Half-life

1. Overview

Radioactive decay is a random process, meaning we cannot predict when a single nucleus will decay. However, because samples contain billions of atoms, we can use the concept of half-life to predict how the overall activity of a sample decreases over time. Understanding half-life is crucial for using radiation safely in medicine, industry, and the home.

Key Definitions

• Half-life ($t_{1/2}$): The time taken for half the nuclei of a specific isotope in a sample to decay. Alternatively: the time taken for the activity (or count rate) of a sample to decrease to half its initial value.
• Activity: The rate at which a source of unstable nuclei decays, measured in Becquerels (Bq). 1 Bq = 1 decay per second.
• Isotope: Atoms of the same element with the same number of protons but different numbers of neutrons.
• Count-rate: The number of decays detected per second by a device (like a Geiger-Müller tube).
• Background Radiation: The low-level radiation present at all times from natural (e.g., rocks, cosmic rays) and man-made sources.

Core Content

Understanding Half-life Since decay is exponential, a substance never "runs out" of radioactivity linearly. Instead, it halves every set period of time.

• After 1 half-life: 50% remains (1/2)
• After 2 half-lives: 25% remains (1/4)
• After 3 half-lives: 12.5% remains (1/8)

Using Decay Curves A decay curve is a graph showing Activity (y-axis) against Time (x-axis).

To find half-life from a graph:

1. Pick a starting activity on the y-axis.
2. Find the time value for that activity.
3. Divide the starting activity by 2.
4. Find the new time value for this halved activity.
5. The difference between these two times is the half-life.

Worked Example (Core) Question: A sample has an initial activity of 1200 Bq. Its half-life is 5 minutes. What is the activity after 15 minutes?

1. Calculate the number of half-lives: $15 \div 5 = 3$ half-lives.
2. Halve the activity three times:
   • 1200 $\rightarrow$ 600 (1st half-life)
   • 600 $\rightarrow$ 300 (2nd half-life)
   • 300 $\rightarrow$ 150 Bq (3rd half-life)

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Extended Content (Extended Curriculum Only)

Calculating Half-life with Background Radiation In real experiments, a Geiger-Müller tube detects both the source and the background radiation. To find the true half-life, you must subtract the background radiation first.

Formula: $Corrected Activity = Measured Activity - Background Radiation$

Worked Example (Extended) Question: The background radiation is 20 counts/min. A source measures 180 counts/min initially. After 4 hours, the source measures 60 counts/min. Calculate the half-life.

1. Find initial corrected activity: $180 - 20 = 160$ counts/min.
2. Find final corrected activity: $60 - 20 = 40$ counts/min.
3. Determine half-lives: $160 \rightarrow 80 \rightarrow 40$ (This is 2 half-lives).
4. Calculate duration: 2 half-lives = 4 hours, so 1 half-life = 2 hours.

Applications of Isotopes The choice of isotope depends on its penetrating power and its half-life.

Application	Radiation Type	Preferred Half-life	Reasoning
Smoke Alarms	Alpha	Long (years)	Alpha is easily blocked by smoke; long half-life means you don't need to replace the source often.
Irradiating Food	Gamma	Long	Gamma kills bacteria/mold throughout the food; long half-life provides a consistent dose over time.
Sterilising Equipment	Gamma	Long	Gamma penetrates packaging to kill bacteria on medical tools; long half-life ensures the machine stays effective.
Thickness Control	Beta	Long	Alpha is too weak (blocked by paper); Gamma is too strong. Beta absorption changes based on thickness; long half-life ensures changes in count-rate are due to metal thickness, not source decay.
Cancer Treatment	Gamma	Short/Long	Focused beams kill tumors. If injected, a short half-life is needed to limit the dose to the patient.
Medical Diagnosis	Gamma	Very Short (hours)	Must be penetrating to leave the body; short half-life ensures the patient isn't radioactive for long.

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Key Equations

1. Number of half-lives ($n$): $n = \frac{\text{Total Time}}{\text{Half-life duration}}$
2. Remaining Activity ($A$ ): $A = A_0 \times (\frac{1}{2})^n$ (where $A_0$ is initial activity)
3. Corrected Activity: $A_{corrected} = A_{measured} - A_{background}$

• Units: Activity = Becquerels (Bq); Time = Seconds, minutes, hours, or years.

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Common Mistakes to Avoid

• ❌ Wrong: Assuming a substance decays at a constant linear rate (e.g., if 100g decays to 50g in 10 mins, it will all be gone in 20 mins).
• ✓ Right: Decay is exponential. After 20 minutes, 25g would remain (half of 50g).
• ❌ Wrong: Calculating half-life by simply dividing the total time by the starting activity.
• ✓ Right: Use the total time to find how many "halving cycles" occurred.
• ❌ Wrong: Forgetting to subtract background radiation before halving the numbers in extended questions.
• ✓ Right: Subtract background radiation from all readings first, then perform the half-life calculation.

