Specification: AQA A-level Physics (7408) 3.10.6.2 Medical physics option | Paper: 3 Section B | Data and formulae booklet: 1/TE = 1/TB + 1/TP is given in the exam; N = N0e−λt, A = λN and T½ = ln 2/λ are in the radioactivity section
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📌 KEY DEFINITIONS – learn them word for word
Physical half-life (TP): the time taken for half the nuclei of a radioactive isotope to decay.
Biological half-life (TB): the time taken for half of a substance to be removed from the body by biological processes such as excretion.
Effective half-life (TE): the time taken for the activity in the body to halve because of both radioactive decay and biological removal. 1/TE = 1/TB + 1/TP, so TE is always shorter than both.
The equation
1 ÷ effective half-life = 1 ÷ biological half-life + 1 ÷ physical half-life 1/TE = 1/TB + 1/TP
| Quantity | Symbol | Unit |
|---|---|---|
| effective half-life | TE | seconds (s), or any time unit used consistently |
| biological half-life | TB | seconds (s), or any time unit used consistently |
| physical half-life | TP | seconds (s), or any time unit used consistently |
Rearranged: TE = TBTP ÷ (TB + TP) | 1/TB = 1/TE − 1/TP | activity in the body after time t: A = A0(½)t/TE | All three half-lives must be in the same unit. Remember to take the reciprocal at the end.
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) Technetium-99m has a physical half-life of 6.0 h. In a patient, its biological half-life is 24 h. Calculate the effective half-life.
Show solution
F: 1/TE = 1/TB + 1/TP
I: 1/TE = 1/24 + 1/6.0 = 0.04166… + 0.1666… = 0.2083… h⁻¹
F: TE = 1 ÷ 0.2083…
A: 4.8 h
Q2 (Core) Iodine-131 has a physical half-life of 8.0 days. In the thyroid its biological half-life is 80 days. Calculate the effective half-life, to 2 significant figures.
Show solution
F: 1/TE = 1/TB + 1/TP
I: 1/TE = 1/80 + 1/8.0 = 0.0125 + 0.125 = 0.1375 day⁻¹
F: TE = 1 ÷ 0.1375 = 7.27… days
A: 7.3 days (2 s.f.)
When TB is much longer than TP, TE is only a little shorter than TP.
Q3 (Core) A tracer containing technetium-99m (TP = 6.0 h) has a biological half-life of 1.5 days. Calculate the effective half-life in hours and in seconds, each to 2 significant figures.
Show solution
Convert first: 1.5 days = 1.5 × 24 = 36 h
F: 1/TE = 1/TB + 1/TP
I: 1/TE = 1/36 + 1/6.0
F: TE = (36 × 6.0) ÷ (36 + 6.0) = 216 ÷ 42 = 5.142… h
A: 5.1 h (2 s.f.); 5.142… × 3600 = 18 514… s = 1.9 × 104 s (2 s.f.)
Watch out: mixing days and hours. Convert to a single unit before adding reciprocals.
Q4 (Core) A technetium-99m tracer (TP = 6.0 h) has an effective half-life of 4.0 h in a patient. Calculate its biological half-life.
Show solution
F: 1/TE = 1/TB + 1/TP
I: 1/4.0 = 1/TB + 1/6.0
F: 1/TB = 0.25 − 0.1666… = 0.08333… h⁻¹
A: TB = 1 ÷ 0.08333… = 12 h
Watch out: subtracting the half-lives (6.0 − 4.0 = 2.0 h). You must subtract the reciprocals.
Q5 (Exam) A patient is injected with technetium-99m of activity 400 MBq. TP = 6.0 h and TB = 24 h. Calculate the activity in the patient’s body 12 h later, to 2 significant figures.
Show solution
Step 1 – effective half-life
F: 1/TE = 1/TB + 1/TP
I: 1/TE = 1/24 + 1/6.0, so TE = 4.8 h
Step 2 – activity remaining
number of effective half-lives = 12 ÷ 4.8 = 2.5
A = A0(½)t/TE = 400 × (½)2.5 = 70.71… MBq
A: 71 MBq (2 s.f.)
Watch out: using the physical half-life gives 400 × (½)² = 100 MBq. Activity in the body falls with the effective half-life.
Q6 (Exam) For the injection in Q5 (400 MBq of technetium-99m, TP = 6.0 h, TE = 4.8 h): (a) calculate the effective decay constant in s⁻¹, to 2 significant figures; (b) calculate the number of technetium-99m nuclei injected, to 2 significant figures.
