Answer to Question #108731 in General Chemistry for Max

Question

The half life of 125-I is 60 days. What percent of original radioactivity would be present after 360 days?

Answer 1

$$N = N_0(0.5)^{{t \over T}}$$

$$0.5^{{t \over T}} = {N \over N_0}$$

$${N \over N_0} = 0.5^{360 \over 60} = 0,015625 = 15.625\%$$