Radioactive decay
Three kinds of radiation
- An unstable nucleus settles down by throwing out radiation.
- There are three kinds — alpha, beta and gamma — and they behave very differently.
- Telling them apart is the heart of this topic.
What each one is
- α: a helium nucleus ($^{4}_{2}\alpha$), charge $+2e$, mass $\approx 4\ \text{u}$.
- β: a fast electron ($\beta^{-}$) or positron ($\beta^{+}$), charge $\mp e$, tiny mass.
- γ: a high-energy photon, no charge, no mass.
An α-particle is:
An α-particle is $^{4}_{2}\alpha$ — a helium-4 nucleus with charge $+2e$.
Penetration and ionising
- α: stopped by paper; strongly ionising.
- β: stopped by a few mm of aluminium.
- γ: only reduced by thick lead; weakly ionising.
Match each radiation to what stops it.
Penetration goes α < β < γ; ionising power goes the other way (α is the strongest ioniser).
Of the three, α radiation is the most strongly ionising.
Yes — α is the strongest ioniser (and so the least penetrating); γ is the weakest ioniser (and most penetrating).
Antiparticles and neutrinos
- Every particle has an antiparticle — same mass, opposite charge (the positron is the electron's).
- β-decay always emits a (anti)neutrino too: nearly massless, no charge, very hard to detect.
The positron is the ____ of the electron.
An antiparticle has the same mass but opposite charge — the positron is $+e$ to the electron's $-e$.
Why β is continuous
- α-decay shares energy between two particles → the α has one fixed energy (discrete).
- β-decay shares energy between three (with the neutrino) → β energies range from zero up to a maximum (continuous).
β-particles come out with a continuous range of energies because the energy is shared among:
With three particles sharing, the β can take any share up to a maximum. α-decay shares between two, fixing the α energy.
Decay equations
- α: $^{A}_{Z}\text{X} \to {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\alpha$.
- β$^{-}$: $^{A}_{Z}\text{X} \to {}^{A}_{Z+1}\text{Y} + {}^{0}_{-1}\beta + \bar{\nu}_{e}$ (a neutron becomes a proton).
In α-decay, by how much does the nucleon number $A$ decrease?
The α carries away 4 nucleons (2 protons + 2 neutrons), so $A$ falls by 4 (and $Z$ by 2).
In β$^{-}$ decay, the proton number $Z$:
A neutron turns into a proton, so $Z$ rises by 1 while $A$ stays the same.
You've got it
- α = He nucleus (paper-stopped, strong ioniser); β = electron/positron (Al-stopped); γ = photon (lead)
- β-decay emits a (anti)neutrino, which is why its energy spectrum is continuous
- α: $A{-}4, Z{-}2$; β$^{-}$: same $A$, $Z{+}1$