Answer:
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
