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The radioactivity of an isotope of radon was measured each day for a week and the following data were recorded - Leaving Cert Physics - Question b - 2018

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The radioactivity of an isotope of radon was measured each day for a week and the following data were recorded. Time (days) 0 1 2 3 4 5 6 7 Ac... show full transcript

Worked Solution & Example Answer:The radioactivity of an isotope of radon was measured each day for a week and the following data were recorded - Leaving Cert Physics - Question b - 2018

Step 1

What is meant by radioactivity?

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Answer

Radioactivity refers to the spontaneous disintegration of a nucleus, a process through which unstable atomic nuclei lose energy by emitting radiation. This emission can take the form of particles (like alpha or beta particles) or electromagnetic waves (such as gamma rays). It is a natural process that occurs in certain isotopes, leading to the transformation of one element into another over time.

Step 2

On graph paper, draw a decay curve (a graph of activity against time).

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  1. Plot the given activity data against time on graph paper.

    • X-axis: Time (days)
    • Y-axis: Activity (MBq)
  2. Mark the points: (0, 600), (1, 490), (2, 400), (3, 330), (4, 270), (5, 220), (6, 180), (7, 150).

  3. Draw a smooth decay curve that connects these points, showing the exponential decrease in activity over time.

Step 3

Use the decay curve to determine the half-life of the isotope.

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From the graph, the half-life is observed by identifying the time taken for the activity to drop from one-half of its initial value (300 MBq). Based on the plotted curve, it can be estimated that the half-life is approximately 3.3 days.

Step 4

Calculate the number of nuclei in the sample at the beginning of the investigation.

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Using the decay constant λˊ\'\lambda = \frac{\ln 2}{T_{1/2}}wherewhereT_{1/2} = 3.3$ days:

  1. Calculate λˊ\'\lambda: λ=ln23.30.21 days1\lambda = \frac{\ln 2}{3.3} \approx 0.21 \text{ days}^{-1}

  2. Use the initial activity: A=λNA = \lambda N Where A=600A = 600 MBq, therefore: 600=0.21NN=6000.212857.14600 = 0.21 N \\ N = \frac{600}{0.21} \approx 2857.14

Thus, the number of nuclei in the sample at the beginning of the investigation is approximately 2.86×1032.86 \times 10^3.

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