Atomic Nuclei and Radioactive Decay

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Senior 6 · Physics

Atomic Nuclei and Radioactive Decay

5 lessons

Lesson 3 of 5

Radioactive Decay: Alpha, Beta and Gamma Radiation

25 min

A photographic plate fogged in the dark

In 1896 Henri Becquerel left uranium salts on a photographic plate wrapped in black paper, in a drawer. When he developed it, the plate showed the outline of the salts: the uranium was giving out something that passed through paper, without any light falling on it. Marie and Pierre Curie soon found two new elements, polonium and radium, that did the same far more strongly. Marie Curie named the effect radioactivity. Within a few years Ernest Rutherford and others showed that the radiation was of three kinds, which they named alpha, beta and gamma. In Lesson 2 you saw WHY some nuclei are unstable. This lesson is about WHAT they emit, how to write the change as an equation, and how the radiation is detected.

Three kinds of radiation

Pass the radiation from a source through a magnetic field and it splits into three beams (Figure 3.1). The direction of the force on a moving charge (Fleming's left-hand rule, or $\vec F = q\vec v \times \vec B$ ) shows that alpha radiation is positive and beta radiation is negative, while gamma radiation carries no charge at all.

Figure 3.1: Radiation in a magnetic field (not to scale). In a 0.10 T field a 4.8 MeV alpha particle moves on a circle of radius about 3 m, while a 1 MeV beta particle's radius is under 5 cm.

Figure 3.2: The three radiations differ greatly in how far they penetrate matter.

Property	alpha (α)	beta-minus (β⁻)	gamma (γ)
nature	helium-4 nucleus, ${}^{4}_{2}\mathrm{He}$	fast electron, ${}^{0}_{-1}\mathrm{e}$	high-energy photon (electromagnetic wave)
charge	$+ 2 e$	$- e$	0
typical speed	about 5% of $c$	up to about 99% of $c$	$c$
energies	a few MeV, at one or a few sharp values for each nuclide	a continuous spread up to a maximum	sharp values, keV to MeV
ionising power	very strong	moderate	weak
range in air	a few cm	up to a few metres	very long; intensity falls with distance
stopped by	a sheet of paper, or the dead outer layer of skin	a few mm of aluminium	never fully stopped; reduced by cm of lead or m of concrete
deflection in E and B fields	small	large, opposite way	none

The penetration and the ionisation are two sides of one fact. An alpha particle, slow and doubly charged, knocks electrons off almost every atom it passes, so it loses its energy in a few centimetres of air. A gamma photon rarely interacts at all, so it travels far and ionises little.

Definition

Radioactive decay is the spontaneous change of an unstable nucleus into a more stable one, with the emission of radiation. The original nucleus is the parent; the one formed is the daughter. Ionising radiation is radiation with enough energy to remove electrons from atoms, forming ions. Example: radium-226 (parent) decays by alpha emission to radon-222 (daughter).

Writing nuclear equations

In every nuclear change the total nucleon number $A$ and the total charge (the sum of the $Z$ values) are conserved. So the top numbers balance, and the bottom numbers balance.

Alpha decay removes 2 protons and 2 neutrons:

{}^{A}_{Z}\mathrm{X} \rightarrow {}^{A-4}_{Z-2}\mathrm{Y} + {}^{4}_{2}\mathrm{He} \qquad \text{e.g. } {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\mathrm{He}

Beta-minus decay: a neutron in the nucleus turns into a proton, an electron and an electron antineutrino, $\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e^-} + \bar{\nu}_e$ . The electron and antineutrino leave:

{}^{A}_{Z}\mathrm{X} \rightarrow {}^{A}_{Z+1}\mathrm{Y} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \qquad \text{e.g. } {}^{14}_{6}\mathrm{C} \rightarrow {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e

Beta-plus decay: a proton turns into a neutron, a positron (the electron's antiparticle) and a neutrino, $\mathrm{p} \rightarrow \mathrm{n} + \mathrm{e^+} + \nu_e$ :

{}^{A}_{Z}\mathrm{X} \rightarrow {}^{A}_{Z-1}\mathrm{Y} + {}^{0}_{+1}\mathrm{e} + \nu_e \qquad \text{e.g. } {}^{11}_{6}\mathrm{C} \rightarrow {}^{11}_{5}\mathrm{B} + {}^{0}_{+1}\mathrm{e} + \nu_e

Electron capture has the same effect on $Z$ : the nucleus captures one of its own inner electrons, ${}^{40}_{19}\mathrm{K} + {}^{0}_{-1}\mathrm{e} \rightarrow {}^{40}_{18}\mathrm{Ar} + \nu_e$ .

