Radioactive decay and half-life: HSC Physics Module 8

Syllabus dot point

“Examine the radioactive decay of atomic nuclei (alpha, beta, gamma) and represent these decays as nuclear equations; use the decay law N = N_0 e^(-lambda t) and the concept of half-life T_1/2”

14 min readUpdated 1 Jul 2026

Quick answer

Radioactive decay proceeds via alpha (loss of $^4_2$ He), beta (n to p with $e^-$ and $\bar{\nu}_e$ , or the reverse with $e^+$ and $\nu_e$ ), and gamma (photon from de-excitation) processes; the number of undecayed nuclei follows $N = N_0 e^{-\lambda t}$ , with half-life $T_{1/2} = (\ln 2)/\lambda$ .

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What this dot point is asking

NESA wants you to describe the three principal types of radioactive decay (alpha, beta, gamma), balance nuclear equations using conservation of mass number and atomic number, use the decay law $N = N_0 e^{-\lambda t}$ along with the equivalent half-life form $N = N_0 (1/2)^{t/T_{1/2}}$ , and connect $\lambda$ and $T_{1/2}$ via $\lambda T_{1/2} = \ln 2$ .

The answer

What radioactive decay is

A radioactive nucleus is one that spontaneously transforms into another nuclear state, releasing energy as kinetic energy of the products and/or as electromagnetic radiation. Decay is a random process for any individual nucleus, but the statistics for large samples follow a predictable exponential law. The probability per unit time that a given nucleus decays is the decay constant $\lambda$ , independent of how long the nucleus has existed.

Alpha decay

A heavy nucleus emits an alpha particle ( $^4_2$ He, two protons and two neutrons). The atomic number decreases by 2, mass number by 4.

^A_Z X \to ^{A-4}_{Z-2} Y + ^4_2 \text{He}

Example:

^{238}_{92}\text{U} \to ^{234}_{90}\text{Th} + ^4_2 \text{He}

Alpha decay typically occurs in heavy nuclei ( $Z > 82$ ) where the Coulomb repulsion between protons becomes hard for the strong force to overcome. The alpha particle escapes by quantum tunnelling. Alphas have short range (a few cm in air, stopped by paper) but cause heavy ionisation per unit path.

Beta-minus decay

A neutron in the nucleus converts to a proton, emitting an electron (the beta particle) and an electron antineutrino. Atomic number increases by 1, mass number unchanged.

n \to p + e^- + \bar{\nu}_e

^A_Z X \to ^{A}_{Z+1} Y + e^- + \bar{\nu}_e

Example:

^{14}_{6}\text{C} \to ^{14}_{7}\text{N} + e^- + \bar{\nu}_e

Beta-minus decay tends to occur in neutron-rich nuclei. The continuous energy spectrum of beta particles was the historical clue that a third particle (the antineutrino) carries away the missing energy.

Beta-plus decay

A proton in the nucleus converts to a neutron, emitting a positron and an electron neutrino. Atomic number decreases by 1, mass number unchanged.

p \to n + e^+ + \nu_e

Example: $^{22}_{11}\text{Na} \to ^{22}_{10}\text{Ne} + e^+ + \nu_e$ . Beta-plus occurs in proton-rich nuclei. The competing process is electron capture, in which a proton absorbs an inner-shell electron and converts to a neutron plus a neutrino.

Gamma decay

The nucleus is left in an excited state after an alpha or beta decay (or after a nuclear reaction). It drops to a lower state by emitting a high-energy photon (gamma ray). No change in $Z$ or $A$ .

^A_Z X^{\ast} \to ^A_Z X + \gamma

Example: $^{60}_{27}\text{Co}^{\ast} \to ^{60}_{27}\text{Co} + \gamma$ . Gamma rays are penetrating (centimetres of lead required to attenuate) but cause less local ionisation than alpha or beta.

Balancing nuclear equations

In any decay equation, two conservation laws must hold:

mass number $A$ balances on both sides,
charge $Z$ balances on both sides (counting an electron as $-1$ and a positron as $+1$ ).

