Answer to Question #345879 in Physics for Thel

Question

Consider cobalt-60 (⁶⁰CO) with a half-line of 5.25 years. How many years will it take for its activity to decrease to one-fourth of its original amount?

Answer 1

$$A=A_0*2^{-t/t_{1/2}}$$

$$\frac{1}{4}A_0=A_0*2^{-t/t_{1/2}}$$

$$t=2t_{1/2}=2*5.25\:\rm yr=10.5\:yr$$