A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has decayed after 3500 years?

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ = > ln (A/A₀) = ln (e⁻ᵏᵗ) = > lnA - lnA₀ = - kt = > lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 ( = 100%)

lnA = lnA₀ - kt

lnA = ln (1) - (6.93 x 10ˉ⁴yrsˉ¹) (3500yrs) = - 2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 - 0.0884 = 0.9116 or 91.16% has decayed.