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An atom has a tiny, dense nucleus of protons and neutrons, surrounded by orbiting electrons.
A nuclide is written AZX, where A is the nucleon number and Z the proton number; the neutron number is N = A − Z.
Specific charge is the charge-to-mass ratio of a particle:
The electron has the largest specific charge (small mass); the neutron's is zero (no charge).
Find the specific charge of a proton (charge 1.6×10−19 C, mass 1.673×10−27 kg).
Isotopes have the same proton number but different neutron numbers; they are chemically identical.
An ion forms when an atom gains or loses electrons, giving it a net negative or positive charge (a multiple of 1.6×10−19 C).
Chlorine's relative atomic mass of 35.5 arises from a mixture of 3517Cl (about 75%) and 3717Cl (about 25%).
Electromagnetic radiation is emitted in packets (photons) of energy:
The electronvolt (eV) is the energy transferred when an electron moves through a p.d. of 1 V: 1 eV = 1.6×10−19 J.
Calculate the energy of a photon of frequency 5.0×1014 Hz (h = 6.63×10−34 J s).
Every particle has a corresponding antiparticle with the same mass and rest energy but opposite charge, baryon number, lepton number and strangeness.
When a particle meets its antiparticle they annihilate, converting their total rest energy into two gamma photons.
The pair approach with equal and opposite momenta, so the total momentum beforehand is zero. The two photons must therefore travel in exactly opposite directions — which is why two photons are produced and never one.
The energy locked up in a particle's mass is its rest energy, given by Einstein's mass–energy equation:
Calculate the rest energy of an electron (mass 9.11×10−31 kg) in J and in MeV.
Pair production is the reverse of annihilation: a high-energy photon creates a particle-antiparticle pair, usually near a nucleus (which recoils to conserve momentum).
The photon must have at least the total rest energy of the pair: Emin = 2 m c2.
The three quarks at A-level are up (u), down (d) and strange (s). Antiquarks have the opposite sign of charge, baryon number and strangeness.
A baryon is three quarks (qqq). The two you must know are the proton and the neutron:
Quarks carry fractional charges, but no free particle has ever been observed with a fractional charge. Every baryon and every meson must therefore have a whole-number charge — 0, ±1e or ±2e. This is a useful check: if a proposed combination of quarks does not come to an integer, it cannot be a real hadron.
The properties of the quarks are given to you in the exam:
| Quark | Charge | Baryon No. | Strangeness | Charmness | Bottomness | Topness |
|---|---|---|---|---|---|---|
| d | −⅓ | +⅓ | 0 | 0 | 0 | 0 |
| u | +⅔ | +⅓ | 0 | 0 | 0 | 0 |
| s | −⅓ | +⅓ | −1 | 0 | 0 | 0 |
| c | +⅔ | +⅓ | 0 | +1 | 0 | 0 |
| b | −⅓ | +⅓ | 0 | 0 | −1 | 0 |
| t | +⅔ | +⅓ | 0 | 0 | 0 | +1 |
All hadrons feel the strong interaction and are made of quarks. They come in exactly two kinds:
A meson is always one quark and one antiquark (qq̅). The two families you need are the pions and the kaons.
Every hadron has an antiparticle, built from the corresponding antiquarks. An antibaryon is three antiquarks (q̅q̅q̅).
Note that the antineutron is not the same as the neutron, even though both are uncharged: their baryon numbers are −1 and +1, so they are genuinely different particles.
Because a meson already contains an antiquark, the antiparticle of a meson is another meson. Replacing every quark by its antiquark turns π⁺ (ud̅) into π⁻ (u̅d), so the two charged pions are antiparticles of each other. The same is true of K⁺ and K⁻. The π⁰ is its own antiparticle.
Forces between particles are transmitted by exchange particles (gauge bosons).
| Interaction | Exchange particle | Range | Acts on | What it governs |
|---|---|---|---|---|
| Gravity | graviton (hypothetical) | infinite | anything with mass | holds planets, stars and galaxies together; negligible between particles |
| Electromagnetic | virtual photon (γ) | infinite | charged particles | binds electrons to nuclei; all of chemistry; friction and contact forces |
| Weak nuclear | W⁺, W⁻ (and Z⁰) | ~10−18 m | all quarks and leptons (leptons are met next lesson) | beta decay and electron capture; the only force that changes quark flavour |
| Strong nuclear | gluon (quarks) pion (nucleons) | ~3 fm | quarks and hadrons | binds quarks into hadrons and holds the nucleus together against proton repulsion |
Protons in the nucleus repel electrostatically, yet the nucleus holds together because of the strong nuclear force.
