CIE AS/A Level Physics 23.2 Radioactive decay Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 23.2 Radioactive decay Study Notes – New Syllabus
CIE AS/A Level Physics 23.2 Radioactive decay Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- understand that fluctuations in count rate provide evidence for the random nature of radioactive decay
- understand that radioactive decay is both spontaneous and random
- define activity and decay constant, and recall and use \( A = \lambda N \)
- define half-life
- use \( \lambda = 0.693 / t_{1/2}\)
- understand the exponential nature of radioactive decay, and sketch and use the relationship \(\displaystyle x = x_0 e^{-\lambda t}\), where ( x ) could represent activity, number of undecayed nuclei or received count rate
Random Nature of Radioactive Decay and Fluctuations in Count Rate
Radioactive decay is a fundamentally random process. Even though large numbers of nuclei decay according to predictable laws (like half-life), it is impossible to predict exactly which nucleus will decay and when.
Because each nucleus decays independently and unpredictably, any measurement of radioactive counts shows natural fluctuations.
Why Count Rate Fluctuates
- Each nucleus decays at a random moment.
- The number of decays in any short time interval varies unpredictably.
- A detector (e.g., Geiger–Müller tube) therefore records slightly different counts each time.
- These variations occur even when the source intensity is constant.
This randomness is strong evidence that radioactive decay is a spontaneous, unpredictable process.
Typical Observation
If you measure the count rate for a radioactive source over multiple 10-second intervals, you might get readings such as:
- 156 counts
- 149 counts
- 161 counts
- 152 counts
- 158 counts
The source did not change; the fluctuations arise purely from the random nature of decay.
Example
A student measures the count rate from a radioactive source for several identical time intervals. The readings are all slightly different. What does this show?
▶️ Answer / Explanation
This shows that radioactive decay is random — each nucleus decays independently and unpredictably, causing fluctuations in count rate.
Example
A student records the following counts in five identical 20-second intervals: 120, 134, 128, 131, 123. Explain why the values are not the same even though the activity of the source is constant.
▶️ Answer / Explanation
Radioactive decay is spontaneous and random. The probability that a nucleus decays in any interval is fixed, but the actual number of decays varies unpredictably. Therefore, the count rate fluctuates even for a stable source.
Example
Explain why keeping the detector, source, and timing interval constant does not eliminate fluctuations in the measured count rate.
▶️ Answer / Explanation
The measurement equipment may be constant, but the decay process itself is inherently random. Each nucleus has a certain probability of decaying per unit time, but whether it actually decays is unpredictable. Because the number of decaying nuclei in any time interval varies randomly, the count rate also fluctuates.
This randomness is a fundamental property of quantum processes and cannot be removed by experimental control.
Spontaneous and Random Nature of Radioactive Decay
Radioactive decay is a fundamental nuclear process in which unstable nuclei transform into more stable nuclei. This process has two key characteristics: it is spontaneous and random.
What “Spontaneous” Means
A radioactive nucleus decays without any external influence. No external factors such as temperature, pressure, or chemical state can change the likelihood of decay.
- The nucleus decides “on its own” when to decay.
- Decay cannot be forced or stopped by physical or chemical means.
- It happens naturally as part of nuclear instability.
What “Random” Means
It is impossible to predict exactly when a particular nucleus will decay. Only the probability of decay can be known.
- Each nucleus behaves independently of others.
- The decay time of each nucleus is unpredictable.
- Count rate fluctuates due to this randomness.
- Only in large samples do we see predictable behaviour such as half-life.
Together:
Radioactive decay occurs without external cause (spontaneous) and cannot be predicted for individual nuclei (random).
Example
Why can’t we predict the exact moment when a particular radioactive nucleus will decay?
▶️ Answer / Explanation
Because radioactive decay is random — each nucleus has an unpredictable decay time, even though the overall probability of decay is fixed.
Example
Explain why heating a radioactive sample does not speed up its decay rate.
▶️ Answer / Explanation
Radioactive decay is spontaneous and is determined by nuclear stability, not external factors. Temperature affects electrons, not the nucleus, so heating cannot alter decay probability.
Example
A sample of a radioactive element shows fluctuating count rates in repeated measurements. Explain how this demonstrates both spontaneity and randomness of decay.
▶️ Answer / Explanation
Random: Each reading differs because the number of nuclei that decay in any given interval varies unpredictably. Each nucleus decays independently with no pattern.
Spontaneous: No external changes were made to the sample, yet decay continues. This shows decay occurs by itself without any external cause or triggering.
Activity, Decay Constant, and the Relation \( \mathrm{A = \lambda N} \)
Radioactive decay is described using two key quantities: the activity of a sample and its decay constant. These describe how quickly nuclei in a radioactive material decay.
Activity (A)
Activity is the number of nuclear decays per unit time.
- Symbol: \( \mathrm{A} \)
- Unit: becquerel (Bq)
- \( \mathrm{1\ Bq = 1\ decay\ per\ second} \)
- Activity decreases with time as the number of undecayed nuclei falls.
