uud	+1	\frac{1}{2}	+1
Neutron n	ddu	0	\frac{1}{2}	+1
Omega \Omega	sss	-1	\frac{3}{2}	+1
Delta \Delta	uuu	+2	\frac{3}{2}	+1

Mesons

A meson is made of one quark and one antiquark. It is a boson, with integer spin, and its baryon number is 0. The pion, kaon and rho are common mesons (Table 14.3).

Table 14.3: some mesons with their properties
Particle	Quarks	Charge	Spin
Pion \pi^{+}	u\bar d	+1	0
Kaon K^{-}	s\bar u	-1	0
Kaon K^{0}	d\bar s	0	0
Rho \rho^{+}	u\bar d	+1	1

Leptons

Leptons are elementary fermions that do not take part in the strong force, and they are not made of quarks. The electron, muon and tau have charge -1 and the three neutrinos have no charge. Leptons have lepton number +1 and antileptons -1, and lepton number is conserved in every reaction (Section 14.5).

Table 14.4: leptons with their properties
Particle	Charge	Mass	Lepton number
Electron e^{-}	-1	0.511\,\mathrm{MeV}/c^{2}	+1
Muon \mu	-1	105.66\,\mathrm{MeV}/c^{2}	+1
Tau \tau	-1	1776.8\,\mathrm{MeV}/c^{2}	+1
Electron neutrino \nu_e	0	<1.0\,\mathrm{eV}/c^{2}	+1
Muon neutrino \nu_\mu	0	<0.17\,\mathrm{MeV}/c^{2}	+1
Tau neutrino \nu_\tau	0	<18.2\,\mathrm{MeV}/c^{2}	+1

Hadrons are made of quarks and feel the strong force. Leptons are elementary particles that do not feel the strong force.

Note. In the book's Tables 14.2 to 14.4 the mass column is headed GeV/c² (Table 14.2), but the values for the proton (938.3) and omega (1672) are in MeV/c², so the header is a misprint for MeV/c².

What are the force carriers? Explain.

Solution

The four fundamental forces are the strong nuclear, weak nuclear, electromagnetic and gravitational forces. Each acts through the exchange of force carriers, which are bosons (Section 14.6). The Standard Model describes three of these forces and excludes gravity, so it lists no carrier for gravity.

Force carriers (gauge and scalar bosons) from Figure 14.13
Carrier	Force	Mass (GeV/c^{2})	Spin	Charge
Photon \gamma	Electromagnetic	0	1	0
Gluon g	Strong nuclear	0	1	0
W boson W^{\pm}	Weak nuclear	80.360	1	\pm1
Z boson Z^{0}	Weak nuclear	91.19	1	0
Higgs boson H	Gives mass to particles	124.97	0	0

Photon

The photon mediates the electromagnetic force. It has no rest mass, travels at c, and has spin 1. In Figure 14.14 two approaching electrons repel each other by exchanging a photon, which is the carrier of the electromagnetic force in that interaction.

Gluon

The gluon binds quarks together into protons and neutrons. It is massless, has spin 1 and carries colour charge, so it can exchange colour between quarks. Gluons are the quanta of the strong force and are short range (Question 6 of the short questions).

W and Z bosons

The W and Z bosons mediate the weak nuclear force, which causes the decays and reactions that power stars. The W bosons have charge \pm1 and the Z boson is neutral. The Z boson was discovered at CERN in 1983. These bosons are very short lived, with lifetimes of the order of 10^{-25}\,\mathrm{s}.

Higgs boson

The Higgs boson is a scalar particle with zero spin. It is produced by the quantum excitation of the Higgs field, which fills the universe and gives mass to the elementary particles. Quarks gain large masses from their interaction with the Higgs field, electrons gain small masses, and photons do not interact with it. The Higgs boson was confirmed at CERN in July 2012.

Note. Section 14.6.4 of the book gives the W mass as 80.360 GeV/c² and also as 91.2 GeV/c². The 91.2 value is the mass of the Z boson, so the second figure is a misprint.

What is particle annihilation? Discuss its two cases.

