Radioactive decay

A **helium nucleus** — 2 protons + 2 neutrons (⁴₂He), charge **+2**.

A **fast electron** emitted from the nucleus, charge **−1**.

A **high-energy photon** (electromagnetic wave), charge **0**, no mass.

To **knock an electron off it**, leaving a charged ion. More ionising = more damage but shorter range.

**Alpha < beta < gamma** — paper, then a few mm of aluminium, then thick lead/concrete.

**Alpha > beta > gamma** — the opposite order to penetration.

α: paper / a few cm of air / skin. β⁻: a few mm of aluminium. γ: thick lead or concrete.

**Gamma** — it is a neutral photon (charge 0), so a field cannot push it. α and β are charged and do deflect.

Its **+2 charge** makes it interact strongly with atoms, so it ionises heavily and loses its energy in a short distance.

**Outside:** the skin stops it. **Inside** (breathed in/swallowed): its strong ionising power damages tissue with no skin to shield it.

Alpha is the least penetrating: a few cm of air, the casing and skin all stop it, and the sealed source is very weak.

$E = mc^{2}$ — the lost mass (mass defect) times the speed of light squared.

In **opposite** directions (opposite charges); beta much **more**, because its mass is tiny.

**Beta**: paper absorbs some of it, so the count rate depends on the thickness (alpha would be stopped, gamma barely affected).

Gamma gets **out of the body** to the detector and ionises weakly; a short half-life means little dose remains afterwards.

Alpha particles and gamma photons from one nuclide have only **certain energies**: line spectra, like atomic spectra.

The **nucleon number A** (top numbers balance) and the **proton number Z** (bottom numbers balance).

${}^{4}_{2}\alpha$ — a **helium-4 nucleus** (2 protons + 2 neutrons).

${}^{\;\;0}_{-1}e$ — an **electron** (created when a neutron turns into a proton). An antineutrino is emitted with it.

A **falls by 4** and Z **falls by 2** (A → A − 4, Z → Z − 2).

A is **unchanged**; Z **rises by 1** (A → A, Z → Z + 1).

A **neutron becomes a proton**, so there is one more proton. The emitted electron's −1 charge forces the daughter's Z up by 1 to balance.

${}^{A}_{Z}X \to {}^{A-4}_{Z-2}Y + {}^{4}_{2}\alpha$.

${}^{A}_{Z}X \to {}^{\;\;A}_{Z+1}Y + {}^{\;\;0}_{-1}e + \bar{\nu}$.

Z + 1 = 83 + 1 = **84** (polonium). A is unchanged.

A = 226 − 4 = **222**, Z = 88 − 2 = **86** (radon-222).

Apply the changes **one emission at a time**, updating A and Z after each step.

Find the daughter's A and Z first, then use **N = A − Z** (nucleon number − proton number).

A is **unchanged**, Z **falls by 1**: a proton becomes a neutron, emitting a positron and a neutrino.

**Neither changes**: the nucleus only drops from an excited energy level to a lower one.

Beta-minus: an **antineutrino**. Beta-plus: a **neutrino**.

With only two products the beta energy would be fixed; a **range** of energies means a third particle shares the energy.

How much **lighter** the products are than the parent nucleus: Δm = parent mass − total product mass.

The energy the **mass defect** turns into, shared as kinetic energy of the products. Found from E = mc².

$E = mc^{2}$ — mass-energy equivalence (given in the data booklet). Here m is the mass defect.

Multiply Δm (in u) by **931.5**, because 1 u = 931.5 MeV c⁻².

The defect is a **tiny** difference of large numbers — rounding early loses the answer entirely.

**Equal and opposite** (same size p), so the total momentum stays zero — conservation of momentum.

Same momentum p, and KE = p²/2m, so the **smaller** mass gives the **bigger** kinetic energy.

KE_{alpha} : KE_{daughter} = m_{daughter} : m_{alpha}. The alpha's share = m_{daughter} ÷ (m_{daughter} + m_{alpha}).

Almost all of it — around **98%** — because the heavy daughter barely recoils.

E = 0.0052 × 931.5 ≈ **4.8 MeV** (about 5 MeV).

1) mass defect Δm = parent − products; 2) E = mc² (or Δm × 931.5 for MeV); 3) the light product carries most of the energy.

Parent mass − total mass of the products; it becomes the energy released, E = Δm c² (Δm in u × 931.5 MeV).

Equal and opposite momenta and KE = p²/2m: the **smaller mass** gets the larger kinetic energy.

The **positron's mass**, 0.000549 u (and the daughter nucleus).

The **time** for the activity (or count rate, or number of undecayed nuclei) to fall to **half** its value.

The number of nuclei that **decay each second**. Unit: the **becquerel (Bq)**, where 1 Bq = 1 decay per second.

How many decays a **detector records each second** (clicks per second). It is always ≤ the activity.

Radiation a detector picks up **even with no source** (from rocks, soil, cosmic rays). It must be **subtracted** to get the true source count.

**Measured count rate − background count rate**. Always correct for background **before** halving.

Start value **× (1/2)ⁿ**. So 1, 2, 3 half-lives leave 1/2, 1/4, 1/8 of the start.

**n = total time ÷ half-life.** Then halve the start value n times.

No — the count rate keeps **halving** and flattens out, but in theory never reaches zero.

It **stays the same** — both halve by the same factor each half-life, so the ratio is unchanged.

Source 84 − 4 = 80; 4 h = 2 half-lives → 80 → 40 → 20; add background → **24 counts s⁻¹**.

You **cannot predict** when any one nucleus will decay; only the **average** behaviour (the half-life) is fixed.

Subtract the **background** first, then read the time for the corrected rate to halve; check it with a second halving.

Random: no way to predict which nucleus decays when. Spontaneous: not affected by temperature, pressure or chemistry.

(72 − b)/2 + b = 40, so **b = 8 s⁻¹**.

$N = N_0 e^{-\lambda t}$ — exponential decay of the un-decayed nuclei.

$A = A_0 e^{-\lambda t}$, and $A = \lambda N$. Activity follows the same curve as N.

The **probability per unit time** that a given nucleus decays. Unit: **s⁻¹** (or day⁻¹, hour⁻¹).

$\lambda = \dfrac{\ln 2}{t_{1/2}}$ (ln 2 ≈ 0.693).

Because $A = \lambda N$ — activity is just λ times N, so both decay by the same factor $e^{-\lambda t}$.

$\left(\tfrac{1}{2}\right)^{n}$ — e.g. after 3 half-lives, ⅛.

$\lambda = \ln 2 / 8.0 = 0.087$ day⁻¹.

24 days = 3 half-lives, so $(½)^3 = ⅛ = 0.125\,N_0$.

Each nucleus is **very likely** to decay each second, so the **half-life is short**.

The **activity A** (decays per second, in **Bq**), not N directly.

t must use the **same time unit** as λ (λ in day⁻¹ ⇒ t in days).

Whole n half-lives ⇒ use $(½)^n$; any other time ⇒ use $e^{-\lambda t}$.

Only when **λt is small**: the time interval is short compared with the half-life.

t = (1/λ) ln(N₀/N), with λ = ln2/T½ in matching units.

N = (m / M) × N_A: moles times the Avogadro constant.