Decay Constant (λ)
The decay constant \( \mathrm{\lambda} \) is the probability per unit time that a nucleus will decay.
- Symbol: \( \mathrm{\lambda} \)
- Unit: \( \mathrm{s^{-1}} \)
- Higher \( \mathrm{\lambda} \) → faster decay → shorter half-life
Key idea: Each nucleus behaves independently, and \( \mathrm{\lambda} \) quantifies its chance of decaying each second.
The Relationship \( \mathrm{A = \lambda N} \)
\( \mathrm{A = \lambda N} \)
- \( \mathrm{A} \) = activity (decays per second)
- \( \mathrm{\lambda} \) = decay constant
- \( \mathrm{N} \) = number of undecayed nuclei
This formula shows that activity is proportional to the number of radioactive nuclei present.
Example
If a radioactive sample has a decay constant of \( \mathrm{0.05\ s^{-1}} \) and contains \( \mathrm{200} \) nuclei, what is its activity?
▶️ Answer / Explanation
Using \( \mathrm{A = \lambda N} \):
\( \mathrm{A = 0.05 \times 200 = 10\ Bq} \)
The sample undergoes 10 decays per second.
Example
A sample has an activity of \( \mathrm{3.0\times10^4\ Bq} \) and a decay constant of \( \mathrm{2.0\times10^{-3}\ s^{-1}} \). Calculate the number of undecayed nuclei in the sample.
▶️ Answer / Explanation
Rearrange \( \mathrm{A = \lambda N} \):
\( \mathrm{N = \dfrac{A}{\lambda}} \)
\( \mathrm{N = \dfrac{3.0\times10^4}{2.0\times10^{-3}} = 1.5\times10^7} \)
There are \( \mathrm{1.5\times10^7} \) undecayed nuclei.
Example
A radioactive isotope initially contains \( \mathrm{5.0\times10^{12}} \) nuclei and has a decay constant of \( \mathrm{4.0\times10^{-4}\ s^{-1}} \). Calculate its initial activity, and explain what happens to this activity over time.
▶️ Answer / Explanation
Initial activity:
\( \mathrm{A = \lambda N = (4.0\times10^{-4})(5.0\times10^{12})} \)
\( \mathrm{A = 2.0\times10^{9}\ Bq} \)
Interpretation:
- The activity starts high: 2 billion decays per second.
- As decay proceeds, \( \mathrm{N} \) decreases → activity decreases.
- The activity falls exponentially according to the decay law.
Half-life
The half-life of a radioactive isotope is a key concept used to describe how unstable nuclei decay over time.
Definition of Half-life
The half-life is the time taken for the number of unstable nuclei in a radioactive sample to fall to half its original value.
It is also equivalent to:
- the time taken for the activity (count rate) of the sample to fall to half its initial value, or
- the time taken for the mass of the radioactive isotope to reduce by half.
Important Notes:
- Half-life is constant for a given isotope.
- It does not depend on temperature, pressure, or chemical reactions.
- Decay is random, but half-life describes predictable statistical behaviour.
Example
A radioactive sample has 800 nuclei initially. After one half-life, how many undecayed nuclei remain?
▶️ Answer / Explanation
After one half-life:
\( \mathrm{N = \dfrac{800}{2} = 400} \)
400 nuclei remain.
Example
A radioactive sample has an activity of \( \mathrm{1200\ counts/s} \). After one half-life, what is its activity?
▶️ Answer / Explanation
Activity is proportional to the number of undecayed nuclei.
New activity = \( \mathrm{\dfrac{1200}{2} = 600\ counts/s} \)
Activity after one half-life = \( \mathrm{600\ counts/s} \)
Example
A sample starts with \( \mathrm{2.0\times10^{6}} \) nuclei. If the sample undergoes two half-lives, how many undecayed nuclei remain?
▶️ Answer / Explanation
After first half-life:
\( \mathrm{N_1 = \dfrac{2.0\times10^{6}}{2} = 1.0\times10^{6}} \)
After second half-life:
\( \mathrm{N_2 = \dfrac{1.0\times10^{6}}{2} = 5.0\times10^{5}} \)
Remaining nuclei = \( \mathrm{5.0\times10^{5}} \)
Using the Decay Constant Formula: \( \mathrm{\lambda = \dfrac{0.693}{t_{1/2}}} \)
Radioactive decay is described by two key quantities:
- Half-life \( \mathrm{t_{1/2}} \)
- Decay constant \( \mathrm{\lambda} \)
These two quantities are linked by the important relation:
\( \mathrm{\lambda = \dfrac{0.693}{t_{1/2}}} \)
Meaning of the Equation
- \( \mathrm{\lambda} \) = probability per unit time that a nucleus will decay.
- \( \mathrm{t_{1/2}} \) = time for half the nuclei to decay.
- The constant 0.693 comes from \( \mathrm{\ln(2)} \), since half-life corresponds to reducing quantity by half.
