Radioactive Decay

SPEC 3.8.1.3 NUCLEAR PHYSICS

Exponential decay, activity, half-life, and graphical analysis of radioactive processes

Random Nature of Decay

Key Principle

Radioactive decay is a random and spontaneous process. It is impossible to predict which particular nucleus will decay next, or when a specific nucleus will decay.

However, for a large sample of identical nuclei, we can define a constant decay probability. Each nucleus has the same fixed probability of decaying per unit time, regardless of how long it has existed or what external conditions are present.

Key Concept

The decay constant

λ

represents the probability that a given nucleus will decay per unit time. Its unit is s⁻¹.

Analogy

Think of rolling 100 dice and removing every six. You cannot predict which specific die will show a six, but you can predict that roughly will be removed each roll. This is how constant decay probability works — each nucleus has the same chance of "being removed" per unit time.

The Decay Equation

Since each nucleus has the same probability

λ

of decaying per unit time, the rate of decay is proportional to the number of undecayed nuclei remaining:

Fundamental Equation

\frac{Δ N}{Δ t} = - λ N

SymbolMeaningUnit

\frac{Δ N}{Δ t}

Rate of change of undecayed nuclei (rate of decay)s⁻¹

λ

Decay constant (probability of decay per unit time)s⁻¹

N

Number of undecayed nuclei remainingdimensionless

Important

The negative sign indicates that N is decreasing with time — nuclei are decaying, so the number of undecayed nuclei falls.

Exponential Decay

Solving the differential equation

\frac{Δ N}{Δ t} = - λ N

gives the exponential decay law:

Exponential Decay Law

N = N_{0} e^{- λ t}

where

N_{0}

= initial number of undecayed nuclei at

t = 0

Exponential decay curve: N decreases by the same fraction in each equal time interval

Key Feature

The curve never reaches zero — it approaches the time axis asymptotically. In each half-life, exactly half the remaining nuclei decay.

Worked Example01

A sample initially contains 5.0 × 10²⁰ undecayed nuclei. The decay constant is 3.0 × 10⁻⁴ s⁻¹. Calculate the number of undecayed nuclei remaining after 2.0 hours.

1

Convert time to seconds:

t = 2.0 \times 3600 = 7200 s

2

Apply exponential decay:

N = N_{0} e^{- λ t}

3

Substitute:

N = 5.0 \times 1 0^{20} \times e^{- (3.0 \times 1 0^{- 4}) (7200)}

4

Calculate exponent:

λ t = 3.0 \times 1 0^{- 4} \times 7200 = 2.16

5

N = 5.0 \times 1 0^{20} \times e^{- 2.16} = 5.0 \times 1 0^{20} \times 0.1153

Answer: $N = 5.8 \times 1 0^{19}$ nuclei

Activity

Definition

Activity (A) is the rate at which nuclei decay — the number of decays per unit time. It is measured in becquerels (Bq), where 1 Bq = 1 decay per second.

Since activity equals the magnitude of the rate of decay:

Activity Equations

A = λ N

A = A_{0} e^{- λ t}

where

A_{0} = λ N_{0}

is the initial activity

Tip

Activity is directly proportional to the number of undecayed nuclei. As N falls exponentially, so does A. The same exponential decay law applies to both N and A.

Important

In exam questions, always check whether you are given N (number of nuclei) or A (activity). Both follow the same exponential pattern, but make sure you use the correct starting value (

N_{0}

or

A_{0}

).

Worked Example02

A radioactive source contains 8.0 × 10¹⁸ undecayed nuclei and has a decay constant of 1.5 × 10⁻³ s⁻¹. Calculate (a) the initial activity and (b) the activity after 20 minutes.

1

(a) Initial activity:

A_{0} = λ N_{0} = 1.5 \times 1 0^{- 3} \times 8.0 \times 1 0^{18}

2

A_{0} = 1.2 \times 1 0^{16} Bq

3

(b) Convert time:

t = 20 \times 60 = 1200 s

4

A = A_{0} e^{- λ t} = 1.2 \times 1 0^{16} \times e^{- (1.5 \times 1 0^{- 3}) (1200)}

5

λ t = 1.8

, so

A = 1.2 \times 1 0^{16} \times e^{- 1.8} = 1.2 \times 1 0^{16} \times 0.1653

Answer: (a) $1.2 \times 1 0^{16}$ Bq (b) $2.0 \times 1 0^{15}$ Bq

Half-Life

Definition

Half-life ( $T_{\frac{1}{2}}$ ) is the time taken for the number of undecayed nuclei (or the activity) to fall to half its original value.

