Specification: AQA A-level Physics (7408) 3.8.1.3 | Paper: 2 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Activity: the activity A of a radioactive source is the number of nuclei that decay per unit time. It is measured in becquerels (Bq), where 1 Bq = 1 decay per second.
Decay constant: the decay constant λ is the probability that an individual nucleus will decay per unit time. It is measured in s⁻¹ (or year⁻¹, day⁻¹, etc.). Radioactive decay is random (we cannot predict which nucleus will decay next) and spontaneous (it is not affected by external conditions such as temperature or pressure).
Rate of decay: the rate of decay is proportional to the number of undecayed nuclei, ΔN/Δt = −λN. The minus sign shows that N decreases with time.
The equation
activity = decay constant × number of undecayed nuclei A = λN and rate of change of N ΔN/Δt = −λN
| Quantity | Symbol | Unit |
|---|---|---|
| activity | A | becquerels (Bq), equivalent to s⁻¹ |
| decay constant | λ | per second (s⁻¹) |
| number of undecayed nuclei | N | no unit (a number) |
| change in number of undecayed nuclei | ΔN | no unit (a number) |
| time interval | Δt | seconds (s) |
| mass of sample | m | kilograms (kg) |
| atomic mass unit | u | 1 u = 1.661 × 10−27 kg |
| Avogadro constant | NA | 6.02 × 1023 mol⁻¹ |
Rearranged: λ = A ÷ N | N = A ÷ λ | number of nuclei N = m ÷ (mass number × u), or N = (mass in g ÷ molar mass in g mol⁻¹) × NA | For A in Bq, λ must be in s⁻¹. Convert year⁻¹ to s⁻¹ by dividing by the number of seconds in a year (3.15 × 107 s).
Use FIFA for every answer
- 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.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A radioactive sample contains 6.0 × 1020 undecayed nuclei. The decay constant of the isotope is 5.0 × 10−12 s⁻¹. Calculate the activity of the sample.
Show solution
F: A = λN
I: A = 5.0 × 10−12 × 6.0 × 1020
F: A is already the subject
A: 3.0 × 109 Bq
Q2 (Core) A source has an activity of 4.5 × 105 Bq and contains 2.0 × 1015 undecayed nuclei. Calculate the decay constant, to 2 significant figures.
Show solution
F: A = λN
I: 4.5 × 105 = λ × 2.0 × 1015
F: λ = 4.5 × 105 ÷ 2.0 × 1015
A: 2.25 × 10−10 s⁻¹ = 2.3 × 10−10 s⁻¹ (2 s.f.)
Q3 (Core) The source in a smoke alarm is americium-241, which has a decay constant of 1.60 × 10−3 year⁻¹. The source contains 7.3 × 1014 undecayed nuclei. Calculate its activity in Bq, to 2 significant figures. (1 year = 3.15 × 107 s)
Show solution
Convert first: λ = 1.60 × 10−3 ÷ 3.15 × 107 = 5.079… × 10−11 s⁻¹
F: A = λN
I: A = 5.079… × 10−11 × 7.3 × 1014
F: A is already the subject
A: 37 079… Bq = 3.7 × 104 Bq (2 s.f.)
Watch out: using λ in year⁻¹ gives decays per year, not becquerels. A becquerel is one decay per second.
Q4 (Core) A hospital sample contains 1.0 μg of iodine-131. The decay constant of iodine-131 is 1.00 × 10−6 s⁻¹. Calculate the activity of the sample, to 2 significant figures. (mass of one iodine-131 nucleus = 131 u; 1 u = 1.661 × 10−27 kg)
Show solution
Convert first: 1.0 μg = 1.0 × 10−9 kg
Step 1 – number of nuclei
mass of one nucleus = 131 × 1.661 × 10−27 = 2.175… × 10−25 kg
N = 1.0 × 10−9 ÷ 2.175… × 10−25 = 4.595… × 1015
Step 2 – activity
F: A = λN
I: A = 1.00 × 10−6 × 4.595… × 1015
A: 4.595… × 109 Bq = 4.6 × 109 Bq (2 s.f.)
