The half life for the radioactive decay of rubidium-87 to strontium-87 is ×4.881010 years.Suppose nuclear chemical analysis shows that there is 0.486mmol of strontium-87 for every 1.000mmol of rubidium-87 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 rubidium-87 to strontium-87 is ×4.881010 years.Suppose nuclear chemical analysis shows that there is 0.486mmol of strontium-87 for every 1.000mmol of rubidium-87 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 need to use the formula for radioactive decay, which is:
N = N0 * (1/2)^(t/T)
where: N is the final quantity of the substance, N0 is the initial quantity of the substance, t is the time that has passed, and T is the half-life of the substance.
In this case, we know that the ratio of Sr-87 to Rb-87 is 0.486, which means that the initial quantity of Rb-87 (N0) has decreased to 1.000 - 0.486 = 0.514 mmol. This is our final quantity (N).
We also know that the half-life (T) of Rb-87 is 4.88 x 10^10 years.
Substituting these values into the formula, we get:
0.514 = 1 * (1/2)^(t/4.88 x 10^10)
To solve for t, we can take the natural logarithm of both sides:
ln(0.514) = ln((1/2)^(t/4.88 x 10^10))
This simplifies to:
ln(0.514) = (t/4.88 x 10^10) * ln(1/2)
Solving for t gives:
t = (ln(0.514) / ln(1/2)) * 4.88 x 10^10
Calculating this gives:
t ≈ 3.55 x 10^10 years
So, the age of the rock is approximately 35.5 billion years, rounded to two significant digits.
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