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

the inner core.

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