Solution

Particle annihilation happens when a particle meets its antiparticle. They destroy each other, and their mass is converted to energy, which can be released as photons or used to create new particles (Section 14.7).

Case 1: annihilation into energy

An electron and a positron annihilate to give two gamma photons: e^{-}+e^{+}\rightarrow\gamma+\gamma (Figure 14.17). Two photons are needed because a single photon could not conserve momentum when the pair is at rest. The energy of the pair is the sum of their rest energies, and it is all carried away by the photons.

Derivation Energy of the photons in electron-positron annihilation

Aim. Find the energy of each photon when an electron and a positron annihilate from rest.

Assumptions.

Both particles are at rest before the annihilation.
The two photons leave in opposite directions, so they have equal energies.

Figure 14.17: Electron-positron annihilation: the pair meets head on and two photons leave in opposite directions.

The total energy before is the sum of the two rest energies.
E_{\text{total}}=2m_{e}c^{2}
Substitute m_{e}c^{2}=0.511\,\mathrm{MeV}.
E_{\text{total}}=2\times0.511=1.02\,\mathrm{MeV}
The total momentum is zero, so the two photons share the energy equally.
E_{\gamma}=\dfrac{E_{\text{total}}}{2}=0.511\,\mathrm{MeV}

ResultE_{\gamma}=0.511\,\mathrm{MeV}\ \text{for each photon, total }1.02\,\mathrm{MeV}

All the mass of the pair is converted to energy, as Einstein's relation predicts.

Case 2: annihilation into new particles

If a quark meets its own antiquark, they annihilate and produce energy. That energy can produce a new quark and antiquark pair. In Figure 14.18 an up quark and an anti-up quark annihilate, form a gluon, and the gluon produces a top quark and an anti-top quark.

At the Large Hadron Collider and at Fermilab, the energy released in annihilations has been converted into matter particles that were not previously discovered (Section 14.7).

Explain the particle accelerators. Also, give working of its three types.

Solution

Why accelerators are needed

Particle accelerators increase the speed and energy of charged particles, so that the particles can be collided and studied (Section 14.8). The book describes two basic types, linear accelerators and circular accelerators, and the circular type includes the synchrotron and the cyclotron. The question asks for three types: linear, cyclotron and synchrotron.

1. Linear accelerator (LINAC)

A linear accelerator is a set of cylindrical drift tubes of increasing length, joined alternately to an oscillator in a vacuum chamber. The tubes alternate in polarity, so the particle is accelerated in each gap between tubes. Inside a tube the particle drifts with no net acceleration. An ion source produces the ions, and the accelerated beam leaves the last tube and strikes the target (Figure 14.19).

Figure 14.19: Linear accelerator: the drift tubes alternate in polarity, and the beam is accelerated across each gap.

The tubes lengthen because the particle speeds up, so each one must be longer to keep the alternation in step with the oscillator. The SLAC LINAC is about 2 miles long and accelerates particles to about 50 billion eV.

2. Cyclotron

A cyclotron has two hollow metal dees connected to a radio-frequency oscillator of frequency 10^{7}\,\mathrm{Hz}. An alternating potential of 10\,\mathrm{kV} is applied between the dees, and a large electromagnet bends the path into a circle. The ions start at the centre and spiral outward, gaining energy each time they cross the gap between the dees. The cyclotron reaches 520\,\mathrm{MeV} (Figure 14.20 (b)).

Figure 14.20 (b): Cyclotron: ions start at the ion source in the centre and spiral outward between two dees, gaining energy at each crossing of the gap.

3. Synchrotron

A synchrotron has a closed ring of vacuum pipe. Particles are injected into the ring and move repeatedly round it. Radio-frequency cavities placed at intervals around the ring increase their speed, and several electromagnets create the fields that bend the beam round the ring. The synchrotron is the highest energy accelerator: the LHC at CERN reached 13.6\,\mathrm{TeV} in proton collisions (Figure 14.20 (a)).

Figure 14.20 (a): Synchrotron: the beam circulates in a ring, is bent by electromagnets and is speeded up by radio-frequency cavities.

