Radioactive Decay Simulation Experiment.Docx

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Radioactive Decay Simulation Experiment quantitative data Test 1 Dice: Roll Number | Number of Dice left | 0 | 100 | 1 | 77 | 2 | 60 | 3 | 53 | 4 | 43 | 5 | 35 | 6 | 29 | 7 | 26 | 8 | 20 | 9 | 14 | 10 | 11 | 11 | 6 | Test 2 Dice: Roll Number | Number of Dice | 0 | 100 | 1 | 84 | 2 | 68 | 3 | 58 | 4 | 51 | 5 | 46 | 6 | 38 | 7 | 33 | 8 | 28 | 9 | 26 | 10 | 20 | 11 | 19 | 12 | 16 | 13 | 14 | 14 | 8 | Test 1 Coins: Roll Number | Number of coins | 0 | 100 | 1 | 49 | 2 | 23 | 3 | 13 | 4 | 6 | 5 | 4 | 6 | 3 | 7 | 2 | 8 | 1 | 9 | 0 | Test 2 Coins: Roll Number | Number of coins | 0 | 100 | 1 | 43 | 2 | 25 | 3 | 14 | 4 | 8 | 5 | 3 | In the graphs above show the roll number as the time period and the number of coins left as the particles remaining, therefore simulating radioactive decay on 4 instances. Qualitative data As this is only a simulation there are limits as to how accurate the results can be. The results are not consistent as the probably of decay in dice is 1 in 6 where as with the coins the probability of decay is 1 in 2. The box shaken in this experiment was not of a consistent amount of times or force due to different people shaking it. Some of the dice coins landed on top of one another and therefore had to be moved to show all the dice, this would reduce the randomness of shaking the box. The error bars in this experiment are too unpredictable and large to be included. This is because the process of radioactive decay is random and spontaneous. Theory Using the radioactive equation: N=Nₒe-λt Where: * N= number of nuclei * No = initial number of nuclei * λ = decay constant * t = time We can rearrange this to: InN=ln⁡(Nₒe-λt) InN=lnNₒ+lne-λt InN=lnNₒ-λt Therefore in the form y=mx+c The decay constant becomes the slope

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