Answer:
C) 10,500 years.
Step-by-step explanation:
- The decay of cobalt-60 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5272 year) = 1.314 x 10⁻⁴ year⁻¹.
The integration law of a first order reaction is:
kt = ln [A₀]/[A]
k is the rate constant = 1.314 x 10⁻⁴ year⁻¹.
t is the time = ??? years.
[A₀] is the initial percentage of cobalt-60 = 25.0 g.
[A] is the remaining percentage of cobalt-60 = 6.25 g.
∵ kt = ln [Ao]/[A]
∴ (1.314 x 10⁻⁴ year⁻¹)(t) = ln (25.0 g)/(6.25 g)
(1.314 x 10⁻⁴ year⁻¹)(t) = 1.386.
∴ t = 1.386/(1.314 x 10⁻⁴ year⁻¹) = 10,550 years ≅ 10,500 years.