Give in detail the uses of linear and circular particle accelerators.

Solution

Linear accelerators

They are used to produce medical isotopes and to create beams of radiation for cancer treatment.
The electron LINAC is the most common type used for cancer therapy. It generates X-rays and high-energy electrons for radiation therapy.
They serve as injectors, supplying particles to the higher energy accelerators.

Synchrotron accelerators

They produce high-energy environments to study the building blocks of the universe. The Higgs boson was discovered using these accelerators.
In medicine they are used for cell biology, crystallography, high-resolution imaging and cancer radiation therapy.
In environmental science they are used in toxicology, clean combustion and atmospheric research, and also in agriculture, mineral exploration, engineering and forensics.

Cyclotron accelerators

They produce large numbers of particles such as muons and neutrinos.
They are also used to produce medical isotopes.
They are the best source of high-energy beams for particle physics experiments.
They are used in particle therapy to treat cancer.

Accelerators are therefore used both in research and in medicine, and the uses follow from the energies and kinds of particle each type can produce (Section 14.8.1).

Numerical problemsp. 283

A uranium-238 nucleus undergoes alpha decay, emitting an alpha particle (helium nucleus) with a mass of 4.00151\,\mathrm{u}. Calculate the energy released in this alpha decay process. The speed of light c is approximately 3.00\times10^{8}\,\mathrm{m\,s^{-1}}, and 1\,\mathrm{u} is approximately 931.5\,\mathrm{MeV}/c^{2}.

Solution

Given

m_{\text{U}} = 238.05079\,\mathrm{u} mass of the uranium-238 atom, standard table value, not in the question
m_{\text{Th}} = 234.04360\,\mathrm{u} mass of the thorium-234 atom, standard table value, not in the question
m_{\alpha} = 4.00151\,\mathrm{u} alpha particle mass as given in the question
1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^{2}

To find

the energy released Q

The energy released is the mass lost, multiplied by c^{2}. Use the atomic mass of helium-4 for the alpha particle, to match the atomic masses of the parent and daughter (see the note).
Q=\left(m_{\text{U}}-m_{\text{Th}}-m_{\text{He}}\right)c^{2}
Substitute the masses, with m_{\text{He}}=4.00260\,\mathrm{u}.
\Delta m=238.05079-234.04360-4.00260=0.00459\,\mathrm{u}
Convert the mass defect to energy with 1\,\mathrm{u}=931.5\,\mathrm{MeV}/c^{2}.
Q=0.00459\times931.5\,\mathrm{MeV}=4.27\,\mathrm{MeV}

Q\approx4.27\,\mathrm{MeV}

Book answer. 4.25 MeV

Note. The question gives only the alpha mass, so the parent and daughter masses come from standard tables. The book's 4.25\,\mathrm{MeV} corresponds to a mass difference of 0.00456\,\mathrm{u}; our value is about 0.5 per cent higher, which comes from the masses used. The bare-nucleus alpha mass 4.00151\,\mathrm{u} combined with atomic masses would give about 5.3\,\mathrm{MeV}, which mixes two different mass definitions, so the atomic mass of helium is used instead.

A carbon-14 nucleus undergoes beta-minus (\beta^{-}) decay, emitting a beta particle (electron) with a negligible mass. Calculate the energy released in this beta-minus decay process. The speed of light c is approximately 3.00\times10^{8}\,\mathrm{m\,s^{-1}}.

Solution

Given

m_{\text{C}} = 14.003242\,\mathrm{u} atomic mass of carbon-14, standard table value, not in the question
m_{\text{N}} = 14.003074\,\mathrm{u} atomic mass of nitrogen-14, standard table value, not in the question
1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^{2}

To find

the energy released Q

Carbon-14 decays to nitrogen-14 with the emission of an electron and an antineutrino, ^{14}\mathrm{C}\rightarrow{}^{14}\mathrm{N}+e^{-}+\bar\nu_e. The energy released is the mass difference between the atoms, multiplied by c^{2}. The electron is included in the nitrogen atom, so its mass cancels.
Q=\left(m_{\text{C}}-m_{\text{N}}\right)c^{2}
Substitute the atomic masses.
\Delta m=14.003242-14.003074=0.000168\,\mathrm{u}
Convert with 1\,\mathrm{u}=931.5\,\mathrm{MeV}/c^{2}.
Q=0.000168\times931.5\,\mathrm{MeV}=0.156\,\mathrm{MeV}

