A radioactive isotope has a half life of 25 years. There was 100 mg of the isotope in 2005.
(a) Find an expression for the amount of the isotope, A(t), that is still active using time t, measured in years since 2005. Your expression must be of the form A(t)=Pect (base "e").
(b) How much of the isotope do we have at the present time (2018) to 1 decimal place?
(c) Determine an expression for the instantaneous rate of change, A'(t), using the limh→0A(t+h)−A(t)h. You will need to use a spreadsheet again and compare the limit value to your c value from part (a).
(d) What is the instantaneous rate of decay in 2018 to 1 decimal place? Be sure to include proper units and an interpretation of this rate in the context of the problem.
so ive gotten all the way up to c but the spread sheet doesnt come out right and i dont know how to do d can someone help?
3 answers
Now 1/2 = e^(-ln2), so
A(t) = 100* e^(-ln2*t/25) = 100e^(-0.0277t)
(b) A(13)
(c) A'(t) = -0.0277 A(t)
I assume you've seen the proof that d/dt e^t = e^t
(d) A'(13)
1 answer