On Fig 4.1 a force above the axis is a repulsion and a force below it is an attraction. The curve crosses the axis at about 0.5 fm: closer than this the nucleons repel, which is what stops the nucleus collapsing.
Between quarks the strong interaction is carried by the gluon. Between whole nucleons inside a nucleus, AQA requires you to know that the exchange particle is the pion (π).
This is the same argument used for the W bosons: a massive exchange particle gives a short-range force, while the massless photon gives the electromagnetic interaction infinite range.
Leptons are fundamental particles: they are not made of quarks and they do not feel the strong interaction. At A-level you need the electron and the muon, together with their neutrinos and all of their antiparticles.
Lepton number is conserved in every interaction. Beta-minus decay shows why the antineutrino has to be there:
Before the decay the lepton number is 0. The electron created carries Le = +1, so an electron antineutrino with Le = −1 must be emitted alongside it for the total to stay at 0.
| Lepton | Charge | Le | Lμ |
|---|---|---|---|
| electron e⁻ | −1 | +1 | 0 |
| muon μ⁻ | −1 | 0 | +1 |
| positron e⁺ | +1 | −1 | 0 |
| antimuon μ⁺ | +1 | 0 | −1 |
| Neutrino | Charge | Le | Lμ |
|---|---|---|---|
| νe | 0 | +1 | 0 |
| νμ | 0 | 0 | +1 |
| ν̅e | 0 | −1 | 0 |
| ν̅μ | 0 | 0 | −1 |
Four quantities are conserved in every particle interaction:
1 Charge — always conserved, in every interaction without exception.
2 Baryon number — always conserved; this is why the proton cannot decay.
3 Lepton number — conserved for the electron and muon families.
4 Energy and momentum — conserved, as in all of physics.
To test whether a proposed interaction is possible, work out each quantity on the left-hand side and on the right-hand side. If any one of them fails to balance, the interaction cannot happen.
Strangeness is a quantum number carried by particles that contain a strange quark. It behaves differently from the other conserved quantities:
Because strangeness is conserved in the strong interaction, strange particles are always created in pairs — one with strangeness +1 alongside one with strangeness −1. They then decay by the weak interaction, in which strangeness may change, e.g. K⁺ → π⁺ + π⁰. This is why strange particles live far longer than their production suggests.
A Feynman diagram is a space–time diagram of an interaction. It shows what goes in, what comes out, and the exchange particle that carries the force.
The three weak processes you must be able to draw all exchange a W boson. You will study them in full next lesson; here they are, so that you can draw them:
In every case the charge of the W follows from conserving charge at the first vertex.
At the first vertex, work out the change in charge of the hadron. In β⁻ decay a neutron (0) becomes a proton (+1), so the emitted boson must carry away −1: a W⁻. In β⁺ decay a proton (+1) becomes a neutron (0), so a W⁺ is emitted.
The electromagnetic interaction acts between charged particles and is carried by a virtual photon. It is virtual because it exists only for the duration of the interaction and cannot be detected.
The same shape describes any electromagnetic interaction between charged particles — two protons repelling, or an electron and a proton attracting. Neither particle changes identity: the virtual photon transfers only energy and momentum, which is why both lines keep the same label from bottom to top.
So far the hadron has been drawn as a single line. A fuller diagram shows all three quarks, because only one of them actually changes flavour — the other two carry straight on through.
This is why beta decay is a weak interaction and not a strong one: changing a quark's flavour is something only the weak interaction can do.
If a nucleus emitted only an electron in beta decay, every electron would carry away exactly the same energy. Measurement shows something different:
The beta particle shares the energy released with a second, almost undetectable particle, so it emerges with any energy from nearly zero up to Emax.
Beta decay therefore appeared to lose energy. Pauli proposed the neutrino — a neutral, almost massless particle — to conserve energy and momentum.
So a third particle has to appear in every beta decay equation, as you will see in the next Part: beta-minus emits an antineutrino (ν̅e), and beta-plus emits a neutrino (νe).
In radioactive decay, nucleon number and proton number (charge) are conserved.
The weak interaction is the only interaction that can change one type of quark into another. At A-level it is limited to four processes, and each one can be drawn as a Feynman diagram using the rules from last lesson.
The W bosons have a large mass, so they can exist only for a very short time. This is why the weak interaction has such a short range (about 10−18 m).
Beta decay changes one quark flavour within a nucleon, mediated by the weak interaction:
This is why beta decay is a weak interaction: only the weak force can change quark flavour.