Lepton number is also conserved, which is why an electron emitted in beta-minus decay is accompanied by an antineutrino, and a positron in beta-plus is accompanied by a neutrino.

The decay law

If $N(t)$ is the number of undecayed nuclei at time $t$ , the rate of decay is proportional to $N$ :

\frac{dN}{dt} = -\lambda N

Solving with $N(0) = N_0$ :

\boxed{N(t) = N_0 \, e^{-\lambda t}}

The activity (number of decays per unit time) is $A(t) = \lambda N(t)$ , measured in becquerels (Bq, 1 decay per second).

Half-life

The half-life $T_{1/2}$ is the time for half the sample to decay. From $N(T_{1/2}) = N_0 / 2$ :

\frac{1}{2} = e^{-\lambda T_{1/2}}, \quad \lambda T_{1/2} = \ln 2 \approx 0.693

So $\lambda = \ln 2 / T_{1/2}$ .

The decay law in terms of half-life:

N(t) = N_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}

The must-know equations

Decay law: $N(t) = N_0 e^{-\lambda t}$ , equivalently $N(t) = N_0 \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$ .

Decay constant and half-life: $\lambda T_{1/2} = \ln 2 \approx 0.693$ , so $\lambda = \dfrac{\ln 2}{T_{1/2}}$ and $T_{1/2} = \dfrac{\ln 2}{\lambda}$ .

Activity: $A(t) = \lambda N(t)$ , measured in becquerels ( $1\ \text{Bq} = 1$ decay per second).

Alpha decay: $^A_Z X \to {}^{A-4}_{Z-2} Y + {}^4_2\text{He}$ . Beta-minus: $^A_Z X \to {}^{A}_{Z+1} Y + e^- + \bar{\nu}_e$ . Beta-plus: $^A_Z X \to {}^{A}_{Z-1} Y + e^+ + \nu_e$ . Gamma: $^A_Z X^{\ast} \to {}^A_Z X + \gamma$ .

Half-lives of common isotopes:

Isotope	Half-life	Decay mode
$^{14}$ C	5730 y	beta-minus
$^{40}$ K	$1.25 \times 10^9$ y	beta-minus, electron capture
$^{60}$ Co	5.27 y	beta-minus then gamma
$^{99\text{m}}$ Tc	6.0 h	gamma (medical imaging)
$^{131}$ I	8.0 d	beta-minus then gamma
$^{238}$ U	$4.47 \times 10^9$ y	alpha
$^{222}$ Rn	3.82 d	alpha

Applications

Radiometric dating. $^{14}$ C for organic material up to 50000 y; $^{238}$ U/ $^{206}$ Pb for rocks up to billions of years; $^{40}$ K/ $^{40}$ Ar for igneous rocks.
Nuclear medicine. $^{99\text{m}}$ Tc for diagnostic imaging; $^{131}$ I for thyroid therapy; positron emitters for PET.
Smoke detectors. $^{241}$ Am alpha source ionises air; smoke disrupts the ion current and triggers the alarm.
Industrial gauging. Penetrating gammas measure thickness of steel sheets without contact.

Try it: Radioactive decay calculator to find remaining activity, time elapsed, or decay constant from half-life.

Worked examples

Model 1 - dating a wooden artefact. A wooden bowl contains $^{14}\text{C}$ at $25\%$ of the modern atmospheric ratio. The half-life of $^{14}\text{C}$ is $5730\ \text{y}$ . Estimate the age of the bowl.

By counting half-lives: $\dfrac{N}{N_0} = 0.25 = \left(\dfrac{1}{2}\right)^2$ , so the bowl is $2$ half-lives old.

$t = 2 \times 5730 = 11460\ \text{y} \approx 1.1 \times 10^4\ \text{y}$ .

