10 GCSE Physics questions about the smoke alarm, from Foundation to Higher, each with a full worked solution.
Many smoke alarms contain a tiny amount of radioactive americium-241. It emits alpha particles that ionise the air, so a small current flows between two plates. When smoke gets in, it absorbs the alpha particles, the current drops – and the alarm sounds.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Atomic structure · Electricity · Waves
📌 KEY DEFINITIONS – learn them word for word
Alpha particle: two protons and two neutrons – the same as a helium nucleus. It is highly ionising but has a short range in air (a few centimetres).
Half-life: the time it takes for the number of nuclei of a radioactive isotope in a sample to halve, or the time for the count rate (or activity) to fall to half.
Activity: the rate at which a source of unstable nuclei decays, measured in becquerels (Bq).
Equations you’ll use
| Quantity | Symbol | Unit |
|---|---|---|
| mass number | A | — |
| atomic number | Z | — |
| activity | — | becquerels (Bq) |
| charge flow | Q | coulombs (C) |
| current | I | amps (A) |
| time | t | seconds (s) |
| energy transferred | E | joules (J) |
| potential difference | V | volts (V) |
| wave speed | v | metres per second (m/s) |
| frequency | f | hertz (Hz) |
| wavelength | λ | metres (m) |
Use FIFA for every calculation
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Foundation: aim for Q1–8. Higher: aim for Q4–10.
Questions
Q1 (F) A nucleus of americium-241 contains 95 protons. Calculate the number of neutrons in the nucleus.
Show solution
F: neutrons = mass number − atomic number
I: neutrons = 241 − 95
A: 146 neutrons
Q2 (F) Explain why an alpha source is used in a smoke alarm rather than a beta or gamma source. [3 marks]
Show answer
• Alpha particles are strongly ionising, so they ionise the air well and a current can flow.
• They are easily absorbed by smoke particles, so the current drops when smoke is present.
• They have a short range and are stopped by the plastic case, so they do not reach people in the room.
Q3 (F) Americium-241 has a half-life of 432 years. What fraction of the americium will be left after 864 years?
Show solution
Step 1 – number of half-lives: 864 ÷ 432 = 2
Step 2: 1 → ½ → ¼
A: ¼ is left
🛒 All the equations in one place: the Every Equation PowerPoint has every AQA GCSE Physics equation with worked examples and 102 practice questions, for £8.
Q4 (F/H) A new smoke alarm source has an activity of 36 000 Bq. Calculate how long it takes for the activity to fall to 4500 Bq. (half-life = 432 years)
Show solution
Step 1 – halve until you reach 4500 Bq: 36 000 → 18 000 → 9000 → 4500, which is 3 half-lives
Step 2: time = 3 × 432
A: 1296 years
Q5 (F/H) Americium-241 (mass number 241, atomic number 95) decays by alpha emission into neptunium (Np). Write down the mass number and atomic number of the neptunium nucleus.
Show answer
• An alpha particle has mass number 4 and atomic number 2.
• Mass number: 241 − 4 = 237
• Atomic number: 95 − 2 = 93
So the equation is: 24195Am → 23793Np + 42He
Q6 (F/H) Smoke alarms are usually replaced after 10 years. Explain why this is not because the radioactive source has run out. [2 marks]
Show answer
• The half-life of americium-241 (432 years) is much longer than 10 years.
• So after 10 years the activity has hardly changed; alarms are replaced because the electronics and sensor wear out.
Q7 (F/H) When the alarm sounds, a current of 0.030 A flows from the battery for 15 s. Calculate the charge that flows.
Show solution
F: Q = I t
I: Q = 0.030 × 15
F: Q is already the subject
A: 0.45 C
Q8 (F/H) The battery has a potential difference of 9.0 V. Calculate the energy transferred when 0.45 C of charge flows.
Show solution
F: E = Q V
I: E = 0.45 × 9.0
F: E is already the subject
A: 4.05 J (4.1 J)
Q9 (H) The alarm beeps at a frequency of 3200 Hz. Sound travels at 340 m/s in air. Calculate the wavelength of the sound.
Show solution
F: v = f λ
I: 340 = 3200 × λ
F: λ = 340 ÷ 3200
A: 0.11 m (0.106 m)
Q10 (H · HT) The activity of the source falls from 36 000 Bq to 4500 Bq over 3 half-lives. Calculate the net decline in activity and express it as a ratio of the starting activity.
Show solution
Step 1 – net decline: 36 000 − 4500 = 31 500 Bq
Step 2 – ratio: 31 500 : 36 000
Step 3 – simplify (divide both by 4500): 7 : 8
A: net decline = 31 500 Bq, a ratio of 7 : 8 (after 3 half-lives, ⅞ of the activity has gone)
⚠️ Where students lose marks
- Adding the protons to the mass number – the mass number already includes the protons.
- Saying alpha is used because it is “the most dangerous” – the key points are that it is highly ionising and has a short range.
- Counting the number of halvings wrongly – write out each step (36 000 → 18 000 → …).
- In nuclear equations, forgetting that alpha emission lowers the mass number by 4 and the atomic number by 2.
More devices: all 40 everyday devices · Atomic structure · Electricity
🧮 Practise every equation, on paper or on screen
Equations Workbook (£12): printable practice for all the AQA GCSE Physics equations, with 237 questions, 109 worked examples and full answers. Includes a Core booklet with the Higher-only equations removed.
Every Equation PowerPoint (£8): every equation with worked examples, 102 questions and 15 multi-step problems. Good for revising on screen or teaching from the board.