We can derive the half-life equation by setting

N = \frac{N _{0}}{2}

in the decay law:

\frac{N _{0}}{2} = N_{0} e^{- λ T_{1/2}}

\frac{1}{2} = e^{- λ T_{1/2}}

ln (\frac{1}{2}) = - λ T_{1/2}

- ln 2 = - λ T_{1/2}

Half-Life Equation

T_{\frac{1}{2}} = \frac{l n 2}{λ}

where

ln 2 \approx 0.693

Key Concept

A large decay constant means a short half-life (rapid decay). A small decay constant means a long half-life (slow decay). They are inversely proportional.

Graphical method: ① Draw horizontal line at N₀/2 → ② Drop vertically to time axis → Read off T½

Exam Technique

To find the half-life graphically from a decay curve: (1) read the initial value from the y-axis, (2) halve it, (3) draw a horizontal line at this half-value across to the curve, (4) drop a vertical line down to the time axis, and (5) read off the time. You can verify by repeating for a second halving — the time interval should be the same.

Count Rate & Background

Key Distinction

Activity (A) is the total number of decays per second from a source, measured in becquerels (Bq).

Count rate is the number of decays detected per second by a detector (e.g. a Geiger–M ller tube). Count rate is always less than the activity because no detector captures every emitted particle.

Background Radiation

Background radiation is always present from natural sources (cosmic rays, rocks, radon gas, medical sources). When measuring the count rate of a source, you must first measure the background count rate and subtract it to find the corrected count rate due to the source alone.

Corrected count rate = Measured count rate − Background count rate

Alternative Half-Life Formula

A = \frac{A _{0}}{2 ^{n}} where n = \frac{t}{T _{1/2}}

This works for any value of

n

, not just whole numbers.

Exam Tip

For whole-number half-lives, repeated halving is quickest (e.g. 3 half-lives → divide by 2³ = 8). For non-integer half-lives, either use

A = A_{0} / 2^{n}

or the full exponential form

A = A_{0} e^{- λ t}

. Both give the same answer — choose whichever suits the question.

Worked Example03

Strontium-90 has a half-life of 28.8 years. Calculate its decay constant in s⁻¹.

1

Convert half-life to seconds:

T_{1/2} = 28.8 \times 365.25 \times 24 \times 3600 = 9.09 \times 1 0^{8} s

2

Rearrange:

λ = \frac{l n 2}{T _{1/2}}

3

λ = \frac{0.693}{9.09 \times 1 0 ^{8}}

Answer: $λ = 7.62 \times 1 0^{- 10} s^{- 1}$

Worked Example04

A radioactive isotope has a half-life of 6.0 hours. A sample initially has an activity of 4800 Bq. Calculate the activity after (a) 18 hours and (b) 10 hours.

1

(a) 18 hours = 3 half-lives. After each half-life, activity halves:

2

A = 4800 \times (\frac{1}{2})^{3} = 4800 \times \frac{1}{8} = 600 Bq

3

(b) 10 hours is not a whole number of half-lives, so use the exponential form.

4

Find

λ

:

λ = \frac{l n 2}{T _{1/2}} = \frac{0.693}{6.0 \times 3600} = 3.21 \times 1 0^{- 5} s^{- 1}

5

t = 10 \times 3600 = 36000 s

6

A = 4800 \times e^{- (3.21 \times 1 0^{- 5}) (36000)} = 4800 \times e^{- 1.155} = 4800 \times 0.3150

Answer: (a) 600 Bq (b) 1510 Bq

Graphical Methods

There are two key graphs used to analyse radioactive decay data:

N vs t: Exponential curve. Read half-life directly from graph.

ln N vs t: Straight line with gradient = −λ and y-intercept = ln N

Why Use ln Graphs?

