Question

Iron-59 has a half-life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron?   Show all calculations leading to a solution.

Answer 1

First order kinetics is followed by all radio decay.

First order decay equation is given by,

A= A₀e^-kt  , where A= final activity, A₀= initial activity , k=rate constant and t=decay time

Now, the rate constant, k can be determined using the equation of as a function of half life as---

k.t^1/2 = 0.693------- equation (1)

Given in the question,

t^1/2 =45.1 days

=> k = (0.693 / 45.1) days^-1 ----- using equation (1)

=> k =0.0154 days^-1

Again,

from first order decay equation we have,

A= A₀e^-kt

=> ln (A/A₀) = -kt

=> t = [ln(A/A₀) /-k]

=> t= ln [(25/100) / -0.0154 ] days

=> t= (ln 0.25 ) / -0.0154 days

=> t= -1.386 /-0.0154 days

=> t= 90 days

Thus, the iron nail is 90 days old.