Checking with the decay law: $\lambda = \dfrac{\ln 2}{5730} = 1.21 \times 10^{-4}\ \text{y}^{-1}$ , and $0.25 = e^{-\lambda t}$ gives $t = \dfrac{\ln 4}{\lambda} = \dfrac{1.386}{1.21 \times 10^{-4}} = 1.15 \times 10^4\ \text{y}$ , consistent with the half-life count.

Marker's note: the fastest full-marks method is recognising $0.25 = (1/2)^2$ and multiplying by $2$ half-lives; showing the decay-law check earns extra credit but is not required when the fraction is an exact power of $\dfrac{1}{2}$ .

Model 2 - initial activity of a medical tracer. A hospital receives a sample of $^{131}\text{I}$ ( $T_{1/2} = 8.0\ \text{d}$ ) containing $N_0 = 3.0 \times 10^{18}$ nuclei. Calculate (a) the decay constant in $\text{s}^{-1}$ and (b) the activity $6.0$ days later.

(a) $T_{1/2} = 8.0 \times 86400 = 6.91 \times 10^5\ \text{s}$ , so $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{6.91 \times 10^5} = 1.00 \times 10^{-6}\ \text{s}^{-1}$ .

(b) Number remaining after $t = 6.0\ \text{d} = 5.18 \times 10^5\ \text{s}$ : $N = N_0 e^{-\lambda t} = (3.0 \times 10^{18}) e^{-(1.00 \times 10^{-6})(5.18 \times 10^5)} = (3.0 \times 10^{18})(0.596) = 1.79 \times 10^{18}$ .

Activity: $A = \lambda N = (1.00 \times 10^{-6})(1.79 \times 10^{18}) = 1.79 \times 10^{12}\ \text{Bq}$ .

Marker's note: convert every time to the same unit as $\lambda$ before substituting into $e^{-\lambda t}$ - mixing days and seconds is the single most common cause of a wrong exponent here.

Common traps

Forgetting the antineutrino in beta decay: Charge balances without it, but lepton number does not. Always include $\bar{\nu}_e$ for beta-minus and $\nu_e$ for beta-plus.
Confusing $\lambda$ and $T_{1/2}$: They are related by $\lambda T_{1/2} = \ln 2$ , not $\lambda T_{1/2} = 1$ . The numerical factor matters.
Using arithmetic decay (subtracting a fixed amount) instead of exponential: Each half-life removes half the remaining sample, not half the initial sample. After two half-lives the sample is 25% of the original, not 0%.
Confusing $A$ and $Z$ in alpha decay: Alpha decay reduces $A$ by 4 and $Z$ by 2. Both must change together.
Saying gamma decay changes the element: Gamma decay changes the nuclear excitation level only. The element is unchanged.

Exam technique

State the relationship before substituting: write " $\lambda = \ln 2 / T_{1/2}$ " or " $N = N_0 e^{-\lambda t}$ " as its own line before you plug in numbers - a stated formula earns a method mark even if the arithmetic slips.

Watch your units. $\lambda$ and $t$ must be in the same time unit before you evaluate $e^{-\lambda t}$ (both in seconds, or both in years); a half-life given in days must be converted to seconds if $\lambda$ is required in $\text{s}^{-1}$ .

When a fraction remaining is an exact power of $\dfrac{1}{2}$ (such as $12.5\%$ or $6.25\%$ ), the fastest full-marks route is counting half-lives ( $t = n \times T_{1/2}$ ), not the full exponential equation - but show the count explicitly.

Reading a decay graph: the half-life is the constant horizontal spacing between the $N_0$ , $N_0/2$ , $N_0/4$ , ... levels, not the spacing between arbitrary points on the curve. Always read off a level that is an exact power of $\dfrac{1}{2}$ of $N_0$ .

Command words: "write the nuclear equation" wants the balanced $^A_Z X$ form with the ejected particle; "calculate" wants full substitution with units; "explain" (e.g. why beta decay has a continuous energy spectrum) wants the physical reasoning, not just a restated fact.

