Medical physics unit.
Revision question for an upcoming assignment. Need to know basis of theory mostly, answers not all that important as long as I know how to get the correct one!
0.5 gram sample of ancient wood has an activity count of 6.5 counts per minute.
The activity is due to the disintegration of Carbon 13 atoms.
The half-life of Carbon 14 is 5568 years.
Calculate the age of the ship...
Really don't know how to tackle this one. 5 stars best answer!
Revision question for an upcoming assignment. Need to know basis of theory mostly, answers not all that important as long as I know how to get the correct one!
0.5 gram sample of ancient wood has an activity count of 6.5 counts per minute.
The activity is due to the disintegration of Carbon 13 atoms.
The half-life of Carbon 14 is 5568 years.
Calculate the age of the ship...
Really don't know how to tackle this one. 5 stars best answer!
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I assume you mean carbon 14, since carbon 13 is stable.
As I learned when trying to answer the question in the reference, an awful lot comes down what you think the original activity of the sample is. Since 15.3 counts/minute/gram worked in that question, that's what I'll use here. But if your textbook uses some other value, that's what you should use.
Relying heavily on the reference:
The decay rate of carbon 14 is 15.3 decays/min/gram of carbon. Your sample has
(6.5 decays/min) / (0.5 gram) = 13.0 decays/min/gram
Letting the current decay rate be r, and the original decay rate be R, in terms of half-lives n, then,
r = R(1/2)^n
n = ln(r / R) / ln(1/2) = ln(13.0 / 15.3) / (ln(1/2) = 0.235
Since the half-life of carbon 14 is 5568 years, the sample is
(0.235)(5568 years) = 1310 years old.
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Edit: It has to be a typo. C-13 is stable, so the activity due to C-13 atoms must be zero.
As I learned when trying to answer the question in the reference, an awful lot comes down what you think the original activity of the sample is. Since 15.3 counts/minute/gram worked in that question, that's what I'll use here. But if your textbook uses some other value, that's what you should use.
Relying heavily on the reference:
The decay rate of carbon 14 is 15.3 decays/min/gram of carbon. Your sample has
(6.5 decays/min) / (0.5 gram) = 13.0 decays/min/gram
Letting the current decay rate be r, and the original decay rate be R, in terms of half-lives n, then,
r = R(1/2)^n
n = ln(r / R) / ln(1/2) = ln(13.0 / 15.3) / (ln(1/2) = 0.235
Since the half-life of carbon 14 is 5568 years, the sample is
(0.235)(5568 years) = 1310 years old.
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Edit: It has to be a typo. C-13 is stable, so the activity due to C-13 atoms must be zero.
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I'm glad to help.
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Carbon 14 is an isotope of carbon which is formed naturally in the atmosphere. All plants and animals have a regular intake of carbon while they are alive. When an animal or plant dies it no longer takes in carbon of any form. That plant or animal most often decays and its constituent parts break down and are eaten by scavengers, or they decay,or are washed away etc. Once in a great while under catastrophic circumstances the animal or plant is encased in some sediment, or frozen, or desicated, or otherwise preserved quickly so that its structure and constituent parts are preserved.
That is about all I know. I am sorry.
That is about all I know. I am sorry.