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Radioactivity and half-life: WACE Year 12 Physics Unit 4

Syllabus dot point

“Describe alpha, beta and gamma decay and apply the concept of half-life to radioactive decay”

8 min read

Quick answer

Unstable nuclei decay by emitting alpha particles (a helium nucleus), beta particles (an electron from a neutron turning into a proton) or gamma rays (a high-energy photon). Decay is random for any single nucleus, but a large sample halves in a fixed time, the half-life. After nn half-lives the remaining amount is (12)n(\tfrac{1}{2})^n of the original. Nuclear equations must balance both mass number and atomic number.

What this dot point is asking

WACE wants you to describe the three decay modes, write balanced nuclear equations, and use half-life to calculate how much of a sample remains. The key statistical idea is that decay is random per nucleus but predictable in bulk.

The three decay modes

Alpha decay emits an alpha particle, a helium-4 nucleus (24He^4_2\text{He}), reducing the mass number by 44 and the atomic number by 22. Beta-minus decay emits an electron when a neutron becomes a proton, raising the atomic number by 11 with the mass number unchanged. Gamma decay emits a high-energy photon as an excited nucleus settles to a lower energy state, changing neither the mass number nor the atomic number. Alpha is the least penetrating and gamma the most.

Balancing nuclear equations

In any nuclear equation the total mass number (top) and the total atomic number (bottom) must be the same on both sides. For example, alpha decay of radium-226:

88226Ra→86222Rn+24He.^{226}_{88}\text{Ra}\rightarrow{}^{222}_{86}\text{Rn}+{}^4_2\text{He}.

The mass numbers (226=222+4226=222+4) and atomic numbers (88=86+288=86+2) both balance, which lets you identify an unknown product.

Randomness and half-life

You cannot predict when a given nucleus will decay; the process is random and spontaneous. Yet across a large sample a fixed fraction decays each second, so the number remaining falls exponentially. The half-life t1/2t_{1/2} is the time for half the nuclei (and half the activity) to decay, and it is a fixed property of each isotope, unaffected by temperature, pressure or chemical state.

Calculating with half-lives

After nn whole half-lives, the fraction remaining is

NN0=(12)n,n=tt1/2.\frac{N}{N_0}=\left(\frac{1}{2}\right)^n,\qquad n=\frac{t}{t_{1/2}}.

Activity (decays per second) falls in exactly the same proportion, since it is proportional to the number of undecayed nuclei. This produces the familiar exponential decay curve, which never quite reaches zero but halves each half-life.

Remaining sample after several half-lives

A sample after a known time

A radioactive isotope has a half-life of 8.08.0 days. A sample starts with an activity of 4800 Bq4800\ \text{Bq}. Find the activity after 2424 days.

Number of half-lives elapsed:

n=tt1/2=248.0=3.n=\frac{t}{t_{1/2}}=\frac{24}{8.0}=3.

Activity remaining:

A=A0(12)n=4800×(12)3=4800×18=600 Bq.A=A_0\left(\frac{1}{2}\right)^n=4800\times\left(\frac{1}{2}\right)^3=4800\times\frac{1}{8}=600\ \text{Bq}.

So after three half-lives the activity has fallen to one eighth of the original.

Penetrating power and the decay particles

WACE expects you to rank the three radiations by penetrating and ionising power. Alpha particles are large and highly charged, so they ionise strongly but are stopped by a sheet of paper or a few centimetres of air. Beta particles are lighter and faster, ionise less, and are stopped by a few millimetres of aluminium. Gamma rays are uncharged photons, ionise weakly, and are the most penetrating, needing thick lead or concrete to attenuate significantly. There is an inverse relationship between ionising power and penetration: the most strongly ionising radiation (alpha) is the least penetrating, because it deposits its energy quickly. This determines the hazard each presents depending on whether the source is outside or inside the body.

Activity, decay constant and dating

Activity is the number of decays per second, measured in becquerels, and it is proportional to the number of undecayed nuclei present, so it falls exponentially with exactly the same half-life as the nuclei themselves. This is why a measured activity can be used in place of a count of nuclei in half-life problems. The exponential fall underpins radioactive dating: in carbon-14 dating, a once-living sample stops taking in carbon-14 at death, so the ratio of carbon-14 to stable carbon then halves every 57305730 years. Measuring how much carbon-14 remains (or its current activity) and comparing with the living value tells you how many half-lives have passed and hence the age. The same principle, with longer-lived isotopes such as uranium, dates rocks over billions of years, which is a frequent application question.

Working with whole and partial half-lives

When the elapsed time is a whole number of half-lives, halving repeatedly is quickest. For non-whole numbers, use N=N0(12)t/t1/2N=N_0(\tfrac{1}{2})^{t/t_{1/2}} directly. Always check your nuclear equations balance before trusting an identified daughter nucleus.

Halving by time instead of by half-life

Students sometimes halve the activity once per unit of time rather than once per half-life, or treat 2424 days as 2424 halvings. The number of halvings is n=t/t1/2n=t/t_{1/2}, so 2424 days with an 88-day half-life is 33 halvings, not 2424. Compute nn first, then apply (12)n(\tfrac{1}{2})^n.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2023-style6 marks
A radioactive source contains 6.4×10206.4\times10^{20} undecayed nuclei and has a half-life of 5.05.0 years. (a) Calculate the number of undecayed nuclei remaining after 2020 years. (b) Calculate the fraction of the original nuclei that has decayed in that time. (c) Explain why the half-life is unaffected by heating the sample.
Show worked answer →

A 6 mark calculation rewards the half-life count, the remaining nuclei, the decayed fraction and a nuclear-property statement.

(a) Nuclei remaining
Number of half-lives n=205.0=4n=\dfrac{20}{5.0}=4. Remaining N=N0(12)n=6.4×1020×116=4.0×1019N=N_0\left(\tfrac{1}{2}\right)^n=6.4\times10^{20}\times\dfrac{1}{16}=4.0\times10^{19}.
(b) Fraction decayed
Fraction remaining =(12)4=116=\left(\tfrac{1}{2}\right)^4=\dfrac{1}{16}, so fraction decayed =1−116=1516=0.94=1-\dfrac{1}{16}=\dfrac{15}{16}=0.94.
(c) Why temperature has no effect
Radioactive decay is a nuclear process driven by the instability of the nucleus, not by the electrons or the chemical or thermal state. Heating affects the motion of atoms and electrons but not the nucleus, so the half-life is unchanged.

Markers reward n=4n=4, N=4.0×1019N=4.0\times10^{19}, the decayed fraction 15/1615/16 and the nuclear-process reason for temperature independence.

WACE 2020-style5 marks
(a) Write a balanced nuclear equation for the beta-minus decay of carbon-14 (614C^{14}_6\text{C}). (b) Explain what happens within the nucleus during beta-minus decay, and state how the mass number and atomic number change.
Show worked answer →

A 5 mark answer rewards a balanced equation and a description of the underlying process.

(a) Equation. 614C→714N+−1 0e^{14}_6\text{C}\rightarrow{}^{14}_7\text{N}+{}^{\,0}_{-1}\text{e} (with an antineutrino also emitted). Mass numbers balance (14=14+014=14+0) and charge balances (6=7−16=7-1).

(b) The process. In beta-minus decay a neutron in the nucleus changes into a proton, emitting a fast electron (the beta particle). The mass number stays the same (a neutron is replaced by a proton, both nucleons), while the atomic number increases by one because there is now an extra proton, changing the element from carbon to nitrogen.

Markers reward the balanced equation including the emitted electron, the neutron-to-proton conversion, unchanged mass number and atomic number increasing by one.

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