In plain English

Imagine a huge bowl of popcorn kernels popping in a microwave, where each kernel has the same tiny, random chance of popping in the next second, no matter how long it has already sat there. You cannot predict which kernel pops next, but with millions of kernels the overall popping rate is completely predictable: it starts fast (lots of kernels, lots of chances to pop) and slows down smoothly as fewer kernels remain, because there are simply fewer kernels left to pop. Radioactive decay is exactly this at the level of atomic nuclei - each nucleus has a fixed, unpredictable chance per second of decaying (alpha, beta or gamma), but a large sample follows a smooth, predictable curve: $N = N_0 e^{-\lambda t}$ . The "half-life" is just a convenient checkpoint on that curve - the time for exactly half of whatever is left to decay, whether you start counting at the beginning or halfway through. That is why half-life is so useful for dating things: if you know how much of a radioactive isotope is left compared with how much you started with, you can work backward to find out how long it has been decaying.

Examples in context

Example 1. Lucas Heights $^{99m}$ Tc supply chain for NSW hospitals. $^{99m}$ Tc has a half-life of $T_{1/2} = 6.0 \text{ h}$ . ANSTO's OPAL reactor extracts $^{99m}$ Tc from $^{99}$ Mo generators and ships it overnight to Sydney hospitals. Starting with $N_0 = 4.0 \times 10^{15}$ atoms at $6$ am, after $t = 12 \text{ h}$ (two half-lives), $N = N_0 (\tfrac{1}{2})^2 = 1.0 \times 10^{15}$ , only $25\%$ remaining. The decay constant is $\lambda = \ln 2 / T_{1/2} = 0.693 / (6 \times 3600) = 3.21 \times 10^{-5} \text{ s}^{-1}$ . Activity $A = \lambda N$ . To deliver $1 \text{ GBq}$ at noon, the morning shipment must contain $A_0 / (\tfrac{1}{2})^1 = 2 \text{ GBq}$ at $6$ am.

Example 2. Carbon-14 dating of an Aboriginal artefact at the Australian Museum. A charcoal sample from a Sydney rock shelter contains $^{14}$ C activity of $A = 5.0 \text{ Bq/g}$ , compared to the present-day biosphere standard $A_0 = 13.6 \text{ Bq/g}$ . The half-life is $T_{1/2} = 5730 \text{ yr}$ , so $\lambda = \ln 2 / 5730 = 1.21 \times 10^{-4} \text{ yr}^{-1}$ . The age is $t = (1/\lambda) \ln(A_0/A) = (5730/0.693) \ln(13.6/5.0) = 8270 \ln(2.72) = 8270 \text{ yr}$ . The artefact dates to roughly 8300 years before present, placing it in the early Holocene period of Aboriginal occupation of the Sydney basin.

Exam-style questions

Questions in the style of NESA 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.

2023 HSC-style5 marks

A sample of carbon-14 has a half-life of 5730 years. (a) Calculate the decay constant. (b) What fraction of the original carbon-14 remains in a sample after 17190 years? (c) An archaeological artefact contains 12.5% of the carbon-14 expected for a living organism of the same mass. Estimate its age.

Show worked answer →

(a) Decay constant:

$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4}$ y $^{-1}$ .

(b) After 17190 years, that is $n = 17190 / 5730 = 3$ half-lives.

$\frac{N}{N_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8} = 12.5\%$ .

$t = 3 \times 5730 = 17190$ years.

Markers reward $\lambda$ from $\ln 2 / T_{1/2}$ , the fraction by counting half-lives, and the age determination by recognising 12.5% = $(1/2)^3$ .

2021 HSC-style4 marks

Write balanced nuclear equations for: (a) the alpha decay of uranium-238, (b) the beta-minus decay of carbon-14, (c) the gamma decay of an excited cobalt-60 nucleus.

Show worked answer →

(a) Alpha decay: parent loses an alpha particle ( $^4_2$ He), so $Z$ decreases by 2 and $A$ decreases by 4.

