If decay is random, how can half-life be constant?

Each nucleus has a fixed probability of decaying. With enormous numbers, the fraction decaying in a given time becomes extremely predictable.

What is the difference between activity and count rate?

Activity is the total decays per second. Count rate is what a detector records, which is only a fraction of the activity and includes background.

Why plot A against time?

It turns the exponential into a straight line with gradient -, which is easy to fit and read accurately.

Why does carbon dating only work up to about 50 000 years?

After roughly nine half-lives, so little carbon-14 remains that its activity cannot be distinguished from background.

Back to course

168–170, 172–173Nuclear PhysicsStretch

Radioactive Decay and Half-Life

You cannot predict when a single unstable nucleus will decay. Yet a sample of billions decays with such regularity that its activity can date a 4000-year-old skeleton. $\textbf{Radioactive decay}$ is random for individual nuclei and exponential for large numbers.

40 min Topic 11 · Nuclear Radiation

What you'll be able to do

Describe alpha, beta and gamma radiation, and write balanced decay equations
Explain the random and spontaneous nature of radioactive decay
Use $A = \lambda N$ and $N = N_{0}e^{-\lambda t}$
Define half-life and use $\lambda t_{1/2} = \ln 2$

Types of radiation

$\textbf{Alpha}$ ( $\alpha$ ) particles are helium nuclei, $^{4}_{2}\mathrm{He}$ . They are strongly ionising, travel a few centimetres in air and are stopped by paper.

$\textbf{Beta-minus}$ ( $\beta^{-}$ ) particles are fast electrons emitted when a neutron becomes a proton. They are moderately ionising, travel about a metre in air and are stopped by a few millimetres of aluminium.

$\textbf{Gamma}$ ( $\gamma$ ) rays are high-energy electromagnetic radiation emitted when a nucleus drops to a lower energy state. They are weakly ionising and are reduced, not stopped, by thick lead.

In a decay equation, both nucleon number and proton number must balance. Alpha decay reduces $A$ by 4 and $Z$ by 2; $\beta^{-}$ decay leaves $A$ unchanged and increases $Z$ by 1.

^{226}_{88}\mathrm{Ra} \rightarrow\ ^{222}_{86}\mathrm{Rn} +\ ^{4}_{2}\alpha

Alpha decay.

^{14}_{6}\mathrm{C} \rightarrow\ ^{14}_{7}\mathrm{N} +\ ^{0}_{-1}\beta + \bar{\nu}_{e}

Beta-minus decay.

Tip — Background radiation is always present. Measure it first and subtract it from every count before analysing a source.

Random decay and activity

Decay is $\textbf{random}$ — you cannot predict which nucleus decays next or when — and $\textbf{spontaneous}$ — it is unaffected by temperature, pressure or chemical state.

The $\textbf{decay constant}$ $\lambda$ is the probability per unit time that a given nucleus decays, in $\mathrm{s^{-1}}$ .

The $\textbf{activity}$ $A$ is the number of decays per second, in becquerels (Bq). With $N$ undecayed nuclei, $A = \lambda N$ .

A = \lambda N = -\dfrac{\mathrm{d}N}{\mathrm{d}t}

Activity.

Example 1A sample contains

3.0\times10^{15}

nuclei with decay constant

2.0\times10^{-9}\ \mathrm{s^{-1}}

. Find its activity.

A = \lambda N = 2.0\times10^{-9} \times 3.0\times10^{15}

Answer

6.0\times10^{6}\ \mathrm{Bq}

The exponential law

Because the decay rate is proportional to $N$ , the number of undecayed nuclei falls exponentially. Activity, being proportional to $N$ , follows the same curve.

Taking natural logs, $\ln N = \ln N_{0} - \lambda t$ . A graph of $\ln N$ or $\ln A$ against $t$ is a straight line of gradient $-\lambda$ — the standard way to find the decay constant from experimental data.

N = N_{0}e^{-\lambda t}

Number of undecayed nuclei.

A = A_{0}e^{-\lambda t}

Activity.

Example 1A source has initial activity 800 Bq and

\lambda = 0.050\ \mathrm{h^{-1}}

. Find its activity after 24 hours.

A = 800e^{-0.050 \times 24} = 800e^{-1.2}

= 800 \times 0.3012

Answer

240\ \mathrm{Bq}

Tip — Keep $\lambda$ and $t$ in matching units. Per hour with hours is fine; you only need $\mathrm{s^{-1}}$ when finding activity in Bq from $\lambda N$ .

Half-life

The $\textbf{half-life}$ $t_{1/2}$ is the time for the number of undecayed nuclei — or the activity — to halve. It is constant for a given isotope, whatever the starting amount.

Setting $N = N_{0}/2$ in the exponential law gives $e^{-\lambda t_{1/2}} = \tfrac12$ , so $\lambda t_{1/2} = \ln 2$ .

After $n$ half-lives, the fraction remaining is $(\tfrac12)^{n}$ . Medical tracers need half-lives of hours so they are detectable but do not linger; dating needs half-lives comparable to the age being measured.

\lambda t_{1/2} = \ln 2

Half-life and decay constant.

Example 1Carbon-14 has a half-life of 5730 years. A bone has 22% of the carbon-14 activity of living tissue. Estimate its age.

\lambda = \ln 2/5730 = 1.210\times10^{-4}\ \mathrm{yr^{-1}}

0.22 = e^{-\lambda t}

, so

t = \dfrac{-\ln 0.22}{\lambda} = \dfrac{1.514}{1.210\times10^{-4}}

AnswerAbout

1.25\times10^{4}

years.

Equation recap

A = \lambda N

Activity.

N = N_{0}e^{-\lambda t}

Exponential decay.

\lambda t_{1/2} = \ln 2

Half-life.

Common mistakes to avoid

Thinking after two half-lives none remains.

A quarter remains; decay never reaches exactly zero.

Mixing units of

\lambda

and

t

Use consistent time units in the exponent.

Saying heating a source speeds up decay.

Decay is spontaneous and unaffected by physical conditions.

Unbalanced nucleon or proton numbers in decay equations.

Check both top and bottom numbers sum equally on each side.

Key takeaways

Alpha, beta and gamma differ in nature, ionising power and penetration.
Decay is random and spontaneous; activity $A = \lambda N$ .
$N = N_{0}e^{-\lambda t}$ ; a $\ln N$ – $t$ graph has gradient $-\lambda$ .
Half-life is constant, with $\lambda t_{1/2} = \ln 2$ .
After $n$ half-lives, a fraction $(\tfrac12)^{n}$ remains.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 4 lessons