Ytterbium-175 is a $\beta$-emitter with a half-life of 101 hours. How long would it take for the activity in a sample of this isotope to decay to exactly one-fifth of its original value?

Question

Quizplus · Accepted Answer

To decay to one-fifth of its original value, the sample must go through log(5)/log(2) ≈ 2.32 half-lives. Multiplying 2.32 by the half-life of 101 hours gives approximately 235 hours.

Ytterbium-175 Is A β\betaβ -Emitter with a Half-Life of 101 Hours

Ytterbium-175 Is A $\beta$ -Emitter with a Half-Life of 101 Hours