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A mummy discovered in Egypt has lost 46% of its carbon-14. Determine its age
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2 Answers

A = P * (1/2)^(t/5730)
P = 100% = 1
A = 100% - 46% = 0.54
0.54 = 1*(1/2)^(t/5730)
Taking the log of both sides (base 10, but could do ln):
log(0.54) = log((1/2)^(t/5730))
Using power rule of logs:
log(0.54) = (t/5730)*log(0.5)
t = 5730 * log(0.54) / log(0.5)
t ≈ 5,094 years old
First, it is important to note that the half life of carbon-14 is 5730 years.  This says that the amount of carbon is exponentially related to the length since decay began.  To start, note that half life is represented by equation 1 below:
Equation 1:
This says that the amount left is equal to the initial amount multiplied by e raise to the power of the k (the decay constant) and the number of half lives, or simply t/t1/2; where t is the time (age) and t1/2 is the half lift of the subject.
If you are unfamiliar with the k constant for carbon-14, then it can be solved by plugging in a known value: the time of the half life, or t/t1/2=1.  Since the time you plug in at this point is the half life of the substance, then A(t)=(1/2)*A.
Then you have:
solving you get:
To find k, you simply take the natural log (ln) of both sides.
Now that you have k, it is a matter of substituting the amount of carbon-14 left at the age you would like to calculate.
Notice that the A's cancel out and you get:
.46=e-.693t/5730    or .46=e-.00012t
Take the natural log (ln) of both sides to eliminate e and you have
Solving for t you get:
t=-ln(.46)/.00012  = 6471 years