Answer:
t(9.32% remaining) = 203 yrs (3 sig. figs.)
Explanation:
All radioactive decay follows a 1st order decay profile. This is defined by the expression ...
A =A₀e^-k·t
A = final activity = 9.32%
A₀ = initial activity = 100%
e = base of natural logs
k = rate constant = 0.693/t(1/2) = (0.693/30.2) yrs⁻¹ = 0.023 yrs⁻¹
t = time of decay = ln(A/A₀)/-k = ln(9.32%/100%)/-0.023 yrs⁻¹
= 203.286637 yrs (calc. ans.)
≅ 203 yrs (3 sig. figs.)
