What is the half year life of a radioactive material?

Question

Find the half-life of a radioactive material if after 1 year 99.69% of the initial amount remains. (Round your answer to one decimal place.)

Answer 1

Hello William,

We will assume that the amount of radioactive material in the sample at any given time t follows an exponential decay model. That is, if y is the amount of material (measured in any convenient units) at time t (in years), then

(Eq.1) y = Ce^kt,

where C and k are constants. For exponential decay, the constant k is negative. It is easy to see that when t = 0, y = C, so C is the initial amount of material present. The half-life is the length of time that it takes for the radioactive substance to decay to one-half of the original amount. Let us denote the half-life by T. Thus, when t = T, y = (1/2)C. Substitute these values into the original equation, then solve for T.

(1/2)C = Ce^kT.

1/2 = e^kT (dividing both sides by C.)

Now take the natural log of each side to obtain

Ln(1/2) = Ln(e^kT)

Ln(1/2) = kT, and hence,

Eq.2 T = [Ln(1/2)]/k

This gives the half-life in terms of the decay constant k. To find k, we use the fact that after 1 year, 99.69% of the original amount remains. That is, when t = 1, y = .9969C. Substitute this information into Eq. 1 and solve for k.

.9969C = Ce^k(1)

.9969 = e^k

Ln(.9969) = Ln(e^k)

Ln(.9969) = k.

Substitute this into the result for half-life (Eq. 2), and obtain

T = [Ln(1/2)]/Ln(.9969)

T ≅ 223.2 years

Hope that helps. Let me know if you need me to clarify anything.

William