Q\approx0.16\,\mathrm{MeV}

Book answer. 0.17 MeV

Note. With the standard atomic masses the energy is 0.156\,\mathrm{MeV}, which rounds to 0.16\,\mathrm{MeV}, not 0.17\,\mathrm{MeV}. The book's value needs a mass difference of about 0.000182\,\mathrm{u}, which the standard masses do not give. The difference is about 9 per cent, so the book's figure is recorded here as a disagreement.

A nucleus undergoes a gamma decay, emitting a gamma ray photon with energy of 2.00\,\mathrm{MeV}. Calculate the frequency and wavelength of this gamma ray.

Solution

Given

E = 2.00\,\mathrm{MeV}=2.00\times10^{6}\times1.60\times10^{-19}\,\mathrm{J}=3.20\times10^{-13}\,\mathrm{J}
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant
c = 3.00\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

the frequency f
the wavelength \lambda

A photon's energy is E=hf, so its frequency is found from the energy.
f=\dfrac{E}{h}=\dfrac{3.20\times10^{-13}}{6.63\times10^{-34}}
Evaluate.
f=4.83\times10^{20}\,\mathrm{Hz}
The wavelength follows from c=f\lambda.
\lambda=\dfrac{c}{f}=\dfrac{3.00\times10^{8}}{4.83\times10^{20}}
Evaluate.
\lambda=6.21\times10^{-13}\,\mathrm{m}

f\approx4.83\times10^{20}\,\mathrm{Hz},\qquad \lambda\approx6.21\times10^{-13}\,\mathrm{m}

Book answer. 3.02\times10^{20}\,\mathrm{Hz}, 9.94\times10^{-13}\,\mathrm{m}

Note. The book's values do not follow from 2.00\,\mathrm{MeV}. They are equal to E/h with E=2.00\times10^{-13}\,\mathrm{J}, which is 1.25\,\mathrm{MeV}, so the factor 1.60\times10^{-19}\,\mathrm{J} per electronvolt seems to have been dropped. Our frequency is about 60 per cent higher and our wavelength about 1.6 times shorter than the book's; the working above uses the energy as given.

If the negative solution of Dirac equation is for electron, then prove that the positive solution for same equation is not for a proton of the same momentum.

Solution

Given

m_e = 9.11\times10^{-31}\,\mathrm{kg} mass of the electron
m_p = 1.67\times10^{-27}\,\mathrm{kg} mass of the proton
p = \text{the same momentum for both}

To find

show that the positive solution with the proton's mass does not equal the electron's positive solution at the same momentum

Dirac's relation has two solutions, E=\pm\sqrt{m^{2}c^{4}+p^{2}c^{2}}. The negative solution is taken for the electron, so the equation with the electron's mass gives E_{-}=-\sqrt{m_e^{2}c^{4}+p^{2}c^{2}}. The positive solution of the same equation has the same magnitude, E_{+}=+\sqrt{m_e^{2}c^{4}+p^{2}c^{2}}.
E_{+}^{2}=m_{e}^{2}c^{4}+p^{2}c^{2}
A proton of the same momentum has mass m_p, so its positive solution would be E_{p}=+\sqrt{m_p^{2}c^{4}+p^{2}c^{2}}.
E_{p}^{2}=m_{p}^{2}c^{4}+p^{2}c^{2}
Subtract the two squared energies. The momentum terms cancel.
E_{p}^{2}-E_{+}^{2}=\left(m_{p}^{2}-m_{e}^{2}\right)c^{4}
Compare the masses. The proton is heavier, so m_p^{2}>m_e^{2}, and the difference is positive.
m_{p}=1.67\times10^{-27}\,\mathrm{kg}\approx1.83\times10^{3}\,m_{e}\quad\Rightarrow\quad E_{p}^{2}>E_{+}^{2}
Since both energies are positive, E_{p}>E_{+} for every momentum p. The positive solution for the electron's mass therefore cannot equal the energy of a proton with the same momentum.
E_{p}\neq E_{+}

E_{p}>E_{+}\ \text{for every }p,\ \text{so the positive solution is not that of a proton of the same momentum}

Note. The energy of a free particle depends on its mass as well as its momentum, so the mass in the equation identifies the particle. This is the proof asked for; the book prints no answer.