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Exam Tips

1. Read the Graph Axes: Ensure you are looking at "Activity" or "Count-rate" and check the units for time (hours vs. days).
2. The "Half-way" Check: If a question asks for the activity after a certain time, always draw a quick arrow diagram ($100 \rightarrow 50 \rightarrow 25$) to keep track of how many half-lives have passed.
3. Application Logic: When asked why a specific isotope is used, always mention two things: its penetrating power (can it get through the material?) and its half-life (safety vs. convenience).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A sample of radioactive iodine-131 is used in a hospital to treat thyroid cancer. Iodine-131 has a half-life of 8.0 days.

(a) Define the term half-life. [2]

(b) A sample of iodine-131 has an initial activity of 800 Bq. Calculate the activity of the sample after 24 days. [3]

Worked Solution:

(a)

1. The time taken for half of the radioactive nuclei in a sample to decay. [Definition of half-life]

How to earn full marks:

• Correctly state "time taken", "half", and "decay" or "disintegrate" in the definition.
• Mention "nuclei" or "atoms" or "original sample" in the definition.

(b)

1. Calculate the number of half-lives in 24 days. $24 \text{ days} / 8.0 \text{ days/half-life} = 3 \text{ half-lives}$ [Determines the number of half-lives]
2. Calculate the activity after 3 half-lives. $800 \text{ Bq} / 2 / 2 / 2 = 100 \text{ Bq}$ [Reduces the activity by a factor of 2 for each half-life]

How to earn full marks:

• Correctly calculate the number of half-lives (3).
• Divide the initial activity by $2^3$ or 8.
• State the final answer with the correct unit. $\boxed{100 \text{ Bq}}$

Common Pitfall: A common mistake is to assume that the activity decreases linearly with time. Remember that radioactive decay is an exponential process, so the activity halves after each half-life, not by a constant amount. Also, be sure to include the correct unit (Bq) in your final answer.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

Americium-241 is a radioactive isotope used in smoke detectors. It emits alpha particles.

(a) State two properties of alpha particles. [2]

(b) Explain why a radioactive isotope that emits alpha particles is suitable for use in a smoke detector. [2]

(c) The half-life of Americium-241 is 432 years. Suggest why it is suitable for use in a smoke detector that is expected to last for 10 years. [2]

Worked Solution:

(a)

1. Alpha particles are helium nuclei. [States the composition of alpha particles]
2. Alpha particles have a positive charge (+2). [States the charge of alpha particles]

How to earn full marks:

• Any two of: helium nucleus, positive charge, strongly ionising, short range in air, stopped by paper, etc.

(b)

1. Alpha particles ionise the air inside the detector. [Explains the ionization process]
2. Smoke particles absorb the alpha particles, reducing the ionisation and triggering the alarm. [Explains how smoke affects the alpha particles and activates the alarm]

How to earn full marks:

• Mention that alpha particles ionise the air.
• Mention that smoke particles absorb/block the alpha particles, reducing ionisation and triggering the alarm.

(c)

1. After 10 years, a negligible amount of the Americium-241 will have decayed. [States the decay rate is very slow]
2. The activity of the Americium-241 will remain approximately constant, ensuring the detector functions correctly for its expected lifespan. [Explains why the slow decay rate is beneficial]

How to earn full marks:

• State that a negligible amount decays in 10 years.
• Explain that the activity remains nearly constant, so the detector works effectively.

Common Pitfall: Students sometimes struggle to explain the connection between a long half-life and the suitability for smoke detectors. The key is to realize that a very long half-life means the activity remains nearly constant over the device's lifespan, ensuring reliable operation. Don't just say "it lasts a long time"; explain why that's important.

Exam-Style Question 3 — Extended Response [7 marks]

Question:

A scientist is investigating the use of a radioactive isotope to measure the thickness of aluminium sheets produced in a factory. The isotope emits beta particles.

