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The Half-life of dice Decay investigation

Extracts from this document...

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

...read more.

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.

image00.png

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

...read more.

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.

image00.png

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.

...read more.

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