$^{238}_{92}\text{U} \to ^{234}_{90}\text{Th} + ^4_2\text{He}$

(b) Beta-minus decay: a neutron becomes a proton, electron and antineutrino, so $Z$ increases by 1 and $A$ is unchanged.

$^{14}_{6}\text{C} \to ^{14}_{7}\text{N} + e^- + \bar{\nu}_e$

(c) Gamma decay: the nucleus drops from an excited state to a lower state, emitting a gamma photon. $A$ and $Z$ unchanged.

$^{60}_{27}\text{Co}^* \to ^{60}_{27}\text{Co} + \gamma$

Markers reward correct daughter nuclei with correct $A$ and $Z$ , the alpha particle as $^4_2$ He, an antineutrino in the beta decay, and the gamma photon with no change in $A$ or $Z$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Write the balanced nuclear equation for the alpha decay of radium-226, and state the change in mass number and atomic number.

Show worked solution →

Alpha decay removes $^4_2\text{He}$ , so $A$ decreases by $4$ and $Z$ decreases by $2$ :

$^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^4_2\text{He}$ .

Marks: one for a correctly balanced equation (mass numbers $226 = 222 + 4$ , atomic numbers $88 = 86 + 2$ ), one for stating $A$ falls by $4$ and $Z$ falls by $2$ .

foundation3 marks

Radon-222 has a half-life of

3.82

days. Calculate its decay constant

\lambda

\text{s}^{-1}

Show worked solution →

Convert the half-life to seconds: $T_{1/2} = 3.82 \times 86400 = 3.30 \times 10^5\ \text{s}$ .

$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{3.30 \times 10^5} = 2.10 \times 10^{-6}\ \text{s}^{-1}$ .

Marks: one for converting days to seconds, one for the formula $\lambda = \ln 2 / T_{1/2}$ , one for the answer to three significant figures with the correct unit.

foundation2 marks

A sample of iodine-131 (

T_{1/2} = 8.0

days) is left for

32

days. How many half-lives have elapsed, and what fraction of the original iodine-131 remains?

Show worked solution →

Number of half-lives: $n = \dfrac{32}{8.0} = 4$ .

Fraction remaining: $\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16} = 6.25\%$ .

Marks: one for $n = 4$ half-lives, one for the fraction $1/16$ (or $6.25\%$ ).

core4 marks

**Data.** The graph shows the number of undecayed nuclei

N

(relative to the initial number

N_0

) against time

t

for a sample of technetium-99m used in a hospital scan. **(a)** Using the graph, state the half-life of

^{99\text{m}}

Tc. **(b)** Calculate the decay constant

\lambda

\text{s}^{-1}

. **(c)** Using the graph, state the fraction of the sample remaining after

3

time units, and confirm it using

N = N_0(1/2)^{t/T_{1/2}}

Show worked solution →

(a) The curve falls from $N_0$ to $N_0/2$ over one time unit on the graph, and the labelled tick marks on the horizontal axis correspond to $6.0$ hours each, so $T_{1/2} = 6.0\ \text{h}$ (matching the known half-life of $^{99\text{m}}$ Tc).

(b) Convert to seconds: $T_{1/2} = 6.0 \times 3600 = 2.16 \times 10^4\ \text{s}$ .

$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{2.16 \times 10^4} = 3.21 \times 10^{-5}\ \text{s}^{-1}$ .

(c) Reading the graph at $t = 3$ time units, the curve sits at $N_0/8$ , i.e. $12.5\%$ remaining. Checking: $N = N_0\left(\dfrac{1}{2}\right)^{3} = N_0 \times \dfrac{1}{8} = 0.125 N_0$ , which matches the graph.

Marks: one for correctly reading $T_{1/2} = 6.0\ \text{h}$ from the graph, one for the unit conversion to seconds, one for $\lambda = 3.21 \times 10^{-5}\ \text{s}^{-1}$ , one for the fraction $1/8$ read from the graph and confirmed algebraically.

core4 marks

A sample of cobalt-60 (

T_{1/2} = 5.27

years) contains

2.5 \times 10^{19}

undecayed nuclei. Calculate (a) the decay constant in

\text{s}^{-1}

and (b) the initial activity of the sample in becquerels.