Show solution
Convert first: TE = 4.8 × 3600 = 17 280 s; TP = 6.0 × 3600 = 21 600 s
Step 1 – (a) effective decay constant
λE = ln 2 ÷ TE = ln 2 ÷ 17 280 = 4.01… × 10−5 s⁻¹
A: 4.0 × 10−5 s⁻¹ (2 s.f.)
Step 2 – (b) number of nuclei
activity is the rate of radioactive decay, so it uses the physical decay constant: λP = ln 2 ÷ 21 600 = 3.20… × 10−5 s⁻¹
F: A = λPN
I: 4.00 × 108 = 3.20… × 10−5 × N
F: N = 4.00 × 108 ÷ 3.20… × 10−5 = 1.24… × 1013
A: 1.2 × 1013 nuclei (2 s.f.)
Watch out: using λE in A = λN. Biological removal takes nuclei out of the body but does not make them decay.
Q7 (Exam) Technetium-99m in a patient has an effective half-life of 4.8 h. Calculate how long it takes for the activity in the body to fall to 10% of its initial value, to 2 significant figures.
Show solution
F: A = A0(½)t/TE
I: 0.10 = (½)t/4.8
F: take logs: t/4.8 = ln 10 ÷ ln 2 = 3.321…, so t = 4.8 × 3.321… = 15.94… h
A: 16 h (2 s.f.)
Q8 (Exam) A patient is injected with technetium-99m (TP = 6.0 h). Measurements show that the activity in the body falls from 640 MBq to 80 MBq in 13.5 h. Calculate the biological half-life, to 2 significant figures.
Show solution
Step 1 – effective half-life from the data
640 → 320 → 160 → 80 MBq is 3 halvings, so 3TE = 13.5 h and TE = 4.5 h
Step 2 – biological half-life
F: 1/TE = 1/TB + 1/TP
I: 1/4.5 = 1/TB + 1/6.0
F: 1/TB = 0.2222… − 0.1666… = 0.05555… h⁻¹
A: TB = 1 ÷ 0.05555… = 18 h (2 s.f.)
Q9 (Stretch) Iodine-123 (TP = 13 h) is used to image the thyroid. In a patient, TB = 24 h. (a) Calculate TE, to 2 significant figures. (b) The scan is done 6.0 h after injection and needs an activity of 15 MBq in the body. Calculate the activity that must be injected, to 2 significant figures. (c) Show that about 65% of the injected nuclei decay inside the body rather than being excreted.
Show solution
Step 1 – (a) effective half-life
TE = (24 × 13) ÷ (24 + 13) = 312 ÷ 37 = 8.432… h
A: 8.4 h (2 s.f.)
Step 2 – (b) activity to inject
A = A0(½)t/TE, so A0 = A × 2t/TE = 15 × 26.0/8.432… = 15 × 1.637… = 24.56… MBq
A: 25 MBq (2 s.f.)
Step 3 – (c) fraction decaying in the body
at any instant, nuclei decay at a rate λPN and are removed at a rate λBN, so the fraction that decays is λP ÷ (λP + λB) = λP/λE = TE/TP
I: TE/TP = 8.432… ÷ 13 = 0.648…
A: about 65% (2 s.f.), as required
Q10 (Stretch) A patient is injected with 400 MBq of technetium-99m (TP = 6.0 h, TB = 24 h). Each decay emits one 140 keV gamma photon. (a) Using the method of Q9(c), calculate the total number of decays that happen inside the patient’s body, to 2 significant figures. (b) Calculate the total gamma energy emitted inside the body, to 2 significant figures. (e = 1.60 × 10−19 C)
Show solution
Step 1 – effective half-life and nuclei injected
TE = 4.8 h (as in Q1); from Q6, N0 = A0/λP = 1.24… × 1013
Step 2 – (a) decays inside the body
fraction decaying inside = TE/TP = 4.8 ÷ 6.0 = 0.80
number of decays = 0.80 × 1.24… × 1013 = 9.97… × 1012 (this equals A0/λE)
A: 1.0 × 1013 decays (2 s.f.)
Step 3 – (b) energy emitted
energy per photon = 140 × 103 × 1.60 × 10−19 = 2.24 × 10−14 J
total energy = 9.97… × 1012 × 2.24 × 10−14 = 0.2233… J
A: 0.22 J (2 s.f.); most of these gamma photons leave the body, which is what lets the gamma camera detect them
Watch out: multiplying the initial activity by the effective half-life. Activity is not constant, so you cannot use A0 × time.
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