Question 470449: Will you please answer this question:
An artifact was found and tested for its carbon-14 content. If 73% of the original carbon-14 was still present, what is its probable age (to the nearest 100 years)? Use that carbon-14 has a half-life of 5,730 years.
Thank you
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 73% of the original carbon-14 was still present.
what is it's probably age to the nearest 100 years.
carbon-14 has a half life of 5,730 years.
here's a reference for half life formulas:
http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm
the half life decay formula is:
f = p * e^(-kt) where:
f is the future value
p is the present value
e is the scientific base of the constant e whose value is 2.718281828.
k is a constant whose value is positive which makes -k negative.
t is the amount of time in years.
in your problem, the half life is 5730 years.
we use this half life to find the value of k.
the formula is:
1/2 = 1 * e^(-kt)
f is 1/2
p is 1
t is 5730
k is what we are trying to find.
our formula becomes:
.5 = e^(-5730k)
to solve this problem, we use logarithms.
in fact, this problem lends itself to natural logarithms, so we'll take the natural log of both sides of this equation to get:
ln(.5) = ln(e^(-5730k)
since ln(x^a) = a*ln(x), our formula becomes:
ln(.5) = -5730k * ln(e)
since ln(e) is equal to 1, our equation becomes:
ln(.5) = -5730k
we divide both sides of this equation by -5730 to get:
ln(.5)/-5730 = k
we solve for k to get:
k = .000120968
we now have the value of k which we can use to solve the problem.
we use our formula again, only this time we replace k with .000120968.
f is the future value which is .73 times the original value which we assign as 1, which means the original value of 100% of the original value. our future value of .73 is 73% of the original value of 1.
so our equation is:
f = p * e^(-kt)
this time we are solving for t which is the number of years.
f = .73
p = 1
k = .000120968
the formula becomes:
.73 = 1 * e^(-.000120968*t)
this becomes:
.73 = e^(-.00012068*t)
since we want to solve for t which is in the exponent, we use natural logs again.
we get:
ln(.73) = ln(e^(-.00012068*t) which becomes:
ln(.73) = -.00012068*t*ln(e) which becomes:
ln(.73) = -.00012068*t because ln(e) is equal to 1.
we divide both sides of this equation by -.00012068 to get:
t = ln(.73) / -.00012068 to get:
t = 2601.601245 years.
we round this to the nearest hundred years to get:
t = 2600 years.
we can test our half life formula to see if it is accurate, by simply replacing .73 with .5 in that final equation to get:
t = ln(.5) / -.00012068 to get:
t = 5730.
the equation is accurate and so we're good.
the answer is that the carbon-14 is approximately 2600 years old.
Note that in the reference they used N and N[0] and they used k.
Their N is equivalent to my f.
Their N[0] is equivalent to my p.
Their k is equivalent to my -k because they state that, in the decay formula, the value of k is negative.
My k is positive but it has a - sign in front of it, making the value negative.
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