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Why are some nuclei unstable, and how do they decay and change over time?

Describe alpha, beta and gamma decay, balance nuclear equations, and apply the concept of half-life.

The structure of the nucleus, the three types of radioactive decay, writing balanced nuclear equations, and using half-life to describe decay over time, with worked examples.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Nuclear structure
  3. The three types of decay
  4. Balancing nuclear equations
  5. Half-life
  6. Penetration and uses
  7. How SACE assesses this

What this dot point is asking

You need to describe the three decay types, write balanced nuclear equations, and apply the concept of half-life.

Nuclear structure

A nucleus contains protons and neutrons (nucleons). The atomic number ZZ is the number of protons, and the mass number AA is the total number of nucleons. A nuclide is written ZAX^A_Z X. Isotopes of an element have the same ZZ but different AA. A nucleus is unstable (radioactive) when the balance of protons and neutrons, or its total energy, makes it unable to hold together indefinitely.

The three types of decay

Balancing nuclear equations

In every nuclear equation, the totals of mass number and atomic number must be the same on both sides.

For alpha decay of uranium-238:

92238U 90234Th+ 24He.^{238}_{\,92}\text{U} \rightarrow\ ^{234}_{\,90}\text{Th} +\ ^4_2\text{He}.

Check: 238=234+4238 = 234 + 4 and 92=90+292 = 90 + 2. Balanced.

For beta-minus decay of carbon-14:

614C 714N+ 10e.^{14}_{\,6}\text{C} \rightarrow\ ^{14}_{\,7}\text{N} +\ ^{\,0}_{-1}e.

Check: 14=14+014 = 14 + 0 and 6=7+(1)6 = 7 + (-1). Balanced - the neutron count drops by one and the proton count rises by one.

Half-life

Radioactive decay is random for any single nucleus, but for a large sample it follows a predictable pattern described by the half-life.

The activity (decays per second, in becquerel) halves over each half-life, giving an exponential decrease.

Penetration and uses

The differing penetration of the three radiations underpins their uses and hazards: alpha is dangerous if ingested but stopped by skin; beta is used in thickness gauges; gamma is used in medical imaging and sterilisation. Half-life governs how long a radioactive source remains active and is the basis of radioactive dating (e.g. carbon-14 dating of once-living material).

How SACE assesses this

SACE Stage 2 questions on this dot point split into two styles. The first asks you to complete or write a balanced nuclear equation, where the reliable method is to balance the mass number (top) and atomic number (bottom) separately on each side; remember that β\beta^- decay raises ZZ by one while leaving AA unchanged, and α\alpha decay lowers AA by four and ZZ by two. The second style is a half-life calculation: find the number of half-lives n=tt1/2n = \dfrac{t}{t_{1/2}}, then multiply the activity or number of nuclei by (12)n\left(\tfrac{1}{2}\right)^n. The common slip is treating the decay as linear rather than exponential, so the sample never reaches zero; three half-lives leave one-eighth, not nothing. Track whether the question asks for activity (Bq) or number of nuclei, since both halve each half-life.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20243 marksA radioactive source has an activity of 800 Bq800\ \text{Bq} and a half-life of 6.0 hours6.0\ \text{hours}. (a) Calculate the activity after 24 hours24\ \text{hours}. (b) State the fraction of the original radioactive nuclei remaining at that time.
Show worked answer →

(a) Number of half-lives: n=246.0=4n = \dfrac{24}{6.0} = 4. The activity halves each half-life, so

A=800×(12)4=800×116=50 Bq.A = 800\times\left(\tfrac{1}{2}\right)^4 = 800\times\dfrac{1}{16} = 50\ \text{Bq}.
(2 marks)

(b) The fraction of original nuclei remaining is (12)4=116\left(\tfrac{1}{2}\right)^4 = \dfrac{1}{16}. (1 mark) Markers reward computing the number of half-lives and applying the halving correctly; the sample does not reach zero.

SACE 20223 marks(a) Complete the alpha decay equation: 92238U ZATh+ 24He^{238}_{\,92}\text{U} \rightarrow\ ^{A}_{Z}\text{Th} +\ ^4_2\text{He}, stating AA and ZZ. (b) Carbon-14 decays by beta-minus emission to nitrogen. Write the balanced nuclear equation and state what happens to the proton and neutron numbers.
Show worked answer →

(a) Mass number and atomic number must balance. A=2384=234A = 238 - 4 = 234 and Z=922=90Z = 92 - 2 = 90, so the product is 90234Th^{234}_{\,90}\text{Th}. (1 mark)

(b) 614C 714N+ 10e^{14}_{\,6}\text{C} \rightarrow\ ^{14}_{\,7}\text{N} +\ ^{\,0}_{-1}e. Check: 14=14+014 = 14 + 0 and 6=7+(1)6 = 7 + (-1). A neutron becomes a proton, so the proton number rises by one and the neutron number falls by one, with the mass number unchanged. (2 marks) Markers reward the balanced equation and the proton/neutron change.

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