Question

In the formula A(t)=A0e^kt, A is the amount of radioactive material remaining from an initial amount A0 at a given time t, and k is a negative constant determined by the nature of the material. A certain radioactive isotope decays at a rate of 0.175% annually. Determine the half-life of this isotope, to the nearest year.

Answer 1

The half life is the time, T, that will give you a value of A(T) = 1/2 * A0.

Let's plug the information you're given into the formula you're given.

After t = 1 year, you're left with A(1 yr) = 1 - 0.175% of your original amount of isotope, or (1 - 0.00175) * A0 = 0.99825 A0. So,

A(1 yr) = A0 e^(k * 1 yr)
0.99825 A0 = A0 e^(k * 1 yr)

Now, divide by A0.

0.99825 = e^(k * 1 yr)

Take the natural log of both sides to isolate the exponent.

ln 0.99825 = ln [e^(k * 1 yr)]
ln 0.99825 = k * 1 yr

k = ln 0.99825 / 1 yr
k = ln 0.99825 yr^-1

So, now your equation looks like

A(t) = A0 * e^(kt)
A(t) = A0 * e^(ln 0.99825 yr^-1 * t)

Now, what's the half life? In other words, what value of T will give A(T) = 1/2 * A0?

A(T) = A0 * e^(ln 0.99825 yr^-1 * T)
1/2 * A0 = A0 * e^(ln 0.99825 yr^-1 * T)
1/2 = e^(ln 0.99825 yr^-1 * T)

Take the natural log of both sides again.

ln (1/2) = ln [e^(ln 0.99825 yr^-1 * T)]
ln (1/2) = ln 0.99825 yr^-1 * T
T = ln (1/2) / (ln 0.99825 yr^-1)
T = ln (1/2) yr / ln 0.99825

So, that's the half life. If you evaluate it, it comes out to about 396 years.