Asked by hipniblet
An artifact contains one-fourth as much carbon-14 as the atmosphere. How old is the artifact? Use Figure 10-3 to answer this question.
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Answers
Answered by
DrBob222
First, determine k
k = 0.693/half life = 0.693/5730 = ?
I looked up the half life on Google. from the link you listed
Then we must know how much C-14 is there. I looked up CO2 in the atmosphere in 2018 and found 0.04%. So there is 0.04/4 = 0.01 C-14 now (using 2018 as a standard. Does your problem tell you which year that was true? Probably not. Google says C-14 gives about 14 dpm (disintergrations per minute) so this sample must give 0.14 x 0.01 = 0.14.
ln (No/N) = kt
No = 14
N = 0.14
k from above
Solve for t in years.
Post your work if you get stuck.
k = 0.693/half life = 0.693/5730 = ?
I looked up the half life on Google. from the link you listed
Then we must know how much C-14 is there. I looked up CO2 in the atmosphere in 2018 and found 0.04%. So there is 0.04/4 = 0.01 C-14 now (using 2018 as a standard. Does your problem tell you which year that was true? Probably not. Google says C-14 gives about 14 dpm (disintergrations per minute) so this sample must give 0.14 x 0.01 = 0.14.
ln (No/N) = kt
No = 14
N = 0.14
k from above
Solve for t in years.
Post your work if you get stuck.
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