Electrons in an atom occupy discrete energy levels. An electron cannot sit between two levels, so it can only gain or lose energy in the exact amounts that take it from one level to another. A change from one level to another is called a transition.
Level energies are quoted in electronvolts. One electronvolt is the energy transferred when an electron moves through a potential difference of 1 V, so 1 eV = 1.6×10−19 × 1 = 1.6×10−19 J. To convert eV to J, multiply by e; to convert J to eV, divide by e.
Energies are measured from the ionisation level, so a free electron at rest counts as 0 eV. Every bound electron has less energy than that, because energy has to be supplied to pull it out, which is why the ground state and every level above it carry a negative value. The lowest level, the ground state, is the most negative.
There are two ways to excite an atom, and the energy requirement is not the same for the two of them.
When an electron falls from a level of energy E1 to a lower level E2, the energy lost leaves the atom as a single photon:
Take care with the labelling: on the AQA data sheet E1 is the level the electron starts in and E2 the level it ends in, so for emission E1 is the upper level and the difference comes out positive.
An electron in a hydrogen atom falls from the level at −3.4 eV to the ground state at −13.6 eV. Find the wavelength of the emitted photon.
Because the levels are discrete, only certain photon energies can ever be emitted. Split the light from a hot, low-pressure gas with a diffraction grating and you see a set of sharp coloured lines rather than a continuous band: a line emission spectrum.
Every element has its own pattern of levels, so every element has its own set of gaps and therefore its own spectrum. The pattern works like a barcode, which is how the composition of a gas that nobody can sample, in a star or a nebula, can still be identified.
Send white light through a cool gas instead and the same transitions run in reverse. Photons of exactly the right energies are absorbed on the way through and scattered out of the beam, leaving dark lines in an otherwise continuous spectrum: an absorption spectrum.
The dark lines of a gas sit at exactly the wavelengths of its bright ones, because both are produced by the same set of level differences. Emission is an electron falling and giving a photon out; absorption is an electron being lifted by a photon of precisely that energy. Finding the dark lines of an element in starlight is therefore evidence that the element is present in the cooler gas the light passed through.
The fluorescent tube puts excitation, ionisation and de-excitation to work in a single device. The tube holds mercury vapour at low pressure, has an electrode at each end, and is coated on the inside with a phosphor.
Fig 8.4 is steps 4 and 5 drawn on an energy level diagram, because that is where the wavelength changes. One large jump up absorbs the whole ultraviolet photon; the electron then comes back down in two or more smaller jumps, giving out one photon per jump.
Absorbing one photon and re-emitting the energy as several photons of lower energy is called fluorescence. Because each step down is smaller than the original jump up, every emitted photon has a lower frequency, and so a longer wavelength, than the ultraviolet that was absorbed.
At low pressure the accelerated electrons travel a long way between collisions, so they pick up enough kinetic energy to excite a mercury atom rather than merely nudging it. Ionising collisions matter too: each one frees another electron, which is then accelerated in its turn, so the current sustains itself once the tube has struck.
Shine light on a clean metal surface and electrons are emitted from it. What made this observation so awkward was not that it happened, but the way it happened.
Einstein's explanation was that light arrives as photons, and that one photon transfers all of its energy hf to one electron. Some of that energy is used to break the electron free of the surface and whatever is left over becomes kinetic energy:
This is the whole of the argument. Because a photon is absorbed whole and by a single electron, nothing can be accumulated: either one photon carries enough on its own or nothing happens at all. That is what gives a sharp threshold and instant emission, and it is why doubling the number of photons doubles the number of electrons without making any of them faster.
Practical opportunity: demonstrate the photoelectric effect with a photocell, or with a gold-leaf electroscope carrying a clean zinc plate. Charge the plate negatively and a UV lamp discharges it; charge it positively and the leaf stays up, because any emitted electrons are pulled straight back.
At the threshold frequency the photon has just enough energy to free an electron and none left over, so Ek(max) = 0 and the photoelectric equation reduces to:
A metal with a large work function therefore has a high threshold frequency. Caesium, with a small work function, emits in visible light; zinc needs ultraviolet.
Rearranging the equation as Ek(max) = h f − φ puts it in the form y = m x + c. A graph of Ek(max) against f is therefore a straight line of gradient h, crossing the frequency axis at f0 and extrapolating back to an intercept of −φ.
Light of frequency 7.0×1014 Hz hits a metal of work function 3.0×10−19 J. Find the maximum KE of the photoelectrons.
Watch the units. Work functions are usually quoted in electronvolts and must be multiplied by e before being used alongside h f in joules.
