A LevelPhysicsAQARevision NotesNuclear PhysicsRadioactive DecayExponential Decay

Exponential Decay (AQA A Level Physics): Revision Note

Exam code: 7408

Written by: Katie M

Reviewed by: Tim

Updated on 6 October 2026

Guided study available on this topic

Understand this topic with deeper explanations and understanding checks.

Exponential decay

In radioactive decay, the number of undecayed nuclei falls very rapidly, without ever reaching zero
- Such a model is known as exponential decay
The graph of number of undecayed nuclei against time has a very distinctive shape:

The key features of this graph are:
- The steeper the slope, the larger the decay constant $λ$ (and vice versa)
- The decay curves always start on the y-axis at the initial number of undecayed nuclei ( $N_{0}$ )

Equations for radioactive decay

The number of undecayed nuclei N can be represented in exponential form by the equation:

$N = N_{0} e^{- λ t}$

Where:
- $N_{0}$ = the initial number of undecayed nuclei (when $t = 0$ )
- $N$ = number of undecayed nuclei at a certain time $t$
- $λ$ = decay constant (s^-1)
- $t$ = time interval (s)

The number of nuclei can be substituted for other quantities
For example, the activity $A$ is directly proportional to $N$ , so it can also be represented in exponential form by the equation:

$A = A_{0} e^{- λ t}$

Where:
- $A$ = activity at a certain time $t$ (Bq)
- $A_{0}$ = initial activity (Bq)
The received count rate $C$ is related to the activity of the sample, hence it can also be represented in exponential form by the equation:

$C = C_{0} e^{- λ t}$

Where:
- $C$ = count rate at a certain time $t$ (counts per minute or cpm)
- $C_{0}$ = initial count rate (counts per minute or cpm)

The exponential function e

The symbol $e$ represents the exponential constant
- It is approximately equal to $e$ = 2.718
On a calculator it is shown by the button $e^{x}$
The inverse function of $e^{x}$ is $\ln y$ , known as the natural logarithmic function
- This is because, if $e^{x} = y$ , then $x = \ln y$

Worked Example

Strontium-90 decays with the emission of a $β$ -particle to form yttrium-90.

The decay constant of strontium-90 is 0.025 year ^-1.

Determine the activity $A$ of the sample after 5.0 years, expressing the answer as a fraction of the initial activity $A_{0}$ .

[2]

Answer:

Step 1: Write out the known quantities

Decay constant, $λ = 0.025 {year}^{- 1}$
Time interval, $t = 5.0 years$
Both quantities have the same unit, so there is no need for conversion

Step 2: Write the equation for activity in exponential form

$A = A_{0} e^{- λ t}$

Step 3: Rearrange the equation for the ratio between $A$ and $A_{0}$

$\frac{A}{A_{0}} = e^{- λ t}$

Step 4: Calculate the ratio $\frac{A}{A_{0}}$

$\frac{A}{A_{0}} = e^{- (0.025 \times 8)}$ [1 mark]

$\frac{A}{A_{0}} = 0.88$ [1 mark]

Therefore, the activity of strontium-90 decreases by a factor of 0.88, or 12%, after 5 years

Using molar mass & the Avogadro constant

Molar mass

The molar mass, or molecular mass, of a substance is the mass of a substance, in grams, in one mole
- Its unit is g mol^-1
The number of moles from this can be calculated using the equation:

$number of moles (mol) = \frac{mass (g)}{molar mass (g {mol}^{- 1})}$

Avogadro’s constant

Avogadro’s constant $N_{A}$ is defined as:

The number of atoms in one mole of a substance; equal to 6.02 × 10²³ mol^-1

For example, 1 mole of sodium (Na) contains 6.02 × 10²³ atoms of sodium
The number of atoms, or nuclei, can be determined using the equation:

$number of nuclei = \frac{mass \times N_{A}}{molar mass}$

Worked Example

Americium-241 is an artificially produced radioactive element that emits α-particles.

In a smoke detector, a sample of americium-241 of mass 5.1 µg is found to have an activity of 5.9 × 10⁵ Bq. The supplier’s website says the americium-241 in their smoke detectors initially has an activity level of 6.1 × 10⁵ Bq.

Determine:

(a) the number of nuclei in the sample of americium-241 [2]

(b) the decay constant of americium-241 [2]

Answer:

Part (a)

Step 1: Write down the known quantities

$Mass = 5.1 μg = 5.1 \times 10^{- 6} g$
Molecular mass of americium = 241
Avogadro constant, $N_{A} = 6.02 \times 10^{23} {mol}^{- 1}$

Step 2: Write down the equation relating number of nuclei, mass and molecular mass

$N = \frac{mass \times N_{A}}{molar mass}$

Step 3: Calculate the number of nuclei

$N = \frac{(5.1 \times 10^{- 6}) \times (6.02 \times 10^{23})}{241}$ [1 mark]

$N = 1.27394 \times 10^{16} = 1.27 \times 10^{16} (3 s . f .)$ [1 mark]

Part (b)

Step 1: Write down the known quantities

Activity, $A = 5.9 \times 10^{5} Bq$
Number of nuclei, $N = 1.27394 \times 10^{16}$

Step 2: Write the equation for activity

Activity: $A = λ N$

Step 3: Rearrange for decay constant λ and calculate the answer

$λ = \frac{A}{N} = \frac{5.9 \times 10^{5}}{1.27394 \times 10^{16}}$ [1 mark]

$λ = 4.63129 \times 10^{- 11} s^{- 1} = 4.63 \times 10^{- 11} s^{- 1} (3 s . f .)$ [1 mark]

Part (c)

Step 1: Write down the known quantities

Activity, $A = 5.9 \times 10^{5} Bq$
Initial activity, $A_{0} = 6.1 \times 10^{5} Bq$
Decay constant, $λ = 4.63129 \times 10^{- 11} s^{- 1}$

Step 2: Write the equation for activity in exponential form

$A = A_{0} e^{- λ t}$

Step 3: Rearrange for time t

$\frac{A}{A_{0}} = e^{- λ t}$

$\ln (\frac{A}{A_{0}}) = - λ t$

$t = - \frac{1}{λ} \ln (\frac{A}{A_{0}})$ [1 mark]

Step 4: Calculate the age of the smoke detector and convert to years

$t = - \frac{1}{4.63129 \times 10^{- 11}} \times \ln (\frac{5.9 \times 10^{5}}{6.1 \times 10^{5}}) = 7.1981 \times 10^{8} s$ [1 mark]

$t = \frac{7.1981 \times 10^{8}}{24 \times 60 \times 60 \times 365} = 22.8 years (3 s . f .)$ [1 mark]

Therefore, the smoke detector is 22.8 years old

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Exponential Decay (AQA A Level Physics): Revision Note

Exponential decay

Equations for radioactive decay

The exponential function e

Worked Example

Using molar mass & the Avogadro constant

Molar mass

Avogadro’s constant

Worked Example

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Author: Katie M

Reviewer: Tim