state that (electron) antineutrinos are produced during β– decay and (electron) neutrinos are produced during β+ decay

Atoms, Nuclei and Radiation – Cambridge A‑Level Physics 9702

Learning objective

State that electron antineutrinos (̄φe) are emitted in β⁻ decay and that electron neutrinos (φe) are emitted in β⁺ decay. In doing so you will see how the (anti)neutrino guarantees conservation of lepton number, energy, momentum and angular momentum.


1 Radiation types required by the syllabus

RadiationParticle(s) emittedTypical energy rangeRepresentative nuclear equation
α‑radiationα‑particle (42He)4–9 MeV⁽ᴬ⁾ZX → ⁽ᴬ⁻⁴⁾{Z‑2}Y + ^4_2He
β⁻‑radiationElectron (e⁻) + electron antineutrino (̄φe)0.5–3 MeV (shared)⁽ᴬ⁾ZX → ⁽ᴬ⁾{Z+1}Y + e⁻ + ̄φe
β⁺‑radiationPositron (e⁺) + electron neutrino (φe)≥1.022 MeV + kinetic⁽ᴬ⁾ZX → ⁽ᴬ⁾{Z‑1}Y + e⁺ + φe
γ‑radiationPhoton (γ‑ray)0.1–10 MeV (typical)⁽ᴬ⁾ZX* → ⁽ᴬ⁾ZX + γ

Key points required by the syllabus

  • Conservation of nucleon number (A) and charge (Z) in every decay.
  • Use of nuclear notation ⁽ᴬ⁾_ZX (mass number A, atomic number Z).
  • Identification of antiparticles (positron ↔ electron, antineutrino ↔ neutrino).
  • Energy released (Q‑value) is obtained from the mass‑defect and E = mc².


2 Fundamental particles (≈150 words)

All matter is built from two families of elementary particles:

  • Quarks – six flavours (up , down , charm, strange, top, bottom). Up‑type quarks carry charge +2⁄3 e; down‑type quarks carry –1⁄3 e. Protons are uud, neutrons are udd.
  • Leptons – three charged leptons (electron , muon <μ⁻>, tau <τ⁻>) and three neutral partners (electron neutrino <νe>, muon neutrino <νμ>, tau neutrino <ντ>), each with a corresponding antiparticle.

In β‑decay the weak interaction changes a down‑quark to an up‑quark (β⁻) or an up‑quark to a down‑quark (β⁺) and creates the appropriate lepton–neutrino pair.


3 Nuclear notation, isotopes and atomic mass

An isotope is written ⁽ᴬ⁾_ZX, where:

  • A = total number of nucleons (protons + neutrons).
  • Z = number of protons (atomic number).
  • X = chemical symbol.

Atomic masses are given in unified atomic mass units (u), with 1 u = 931.5 MeV c⁻². The mass‑defect Δm is the difference between the sum of the individual nucleon masses and the measured nuclear mass; the binding energy B = Δm c².


4 β⁻ Decay (neutron → proton)

4.1 Particle‑level reaction

n → p + e⁻ + ̄φe

4.2 Nuclear equation

⁽ᴬ⁾ZX → ⁽ᴬ⁾{Z+1}Y + e⁻ + ̄φe

4.3 Worked example – Carbon‑14

⁽¹⁴⁾6C → ⁽¹⁴⁾7N + e⁻ + ̄φe

Atomic masses (to 5 dp):

  • m(14C) = 14.003242 u
  • m(14N) = 14.003074 u
  • me = 0.00054858 u

Q‑value:

\[

\begin{aligned}

Q &= \bigl[m(^{14}\!C)-m(^{14}\!N)-m_e\bigr]c^{2} \\

&= (0.0001194\;\text{u})\times 931.5\;\text{MeV/u} \\

&\approx 0.11\;\text{MeV}.

\end{aligned}

\]

Because the decay is a three‑body process, the kinetic energy is shared continuously between the electron and the antineutrino, producing the characteristic continuous β‑spectrum (0 → Q).