Show that the following particle reactions satisfy the laws of conservation of charge, lepton-number and baryon-number.

Solution

(i) \pi^{-}+p=K^{+}+K^{-}+n

Charge: left side (-1)+(+1)=0; right side (+1)+(-1)+0=0. Conserved.
0=0
Lepton number: the pion, kaons, proton and neutron are all hadrons, so each has lepton number 0 on both sides. Conserved.
0=0
Baryon number: left side 0+1=+1 (pion 0, proton +1); right side 0+0+1=+1 (kaons 0, neutron +1). Conserved.
+1=+1

(ii) \pi^{+}=\mu^{+}+\nu_{\mu}

Charge: left side +1; right side (+1)+0=+1. Conserved.
+1=+1
Lepton number: the pion has 0; on the right, \mu^{+} is an antilepton (-1) and \nu_\mu is a neutrino (+1), total 0. Conserved.
0=(-1)+(+1)=0
Baryon number: all particles are mesons or leptons, so the baryon number is 0 on both sides. Conserved.
0=0

\text{Both reactions conserve charge, lepton number and baryon number.}

Note. Part (ii) is taken as printed, with the muon neutrino \nu_\mu and no bar. With an antineutrino, the lepton number on the right would be -1-1=-2 and the reaction would fail, so the printed form is the one that is consistent.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 151 mark

Proton is made up of:

A2 down and 1 up quark
B1 down and 2 up quarksAnswer
C2 down and 2 up quarks
D4 down and 2 up quarks

Why

The proton consists of two up quarks and one down quark (uud). The charge check: +\tfrac23+\tfrac23-\tfrac13=+1. Option A (udd, with two down quarks) is the neutron, with charge +\tfrac23-\tfrac13-\tfrac13=0. Options C and D have too many quarks, since a baryon has three.

2025 Annual I, Set B, MCQ 171 mark

A hadron may either be a Baryon or a:

ALepton
BGluon
CPhoton
DMesonAnswer

Why

Hadrons are made of quarks. A baryon has three quarks (proton, neutron) and a meson has a quark and an anti-quark (pion, kaon). So a hadron is a baryon or a meson. Option A (lepton) is the tempting wrong one: leptons are elementary particles that do not take part in the strong interaction, so they are not hadrons. Gluons and photons are force carriers.

2026 Annual I, Set B, Q.2 (ii)3 marks

Describe photon as a force carrier.

Solution

Photon as a force carrier (3 marks). The photon is the exchange particle (force carrier) of the electromagnetic force.

According to quantum field theory, charged particles exert the electromagnetic force on each other by exchanging photons. One charged particle emits a photon and another absorbs it, so the photon carries the force between them; hence photons are the mediators of the electromagnetic force.
A photon is the smallest quantum of electromagnetic radiation and the basic unit of light. It travels through vacuum at 3\times10^{8}\,\mathrm{m\,s^{-1}} and has energy E=hf.
It has zero rest mass and zero charge, it is stable, and it has spin 1, so it is a boson (a gauge boson).

In short, the electromagnetic interaction between two charges is the exchange of photons between them.

2025 Annual I, Set B, Q.2 (ix) OR3 marks

If ^{233}_{92}\mathrm{U} decays twice by alpha particle emission, what is the resulting isotope? Show by equation.

Solution

Given

{}^{233}_{92}\mathrm{U} = parent decays twice by alpha emission

To find

the final isotope

In each alpha decay the nucleus emits an alpha particle {}^{4}_{2}\mathrm{He}: the mass number falls by 4 and the atomic number falls by 2. Charge and nucleon number are conserved in each step.