Show worked solution →

(a) Convert the half-life to seconds: $T_{1/2} = 5.27 \times 365.25 \times 86400 = 1.663 \times 10^{8}\ \text{s}$ .

$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{1.663 \times 10^{8}} = 4.17 \times 10^{-9}\ \text{s}^{-1}$ .

(b) Activity $A = \lambda N = (4.17 \times 10^{-9})(2.5 \times 10^{19}) = 1.04 \times 10^{11}\ \text{Bq}$ .

Marks: one for the unit conversion to seconds, one for $\lambda = 4.17 \times 10^{-9}\ \text{s}^{-1}$ , one for the formula $A = \lambda N$ , one for the activity to three significant figures with the correct unit.

core7 marks

Radium-226 decays by alpha emission with

T_{1/2} = 1600

years. **(a)** Write the balanced nuclear equation. **(b)** Calculate the fraction of the original

^{226}

Ra remaining after

4800

years. **(c)** Find the activity of

1.0

g of pure

^{226}

Ra, in becquerels. (

N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}

;

1

year

= 3.15 \times 10^7

s.)

Show worked solution →

(a) $^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$ .

(b) $4800 / 1600 = 3$ half-lives, so $\dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} = 12.5\%$ .

(c) Number of nuclei in $1.0$ g ( $M = 226$ g/mol): $N = \dfrac{1.0}{226} \times 6.022 \times 10^{23} = 2.66 \times 10^{21}$ .

Decay constant: $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{1600 \times 3.15 \times 10^7} = 1.38 \times 10^{-11}\ \text{s}^{-1}$ .

Activity: $A = \lambda N = (1.38 \times 10^{-11})(2.66 \times 10^{21}) = 3.66 \times 10^{10}\ \text{Bq}$ (this defines the curie, $1\ \text{Ci} \approx 3.7 \times 10^{10}\ \text{Bq}$ ).

Marks: one for the balanced equation (mass numbers $226 = 222+4$ , atomic numbers $88 = 86+2$ ), one for identifying $3$ half-lives, one for the fraction $1/8$ , one for $N = 2.66 \times 10^{21}$ from moles, one for $\lambda$ in $\text{s}^{-1}$ with the year-to-second conversion shown, one for $A = \lambda N$ , one for the final activity to three significant figures with the correct unit.

exam6 marks

Analyse how the shape of the

N

t

decay curve, together with the decay law

N = N_0 e^{-\lambda t}

, distinguishes radioactive decay from a process in which a fixed number of nuclei decayed every second. Refer to the concept of half-life in your answer.

Show worked solution →

Band-6 plan. (1) Describe the shape of the graph (exponential fall, never reaching zero, concave up, decreasing gradient). (2) State the decay law and what it predicts about the rate of decay. (3) Contrast with a hypothetical constant (linear) decay. (4) Use half-life to show why the exponential model is correct and the linear model is not, with a concrete numerical check.

Model answer. The $N$ - $t$ graph is a smooth curve that falls steeply at first and flattens as $t$ increases, always approaching but never reaching zero. This shape follows directly from $\dfrac{dN}{dt} = -\lambda N$ : the rate of decay is proportional to the number of nuclei still present, so as $N$ falls, the rate of decay falls too, giving the characteristic exponential curve $N = N_0 e^{-\lambda t}$ .

If instead a fixed number of nuclei decayed every second (a linear model, $N = N_0 - kt$ ), the graph would be a straight line reaching $N = 0$ at a definite, finite time, and the rate of decay would stay constant no matter how few nuclei remained. This is physically implausible once few nuclei remain, since there is no way for exactly $k$ of a tiny remaining population to decay each second.