Watch out: 1 μg is 10−9 kg, not 10−6 kg. 10−6 is the conversion to grams.
Q5 (Exam) A 5.0 g sample of plutonium-239 has an activity of 1.15 × 1010 Bq. The molar mass of plutonium-239 is 239 g mol⁻¹. Calculate the decay constant in s⁻¹ and in year⁻¹, both to 2 significant figures. (NA = 6.02 × 1023 mol⁻¹; 1 year = 3.15 × 107 s)
Show solution
Step 1 – number of nuclei
moles = 5.0 ÷ 239 = 0.02092… mol
N = 0.02092… × 6.02 × 1023 = 1.259… × 1022
Step 2 – decay constant
F: A = λN
I: 1.15 × 1010 = λ × 1.259… × 1022
F: λ = 1.15 × 1010 ÷ 1.259… × 1022
A: 9.131… × 10−13 s⁻¹ = 9.1 × 10−13 s⁻¹ (2 s.f.)
Step 3 – convert to year⁻¹
λ = 9.131… × 10−13 × 3.15 × 107 = 2.876… × 10−5 year⁻¹ = 2.9 × 10−5 year⁻¹ (2 s.f.)
Watch out: to go from s⁻¹ to year⁻¹ you multiply (a nucleus has more chance of decaying in a year than in a second).
Q6 (Exam) A sample contains 3.0 × 1016 undecayed nuclei of an isotope with decay constant 1.8 × 10−9 s⁻¹. (a) Calculate the rate of change of N. (b) Estimate the number of nuclei that decay in 1.0 minute, to 2 significant figures, and explain why it is reasonable to treat N as constant over this time.
Show solution
Step 1 – (a) rate of decay
F: ΔN/Δt = −λN
I: ΔN/Δt = −1.8 × 10−9 × 3.0 × 1016
A: −5.4 × 107 s⁻¹ (N falls by 5.4 × 107 nuclei every second)
Step 2 – (b) decays in 1.0 minute
ΔN = −5.4 × 107 × 60 = −3.24 × 109, so 3.2 × 109 nuclei decay (2 s.f.)
this is only 3.24 × 109 ÷ 3.0 × 1016 = 1.08 × 10−7 of the nuclei, so N (and the activity) hardly changes during the minute
Watch out: ΔN/Δt = −λN only works directly when Δt is very short compared with the half-life. Over longer times use N = N0e−λt.
Q7 (Exam) A smoke alarm contains americium-241 with an activity of 3.7 × 104 Bq. The decay constant of americium-241 is 5.09 × 10−11 s⁻¹. Calculate the mass of americium-241 in the alarm, in kg, to 2 significant figures. (mass of one americium-241 nucleus = 241 u; 1 u = 1.661 × 10−27 kg)
Show solution
Step 1 – number of nuclei
F: A = λN
I: 3.7 × 104 = 5.09 × 10−11 × N
F: N = 3.7 × 104 ÷ 5.09 × 10−11 = 7.269… × 1014
Step 2 – mass
mass of one nucleus = 241 × 1.661 × 10−27 = 4.003… × 10−25 kg
m = 7.269… × 1014 × 4.003… × 10−25 = 2.909… × 10−10 kg
A: 2.9 × 10−10 kg (2 s.f.), about 0.3 μg
Q8 (Exam) Living material contains carbon-14 nuclei: 1.3 × 10−12 of all carbon atoms are carbon-14. The decay constant of carbon-14 is 1.21 × 10−4 year⁻¹. Calculate the activity due to carbon-14 in 1.0 g of carbon from a living tree, to 2 significant figures. (Take the molar mass of carbon as 12 g mol⁻¹; NA = 6.02 × 1023 mol⁻¹; 1 year = 3.15 × 107 s)
Show solution
Convert first: λ = 1.21 × 10−4 ÷ 3.15 × 107 = 3.841… × 10−12 s⁻¹
Step 1 – number of carbon-14 nuclei
carbon atoms in 1.0 g = (1.0 ÷ 12) × 6.02 × 1023 = 5.016… × 1022
N = 5.016… × 1022 × 1.3 × 10−12 = 6.521… × 1010
Step 2 – activity
F: A = λN
I: A = 3.841… × 10−12 × 6.521… × 1010
A: 0.2505… Bq = 0.25 Bq (2 s.f.)
