Calculate the activity of 60Co after 72hr, if you know the activity at time 0 is 35MBq (decay constant=0.001 day-1) ?

The activity of cobalt-60 after 72 hours is 34.895 MBq

Explanation:

A (t) = Ao (0.5) ^t/t1/2

A (t) is the activity of cobalt-60 after time t

Ao is the initial activity of cobalt-60 = 35 MBq

t is time taken to reduce in activity = 72 hours = 72/24 = 3 days

t1/2 is the half-life = 0.693 : decay constant = 0.693 : 0.001/day = 693 days

A (72) = 35 (0.5) ^3/693 = 35 * 0.5^0.00433 = 35 * 0.997 = 34.895 MBq