(a) Describe how the scientist could use the radioactive isotope and a detector to measure the thickness of the aluminium sheets. Include a diagram in your answer. [4]

(b) Explain why an isotope that emits alpha particles would not be suitable for this application. [2]

(c) The scientist measures the count rate detected through an aluminium sheet of a certain thickness. The count rate is 640 counts per minute (cpm). Without the aluminium sheet, the count rate is 5120 cpm. The half-thickness of aluminium for these beta particles is 2.0 mm (the thickness of aluminium that reduces the count rate by half). Calculate the thickness of the aluminium sheet. [1]

Worked Solution:

(a)

1. 📊A source of beta radiation is placed on one side of an aluminium sheet. A detector is placed on the opposite side. The aluminium sheet is between the source and the detector. Label the source, sheet, and detector.
   [Diagram shows the setup]
2. The radioactive source emits beta particles. [States the radiation type]
3. The aluminium sheet absorbs some of the beta particles. [Explains the absorption process]
4. The detector measures the count rate of the beta particles that pass through the sheet. A thicker sheet will absorb more beta particles, resulting in a lower count rate. [Explains how the count rate relates to thickness]

How to earn full marks:

• Diagram must show the source, sheet, and detector in the correct arrangement, and be labelled.
• State that beta particles are emitted.
• State that the aluminium sheet absorbs beta particles.
• Explain that a lower count rate indicates a thicker sheet.

(b)

1. Alpha particles have a very short range in air and are easily absorbed by even thin materials. [States the limited penetration of alpha particles]
2. Alpha particles would be completely absorbed by any aluminium sheet, regardless of its thickness. Therefore, the detector would always read zero, and it would be impossible to measure the thickness. [Explains why alpha particles are unsuitable]

How to earn full marks:

• State that alpha particles have a short range and are easily absorbed.
• Explain that all alpha particles would be absorbed, making thickness measurement impossible.

(c)

1. Determine the number of half-thicknesses required to reduce the count rate from 5120 cpm to 640 cpm. $5120 \text{ cpm} / 2 = 2560 \text{ cpm}$ $2560 \text{ cpm} / 2 = 1280 \text{ cpm}$ $1280 \text{ cpm} / 2 = 640 \text{ cpm}$ This is 3 half-thicknesses.
2. Calculate the total thickness. $3 \times 2.0 \text{ mm} = 6.0 \text{ mm}$

How to earn full marks:

• Correctly determine the number of half-thicknesses (3).
• Multiply the number of half-thicknesses by the half-thickness value (2.0 mm).
• State the final answer with the correct unit. $\boxed{6.0 \text{ mm}}$

Common Pitfall: Many students forget to include a labeled diagram in part (a), even when explicitly asked. Also, in part (c), make sure you understand the concept of "half-thickness" and how it relates to the number of half-lives. Don't just divide the initial count rate by the half-thickness value.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A medical technician is preparing a sample of technetium-99m (Tc-99m) for use in a medical scan. Tc-99m is a radioactive isotope that emits gamma radiation.

(a) State two properties of gamma radiation. [2]

(b) Explain why gamma radiation, rather than alpha or beta radiation, is used for medical scans. [3]

(c) Tc-99m has a half-life of 6.0 hours. The technician needs to prepare a sample with an activity of 40 MBq (megabecquerels) at 08:00. Calculate the activity of the Tc-99m sample that the technician needs to prepare at 02:00 (6 hours earlier). [4]

Worked Solution:

(a)

1. Gamma radiation is an electromagnetic wave. [States the nature of gamma radiation]
2. Gamma radiation has no charge. [States the charge of gamma radiation]

How to earn full marks:

• Any two of: electromagnetic wave, high energy, no mass, no charge, highly penetrating, travels at the speed of light.

(b)

1. Alpha radiation is strongly ionising and has a short range, so it would be absorbed by the body before reaching the detector. [Explains why alpha radiation is unsuitable]
2. Beta radiation is also strongly ionising and has a limited range, so it would also be absorbed by the body. [Explains why beta radiation is unsuitable]
3. Gamma radiation is weakly ionising and highly penetrating, so it can pass through the body and be detected externally, allowing internal organs to be imaged. [Explains why gamma radiation is suitable]

How to earn full marks:

• Explain that alpha radiation is absorbed due to its short range.
• Explain that beta radiation is absorbed due to its limited range.
• Explain that gamma radiation is penetrating and can be detected externally.

(c)

1. Calculate the number of half-lives between 02:00 and 08:00. $8:00 - 2:00 = 6 \text{ hours}$ $6 \text{ hours} / 6.0 \text{ hours/half-life} = 1 \text{ half-life}$ [Determines the number of half-lives]
2. Calculate the required activity at 02:00. $40 \text{ MBq} \times 2 = 80 \text{ MBq}$ [Multiplies the activity by 2 to account for one half-life]

How to earn full marks:

• Correctly calculate the number of half-lives (1).
• Multiply the activity at 08:00 by 2.
• State the final answer with the correct unit. $\boxed{80 \text{ MBq}}$

Common Pitfall: A common mistake in part (c) is to divide the activity by 2 instead of multiplying, because students forget that you're calculating the initial activity needed to reach a certain activity later. Also, remember that MBq stands for megabecquerels, so the unit is already correct and doesn't need any further conversion.