The half life for the radioactive decay of potassium-40 to argon-40 is ×1.26109 years.Suppose nuclear chemical analysis shows that there is 0.313mmol of argon-40 for every 1.000mmol of potassium-40 in a certain sample of rock. Calculate the age of the rock.Round your answer to 2 significant digits.
Question
The half life for the radioactive decay of potassium-40 to argon-40 is ×1.26109 years.Suppose nuclear chemical analysis shows that there is 0.313mmol of argon-40 for every 1.000mmol of potassium-40 in a certain sample of rock. Calculate the age of the rock.Round your answer to 2 significant digits.
Solution
To solve this problem, we will use the formula for radioactive decay, which is:
N = N0 * e^(-λt)
where: N is the final amount of the substance, N0 is the initial amount of the substance, λ is the decay constant, t is the time that has passed.
First, we need to find the decay constant (λ). The decay constant is related to the half-life (T) of a substance by the formula:
λ = ln(2) / T
Substituting the given half-life of potassium-40 (1.26109 years) into the formula, we get:
λ = ln(2) / 1.26109 = 0.5493 per year
Next, we need to find the ratio of the final amount of the substance to the initial amount. This is given by the ratio of argon-40 to potassium-40 in the rock, which is 0.313/1.000 = 0.313.
Substituting these values into the formula for radioactive decay, we get:
0.313 = e^(-0.5493t)
To solve for t, we take the natural logarithm of both sides:
ln(0.313) = -0.5493t
Solving for t gives:
t = ln(0.313) / -0.5493 = 1.17 years
So, the age of the rock is approximately 1.17 years, rounded to two significant digits.
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