Q9 (Stretch) A hospital prepares a dose of technetium-99m with an activity of 4.0 × 108 Bq at the moment of injection. The decay constant of technetium-99m is 3.2 × 10−5 s⁻¹. (a) Calculate the mass of technetium-99m in the dose, to 2 significant figures. (b) Estimate the number of nuclei that decay in the first minute after injection, and show that the activity changes by less than 0.2% in this time. (mass of one technetium-99m nucleus = 99 u; 1 u = 1.661 × 10−27 kg)
Show solution
Step 1 – (a) number of nuclei
F: A = λN
I: 4.0 × 108 = 3.2 × 10−5 × N
F: N = 4.0 × 108 ÷ 3.2 × 10−5 = 1.25 × 1013
Step 2 – (a) mass
mass of one nucleus = 99 × 1.661 × 10−27 = 1.644… × 10−25 kg
m = 1.25 × 1013 × 1.644… × 10−25 = 2.055… × 10−12 kg
A: 2.1 × 10−12 kg (2 s.f.), about 2 nanograms
Step 3 – (b) decays in the first minute
ΔN ≈ AΔt = 4.0 × 108 × 60 = 2.4 × 1010 nuclei
fraction decayed = 2.4 × 1010 ÷ 1.25 × 1013 = 1.92 × 10−3 = 0.192%
A ∝ N, so the activity also falls by only about 0.19%, which is less than 0.2%
Watch out: a tiny mass can have a huge activity. The activity depends on λ and N, not on the mass alone.
Q10 (Stretch) Strontium-90 has a decay constant of 0.0241 year⁻¹ and caesium-137 has a decay constant of 0.0230 year⁻¹. (a) Calculate the activity of 1.0 g of strontium-90 in Bq, to 2 significant figures. (b) Show that 1.0 g of strontium-90 has an activity about 1.6 times that of 1.0 g of caesium-137. (mass of one strontium-90 nucleus = 90 u, mass of one caesium-137 nucleus = 137 u; 1 u = 1.661 × 10−27 kg; 1 year = 3.15 × 107 s)
Show solution
Convert first: 1.0 g = 1.0 × 10−3 kg; λ = 0.0241 ÷ 3.15 × 107 = 7.650… × 10−10 s⁻¹
Step 1 – (a) number of strontium-90 nuclei
mass of one nucleus = 90 × 1.661 × 10−27 = 1.4949 × 10−25 kg
N = 1.0 × 10−3 ÷ 1.4949 × 10−25 = 6.689… × 1021
Step 2 – (a) activity
F: A = λN
I: A = 7.650… × 10−10 × 6.689… × 1021
A: 5.117… × 1012 Bq = 5.1 × 1012 Bq (2 s.f.)
Step 3 – (b) ratio
for equal masses N ∝ 1 ÷ mass number, so A ∝ λ ÷ mass number
ratio = (0.0241 ÷ 90) ÷ (0.0230 ÷ 137) = 0.0241 × 137 ÷ (0.0230 × 90) = 1.595…
A: ratio ≈ 1.6 (2 s.f.); the unit conversions cancel, so year⁻¹ can be used for a ratio
Watch out: the two decay constants are almost equal, but strontium-90 has more nuclei per gram because each nucleus is lighter. Both factors matter.
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