Rhenium186 is a$\beta$-emitter with a half-life of 90.0 hours. How long would it take for the activity in a sample of this isotope to decay to exactly one-third of its original value?

Question

Quizplus · Accepted Answer

The decay of a radioactive isotope follows the formula $N(t) = N_0 \times (1/2)^{t/T_{1/2}}$. To find the time $t$ when the activity is one-third of the original, solve $(1/2)^{t/90} = 1/3$. Taking the natural logarithm of both sides and solving for $t$ gives $t \approx 143$ hours.

Rhenium186 Is A β\betaβ -Emitter with a Half-Life of 90

Rhenium186 Is A $\beta$ -Emitter with a Half-Life of 90