Problema Solution
A certain radioactive isotope has leaked into a small stream. One hundred days after the leak, 2% of the original amount of the substance remained. Determine the half-life of this radioactive isotope.
Answer provided by our tutors
A model that describes the exponential decay is:
A(t) = A0*e^(kt),
A = the amount left after time t
A0 = the initial quantity
k = constant
t = 100 days
A(100) = 0.02A0
first we will find k:
0.02A0 = A0*e^(100k)
e^(100k) = 0.02
k = (1/100)ln(0.02)
let t = half-life time then we have
A(t) = (1/2)A0
(1/2)A0 = A0*e^(((1/100)ln(0.02))*t)
e^(((1/100)ln(0.02))*t) = 1/2
((1/100)ln(0.02))*t = ln(1/2)
t = ln(1/2)/((1/100)ln(0.02))
t = 100*(ln(1/2))/(ln(0.02))
t = 17.72 days
the half life of the radioactive isotope is 17.72 days.