Taking natural logs of

N = N_{0} e^{- λ t}

gives:

ln N = ln N_{0} - λ t

. This is in the form

y = m x + c

, so plotting

ln N

against

t

gives a straight line. The gradient equals

- λ

, and the y-intercept gives

ln N_{0}

.

Worked Example05

A graph of ln(A) against time t gives a straight line. Two points on the line are (0 s, 18.4) and (500 s, 16.2). Determine the half-life of the isotope.

1

The gradient of ln(A) vs t equals

- λ

.

2

Gradient

= \frac{16.2 - 18.4}{500 - 0} = \frac{- 2.2}{500} = - 4.4 \times 1 0^{- 3} s^{- 1}

3

So

λ = 4.4 \times 1 0^{- 3} s^{- 1}

4

T_{1/2} = \frac{l n 2}{λ} = \frac{0.693}{4.4 \times 1 0 ^{- 3}}

Answer: $T_{1/2} = 158 s$ (2 min 38 s)

Calculating N from Mass

Exam questions often give the mass of a radioactive sample rather than the number of nuclei directly. You must be able to convert between the two.

Key Relationship

N = \frac{m}{M} \times N_{A}

where

m

= mass of sample (g),

M

= molar mass (g mol⁻¹),

N_{A} = 6.02 \times 1 0^{23} mol^{- 1}

TermMeaningUnit

m

Mass of the radioactive sampleg (or kg — convert accordingly)

M

Molar mass (numerically equal to nucleon number A)g mol⁻¹

N_{A}

Avogadro constant6.02 × 10²³ mol⁻¹

Tip

For a specific isotope, the molar mass in grams is numerically equal to the nucleon number (mass number). For example, carbon-14 has molar mass 14 g mol⁻¹, and uranium-238 has molar mass 238 g mol⁻¹.

Worked Example06

A 2.0 g sample of iodine-131 ( $T_{1/2} = 8.02$ days) is prepared. Calculate the initial activity of the sample. (Molar mass of I-131 = 131 g mol⁻¹)

1

Find number of nuclei:

N = \frac{m}{M} \times N_{A} = \frac{2.0}{131} \times 6.02 \times 1 0^{23} = 9.19 \times 1 0^{21}

2

Find decay constant:

T_{1/2} = 8.02 \times 24 \times 3600 = 6.93 \times 1 0^{5} s

3

λ = \frac{l n 2}{T _{1/2}} = \frac{0.693}{6.93 \times 1 0 ^{5}} = 1.00 \times 1 0^{- 6} s^{- 1}

4

Activity:

A = λ N = 1.00 \times 1 0^{- 6} \times 9.19 \times 1 0^{21}

Answer: $A = 9.2 \times 1 0^{15} Bq$ (9.2 PBq)

Worked Example07

A sample of wood from an archaeological dig has a carbon-14 activity of 0.19 Bq g⁻¹. Living wood has an activity of 0.25 Bq g⁻¹. The half-life of carbon-14 is 5730 years. Estimate the age of the sample.

1

Use

A = A_{0} e^{- λ t}

, so

\frac{A}{A _{0}} = e^{- λ t}

2

Take natural log:

ln (\frac{A}{A _{0}}) = - λ t

3

t = - \frac{1}{λ} ln (\frac{A}{A _{0}}) = \frac{T _{1/2}}{l n 2} ln (\frac{A _{0}}{A})

4

t = \frac{5730}{0.693} \times ln (\frac{0.25}{0.19}) = 8268 \times ln (1.316)

5

t = 8268 \times 0.2744 = 2269 years

Answer: The sample is approximately 2300 years old

Applications of Half-Life

Carbon-14 Dating

All living organisms absorb carbon-14 (

T_{1/2} = 5730

years) from the atmosphere. When an organism dies, it stops absorbing C-14 and the amount decreases by radioactive decay. By comparing the current activity of a sample with the activity of living material, we can estimate the age of organic objects (wood, bone, cloth) up to around 50 000 years old.