A photocell turns the photoelectric effect into a measurable current. Radiation falls on a photoemissive surface inside an evacuated tube, the emitted electrons cross to a wire anode, and the current round the external circuit is carried entirely by those electrons.
The current tells you how many electrons leave each second, so it follows the intensity. It says nothing about how fast they are going. To measure that, the circuit has to be turned against them.
Connect a variable supply so that its positive terminal goes to the photoemissive surface. The anode is now negative, so the field pushes the emitted electrons back the way they came. Raise the p.d. and the slowest electrons no longer reach the anode, so the current falls. At the stopping potential Vs even the fastest electrons are turned back and the current reaches zero.
An electron turned back at Vs has had all of its kinetic energy converted into electrical potential energy, so:
Substituting this into the photoelectric equation gives a form that connects the stopping potential straight to the frequency of the light:
A photoemissive surface of work function 2.1 eV is illuminated with light of frequency 7.23×1014 Hz. Find the stopping potential.
Because e Vs = Ek(max), a stopping potential of 0.90 V means the fastest electrons left with 0.90 eV of kinetic energy. The two numbers are always the same, which makes the stopping potential a direct reading of Ek(max) in electronvolts.
Einstein had shown that a wave can behave as a particle. In 1923 Louis de Broglie turned the idea round and suggested that all particles have a wave nature, with a wavelength set by their momentum:
This is the de Broglie wavelength, measured in metres. Note what it says: the greater the momentum, the shorter the wavelength. A fast particle, or a heavy one, has a tiny wavelength, which is why wave behaviour never shows up in everyday objects.
Most electrons in the laboratory are speeded up by an accelerating p.d. V, so it is worth being able to go straight from V to λ. The work done on the electron becomes its kinetic energy:
Solving for v and substituting into λ = h/mv gives a single expression for an accelerated electron:
Electrons are accelerated from rest through a p.d. of 2500 V. Find their de Broglie wavelength.
The modern reading of the de Broglie wavelength is not that the electron wobbles along a wavy path. The wave describes the probability of finding the particle at a given point in space, and it is that probability which shows interference.
De Broglie's suggestion was testable: if electrons have a wavelength, they must diffract. In 1927 Davisson and Germer showed exactly that, firing electrons at a single nickel crystal and finding that the scattered intensity peaked at particular angles. In the same year G. P. Thomson sent electrons through a thin film instead, and it is his version that the laboratory tube reproduces.
The regular rows of atoms act as the diffraction grating. A thin film is made of countless tiny crystals lying at every orientation, so the diffracted beams come out at the same angle in every direction at once and the pattern on the screen is a set of complete rings rather than the row of spots a single crystal would give.
Diffraction is only noticeable when the wavelength is comparable to the size of the gap. The atomic spacing in a crystal is about 10−10 m, so electrons must be given a de Broglie wavelength of roughly that size, which an accelerating p.d. of a few hundred volts provides.
Changing the speed of the electrons changes the pattern, and the direction of the change is a standard exam point.
Raising the accelerating p.d. gives the electrons more momentum. Since λ = h/p, their wavelength falls, so there is less diffraction and the rings close in towards the centre. Slowing the electrons down does the opposite: a longer wavelength spreads the pattern out.
Practical opportunity: an electron diffraction tube, in which the accelerating p.d. can be varied while the rings on the fluorescent screen are watched.
The same relationship is a tool rather than a curiosity. To probe something of size d you need a wavelength no larger than d, so investigating structure inside a nucleus demands very small wavelengths and therefore very high momenta. This is why particle accelerators have to reach such enormous energies: they are buying resolution.
Electrons are to be used to investigate a nucleus of diameter 5.0×10−15 m. Find the smallest momentum that will do.
Putting the two halves of this topic together gives wave-particle duality: waves sometimes behave as particles, and particles sometimes behave as waves. Neither model on its own accounts for everything, and which one is useful depends on what is being measured.
The two key equations sit either side of the divide, and it is worth being able to say which is which:
Refraction is often added to the list of wave behaviours, and it is the one to leave off. A particle model accounts for refraction perfectly well, as Newton showed, so it does not decide between the two pictures. Diffraction, interference and polarisation do.
Electron diffraction is the evidence that particles possess wave properties. The photoelectric effect is the evidence that electromagnetic waves have a particulate nature. Each experiment supplies exactly one half of the argument, and examiners expect the right one to be quoted for the right half.
Both constants in those equations are the same h, which is the real point of de Broglie's idea. One quantity governs how much energy a wave of a given frequency carries, and how long a wave a particle of a given momentum has.