Americium-243 has a decay constant of 9.39 × 10^-5 years^-1. How long will it take for a sample of americium to lose one-third of its nuclei?

Question

Quizplus · Accepted Answer

We can start by using the formula for radioactive decay:
N = N0e^(-λt)
where N is the current number of nuclei, N0 is the initial number of nuclei, λ is the decay constant, and t is time.
We want to find how long it takes for the sample to lose one-third of its nuclei, which means the current number of nuclei is 2/3 of the initial number of nuclei:
N = (2/3)N0
We can plug in the values given in the problem:
(2/3)N0 = N0e^(-λt)
Simplifying,
2/3 = e^(-λt)
ln(2/3) = -λt
t = (ln(2/3))/λ
Plugging in the values given in the problem for λ,
t = (ln(2/3))/(9.39 × 10^-5 years^-1)
t = 4.32 × 10^13 years, or approximately 4.32 × 10^3 billion years.
Therefore, the answer is (C) 4.32 × 10^13 years.

Americium-243 Has a Decay Constant of 9