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Radioactive decay lab answers for A Level physics (free PDF download)

The method and the results are on the radioactive decay lab. Do the analysis yourself first, then check the answers below.

Quick answer: Radioactive decay is the random change of an unstable nucleus, and a sample decays exponentially with a fixed half-life. These answers go with the Physics 11-12 lab on radioactive decay: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.

Radioactive decay lab answers for A Level physics (free PDF download)

The model answers use the sample table. Your own counts will differ because the dice are random, but the half-life of the model should come out within about 10 % of 3.8 throws, and the graph of ln N against n should be close to a straight line sloping downwards.

Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB E.3-E.5

Question 1 Why results vary

Whether any one die shows a six is random, so the number removed changes from run to run. In a real sample each nucleus decays at random too, so the number decaying in a short time fluctuates around the average. The average behaviour is predictable even though each event is random.

Question 2 Expected values

Expected N = 120 × (5/6)10 = 120 × 0.1615 = 19.4. The measured 22 is 14 % above the expected value, which is acceptable for such small numbers.

Question 3 Half-life from the graph

Half of 120 is 60. N is 61 at n = 4 and 48 at n = 5, so by interpolation n = 4 + (61 − 60) / (61 − 48) = 4.08 throws.

Question 4 Straight-line plot

The values of ln N are: 4.79, 4.62, 4.50, 4.30, 4.11, 3.87, 3.66, 3.56, 3.50, 3.30, 3.09. A least-squares line gives a gradient of −0.17 per throw. The gradient equals −λ, where λ is the decay constant of the model, so λ = 0.17 per throw.

Question 5 Half-life from the gradient

T½ = ln 2 / λ = 0.693 / 0.17 = 4.07 throws. This agrees with the graph value of 4.08 throws to within a few percent.

Question 6 Prediction from probability

N = N₀ (5/6)n = N₀ e−λn, so λ = −ln(5/6) = ln(6/5) = 0.182 per throw. T½ = 0.693 / 0.182 = 3.80 throws. The measured value from the gradient, 4.07 throws, is 6.9 % different.

Question 7 Improving the reliability

Random counts fluctuate by about √N, so for N = 22 the fluctuation is about √22 = 4.7, which is 21 % of the count. The smaller N becomes, the bigger the fractional uncertainty. Use more dice (for example 500) or repeat the run several times and average N at each throw.

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