- Level: International Baccalaureate
- Subject: Physics
- Word count: 927
The Half-life of dice Decay investigation
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Introduction
The Half-life of dice – Decay investigation
Cheung Yu On Andy 12L
Aim: To investigate the half-life of dice
Qualitative Data
- A total of 100 dice were put into a blue plastic container. The container was then shuffled in random speeds and directions to allow the dice to roll.
- After every shuffled, all the dice in the container with “6” facing up were removed from the container and the number was recorded. The remaining dice were shuffled again and dice with “6” facing up were removed again. The process repeats for 20 shuffles in total.
Data collection
Our group’s data
Time shuffled | No. of dice that show a “6” removed from the container |
0 | 0 |
1 | 16 |
2 | 10 |
3 | 10 |
4 | 10 |
5 | 8 |
6 | 7 |
7 | 10 |
8 | 8 |
9 | 4 |
10 | 2 |
11 | 2 |
12 | 3 |
13 | 0 |
14 | 1 |
15 | 2 |
16 | 2 |
17 | 2 |
18 | 0 |
19 | 2 |
20 | 0 |
Data processing
Time shuffled | No. of dice that show a “6” removed from the container | No. of dice remaining in the container | Percentage error | Numerical error |
0 | 0 | 100 | 0% | 0 |
1 | 16 | 84 | 10% | 8 |
2 | 10 | 74 | 10% | 7 |
3 | 10 | 64 | 9% | 6 |
4 | 10 | 54 | 9% | 5 |
5 | 8 | 46 | 8% | 4 |
6 | 7 | 39 | 8% | 3 |
7 | 10 | 29 | 7% | 2 |
8 | 8 | 21 | 7% | 1 |
9 | 4 | 17 | 6% | 1 |
10 | 2 | 15 | 6% | 1 |
11 | 2 | 13 | 5% | 1 |
12 | 3 | 10 | 5% | 1 |
13 | 0 | 10 | 4% | 0 |
14 | 1 | 9 | 4% | 0 |
15 | 2 | 7 | 3% | 0 |
16 | 2 | 5 | 3% | 0 |
17 | 2 | 3 | 2% | 0 |
18 | 0 | 3 | 2% | 0 |
19 | 2 | 1 | 1% | 0 |
20 | 0 | 1 | 1% | 0 |
Justification of errors
The percentage error is larger at first (10%) since there is more dice involved in the shuffling of the container, so there is a higher possibility that more dice are not rolled probably. The
Middle
3.5
18
0.03
3.5
19
0.01
4.6
20
0.01
4.6
Justification
N/N0 = Number of dice remaining / Initial total number of dice
For 1st shuffle: 84 / 100 = 0.84
-ln(N/N0) = -ln(0.84) = 0.17 (2s.f.)
Note: negative ln is used to produce a trend line with a positive slope
The following table shows the necessary values for the maximum and minimum lines of best fit.
Time shuffled | Maximum | Minimum | Percentage error |
0 | 0.00 | 0.00 | 0.0% |
1 | 0.27 | 0.08 | 9.6% |
2 | 0.40 | 0.21 | 9.5% |
3 | 0.54 | 0.36 | 9.4% |
4 | 0.71 | 0.53 | 9.3% |
5 | 0.87 | 0.69 | 8.7% |
6 | 1.0 | 0.87 | 7.7% |
7 | 1.3 | 1.2 | 6.9% |
8 | 1.6 | 1.5 | 4.8% |
9 | 1.8 | 1.7 | 5.9% |
10 | 2.0 | 1.8 | 6.7% |
11 | 2.1 | 2.0 | 7.7% |
12 | 2.4 | 2.2 | 10.0% |
13 | 2.3 | 2.3 | 0.0% |
14 | 2.4 | 2.4 | 0.0% |
15 | 2.7 | 2.7 | 0.0% |
16 | 3.0 | 3.0 | 0.0% |
17 | 3.5 | 3.5 | 0.0% |
18 | 3.5 | 3.5 | 0.0% |
19 | 4.6 | 4.6 | 0.0% |
20 | 4.6 | 4.6 | 0.0% |
Justification
For 1st shuffle:
Max line of best fit = -ln(N - numerical error /N0) = -ln[(84-8)/100] = 0.27 (2s.f.)
Max line of best fit = -ln(N + numerical error /N0) = -ln[(84+8)/100] = 0.08 (2s.f.)
Note: negative ln is used to produce a trend line with a positive slope
Percentage error = (Max – min) / 2 = (0.27 – 0.08) /2 = 9.6% (2s.f.)
From the equation of the ln line plotted, the value for the decay constant, λ is found to be 0.22.
The half-life, t1/2, of the decay model can be calculated from t1/2 = (ln2/λ), which is derived from N = N0 e –λt.
Data collection
Class data
Time shuffled | No. of dice that show a “6” removed from the container |
0 | 0 |
1 | 21 |
2 | 14 |
3 | 18 |
4 | 7 |
5 | 5 |
6 | 7 |
7 | 4 |
8 | 7 |
9 | 2 |
10 | 4 |
11 | 5 |
12 | 0 |
13 | 1 |
14 | 1 |
15 | 0 |
16 | 1 |
17 | 0 |
18 | 1 |
19 | 0 |
20 | 1 |
Conclusion
The following table shows the necessary values for the maximum and minimum lines of best fit.
Time shuffled | Maximum | Minimum | Percentage error |
0 | 0.00 | 0.00 | 0.0% |
1 | 0.34 | 0.14 | 10% |
2 | 0.54 | 0.33 | 11% |
3 | 0.84 | 0.67 | 8.5% |
4 | 1.0 | 0.82 | 10% |
5 | 1.1 | 0.97 | 8.6% |
6 | 1.3 | 1.2 | 7.2% |
7 | 1.5 | 1.3 | 8.4% |
8 | 1.8 | 1.7 | 5.9% |
9 | 2.0 | 1.8 | 6.7% |
10 | 2.3 | 2.1 | 9.1% |
11 | 2.8 | 2.8 | 0.0% |
12 | 2.8 | 2.8 | 0.0% |
13 | 3.0 | 3.0 | 0.0% |
14 | 3.2 | 3.2 | 0.0% |
15 | 3.2 | 3.2 | 0.0% |
16 | 3.5 | 3.5 | 0.0% |
17 | 3.5 | 3.5 | 0.0% |
18 | 3.9 | 3.9 | 0.0% |
19 | 3.9 | 3.9 | 0.0% |
20 | 4.6 | 4.6 | 0.0% |
Justification
For 1st shuffle:
Max line of best fit = -ln(N - numerical error /N0) = -ln[(79-8)/100] = 0.34 (2s.f.)
Max line of best fit = -ln(N + numerical error /N0) = -ln[(84+8)/100] = 0.14 (2s.f.)
Note: negative ln is used to produce a trend line with a positive slope
Percentage error = (Max – min) / 2 = (0.34 – 0.14) /2 = 10% (2s.f.)
From the equation of the ln line plotted, the value for the decay constant, λ is found to be 0.22. This decay constant calculated from class data is the same as the decay constant calculated from our group’s data. This can possibly due to the large number of shuffles made (20), which evens out the error involved with the differences in the speed and direction of shuffling by different students.
The half-life, t1/2, of the decay model can be calculated again from t1/2 = (ln2/λ), which is derived from N = N0 e –λt.
Since the decay constant calculated from class data is the same as that calculated from our group’s data, the half-life of the decay model is also the same.
This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.
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