Word problem help

Dear Erica,

 

In general

 

A = A₀e-^kt

 

where A is the amount remaining after t years, A₀ is the starting amount and k is the decay constant.

 

Let us rewrite this as

 

A/A₀ = e^-kt = e^{-[(0.00012)*(4000)]} = e^-0.48 = 0.62

 

I think all we can say with certainty is that, during those 4000 years, the amount of C-14 decreased to 62% of its original value, or it lost 38% of its original value (A₀).

N(t)=N(0)e^-kt, where k is the decay rate. I assume k=-0.00012 is measured in year^-1, so that t is in years (in fact, it is the case, I believe, since half-life time is 5730 years for C¹⁴).

 

Then, 

N(4000)/N(0)=e^{-0.00012*4000}=e^-0.48≈0.62

 

So the amount of carbon-14 left at the present moment is 0.62 or 62% of its original quantity. Therefore, 38% has been lost due to decay.

 

Answer: 38%