First alpha decay: A=233-4=229, Z=92-2=90, which is thorium.
{}^{233}_{92}\mathrm{U}\to{}^{229}_{90}\mathrm{Th}+{}^{4}_{2}\mathrm{He}
Second alpha decay: A=229-4=225, Z=90-2=88, which is radium.
{}^{229}_{90}\mathrm{Th}\to{}^{225}_{88}\mathrm{Ra}+{}^{4}_{2}\mathrm{He}
Overall, the two steps combine into one equation. Check: 233=225+2\times4 and 92=88+2\times2.
{}^{233}_{92}\mathrm{U}\to{}^{225}_{88}\mathrm{Ra}+2\,{}^{4}_{2}\mathrm{He}

The resulting isotope is radium-225, {}^{225}_{88}\mathrm{Ra}.

{}^{225}_{88}\mathrm{Ra}

2025 Annual I, Set B, Q.2 (x) OR3 marks

Compare gluons and photons.

Solution

Gluons and photons are both force carriers (gauge bosons), but they belong to different forces.

Property	Photon	Gluon
Force carried	Electromagnetic force	Strong (colour) force
Acts between	Electrically charged particles	Quarks and anti-quarks (colour-charged particles)
Charge it carries	No electric charge, and no colour	No electric charge, but it carries colour charge and exchanges colour
Mass and speed	Zero rest mass, speed c	Massless, speed c
Spin	1 (boson)	1 (vector boson)
Range	Long range	Short range, within a nucleon or nucleus
Role	Quantum of electromagnetic radiation; E=hf	Quantum of the strong force; binds quarks into protons and neutrons

Both are massless, travel at the speed of light and have spin 1. The main difference is that the gluon carries colour charge and acts on quarks, while the photon is colourless and acts on electric charge.

2025 Annual I, Set B, Q.2 (xi)3 marks

What is meant by annihilation? Give one annihilation reaction.

Solution

Meaning (1 mark). Annihilation is the process in which a particle and its anti-particle meet and destroy each other. Their mass is converted into energy, in accordance with E=mc^{2}, usually as gamma-ray photons. A particle and its anti-particle cannot exist in the same place.

Reaction (1 mark). An electron and a positron that come close annihilate and give two gamma-ray photons:

e^{-}+e^{+}\to\gamma+\gamma

Two photons are needed to conserve momentum. The pair at rest has zero total momentum, and two photons moving in opposite directions have momenta that cancel. Each photon carries 0.511\,\mathrm{MeV}, the rest energy of an electron.

Electron-positron annihilation: the pair meets head on and two photons leave in opposite directions.

Second case (1 mark). The released energy may instead form a new particle and anti-particle pair, as when an up quark and an anti-up quark annihilate and produce a top quark and an anti-top quark.

Note. The paper does not say how many reactions to give; one is asked.

2026 Annual I, Set D, MCQ 161 mark

Neutrino is:

AMagnetic particle
BCharged particle
CHeavy particle
DNeutral particleAnswer

Why

The neutrino is a lepton with zero electric charge, so it is a neutral particle. Neutrinos are not magnetic or charged, and their mass is extremely small (less than 1\,\mathrm{eV}/c^{2} for the electron neutrino), so option C (heavy) is wrong. Option B, charged, is the property of the electron, muon and tau, the charged leptons.

2025 Annual I, Set D, MCQ 171 mark

Two up quarks and one down quark make a:

AMeson
BPhoton
CNeutron
DProtonAnswer

Why

Add the charges of the quarks: uud gives +\tfrac23+\tfrac23-\tfrac13=+1, which is a proton. The neutron is udd (two down quarks and one up quark), charge 0. A meson has only two quarks (a quark and an anti-quark), and a photon is a force carrier, not made of quarks.

2026 Annual I, Set D, Q.2 (ii) OR3 marks

State the Higgs Boson as a fundamental particle which is responsible for the particle's mass.

Solution

The Higgs boson (3 marks). The Higgs boson is a fundamental (elementary) particle that is responsible for the mass of particles.

