Lesson 2 · 30 min

Radioactive decay

α, β⁻, β⁺, electron capture and γ: five ways a nucleus moves toward stability. Balance decay equations, count the steps from uranium-238 to lead-206, and watch individual nuclear “dice” add up to the exponential law.

In this lesson

Write equations for α, β⁻, β⁺, electron-capture and γ decay and check that mass and charge numbers balance
Compute the decay energy Q from atomic masses and judge whether a decay is energetically possible
Count the α and β decays in a decay chain and explain secular equilibrium
Distinguish the randomness of a single nucleus from the exponential law obeyed by many

In 1896 Henri Becquerel left uranium salts on a photographic plate wrapped in black paper, locked in a drawer. Days later the developed plate showed the outline of the salts. There had been no light and no chemical reaction: the uranium nuclei themselves were changing and giving off invisible rays. Today we know that in 1 g of uranium-238 about 12 000 nuclei decay every second — and there is no way at all to predict which nucleus will be next.

Predict first

Take 400 nuclei of a nuclide with a half-life of 1 hour. How many are left after 1 hour?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Exactly 200 About 200, differing by ten or so from one run to the next None: they all decay together when the half-life is up 400: nothing happens until the half-life is reached

Five ways to decay

When a nucleus changes, both the mass number A and the charge number Z are conserved: the superscripts on each side add up to the same total, and so do the subscripts. An electron is written $\ce{^{0}_{-1}e}$ , a positron $\ce{^{0}_{+1}e}$ , and an α particle is a helium nucleus, $\ce{^{4}_{2}He}$ .

α decay emits $\ce{^4_2He}$ : A falls by 4 and Z by 2. $\ce{^{238}_{92}U} \mathrel{\ce{->}} \ce{^{234}_{90}Th} + \ce{^{4}_{2}He}$ releases 4.27 MeV, of which the α particle carries off 4.20 MeV.
β⁻ decay: a neutron in the nucleus becomes a proton, emitting an electron and an antineutrino; A is unchanged and Z rises by 1: $\ce{^{14}_{6}C} \mathrel{\ce{->}} \ce{^{14}_{7}N} + \ce{^{0}_{-1}e} + \ce{\bar{\nu}_e}$ .
β⁺ decay: a proton becomes a neutron, emitting a positron and a neutrino; Z falls by 1: $\ce{^{18}_{9}F} \mathrel{\ce{->}} \ce{^{18}_{8}O} + \ce{^{0}_{+1}e} + \ce{\nu_e}$ — the basis of PET imaging.
Electron capture: the nucleus swallows an inner electron, which also lowers Z by 1: $\ce{^{40}_{19}K} + \ce{^{0}_{-1}e} \mathrel{\ce{->}} \ce{^{40}_{18}Ar} + \ce{\nu_e}$ .
γ decay: an excited nucleus emits a high-energy photon; A and Z stay the same: $\ce{^{99m}_{43}Tc} \mathrel{\ce{->}} \ce{^{99}_{43}Tc} + \ce{\gamma}$ releases 142.7 keV, almost all of it as a γ ray of about 140 keV.

Q_\alpha = \left[M_\mathrm{P} - M_\mathrm{D} - M(\ce{^4He})\right]c^2,\quad Q_{\beta^-} = \left[M_\mathrm{P} - M_\mathrm{D}\right]c^2,\quad Q_{\beta^+} = \left[M_\mathrm{P} - M_\mathrm{D} - 2m_\mathrm{e}\right]c^2

With neutral-atom masses; electron capture uses the same expression as β⁻ (parent − daughter). A decay can happen only if Q > 0.

$M_\mathrm{P},\ M_\mathrm{D}$: atomic masses of parent and daughter
$Q$: total energy released, shared among the products
$2m_\mathrm{e}c^2$: the two electron rest energies β⁺ decay must also pay, 1.022 MeV

Mass and charge numbers balance · Q from AME2020 atomic masses

Decay-equation workbench

Parent nuclide

$\ce{^{40}_{19}K -> ^{40}_{20}Ca + ^{0}_{-1}e + \bar{\nu}_e}$

Bookkeeping
Mass number A	40 = 40 + 0
Charge number Z	19 = 20 + (-1)

Energy released Q = 1.311 MeV Q > 0: energetically allowed

measured branching 89.28 %

Q is shared between the electron (or positron) and a neutrino, so β energies form a continuous spectrum reaching up to about Q.

