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.

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11.1Nuclear 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. $Radioactive decay$ 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 $A = λ N$ and $N = N_{0} e^{- λ t}$
Define half-life and use $λ t_{1/2} = ln 2$

Types of radiation

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

$Beta-minus$ ( $β^{-}$ ) 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.

$Gamma$ ( $γ$ ) 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; $β^{-}$ decay leaves $A$ unchanged and increases $Z$ by 1.

_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} α

Alpha decay.

_{6}^{14} C \to_{7}^{14} N +_{- 1}^{0} β + \overset{ν}{ˉ}_{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 $random$ — you cannot predict which nucleus decays next or when — and $spontaneous$ — it is unaffected by temperature, pressure or chemical state.

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

The $activity$ $A$ is the number of decays per second, in becquerels (Bq). With $N$ undecayed nuclei, $A = λ N$ .

A = λ N = - \frac{d N}{d t}

Activity.

A = λ N = 2.0 \times 1 0^{- 9} \times 3.0 \times 1 0^{15}

Answer

6.0 \times 1 0^{6} 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, $ln N = ln N_{0} - λ t$ . A graph of $ln N$ or $ln A$ against $t$ is a straight line of gradient $- λ$ — the standard way to find the decay constant from experimental data.

N = N_{0} e^{- λ t}

Number of undecayed nuclei.

A = A_{0} e^{- λ t}

Activity.

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

= 800 \times 0.3012

Answer

240 Bq

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

Half-life

The $half-life$ $t_{1/2}$ 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 $e^{- λ t_{1/2}} = \frac{1}{2}$ , so $λ t_{1/2} = ln 2$ .

After $n$ half-lives, the fraction remaining is $(\frac{1}{2})^{n}$ . 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.

λ t_{1/2} = ln 2

Half-life and decay constant.

λ = ln 2/5730 = 1.210 \times 1 0^{- 4} y r^{- 1}

0.22 = e^{- λ t}

, so

t = \frac{- ln 0.22}{λ} = \frac{1.514}{1.210 \times 1 0 ^{- 4}}

AnswerAbout

1.25 \times 1 0^{4}

years.

Equation recap

A = λ N

Activity.

N = N_{0} e^{- λ t}

Exponential decay.

λ 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

λ

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 $A = λ N$ .
$N = N_{0} e^{- λ t}$ ; a $ln N$ – $t$ graph has gradient $- λ$ .
Half-life is constant, with $λ t_{1/2} = ln 2$ .
After $n$ half-lives, a fraction $(\frac{1}{2})^{n}$ remains.

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Radioactive Decay and Half-Life

Types of radiation

Random decay and activity

The exponential law

Half-life

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Radioactive Decay and Half-Life

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