Up to 1964 the mass of a particle was thought to be an intrinsic property. In 1964 Peter Higgs proposed the Higgs field, which fills the universe and gives mass to elementary particles. The Higgs boson is the wave (quantum excitation) of that field. It was confirmed at CERN in July 2012, with a mass of 124.97\,\mathrm{GeV}/c^{2}, about 130 times that of a proton.
How strongly a particle interacts with the Higgs field decides its mass. Quarks interact strongly and gain relatively large mass; electrons interact only slightly, so they are very light; photons do not interact with it and have no mass.
The Higgs boson has zero spin, zero electric charge and no strong force interaction.

2025 Annual I, Set D, Q.4 OR2+4 = 6 marks

What is meant by quark? Explain the quark family of particles.

Solution

Quark (2 marks)

A quark is an elementary particle, that is, a particle not composed of other particles, and it is one of the basic building blocks of matter. Quarks combine to form composite particles called hadrons (baryons and mesons), and they carry fractional electric charge (\pm\tfrac13e or \pm\tfrac23e), spin \tfrac12, and a colour charge. The quark theory was proposed in 1964 by Murray Gell-Mann and George Zweig and confirmed in 1968 in deep inelastic scattering at SLAC. Each quark has an anti-quark with the same mass and opposite charge.

The quark family (4 marks)

There are six flavours of quark, arranged in three generations. In each generation one quark has charge +\tfrac23e and the other -\tfrac13e.

Quark	Symbol	Charge	Generation	Note
Up	u	+\tfrac23e	I	lightest, most stable
Down	d	-\tfrac13e	I	stable, a little heavier than up
Charm	c	+\tfrac23e	II	hundreds of times heavier than up
Strange	s	-\tfrac13e	II	very long lived
Top	t	+\tfrac23e	III	heaviest, decays before forming hadrons
Bottom	b	-\tfrac13e	III	decays mostly into charm

Colour charge. Quarks also carry a colour charge (red, blue or green) and anti-quarks carry anti-colours. This has nothing to do with ordinary colour. It is the property on which the strong force, described by quantum chromodynamics, acts, and gluons bind the quarks together.

Hadrons made of quarks. A baryon has three quarks and a meson has a quark and an anti-quark. The proton is uud (charge +1) and the neutron is udd (charge 0); the anti-proton and anti-neutron contain the anti-quarks.

The quark content of the proton and the neutron. The charges of the three quarks add up to the charge of the particle.

Exam focus

Explain the quark family of particles (quark, six flavours, charges, colour). (long)
2025 Set D Q.4 OR (6 marks, 2+4); comprehensive question 7 of the book.
Go to answer
Quark composition of the proton and neutron; check by adding the charges. (mcq)
2026 Set B MCQ 15 (proton) and 2025 Set D MCQ 17 (two up and one down quark).
Go to answer
Compare gluons and photons; photon as a force carrier. (short)
2025 Set B Q.2 (x) OR (comparison) and 2026 Set B Q.2 (ii) (photon as a force carrier).
Go to answer
The Higgs boson and the mass of particles. (short)
2026 Set D Q.2 (ii) OR; an SLO of the unit.
Go to answer
What is annihilation? Give a reaction, and the two cases. (short)
2025 Set B Q.2 (xi); comprehensive question 10 of the book.
Go to answer
Alpha decay: write the equation, and the result of two successive alpha decays. (short)
2025 Set B Q.2 (ix) OR ({}^{233}_{92}\mathrm{U} decaying twice); alpha decay lowers A by 4 and Z by 2.
Go to answer
Hadrons: baryons and mesons, and the lepton as a different family. (mcq)
2025 Set B MCQ 17 (a hadron is a baryon or a meson) and 2026 Set D MCQ 16 (the neutrino is neutral).
Go to answer
Beta decay and the antineutrino; conservation of charge, nucleon number and lepton number. (long)
Not asked in the 2025 or 2026 papers, but comprehensive question 2 and numerical problem 5 of the book and several SLOs cover it.
Go to answer