Question

A radioactive isotope is produced at a rate of 30 grams hourly. If over long periods of time the quantity of the isotope that is present stabilizes near 2400 grams, what is the half-life of the isotope?

Answer 1

Solution :

From Radioactive decay law, we have,

N(t) = N0 exp(–Lamda*t) ----(i)

Differentiate equation (i) w.r.t time, we get,

dN(t)/dt =–Lamda * N0 exp(–Lamda*t)=(–Lamda) * N ----(ii)

=> |dN(t)/dt| = Landa* N [ where Lamda is decay constant]

Given, dN(t)/dt = 30 grams per hour

N=24 grams , to find the value of decay constant substitute given data in equation(ii) we get,

decay constant=Lamda= (1/80) per hour.

So half life = T(1/2) = ln 2 /lamda

=> T(1/2) =55.44 hour

Hence for radioactive isotope, half life is 55.44 hours