Gamma emission: after an alpha or beta decay the daughter is often left in an excited state, marked *. It drops to its ground state by emitting a gamma photon, and $A$ and $Z$ do not change. Cobalt-60 decays by beta-minus emission to an excited nickel-60 nucleus, which emits two gamma photons of 1.17 MeV and 1.33 MeV:

{}^{60}_{27}\mathrm{Co} \rightarrow {}^{60}_{28}\mathrm{Ni}^{*} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e, \qquad {}^{60}_{28}\mathrm{Ni}^{*} \rightarrow {}^{60}_{28}\mathrm{Ni} + \gamma

Why a neutrino? Alpha particles from one nuclide come out with one or a few sharply defined energies, but beta particles come out with a continuous spread of energies. In 1930 Wolfgang Pauli proposed that a third, almost undetectable particle shares the energy: the neutrino, detected in 1956. You will meet it again in Unit 6.

Worked example 1: completing nuclear equations

Identify X in each decay: (a) ${}^{238}_{92}\mathrm{U} \rightarrow \mathrm{X} + {}^{4}_{2}\mathrm{He}$ ; (b) ${}^{90}_{38}\mathrm{Sr} \rightarrow {}^{90}_{39}\mathrm{Y} + \mathrm{X} + \bar{\nu}_e$ ; (c) ${}^{22}_{11}\mathrm{Na} \rightarrow \mathrm{X} + {}^{0}_{+1}\mathrm{e} + \nu_e$ ; (d) ${}^{131}_{53}\mathrm{I} \rightarrow \mathrm{X} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e$ .

Solution (balance $A$ on top, $Z$ below)

(a) $A = 238 - 4 = 234$ , $Z = 92 - 2 = 90$ : X is ${}^{234}_{90}\mathrm{Th}$ (thorium).

(b) $A = 90 - 90 = 0$ , $Z = 38 - 39 = - 1$ : X is ${}^{0}_{-1}\mathrm{e}$ , a beta-minus particle.

(d) $A = 131$ , $Z = 53 + 1 = 54$ : X is ${}^{131}_{54}\mathrm{Xe}$ (xenon). Neutrinos have $A = 0$ and $Z = 0$ , so they do not affect the balance.

Worked example 2: the energy of an alpha decay

Radium-226 decays to radon-222 by alpha emission. The atomic masses are $m(\mathrm{Ra}) = 226.025\,410\,\mathrm{u}$ , $m(\mathrm{Rn}) = 222.017\,578\,\mathrm{u}$ and $m(\mathrm{He}) = 4.002\,603\,\mathrm{u}$ . Find (a) the energy released, $Q$ , and (b) the kinetic energy of the alpha particle, if the radium nucleus was at rest.

Solution

(a) $\Delta m = 226.025\,410 - 222.017\,578 - 4.002\,603 = 0.005\,229\,\mathrm{u}$ (the electrons balance: 88 on each side). $Q = 0.005\,229 \times 931.5 = 4.87\,\mathrm{MeV}$ .

(b) Momentum is conserved, so the alpha particle and the radon nucleus fly apart with equal and opposite momenta $p$ . Since $E_k = \frac{p^2}{2m}$ , the kinetic energy divides in inverse proportion to the masses:

E_{k,\alpha} = Q \times \frac{m_{\text{Rn}}}{m_{\text{Rn}} + m_\alpha} = 4.87 \times \frac{222}{226} = 4.78\,\mathrm{MeV}

The measured alpha energy is 4.78 MeV. The radon nucleus recoils with the other 0.09 MeV. Because the split is fixed, every alpha particle from this decay has the same energy.

Worked example 3: a decay series

Uranium-238 decays through a series of radioactive daughters to stable lead-206. How many alpha and how many beta-minus decays occur?

Solution

Only alpha decay changes $A$ , by 4 each time: $\frac{238 - 206}{4} = 8$ alpha decays.

