### Half-life problems involving carbon-14

Note: If you have not looked at the half-life videos on the Radioactivity menu, there are several which measure the drop off in radioactivity in terms of disintegrations. Included in these are two which use C-14 as the example problem to be solved.

Problem #41: A chemist determines that a sample of petrified wood has a carbon-14 decay rate of 6.00 counts per minute per gram. What is the age of the piece of wood in years? The decay rate of carbon-14 in fresh wood today is 13.6 counts per minute per gram, and the half life of carbon-14 is 5730 years.

Solution:

1) Determine decimal fraction of C-14 remaining:

6.00 / 13.6 = 0.4411765

2) Determine how many half-lives have elapsed:

(1/2)n = 0.4411765

n log 0.5 = log 0.4411765

n = 1.18057

3) Determine length of time elapsed:

5730 yr x 1.18057 = 6765 yr

Problem #42: The carbon-14 decay rate of a sample obtained from a young tree is 0.296 disintegration per second per gram of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of 0.109 disintegration per second per gram of the sample. What is the age of the object?

Solution:

1) Determine decimal fraction of C-14 remaining:

0.109 / 0.296 = 0.368243

2) Determine how many half-lives have elapsed:

(1/2)n = 0.368243

n log 0.5 = log 0.368243

n = 1.441269

3) Determine length of time elapsed:

5730 yr x 1.441269 = 8258 yr

Problem #43: The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees (indicating 70% of the C-14 had decayed). How old is that piece of wood?

Solution:

1) Determine decimal fraction of C-14 remaining:

0.300 (from text of problem)

2) Determine how many half-lives have elapsed:

(1/2)n = 0.300

n log 0.5 = log 0.300

n = 1.737

3) Determine length of time elapsed:

5730 yr x 1.737 = 9953 yr

Problem #44: Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 mg of carbon-14, how much carbon-14 must have been present in the sample 11,430 years ago?

Solution:

1) Determine half-lives elapsed:

11,430 / 5730 = 1.9947644

2) Determine decimal fraction remaining:

(1/2)1.9947644 = x

x = 0.25091

3) Use a ratio and proportion to find C-14 present in the past:

(0.060 mg / 0.25091) = (x / 1)

x = 0.239 mg

Problem #45: Determine the age of a sample of charcoal which is giving off 25 counts per hour, if carbon-14 from a just made piece of charcoal gives off 85 counts per hour. The half life of carbon-14 is 5730 years.

The solution done in a video.

Problem #46: A living plant contains approximately the same isotopic abundance of C-14 as does atmospheric carbon dioxide. The observed rate of decay of C-14 from a living plant is 15.3 disintegrations per minute per gram of carbon. How many disintegrations per minute per gram of carbon will be measured from a 12900-year-old sample? (The half-life of C-14 is 5730 years.)

Solution:

1) Determine half-lives elapsed:

12,900 yr / 5730 yr = 2.2513

2) Determine decmal fraction remaining:

(1/2)2.2513 = x

x = 0.21

3) Determine counts remaining:

15.3 x 0.21 = 3.2

Problem #47: All current plants have a C-14 count of 15.3 cpm. How old is a wooden artifact if it has a count of 9.58 cpm?

Solution:

1) Determine decmal fraction remaining:

9.58 / 15.3 = 0.6261438

2) Determine half-lives elapsed:

(1/2)n = 0.6261438

n log 0.5 = log 0.6261438

n = 0.675434

3) Determine number of years:

5730 years x 0.675434 = 3870 years

Problem #48: Using dendrochronology (using tree rings to determine age), tree materials dating back 10,000 years have been identified. Assuming you had a sample of such a tree in which the number of C-14 decay events was 15.3 decays per minute before decomposition, what would the decays per minute be in the present day?

Solution:

10,000 yr / 5730 yr = 1.7452 half-lives

(1/2)1.7452 = 0.2983 (this is the decimal amount remaining)

15.3 times 0.2983 = 4.56 (rounded off to three sig figs)

Problem #49: You read that a fossil dinosaur skull has been found in Montana and that it has been carbon-14 dated to be 73 million years old. Provide two (2) scientifically-based reasons to explain why C-14 dating cannot do this.

Solution:

1) A common rule of thumb is that a radioactive dating method is good out to about 10 half-lives. Given a C-14 half-life of 5730 years, you can see that C-14 dating is (theoretically) good out to around 60,000 years (more-or-less). In fact, due to fluctuations in the carbon amount in the atmosphere, modern C-14 dating needs to be correlated to dates determined by analysis of tree-ring records (dendrochronology). Here is a brief article about how and why radiocarbon dates must be calibrated.

2) A skull does not have very much (if any) carbon in it after 73 million years. It would not be dated using C-14 dating. In fact, the value of 73 million years is not arrived at by directly testing the skull. Minerals containing radioactive elements are dated and the age of the skull would be assumed to be of the same age as the strata in which it was discovered.

You can find a brief discussion about the techniques here.

Problem #50: A mammoth skeleton has a carbon-14 decay rate of 0.0077 disintegrations per second per gram of carbon. How long ago did the mammoth live? (Assume that living organisms have a carbon-14 decay rate of 0.255 s¯11 and that carbon-14 has a half-life of 5730 y.)

Solution:

0.0077 / 0.255 = 0.030196

(1/2)n = 0.030196

n log 0.5 = 0.030196

n = 5.0495

(5730 y) (5.0495) = 28933.635 y

29000 y seems a reasonable answer to report

Comment: the skeleton itself was not dated by C-14 since no organic material remains in the bones. However, organic material the skeleton was buried in was dated or perhaps food in its stomach was dated (which has happened).

Problem #51: All living things have a steady state C-14 activity of 15.3 atoms min¯11. Suppose a sample of cloth from an archeological site shows an activity of 0.372 atoms min¯11. How old is the cloth?

Solution:

0.372 / 15.3 = 0.0243137 (this is the decimal amount of activity remaining)

(1/2)n = 0.0243137

n log 0.5 = log 0.0243137

n = 5.362 (this is how many half-lies have passed)

(5730 yr) (5.362( = 30724.26 yr

Problem #52: The smallest C-14 activity that can be measured is about 0.20%. If C-14 is used to date an object, the object must have died within how many years?

Solution:

In decimal form, 0.20% is 0.0020. This is the amount remaining.

(1/2)n = 0.0020

n log 0.5 = log 0.0020

n = 8.9658 <--- this is how many half-lives have elapsed

(5730 yr) (8.9658) = 51374 yr

Problem #53: How long will it take for 25% of the C-14 atoms in a sample of C-14 to decay?

Solution:

25% decay means 75% remains

(1/2)n = 0.75

n log 0.5 = log 0.75

n = 0.415

(5730 yr) (0.415) = 2378 yr

Problem #54: