11.1 Atoms, nuclei and radiation

Learning objectives

11.1.1The results of the α-particle scattering experiment the existence and small• infer from the results of the α-particle scattering experiment the existence and small size of the nucleus 11.1.2A simple model for the nuclear atom to include protons, neutrons and orbital• describe a simple model for the nuclear atom to include protons, neutrons and orbital electrons 11.1.3Nucleon number and proton number• distinguish between nucleon number and proton number 11.1.4Isotopes are forms of the same element with different numbers of neutrons in• understand that isotopes are forms of the same element with different numbers of neutrons in their nuclei 11.1.5The notation A Z X for the representation of nuclides• understand and use the notation A Z X for the representation of nuclides 11.1.6Nucleon number and charge are conserved in nuclear processes• understand that nucleon number and charge are conserved in nuclear processes 11.1.7The composition, mass and charge of α-, β- and γ-radiations (both β–• describe the composition, mass and charge of α-, β- and γ-radiations (both β– (electrons) and β+ (positrons) are included)11.1.8An antiparticle has the same mass but opposite charge to the corresponding• understand that an antiparticle has the same mass but opposite charge to the corresponding particle, and that a positron is the antiparticle of an electron 11.1.9That (electron) antineutrinos are produced during β– decay and (electron)• state that (electron) antineutrinos are produced during β– decay and (electron) neutrinos are produced during β+ decay 11.1.10Α-particles have discrete energies but that β-particles have a continuous• understand that α-particles have discrete energies but that β-particles have a continuous range of energies because (anti)neutrinos are emitted in β-decay 11.1.11Α- and β-decay by a radioactive decay equation of the form UT h92 238 90 234• represent α- and β-decay by a radioactive decay equation of the form UT h92 238 90 234 2 4" + α 11.1.12The unified atomic mass unit (u) as a unit of mass• use the unified atomic mass unit (u) as a unit of mass

Most alpha particles passed through thin foil, but a small fraction were deflected through large angles; this implies most atomic volume is empty with mass and positive charge concentrated in a tiny nucleus.

Use the scattering observations as evidence and distinguish the nuclear model from the older diffuse-charge model.

A near head-on deflection is rare because the nucleus occupies a tiny cross-section, yet it reveals strong repulsion from concentrated positive charge.

The experiment did not show electrons orbiting like planets; it established nuclear concentration, not a complete modern atom model.

A simple nuclear model places positively charged protons and neutral neutrons in a tiny nucleus, with negatively charged electrons occupying the surrounding region.

Mass is concentrated mainly in nucleons while atomic size is set by the electron region. This is a model, not a literal set of classical orbits.

Changing the number of electrons can make an ion without changing the nucleus or the element identity.

Electrons are not inside the nucleus in this model, and a neutral atom does not mean it contains no charged particles.

Proton number Z is the number of protons; nucleon number A is total protons plus neutrons, so neutron number is A−Z.

Z determines chemical identity. A can vary between isotopes while Z remains fixed.

A nuclide with A=23 and Z=11 contains 11 protons and 12 neutrons.

Nucleon number is not the number of electrons, and changing A alone does not create a new element.

Isotopes have the same proton number but different numbers of neutrons, so they share chemical identity but differ in mass and nuclear properties.

Compare Z first, then A. Isotopic abundance affects relative atomic mass without changing the element symbol.

Carbon-12 and carbon-14 both have Z=6; carbon-14 has two more neutrons and is radioactive.

Isotopes are not different elements, and isotope differences are not caused by different electron counts in neutral atoms.

$⁽ᴬ⁾₍Z₎X: A = nucleon number, Z = proton number, X = element symbol$

Quantity	Count
protons	Z
neutrons	A − Z
electrons in a neutral atom	Z
electrons in an ion of charge +ne	Z − n
electrons in an ion of charge −ne	Z + n

For ²³⁹₉₄Pu⁺, protons = 94, neutrons = 239 − 94 = 145 and electrons = 94 − 1 = 93. The + sign means one electron has been removed; it does not change the nucleus.

To write a nuclide symbol, use the periodic-table identity to select X and Z, then place total protons plus neutrons as A at upper left and Z at lower left.

Do not swap A and Z. A is not neutron number, and an ionic charge changes electron count only; it does not change A, Z or the isotope.

$ΣA_before = ΣA_after and Σ(q/e)_before = Σ(q/e)_after$

Emitted object	nucleon number A	charge number q/e
α particle	4	+2
β⁻ electron	0	−1
β⁺ positron	0	+1
γ photon or (anti)neutrino	0	0

Write every reactant and product, then total the upper numbers for nucleon number and the signed lower charge numbers independently. A missing particle or wrong daughter is exposed when either ledger fails.

Alpha decay conserves A through A = (A − 4) + 4 and charge through Z = (Z − 2) + 2. In beta-minus decay, the daughter's nuclear proton number rises by one, but total charge still balances: Z = (Z + 1) + (−1).

Proton number and neutron number are not each separately conserved in beta decay: one nucleon changes type. The conserved quantities here are total nucleon number and total electric charge across all products.