Eight alpha decays lower $Z$ by 16, from 92 to 76. Lead has $Z = 82$ , so six beta-minus decays must raise $Z$ by 6. Answer: 8 α and 6 β⁻. Check: $92 - 2 (8) + 6 = 82$ . Radium-226 and radon-222 are two of the members of this series.

Detecting radiation

Every detector uses the fact that the radiation ionises matter.

Geiger–Müller (GM) tube. Radiation entering the tube ionises a gas atom. The strong field near the thin central wire accelerates the freed electron, which ionises more atoms: an avalanche that gives a pulse of current. A counter records the pulses. It counts α (if its window is thin enough), β and γ, but cannot tell them apart or measure their energy.
Cloud chamber. Alcohol vapour cooled below its condensing point forms droplets on the ions left by a particle, making its track visible: thick, straight and short for α; thin and twisting for β; γ leaves hardly any track.
Scintillation counter. Radiation makes a crystal (such as sodium iodide) emit flashes of light, which a photomultiplier turns into pulses whose size measures the energy.
Spark counter. A high voltage across a wire gauze and a plate sparks where α particles ionise the air.
Photographic film. Radiation darkens it, as in Becquerel's discovery; film badges and electronic dosimeters record the exposure of radiation workers.

Figure 3.3: A Geiger–Müller tube and its circuit.

Background radiation

A GM tube clicks even with no source present. This background radiation comes from radon gas seeping out of rocks and soil, from rocks and building materials, from cosmic rays, from radioactive potassium-40 and carbon-14 in food and in our own bodies, and from medical uses. Always measure the background count rate first and subtract it from every reading.

Experiment 3.1: identifying radiation by absorption (teacher demonstration)

Apparatus: a GM tube and counter, sealed school sources, tongs, absorbers: paper, aluminium sheets of 1 to 5 mm, lead sheets.

Method: Record the background count over 5 minutes. Place the source a fixed distance (about 2 cm) from the window and record the count rate with no absorber, then with paper, then with 3 mm of aluminium, then with 2 cm of lead between them. Subtract the background from each reading.

Warning

Safety. In Rwanda, radioactive sources are regulated by the Rwanda Utilities Regulatory Authority (RURA) under Law N° 59 of 2017 governing radiation protection. School sources are sealed and must be handled only by the teacher, with tongs, held at arm's length and pointed away from people. Return each source to its lead-lined box as soon as the reading is taken. Never touch, open or take a source home; wash your hands after the lesson. The three rules, time, distance and shielding, are explained in Lesson 5.

Worked example 4: analysing absorption readings

In Experiment 3.1 the background was 24 counts per minute. With a mixed source the readings were:

Absorber	none	paper	3 mm aluminium	2 cm lead
count rate / min⁻¹	1850	1210	460	150

Which radiations does the source emit?

Solution

Corrected rates: 1826, 1186, 436, 126 per minute.

Paper removes 640 per minute: α is present (only α is stopped by paper).
3 mm of aluminium removes a further 750 per minute: β is present.
After 3 mm of aluminium, 436 per minute remain, and even 2 cm of lead leaves 126 per minute, well above background: γ is present.

The source emits all three. (The lead reduces the γ count without stopping it.)

Warning

Common mistakes. (1) Forgetting the neutrino or antineutrino in beta decay. (2) Writing the beta particle as ${}^{0}_{1}\mathrm{e}$ in β⁻ decay: the electron is ${}^{0}_{-1}\mathrm{e}$ . (3) Thinking gamma emission changes the element: $A$ and $Z$ stay the same. (4) Saying beta particles come from the atom's electron shells: they are created in the nucleus when a neutron changes into a proton. (5) Not subtracting the background count.

Tip

For decay series: alpha changes A, beta does not. Count the alphas from the change in $A$ (divide by 4), then use the change in $Z$ to count the betas.

Penetrating Properties of Radiation (FuseSchool). Watch how paper, aluminium and lead absorb alpha, beta and gamma radiation, and how this links to their ionising power.

