SOLUTION: The radioactive isotope carbon 14 used to date fossils decays with an annual rate of about 0.000124. If a fossil is found which originally has 2 mg of carbon 14, and it now has 0.1

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Question 1196893: The radioactive isotope carbon 14 used to date fossils decays with an annual rate of about 0.000124. If a fossil is found which originally has 2 mg of carbon 14, and it now has 0.18 mg, how old is it?
Found 4 solutions by math_tutor2020, josgarithmetic, ikleyn, greenestamps:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

a = initial value
b = determines if we have growth or decay depending if b > 1 or 0 < b < 1.

x = number of years
y = amount of carbon-14 in mg

Given info:
a = 2
b = 1 + r = 1 - 0.000124 = 0.999876
y = 0.18

y = a*b^x
0.18 = 2*0.999876^x
2*0.999876^x = 0.18
0.999876^x = 0.18/2
0.999876^x = 0.09
log(0.999876^x) = log(0.09)
x*log(0.999876) = log(0.09)
x = log(0.09)/log(0.999876)
x = 19,417.7122011154

Answer: Approximately 19,417 years old

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Here's an approach using the half-life formula

The half-life of carbon 14 is approximately H = 5730 years according to these sources here
https://pubchem.ncbi.nlm.nih.gov/compound/carbon-14
https://en.wikipedia.org/wiki/Carbon-14
Every 5730 years or so, the amount cuts in half.

y = a*0.5^(x/H)
0.18 = 2*0.5^(x/5730)
0.5^(x/5730) = 0.18/2
0.5^(x/5730) = 0.09
log( 0.5^(x/5730) ) = log(0.09)
(x/5730)*log(0.5) = log(0.09)
x/5730 = log(0.09)/log(0.5)
x = 5730*log(0.09)/log(0.5)
x = 19,905.6257091448
x = 19,906
This isn't too far from the 19,417 value calculated earlier.

On the scale of tens of thousands of years, a few hundred years isn't that much of a difference
19906-19417 = 488
489/19906 = 0.0246 = 2.46% error approximately
Though of course the level of precision will depend on what context you're in. If you're casually talking to a friend, then you don't need that much precision. For scientific papers, then you'll definitely need more accuracy.

Here's a calculator to help check your work
https://www.omnicalculator.com/chemistry/carbon-dating
In this case there's 0.18/2 = 0.09 = 9% of the carbon 14 left
The calculator will produce the result of 19,906
This result is approximate due to the fact that the half-life 5730 was approximate.

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
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The radioactive isotope carbon 14 used to date fossils decays with an annual rate of about 0.000124. If a fossil is
found which originally has 2 mg of carbon 14, and it now has 0.18 mg, how old is it?
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"annual rate of about 0.000124. " ??????

Half Life of Carbon 14 is 5730 years.
y, amount present after x years
x, how many years passed
p, original amount
highlight_green%28y=p%281%2F2%29%5E%28x%2F5730%29%29

If given y and p, and want to solve x, then
log%28%28y%29%29=log%28%28p%29%29%2Blog%28%28%281%2F2%29%5E%28x%2F5730%29%29%29
log%28%28y%29%29-log%28%28p%29%29=%28x%2F5730%29log%28%281%2F2%29%29
%28log%28%28y%29%29-log%28%28p%29%29%29%2Flog%28%281%2F2%29%29=x%2F5730
x=5730%28log%28%28y%29%29-log%28%28p%29%29%29%2Flog%28%281%2F2%29%29

Using your given data for the question
x=5730%28log%28%280.18%2F2%29%29%29%2Flog%28%281%2F2%29%29---------compute this in the way you like.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

For your information and for your education 

    It is well known fact form Science that ratio of masses of radioactive Carbon-14
    to the regular stable Carbon-12 in any live organic organism is about  1:10%5E12.


    Therefore, if the mass of radioactive Carbon-14 in a sample is 2 mg, it 
    means that the regular Carbon-12 in the same sample has the mass


        2%2A10%5E12 mg = 2%2A10%5E9 grams = 2%2A10%5E6 kilograms = 2%2A10%5E3 metric tons = 2000 metric tons.


    2000 metric tons is the mass of fully loaded freight train consisting of 100 wagons, 
    or is the mass of two  fully loaded freight trains consisting of 50 wagons each.

About these facts, see this Wikipedia article

https://en.wikipedia.org/wiki/Carbon-14


I want to say that,  when somebody starts create his or her own problems in  Math,
this person should be  (and MUST be)  familiar with a subject,  he or she is writing about.


Otherwise,  the whole ugly picture is obtained.

Ugly picture to such a degree that spreading/distribution such problems in the Internet becomes anti-pedagogic.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Some comments about the problem and the other responses you have received....

(1) Your decay rate of 0.000124 corresponds to a carbon 14 half life of 5590 years. Internet sources place the half life at 5730 plus or minus 30 years.

(2) Radioactive decay is a statistical process; the amount remaining does not follow a smooth decaying exponential curve. Any age obtained using carbon 14 dating can only be treated as approximate. Any answer that gives a carbon 14 age should show no more than about 3 significant figures.