only 20% of the carbon-14 in an small wooden bowl remains. how old is the bowl?

Question

NOTE: THE HALF LIFE OF CARBON-14 IS 5700 YEARS and A=A     e^kt

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Answer 1

The formula for exponential decay should be of the form A=A₀·e^-kt where k is the decay constant. To find the decay constant, plug in the provided half-life:

 

1/2 = 1 · e^{-k(5700 years)}  →  ln(1/2) = -k(5700 years)  →  k = -ln(1/2)/(5700 years) = 1.22E-4 years^-1

 

Plug this decay constant back into the formula with the given percentage remaining of carbon-14:

 

1/5 = 1 · e^{-(1.22E-4 years-1)t}  →  ln(1/5) = -(1.22E-4 years-1)t  →  t = -ln(1/5)/(1.22E-4 years-1) = 13,192 years old

 

This places the bowl's age at a little before the start of the Neolithic Era, which seems like a reasonable answer. 

 

Answer 2

The formula given is not actually correct.  It should be A = A₀e^-kt, but it seems easier to use F = 0.5ⁿ where 

F = fraction remaining = 20% = 0.20 and n = number of half lives that have elapsed = ?

0.2 = 0.5ⁿ

log 0.2 = n log 0.5

-0.699 = -.301n

n = 2.32 half lives have elapsed

5700 yr x 2.32 = 13,237 years old