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
Is this right ??? thanks
To calculate the maximum number of 116In nuclei in the sample, you need to use the concept of radioactive decay and equilibrium.
The radioactive decay of 116In follows an exponential decay process, and its rate is given by the equation N = N0 * e^(-λt), where N is the number of nuclei at time t, N0 is the initial number of nuclei, λ is the decay constant, and e is Euler's number (approximately 2.718).
Given that the half-life of 116In is 54 minutes, you correctly converted it to seconds as 3.24 x 10^3 s.
To determine the equilibrium number of nuclei, you need to find the value of λ. Using the half-life formula T1/2 = ln(2) / λ, you can rearrange it to find λ = ln(2) / T1/2.
Substituting the values, λ = ln(2) / (3.24 x 10^3 s) ≈ 2.14 x 10^(-4) s^(-1).
Now, you can calculate the equilibrium number of nuclei using the formula N = ∆N0 / λ, where ∆N0 is the production rate of nuclei per second. In this case, ∆N0 = 10 s^(-1).
Therefore, N = (10 s^(-1)) / (2.14 x 10^(-4) s^(-1)) ≈ 4.67 x 10^4 nuclei.
So, your calculation is correct. The maximum number of 116In nuclei in the sample, in equilibrium, is approximately 4.67 × 10^4 nuclei.