Question

Potassium-40 is a radioactive isotope that decays into a single argon-40 atom and other particles with a half-life of 1:25 billion years. A rock sample was found that contained 8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay. Date the rock to the time it contained only potassium-40.

Answer 1

Answer:

0.147 billion years = 147.35 million years.

Explanation:

• It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

• Half-life time is the time needed for the reactants to be in its half concentration.

• If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

• Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

• The half-life of Potassium-40 is 1.25 billion years.

• For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

• Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

• At the time needed to be determined:

8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.

• If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

• That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.