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:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
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:
<em>kt = ln([A₀]/[A]),</em>
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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:
<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>
- 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.