✦

Practice

Complete: (a) ${}^{210}_{84}\mathrm{Po} \rightarrow {}^{206}_{82}\mathrm{Pb} + \mathrm{X}$ ; (b) ${}^{32}_{15}\mathrm{P} \rightarrow \mathrm{X} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e$ ; (c) ${}^{18}_{9}\mathrm{F} \rightarrow {}^{18}_{8}\mathrm{O} + \mathrm{X} + \nu_e$ .
Thorium-232 decays to lead-208. How many α and β⁻ decays occur?
Explain why alpha particles are strongly ionising but have a short range.
Use $\vec F = q\vec v \times \vec B$ to explain the directions of the α and β⁻ paths in Figure 3.1.
Polonium-210 (209.982 874 u) decays by α emission to lead-206 (205.974 465 u). Find $Q$ ( $m_{\text{He}} = 4.002\,603\,\mathrm{u}$ ) and the kinetic energy of the α particle.
Why do α particles from one nuclide have discrete (sharply defined) energies, while β particles have a continuous range of energies?
A source gives 530 counts per minute with no absorber, 525 with paper and 36 with 5 mm of aluminium. The background is 30 per minute. What does it emit?
Mutesi says: "A gamma-emitting source is safe behind a sheet of aluminium, because aluminium stops beta particles." Evaluate her statement.

Show answersHide answers

(a) ${}^{4}_{2}\mathrm{He}$ ; (b) ${}^{32}_{16}\mathrm{S}$ ; (c) ${}^{0}_{+1}\mathrm{e}$ , a positron.
$\frac{232 - 208}{4} = 6$ α; $Z$ : $90 - 12 = 78$ , so $82 - 78 = 4$ β⁻.
Its charge $+ 2 e$ and low speed mean it interacts strongly with the electrons of every atom it passes, forming many ions per mm. It loses a little energy in each ionisation, so its energy is used up within a few cm of air.
For α (positive), $\vec v$ up and $\vec B$ into the page give $\vec F$ to the left. β⁻ has negative charge, so the force is to the right. The α path is far straighter because its momentum is much larger (radius $r = \frac{mv}{qB}$ ).
$\Delta m = 209.982\,874 - 205.974\,465 - 4.002\,603 = 0.005\,806\,\mathrm{u}$ ; $Q = 5.41\,\mathrm{MeV}$ ; $E_{k,\alpha} = 5.41 \times \frac{206}{210} = 5.31\,\mathrm{MeV}$ .
An alpha decay has only two products, so momentum conservation fixes how the energy is shared. A beta decay has three products (daughter, electron and antineutrino), which can share the energy in many ways.
Corrected: 500, 495, 6. Paper hardly changes the rate: no α. 5 mm of aluminium removes almost everything: β. The small remainder is close to zero, so there is little or no γ. It is a β emitter.
Wrong. Aluminium stops β but only slightly weakens γ rays, which need centimetres of lead or thick concrete to reduce them substantially and are never stopped completely. Safety also needs distance and short exposure times.

Key points

α: ${}^{4}_{2}\mathrm{He}$ , strongly ionising, stopped by paper. β⁻: electron from the nucleus, stopped by mm of aluminium. γ: photon, weakly ionising, only reduced by lead.
Balance $A$ and $Z$ : α lowers $A$ by 4 and $Z$ by 2; β⁻ raises $Z$ by 1; β⁺ and electron capture lower $Z$ by 1; γ changes neither.
$Q = \Delta m \times 931.5\,\mathrm{MeV}$ ; in α decay the α particle takes the fraction $\frac{m_{\text{daughter}}}{m_{\text{total}}}$ .
Detectors use ionisation: GM tube, cloud chamber, scintillation counter. Always subtract the background.

Sources and credits

10.4: Radioactive Decay, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Atomic masses: AME2020, IAEA Atomic Mass Data Center.
Rwanda's radiation law and regulator (facts quoted): Rwanda National Report for the Joint Convention on the Safety of Spent Fuel Management and on the Safety of Radioactive Waste Management, Government of Rwanda, August 2024, published by the IAEA.
Photo: Particle Tracks in AWAN Expansion Cloud Chamber.jpg by Kebuk awan, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Penetrating Properties of Radiation, FuseSchool - Global Education, embedded from YouTube.

The three diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1A photographic plate fogged in the dark

2Three kinds of radiation

3Writing nuclear equations

4Detecting radiation

5Background radiation

6Experiment 3.1: identifying radiation by absorption (teacher demonstration)

A photographic plate fogged in the dark

Three kinds of radiation

Writing nuclear equations

Detecting radiation

Background radiation

Experiment 3.1: identifying radiation by absorption (teacher demonstration)