The half-life concept exposes the difference cleanly. For exponential decay, the time to go from $N_0$ to $N_0/2$ is the same as the time to go from $N_0/2$ to $N_0/4$ (a constant $T_{1/2} = \ln2/\lambda$ , independent of how much sample remains) - visible on the graph as equal horizontal spacing between the $N_0/2$ , $N_0/4$ and $N_0/8$ levels. For the linear model, the time to halve the sample would keep shrinking as $N$ falls, since a constant number of decays per second represents a larger and larger fraction of a dwindling sample. Because measured half-lives are constant (e.g. every $^{99\text{m}}$ Tc sample takes $6.0$ h to halve, regardless of its size), the exponential model is confirmed and the constant-rate model is ruled out.

Marker's note: the top band explicitly derives the shape from $dN/dt = -\lambda N$ , states the constant-half-life property as the discriminating test, and rejects the linear alternative with a physical reason (not just "it's wrong"). A response that only describes the curve as "decreasing" without the constant-half-life argument caps in the middle band.

exam7 marks

Evaluate the usefulness of radioactive half-life for dating objects of very different ages, such as a

200

-year-old wooden artefact and a

2

-billion-year-old rock, referring to the choice of isotope and the limitations of the technique.

Show worked solution →

Band-6 plan. Thesis: half-life dating is powerful but only within a "useful window" set by the chosen isotope's $T_{1/2}$ , so different isotopes suit different timescales. Argue (1) the maths of why a very short or very long half-life relative to the sample age fails, (2) match isotopes to the two examples ( $^{14}$ C vs $^{238}$ U/ $^{206}$ Pb), (3) name genuine limitations (initial-ratio assumption, contamination, measurement precision), (4) reach a judgement.

Model answer. Radiometric dating relies on comparing the current amount of a radioactive isotope (or its stable daughter) with the known initial amount, using $N = N_0 e^{-\lambda t}$ . This only works well when the object's age is comparable to the isotope's half-life: if the age is much less than $T_{1/2}$ , almost none of the isotope has decayed and the small change cannot be measured precisely; if the age is much greater than $T_{1/2}$ , almost none of the parent isotope remains to measure.

For the $200$ -year-old wooden artefact, carbon-14 ( $T_{1/2} = 5730$ y) is appropriate: $200$ years is a small fraction of one half-life, so a sensitive enough detector can still resolve the small drop in $^{14}$ C activity, though the uncertainty is proportionally larger for very young samples. For the $2$ -billion-year-old rock, $^{14}$ C is useless (after $2 \times 10^9$ y, the fraction remaining $e^{-\lambda t}$ is effectively zero and unmeasurable); instead $^{238}$ U ( $T_{1/2} = 4.47 \times 10^9$ y) decaying to stable $^{206}$ Pb is used, because $2$ billion years is a comparable fraction of its half-life, giving a measurable but not fully-decayed parent-to-daughter ratio.

Limitations apply to both cases. The method assumes the initial isotope ratio is known (for $^{14}$ C, that atmospheric $^{14}$ C/ $^{12}$ C has been roughly constant, corrected using tree-ring calibration; for U-Pb, that no lead was present when the rock crystallised, checked using isotope-ratio methods). Contamination or leaching of parent or daughter isotopes after formation, and the practical precision of activity or mass-spectrometry measurements, both introduce error. Despite these caveats, because the half-life of the chosen isotope is matched to the expected age, radiometric dating gives a reliable, physically grounded age estimate in both cases, with quoted uncertainties reflecting the calibration and measurement limits rather than a flaw in the underlying exponential law.

Marker's note: the top band gives the quantitative reasoning for why half-life must be comparable to age (not just "different isotopes for different ages"), correctly pairs each isotope to each timescale, and names at least two genuine limitations (initial-ratio assumption and contamination/precision) before reaching an overall judgement. A response that only lists isotopes and half-lives without this reasoning caps in the middle band.

Practise this

Sources & how we know this

Physics Stage 6 Syllabus (2017) — NESA
Physics HSC exam papers — NESA