Units:
- If \( \mathrm{t_{1/2}} \) is in seconds → \( \mathrm{\lambda} \) is in \( \mathrm{s^{-1}} \)
- If \( \mathrm{t_{1/2}} \) is in minutes → \( \mathrm{\lambda} \) is in \( \mathrm{min^{-1}} \)
- If \( \mathrm{t_{1/2}} \) is in years → \( \mathrm{\lambda} \) is in \( \mathrm{year^{-1}} \)
Example
A radioactive isotope has a half-life of 20 s. Calculate its decay constant.
▶️ Answer / Explanation
\( \mathrm{\lambda = \dfrac{0.693}{20}} \)
\( \mathrm{\lambda = 0.03465\ s^{-1}} \)
Decay constant = \( \mathrm{3.47\times10^{-2}\ s^{-1}} \)
Example
An isotope has a decay constant \( \mathrm{0.010\ s^{-1}} \). Calculate its half-life.
▶️ Answer / Explanation
Rearrange:
\( \mathrm{t_{1/2} = \dfrac{0.693}{\lambda}} \)
\( \mathrm{t_{1/2} = \dfrac{0.693}{0.010}} \)
\( \mathrm{t_{1/2} = 69.3\ s} \)
Half-life = 69.3 s
Example
An isotope has a half-life of 3.2 hours. Calculate the decay constant in both \( \mathrm{h^{-1}} \) and \( \mathrm{s^{-1}} \).
▶️ Answer / Explanation
Step 1: Decay constant in \( \mathrm{h^{-1}} \)
\( \mathrm{\lambda = \dfrac{0.693}{3.2} = 0.21656\ h^{-1}} \)
Step 2: Convert half-life to seconds
\( \mathrm{3.2\ h = 3.2\times3600 = 11520\ s} \)
Step 3: Decay constant in \( \mathrm{s^{-1}} \)
\( \mathrm{\lambda = \dfrac{0.693}{11520} = 6.01\times10^{-5}\ s^{-1}} \)
Decay constant:
- \( \mathrm{0.217\ h^{-1}} \)
- \( \mathrm{6.0\times10^{-5}\ s^{-1}} \)
Exponential Nature of Radioactive Decay
Radioactive decay is a random process, but for a large number of nuclei it follows a very predictable pattern: a smooth, continuous exponential decrease over time.
The quantity that decreases may be: 
- \( \mathrm{N} \): number of undecayed nuclei
- \( \mathrm{A} \): activity of the sample
- \( \mathrm{R} \): count rate from a detector
All of these decrease in the same exponential way.
The Exponential Decay Equation
\( \mathrm{x = x_0 e^{-\lambda t}} \)
- \( \mathrm{x_0} \): initial value at \( \mathrm{t = 0} \)
- \( \mathrm{x} \): value at time \( \mathrm{t} \)
- \( \mathrm{\lambda} \): decay constant
- \( \mathrm{t} \): time
Interpretation:
- Decay is rapid at first (steep slope).
- The rate slows down over time but never reaches zero.
- After each half-life, the quantity becomes half its previous value.
Qualitative Sketch of Exponential Decay
- Curve starts at \( \mathrm{x_0} \).
- Steep fall initially.
- Gradually flattens but never touches the time-axis.
- Shows half the quantity remaining after every equal time interval.
Example
A radioactive sample has \( \mathrm{N_0 = 1000} \) undecayed nuclei. If the decay constant is \( \mathrm{\lambda = 0.2\ s^{-1}} \), find \( \mathrm{N} \) after 5 s.
▶️ Answer / Explanation
Use:
\( \mathrm{N = N_0 e^{-\lambda t}} \)
\( \mathrm{N = 1000 \, e^{-0.2\times5}} \)
\( \mathrm{N = 1000 \, e^{-1}} = 1000 \times 0.3679 \)
\( \mathrm{N \approx 368} \)
368 nuclei remain after 5 s.
Example
A sample has an initial activity of \( \mathrm{A_0 = 500\ Bq} \). Its decay constant is \( \mathrm{0.005\ s^{-1}} \). Find the activity after 200 s.
▶️ Answer / Explanation
\( \mathrm{A = A_0 e^{-\lambda t}} \)
\( \mathrm{A = 500 \, e^{-0.005\times200}} \)
\( \mathrm{A = 500\, e^{-1}} = 500 \times 0.3679 \)
\( \mathrm{A \approx 184\ Bq} \)
Activity after 200 s ≈ 184 Bq.
Example
A detector initially records a count rate of 2000 counts/min from a radioactive sample. After 10 minutes the count rate is 800 counts/min. Determine the decay constant \( \mathrm{\lambda} \).
▶️ Answer / Explanation
Step 1: Use exponential decay:
\( \mathrm{800 = 2000\, e^{-\lambda(10)}} \)
Divide both sides by 2000:
\( \mathrm{0.4 = e^{-10\lambda}} \)
Take natural log:
\( \mathrm{\ln(0.4) = -10\lambda} \)
So:
\( \mathrm{\lambda = -\dfrac{\ln(0.4)}{10}} \)
\( \mathrm{\lambda = \dfrac{0.9163}{10} = 0.0916\ min^{-1}} \)
Decay constant = \( \mathrm{9.16\times10^{-2}\ min^{-1}} \)