4.4 Lepton‑number balance

  • Electron: L = +1
  • Electron antineutrino: L = –1
  • Total lepton number of products = 0, equal to the initial neutron (L = 0).


5 β⁺ Decay (proton → neutron)

5.1 Particle‑level reaction

p → n + e⁺ + φe

5.2 Nuclear equation

⁽ᴬ⁾ZX → ⁽ᴬ⁾{Z‑1}Y + e⁺ + φe

5.3 Worked example – Sodium‑22

⁽²²⁾11Na → ⁽²²⁾10Ne + e⁺ + φe

Atomic masses:

  • m(22Na) = 21.994437 u
  • m(22Ne) = 21.991385 u
  • me = 0.00054858 u

Q‑value:

\[

\begin{aligned}

Q &= \bigl[m(^{22}\!Na)-m(^{22}\!Ne)-2m_e\bigr]c^{2} \\

&= (0.001994\;\text{u})\times 931.5\;\text{MeV/u} \\

&\approx 1.86\;\text{MeV}.

\end{aligned}

\]

The first 1.022 MeV supplies the rest‑mass energy of the e⁺e⁻ pair; the remaining energy appears as kinetic energy of the positron and the neutrino.

5.4 Lepton‑number balance

  • Positron (antilepton): L = –1
  • Electron neutrino: L = +1
  • Total lepton number of products = 0, matching the initial proton (L = 0).


6 Comparison of β⁻ and β⁺ decay

Featureβ⁻ decayβ⁺ decay
Quark changed → uu → d
Nuclear changen → p (Z + 1)p → n (Z – 1)
Lepton emittedElectron e⁻Positron e⁺
Associated (anti)neutrinoElectron antineutrino ̄φeElectron neutrino φe
Lepton‑number balance+1 + (–1) = 0–1 + (+1) = 0
Typical Q‑value0.5–3 MeV (continuous spectrum)≥1.022 MeV + kinetic
Detectable radiationElectron track in cloud/chamber; antineutrino undetected.Positron track + two 511 keV annihilation γ‑rays; neutrino undetected.


7 Brief overview of α and γ radiation

  • α‑decay: ⁽ᴬ⁾ZX → ⁽ᴬ⁻⁴⁾{Z‑2}Y + ^4_2He. Conserves A and Z; releases 4–9 MeV.
  • γ‑decay: Excited nucleus de‑excites by emitting a photon, ⁽ᴬ⁾ZX* → ⁽ᴬ⁾ZX + γ. No change in A or Z; photon energy equals the difference between nuclear energy levels.


8 Mass‑defect, binding energy and Q‑values

For any nuclear transformation

\[

Q = \bigl[M{\text{initial}}-M{\text{final}}\bigr]c^{2},

\]

where the masses are taken from atomic‑mass tables (including the electrons that belong to neutral atoms).

  • In β⁻ decay the daughter atom already contains the emitted electron, so only one me is subtracted.
  • In β⁺ decay a positron must be created; therefore two me are subtracted (the positron plus the electron that will later annihilate).

Binding energy per nucleon B/A is obtained from the total mass‑defect of the nucleus and helps explain why very light nuclei favour β⁺ decay and very heavy nuclei favour α decay.


9 Antiparticles – quick reminder

  • Every particle has an antiparticle with opposite electric charge and opposite lepton or baryon number.
  • Positron e⁺ ↔ electron e⁻.
  • Electron antineutrino ̄φe ↔ electron neutrino φe.
  • In β⁻ decay the antineutrino carries the “missing” lepton number; in β⁺ decay the neutrino does the same.


10 Key take‑away statement

During β⁻ decay a neutron converts to a proton and emits an electron (e⁻) together with an electron antineutrino (̄φe). During β⁺ decay a proton converts to a neutron and emits a positron (e⁺) together with an electron neutrino (φe). The (anti)neutrino ensures that lepton number, energy, momentum and angular momentum are all conserved.