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

Question

Quizplus · Accepted Answer

We can use the exponential decay formula to find the time it takes for a sample to lose one-third of its nuclei:

N(t) = N₀e^(-λt)

where N(t) is the number of nuclei at time t, N₀ is the initial number of nuclei, λ is the decay constant, and e is the mathematical constant approximately equal to 2.718.

We want to solve for t when N(t) = N₀/3:

N₀/3 = N₀e^(-λt)

1/3 = e^(-λt)

Taking the natural logarithm of both sides, we get:

ln(1/3) = -λt

Solving for t, we get:

t = -ln(1/3)/λ

Plugging in the given values for λ, we get:

t = -ln(1/3)/(9.39 × 10^-5 years^-1)

t = 4320 years

Therefore, the answer is C, 4.32 × 10^3 y.

Americium-243 Has a Decay Constant of 9 10−510^{ - 5}10−5 years −1\text { years } - 1 years −1

Americium-243 Has a Decay Constant of 9 $10^{ - 5}$ $\text { years } - 1$