Method

Measure the current activity

A

of the sample. Compare with the known activity

A_{0}

of living material. Use

A = A_{0} e^{- λ t}

rearranged to find

t

:

t = \frac{1}{λ} ln (\frac{A _{0}}{A})

Medical Tracers

Technetium-99m (

T_{1/2} = 6.0

hours) is the most widely used radioactive tracer in medicine. It is ideal because it is a pure gamma emitter (so gamma rays can be detected outside the body with minimal tissue damage), its half-life is short enough to limit patient exposure but long enough for imaging, and it can be chemically attached to different molecules to target specific organs.

Exam Tip

Exam questions may ask you to justify the choice of isotope for a given use. Think about: half-life (too short = decays before use; too long = prolonged exposure), type of radiation emitted, and whether it can be safely introduced into the body or environment.

Storage & Disposal

Isotopes with very long half-lives (e.g. plutonium-239,

T_{1/2} = 24100

years) remain hazardous for thousands of years. They must be stored securely, for example in steel casks within deep underground repositories, to prevent environmental contamination.

Practice

Question 1: Cobalt-60 has a half-life of 5.27 years. A hospital source has an initial activity of 3.7 × 10¹⁴ Bq. Calculate the decay constant in s⁻¹ and the activity after 2.0 years.

1

T_{1/2} = 5.27 \times 365.25 \times 24 \times 3600 = 1.663 \times 1 0^{8} s

2

λ = \frac{0.693}{1.663 \times 1 0 ^{8}} = 4.17 \times 1 0^{- 9} s^{- 1}

3

t = 2.0 \times 365.25 \times 24 \times 3600 = 6.311 \times 1 0^{7} s

4

A = 3.7 \times 1 0^{14} \times e^{- (4.17 \times 1 0^{- 9}) (6.311 \times 1 0^{7})} = 3.7 \times 1 0^{14} \times e^{- 0.2632}

Answer: $λ = 4.17 \times 1 0^{- 9} s^{- 1}$ ; $A = 2.8 \times 1 0^{14} Bq$

Practice

Question 2:

Time / s0100200300400500

Activity / Bq84054035022514593

1

Plot ln(A) vs t. Calculate ln values: ln(840)=6.73, ln(540)=6.29, ln(350)=5.86, ln(225)=5.42, ln(145)=4.98, ln(93)=4.53

2

Draw best-fit straight line through points.

3

Gradient

= \frac{4.53 - 6.73}{500 - 0} = \frac{- 2.20}{500} = - 4.4 \times 1 0^{- 3} s^{- 1}

4

λ = 4.4 \times 1 0^{- 3} s^{- 1}

5

T_{1/2} = \frac{0.693}{4.4 \times 1 0 ^{- 3}}

Answer: $T_{1/2} \approx 158 s$

Practice

Question 3: Technetium-99m is used in medical imaging and has a half-life of 6.01 hours. A patient is injected with a solution containing 5.0 × 10⁻⁸ g of Tc-99m. Calculate the initial activity of the sample. (Molar mass = 99 g mol⁻¹,

N_{A} = 6.02 \times 1 0^{23} mol^{- 1}

)

1

Number of nuclei:

N = \frac{5.0 \times 1 0 ^{- 8}}{99} \times 6.02 \times 1 0^{23} = 3.04 \times 1 0^{14}

2

Convert half-life:

T_{1/2} = 6.01 \times 3600 = 21636 s

3

λ = \frac{0.693}{21636} = 3.20 \times 1 0^{- 5} s^{- 1}

4

A = λ N = 3.20 \times 1 0^{- 5} \times 3.04 \times 1 0^{14}

Answer: $A = 9.7 \times 1 0^{9} Bq \approx 9.7 GBq$

Summary

Key Equations

EquationUse

\frac{Δ N}{Δ t} = - λ N

Defines constant decay probability

N = N_{0} e^{- λ t}

Number of undecayed nuclei at time t

A = λ N

Activity from number of nuclei

A = A_{0} e^{- λ t}

Activity at time t

T_{1/2} = \frac{l n 2}{λ}

Relationship between half-life and decay constant

N = \frac{m}{M} \times N_{A}

Number of nuclei from mass

Key Exam Skills

Convert time units carefully (years → days → hours → seconds)

Use ln graphs to determine λ from experimental data (gradient = −λ)

Convert between mass and N using the Avogadro constant

For whole numbers of half-lives, use repeated halving; for arbitrary times, use the exponential equation