Question

If the half-life of a radioisotope is 20,000 years, then a sample in which three-quarters of that radioisotope has decayed is ______________ years old. a. 15,000 b. 26,667 c. 30,000 d. 40,000

Short answer

The sample is 40,000 years old.

Step-by-step solution

Understanding Half-Life

The half-life of a radioisotope is the time it takes for half of the radioisotope to decay. In this problem, the half-life is given as 20,000 years.

Determine Decay Fraction

We are asked to find out how long it took for three-quarters (or 75%) of the isotope to decay. If three-quarters of it has decayed, then only one-quarter (or 25%) of the original sample is left.

Use Exponential Decay Formula

The decay of a substance can be described by the formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \] where \( N(t) \) is the remaining quantity of the substance after time \( t \), \( N_0 \) is the initial quantity of the substance, and \( T \) is the half-life. Here, we want to find \( t \) when \( N(t) = \frac{1}{4} N_0 \).

Set Up the Equation

Substitute \( N(t) = \frac{1}{4}N_0 \) and \( T = 20,000 \) into the formula:\[ \frac{1}{4}N_0 = N_0 \left(\frac{1}{2}\right)^{\frac{t}{20,000}} \]

Simplify the Equation

Cancel \( N_0 \) from both sides to get: \[ \frac{1}{4} = \left(\frac{1}{2}\right)^{\frac{t}{20,000}} \]

Solve for t using Logs

Take the logarithm of both sides:\[ \log_{10}\left(\frac{1}{4}\right) = \frac{t}{20,000} \cdot \log_{10}\left(\frac{1}{2}\right) \]Solve for \( t \): \[ t = \frac{\log_{10}(\frac{1}{4})}{\log_{10}(\frac{1}{2})} \times 20,000 \]

Calculate the Value

Compute the values using a calculator:\[ \log_{10}(\frac{1}{4}) = -0.60206 \] \[ \log_{10}(\frac{1}{2}) = -0.30103 \] Plug these into the equation:\[ t = \frac{-0.60206}{-0.30103} \times 20,000 \approx 40,000 \text{ years} \]

Conclusion

Thus, the age of the sample in which three-quarters of the radioisotope has decayed is 40,000 years.

Key concepts

half-life

The concept of half-life in radioactive decay is crucial for understanding how a radioisotope decreases over time. A half-life is the time required for one-half of a sample of a radioactive substance to decay into another element or isotope. If you have a sample of 100 atoms, after one half-life, only 50 of those atoms would remain as the original isotope.

Half-life is a characteristic property of each specific isotope, which means it remains constant and does not change regardless of the starting amount of the substance. In this exercise, the half-life is specified as 20,000 years.

Knowing the half-life of an isotope can help us predict how much of a sample remains after a given period. This invaluable tool is used in various fields, from archaeology, where it's applied in carbon dating, to medicine in tracing isotopes' behavior in the body. It helps calculate timelines and estimates when certain percentages of decay occur—such as the three-quarters decay in the exercise.

exponential decay

Exponential decay describes a process where the quantity of a substance decreases at a rate proportional to its current value.

It is mathematically represented by the equation:

• \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \]
• Where \( N(t) \) is the amount left after time \( t \), \( N_0 \) is the initial amount, and \( T \) is the half-life.

When dealing with radioactive decay, this formula helps determine how much of the original substance remains as time proceeds. In our exercise, this formula is rearranged to solve for the time \( t \) when a particular fraction of the isotope, such as one-quarter, remains.

Exponential decay implies a rapid decrease initially and then a slower reduction as the substance approaches its decay limits. It's in many natural processes and systems, where changes happen gradually over time, showcasing the persistent power of small forces acting over long periods.

logarithm

Logarithms play a vital role when solving exponential decay problems. A logarithm is the opposite of exponentiation, which allows solving equations where variables appear as exponents. Used in mathematical solutions, like in this exercise, they help isolate the time \( t \) in the exponential decay formula.

In the equation:

• \[ \log_{10}\left(\frac{1}{4}\right) = \frac{t}{20,000} \cdot \log_{10}\left(\frac{1}{2}\right) \]
• Logarithms simplify calculations involving exponential decay by making them easier to handle numerically.

The property of logarithms, where the log of a fraction or product can be broken down into simpler components, is core to these calculations. Understanding logarithms provides clearer insight into how to determine unknown variables, such as time, in decay equations, ultimately helping predict how long it would take for a significant part of a substance to diminish.