Half-life (⁴⁰K): 1.25 billion years

The step on the chart of nuclides

Potassium-40 is unusual: it undergoes both β⁻ decay and electron capture. Try each and compare the Q values; then see why β⁺ hardly happens. Switch to carbon-14 and find out why its α and β⁺ Q values are negative.

Decay chains and secular equilibrium

The daughter of ²³⁸U, ²³⁴Th, is itself radioactive, and so is its daughter … it takes 14 decays to reach stable ²⁰⁶Pb. Count them: the mass number falls from 238 to 206, a drop of 32, and only α decay changes A, so there are 32/4 = 8 α decays. Eight α decays lower Z by 16, but Z only falls from 92 to 82, by 10, so 6 β⁻ decays must raise it back by 6.

The uranium in an ore has been there for billions of years. Each member of the chain is produced by the one before it exactly as fast as it decays itself, so every member has the same activity — secular equilibrium. The shorter a member’s half-life, the less of it there is.

A_i = \lambda_i N_i,\qquad \lambda_1 N_1 = \lambda_2 N_2 = \cdots = \lambda_n N_n \quad(\text{secular equilibrium})

Holds once the parent outlives every daughter by far and enough time has passed.

$A_i$: activity of member i (decays per second, Bq)
$\lambda_i$: decay constant, $\lambda = \ln 2 / t_{1/2}$
$N_i$: number of atoms of that member

²³⁸U → ²⁰⁶Pb in 14 decays · NUBASE2020 half-lives

The uranium series: a Bateman-equation solver

Time elapsed · 1 million years

2 of 13 daughters are within 1 % of the parent’s current activity

²⁰⁶Pb/²³⁸U atom ratio: 8.81 × 10⁻⁵

Click a row to highlight that nuclide in the graph.

Activity of each member now (relative to the parent’s initial activity)
Nuclide	Half-life	Relative activity
	4.46 billion years	1
	24.1 days	1
	69.5 s	1
	246 thousand years	0.941
	75.4 thousand years	0.915
	1600 years	0.914
	3.82 days	0.914
	3.1 min	0.914
	27.1 min	0.914
	19.9 min	0.914
	163 µs	0.914
	22.2 years	0.914
	5.01 days	0.914
	138 days	0.914

Start from freshly purified uranium-238 and drag the time: the short-lived ²³⁴Th and ²³⁴ᵐPa catch up within months, ²³⁴U takes hundreds of thousands of years, and radium and everything after it wait for ²³⁰Th to grow in. After about two million years every daughter is within 1 % of the parent’s activity. Then try “a flask of radon-222” and watch its daughters come into equilibrium within a few hours.

Worked example

How many decays from thorium-232 to lead-208?

The thorium series starts at $\ce{^{232}_{90}Th}$ and ends at $\ce{^{208}_{82}Pb}$ . How many α and how many β⁻ decays does it contain?

Concept check

What does ²²²Rn become when it α-decays?

²¹⁸Po ²¹⁸Rn ²²²Fr ²²⁶Ra

Try the calculation

Uranium-238 and radium-226 are in secular equilibrium in an old uranium ore. How many grams of ²²⁶Ra accompany 1000 kg of ²³⁸U? (Half-lives 4.463 × 10⁹ years and 1600 years; molar masses 238.05 and 226.03 g/mol.)

Key takeaways

Nuclear equations conserve A and Z: α decay gives A − 4 and Z − 2, β⁻ gives Z + 1, β⁺ and electron capture give Z − 1, and γ decay changes neither.
A decay can happen only if Q, from atomic masses, is positive; β⁺ must pay an extra 1.022 MeV.
²³⁸U reaches ²⁰⁶Pb through 8 α and 6 β⁻ decays; in an old ore the whole chain is in secular equilibrium, every member with the same activity.
When one nucleus decays cannot be predicted, but many nuclei follow the exponential law precisely, with fluctuations of about √N.

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Data and references

F. G. Kondev et al., “The NUBASE2020 evaluation of nuclear physics properties”, Chinese Physics C 45, 030001 (2021) — half-lives, decay modes and branching ratios.
M. Wang et al., “The AME 2020 atomic mass evaluation (II)”, Chinese Physics C 45, 030003 (2021) — atomic masses for Q-values.
K. S. Krane, Introductory Nuclear Physics, Wiley (1988), chapters 6–10 (decay laws, α, β and γ decay).
G. Friedlander, J. W. Kennedy, E. S. Macias, J. M. Miller, Nuclear and Radiochemistry, 3rd ed., Wiley (1981), chapter 6 (radioactive growth and decay, secular equilibrium).