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
T is the half-life of the substance

In this case, we know that the final quantity of potassium-40 (N) is 1.000mmol - 0.362mmol = 0.638mmol (because the amount of argon-40 represents the amount of potassium-40 that has decayed). The initial quantity (N0) is 1.000mmol, and the half-life (T) is 1.26109 years.

We can plug these values into the formula and solve for t:

0.638mmol = 1.000mmol * (1/2)^(t/1.26109 years)

To isolate t, we can take the natural logarithm of both sides and rearrange the equation:

ln(0.638) = ln(1.000 * (1/2)^(t/1.26109 years))
ln(0.638) = ln(1.000) + (t/1.26109 years) * ln(1/2)
t = 1.26109 years * (ln(0.638) - ln(1.000)) / ln(1/2)

Calculating the right-hand side gives us the age of the rock. Please note that you need to use a calculator that can handle natural logarithms to get the numerical result.

Question

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
T is the half-life of the substance

In this case, we know that the final quantity of potassium-40 (N) is 1.000mmol - 0.362mmol = 0.638mmol (because the amount of argon-40 represents the amount of potassium-40 that has decayed). The initial quantity (N0) is 1.000mmol, and the half-life (T) is 1.26109 years.

We can plug these values into the formula and solve for t:

0.638mmol = 1.000mmol * (1/2)^(t/1.26109 years)

To isolate t, we can take the natural logarithm of both sides and rearrange the equation:

ln(0.638) = ln(1.000 * (1/2)^(t/1.26109 years))
ln(0.638) = ln(1.000) + (t/1.26109 years) * ln(1/2)
t = 1.26109 years * (ln(0.638) - ln(1.000)) / ln(1/2)

Calculating the right-hand side gives us the age of the rock. Please note that you need to use a calculator that can handle natural logarithms to get the numerical result.

Knowee AI · Accepted Answer

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
T is the half-life of the substance

In this case, we know that the final quantity of potassium-40 (N) is 1.000mmol - 0.362mmol = 0.638mmol (because the amount of argon-40 represents the amount of potassium-40 that has decayed). The initial quantity (N0) is 1.000mmol, and the half-life (T) is 1.26109 years.

We can plug these values into the formula and solve for t:

0.638mmol = 1.000mmol * (1/2)^(t/1.26109 years)

To isolate t, we can take the natural logarithm of both sides and rearrange the equation:

ln(0.638) = ln(1.000 * (1/2)^(t/1.26109 years))
ln(0.638) = ln(1.000) + (t/1.26109 years) * ln(1/2)
t = 1.26109 years * (ln(0.638) - ln(1.000)) / ln(1/2)

Calculating the right-hand side gives us the age of the rock. Please note that you need to use a calculator that can handle natural logarithms to get the numerical result.

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 0.362mmol 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.

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