Question
If, initially, there is 20 mg of gold-198 (198Au) (half-life = 2.696 days),
how many days does it take for less than 1 mg to
remain?
how many days does it take for less than 1 mg to
remain?
Answers
Reiny
20(.5)^(t/2.696) < 1, where t is number of days
(.5)^(t/2.696) < .05
take logs of both sides and use log rules
(t/2.696) log .5 = log .05
t/2.696 < log.05/log.5 = 4.3219..
t < 11.65 days
So it takes just less than 12 days
(.5)^(t/2.696) < .05
take logs of both sides and use log rules
(t/2.696) log .5 = log .05
t/2.696 < log.05/log.5 = 4.3219..
t < 11.65 days
So it takes just less than 12 days