13 (a) State what is meant by radioactive decay. (b) The variation with time t of the number N of technetium-101 nuclei in a sample of radioactive material is shown in Fig. 13.1. [Figure 13.1] (i) Use Fig. 13.1 to determine the activity, in Bq, of the sample of technetium-101 at time t = 14.0 minutes. Show your working. activity = .............................................. Bq (ii) Without calculating the half-life of technetium-101, use your answer in (i) to determine the decay constant λ of technetium-101. λ = .............................................. s⁻¹
- (a)State what is meant by radioactive decay.[2]
- (b)(i)Use Fig. 13.1 to determine the activity, in Bq, of the sample of technetium-101 at time t = 14.0 minutes. Show your working. activity = .............................................. Bq[4]
- (b)(ii)Without calculating the half-life of technetium-101, use your answer in (i) to determine the decay constant λ of technetium-101. λ = .............................................. s⁻¹[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [B1] emission of particles/radiation by unstable nucleus
- [B1] spontaneous emission
- [B1] use of graph to determine half-life = 14 minutes
- [C1] hence λ = ln 2 / (14 × 60) (s⁻¹)
- [C1] N at 14 minutes = 4.4 × 10⁷ AND A = λN
- [A1] activity = 4.4 × 10⁷ × ln 2 / (14 × 60) = 3.6 × 10⁴ Bq
- OR
- [(B1)] correct tangent drawn at time t = 14 minutes
- [(C1)] magnitude of gradient of tangent identified as activity
- [(C1)] correct working for gradient leading to activity
- [(A1)] activity = 3.6 × 10⁴ Bq
- [C1] 3.6 × 10⁴ = λ × 4.4 × 10⁷ OR λ = ln 2 / (14.0 × 60)
- [A1] λ = 8.2 × 10⁻⁴ s⁻¹
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