151k views
4 votes
The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years. Suppose C(t) is the amount of carbon-14 present at time t.

(a) Find the value of the constant kk in the differential equation C′=−kC.

User Sweetkorn
by
6.2k points

1 Answer

3 votes

Answer:

k-0.000125

Explanation:

Given that the radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years.


C' = -kC\\(dC)/(C) =-kdt\\ln C = -kt+C_1, where C_1 is arbitrary constant.

Or
C(t) = Ae ^(-kt)

To find K.

C(t) = 1/2 C when t = 5551

i.e. A will become A/2 in 5551 years


A/2 = Ae^(-5551k) \\ln 0.5= -5551k\\k = 0.000125

User Hjaltij
by
6.3k points