Radiation	Composition	Rest mass	Charge
α	helium nucleus: 2 protons + 2 neutrons	about 4 u	+2e
β⁻	electron	9.11 × 10⁻³¹ kg ≈ 5.49 × 10⁻⁴ u	−e
β⁺	positron	same as electron	+e
γ	electromagnetic photon	0	0

Identify radiation from all three requested properties. Positive charge alone is insufficient: both α and β⁺ are positive, but α has charge magnitude 2e and a mass thousands of times larger.

$α = ⁴₂He, β⁻ = ⁰₋₁e, β⁺ = ⁰₊₁e, γ = ⁰₀γ$

Gamma radiation is electromagnetic radiation, not a neutral nucleon. A beta particle has the electron/positron rest mass; its high speed does not make its rest mass zero.

The antiparticle corresponding to a particle has the same rest mass and electric charge of equal magnitude but opposite sign.

Particle	Antiparticle	Mass relation	Charge relation
electron e⁻	positron e⁺	equal	−e and +e
proton p	antiproton p̄	equal	+e and −e

If a top quark has charge +(2/3)e, its antiquark has the same mass and charge −(2/3)e. Reverse the sign; do not change the magnitude or mass.

A positron is specifically the electron's antiparticle. It is not a proton: both carry +e, but their masses and particle identities are different.

Decay	Nucleon change	Charged beta particle	Neutral lepton
β⁻	neutron → proton	electron e⁻	electron antineutrino ν̄ₑ
β⁺	proton → neutron	positron e⁺	electron neutrino νₑ

$n → p + e⁻ + ν̄ₑ p → n + e⁺ + νₑ$

Match the beta sign first: β⁻ is the electron and therefore accompanies the antineutrino; β⁺ is the positron and accompanies the neutrino. Both (anti)neutrinos have zero charge and zero nucleon number.

The (anti)neutrino is an additional emitted particle, not another name for the beta particle. Omitting it makes a beta-decay equation incomplete for this syllabus.

Alpha decay between quantised nuclear levels gives particles with characteristic discrete energies; beta decay shares energy with a neutrino, producing a continuous range.

Interpret line spectra and broad beta spectra as evidence about the number of bodies sharing decay energy.

A detector can show sharp alpha peaks but a broad beta distribution ending at a maximum energy.

A continuous beta spectrum does not mean beta particles are emitted with random violation of energy conservation; total energy is conserved event by event.

$α: ᴬ_ZX → ᴬ⁻⁴_Z₋₂Y + ⁴₂He β⁻: ᴬ_ZX → ᴬ_Z₊₁Y + ⁰₋₁e + ⁰₀ν̄ₑ β⁺: ᴬ_ZX → ᴬ_Z₋₁Y + ⁰₊₁e + ⁰₀νₑ$

Identify the decay type, calculate the daughter A and Z, use Z to identify its element symbol, add every emitted particle, then check that upper numbers and signed lower numbers balance separately.

Decay	Complete example	Check
α	²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He	238 = 234 + 4; 92 = 90 + 2
β⁻	³₁H → ³₂He + ⁰₋₁e + ⁰₀ν̄ₑ	3 = 3; 1 = 2 − 1
β⁺	¹²₇N → ¹²₆C + ⁰₊₁e + ⁰₀νₑ	12 = 12; 7 = 6 + 1

A is unchanged in beta decay. The daughter Z rises for β⁻ and falls for β⁺, and the correct electron (anti)neutrino must be included even though it contributes 0 to both number checks.

One unified atomic mass unit, 1 u, is one twelfth of the mass of a neutral carbon-12 atom. Use 1 u = 1.66 × 10⁻²⁷ kg.

$mass in kg = mass in u × 1.66 × 10⁻²⁷ mass in u = mass in kg ÷ (1.66 × 10⁻²⁷)$

A particle of mass 12.0 u has mass 12.0 × 1.66 × 10⁻²⁷ = 1.99 × 10⁻²⁶ kg.

A nucleus of mass 7.30 × 10⁻²⁶ kg has mass (7.30 × 10⁻²⁶)/(1.66 × 10⁻²⁷) = 44.0 u.

Nucleon number A is an integer count, not an exact nuclear mass in u. Binding energy causes the actual mass of a nucleus to differ from the simple sum of free nucleon masses.

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—11.2 Fundamental particlesRead

11.1 Atoms, nuclei and radiation

Learning objectives

Alpha scattering showed that atoms contain a tiny dense positive nucleus

The nuclear atom contains protons and neutrons in a small nucleus with electrons outside it

Proton number identifies the element, while nucleon number counts protons plus neutrons

Isotopes are atoms of the same element with different neutron numbers

Read nuclide notation and count particles in atoms and ions

Balance nucleon number and electric charge separately

Compare alpha, beta-minus, beta-plus and gamma radiation

An antiparticle has equal mass and opposite charge

Beta-minus emits an antineutrino; beta-plus emits a neutrino

Alpha particles have discrete energies, whereas beta particles have a continuous energy spectrum

Write complete alpha, beta-minus and beta-plus decay equations

Convert particle and nuclear masses using the unified atomic mass unit

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