116In nuclei are produced at the rate of 10 s-1. Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m.
So far I have come up with ...
The half life = 54m x 60 s-1 = 3.24 x 103 s-1
(In2) / λ = In2 / 3.24 x 10^3 s-1 = 2.14 x 10^-4 s-1,
∆N0/∆t = 10 s-1 So, N = 10 / 2.14 x 10^-4 s-1 = 4.67×10^4 nuclei