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
To calculate the maximum number of 116In nuclei in the sample, we can use the concept of radioactive decay and equilibrium.
First, let's calculate the decay constant (λ) for 116In using its half-life (T1/2) of 54 minutes:
λ = ln(2) / T1/2
= ln(2) / (54 minutes x 60 seconds)
≈ 1.274 x 10^(-4) s^(-1)
Now, we have the decay constant for 116In. The rate of production of 116In nuclei is given as 10 s^(-1).
We can use the equation for the rate of change of the number of nuclei (dN/dt), which is equal to the difference between the rate of production (∆N0/∆t) and the rate of decay (λN).
dN/dt = ∆N0/∆t - λN
Since we want to calculate the number of nuclei in equilibrium, the rate of change is zero (dN/dt = 0).
0 = ∆N0/∆t - λN
Rearranging this equation, we get:
N = ∆N0/∆t / λ
Substituting the given values, we have:
N = (10 s^(-1)) / (1.274 x 10^(-4) s^(-1))
≈ 78,525 nuclei
Therefore, the maximum number of 116In nuclei (number in equilibrium) in the sample is approximately 78,525 nuclei.