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When a sample from a meteorite is analyzed, it is determined that 93.8\(\%\) of the original mass of a certain radioactive isotope is still present. Based on this finding, the age of the meteorite is calculated to be \(4.51 \times 10^{9}\) yr. What is the half-life (in yr) of the isotope used to date the meteorite?

Short Answer

Expert verified
The half-life of the isotope is approximately \(4.85 \times 10^9\) years.

Step by step solution

01

Understand the Problem

We need to find the half-life of a radioactive isotope given that 93.8\(\%\) of the isotope remains after \(4.51 \times 10^9\) years.
02

Use the Exponential Decay Formula

Radioactive decay follows the formula \( N = N_0 e^{- rac{t}{\tau}} \), where \( N \) is the remaining quantity, \( N_0 \) is the initial quantity, \( t \) is the time, and \( \tau \) is the decay constant. In this problem, \( N/N_0 = 0.938 \) and \( t = 4.51 \times 10^9 \) years.
03

Convert Percentage to Proportion

We have \( N/N_0 = 93.8\% = 0.938 \). This means \( 0.938 = e^{- rac{t}{\tau}} \).
04

Solve for Decay Constant (\(\tau\))

Use the equation from Step 3: \( 0.938 = e^{- rac{4.51 \times 10^9}{\tau}} \). Take the natural logarithm on both sides to get \( \ln(0.938) = -\frac{4.51 \times 10^9}{\tau} \).
05

Calculate \(\tau\)

Rearrange the equation: \( \tau = -\frac{4.51 \times 10^9}{\ln(0.938)} \). Calculate \( \ln(0.938) \approx -0.0645 \), so \( \tau \approx \frac{4.51 \times 10^9}{0.0645} \approx 6.99 \times 10^{10} \) years.
06

Relate \(\tau\) to Half-Life (\(T_{1/2}\))

The half-life \( T_{1/2} \) is related to \( \tau \) by \( T_{1/2} = \tau \ln(2) \).
07

Calculate the Half-Life

Substitute \( \tau \approx 6.99 \times 10^{10} \) years into \( T_{1/2} = \tau \ln(2) \). We find \( T_{1/2} \approx 6.99 \times 10^{10} \times 0.693 \approx 4.85 \times 10^9 \) years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
The half-life of a radioactive isotope is the time it takes for half of the initial amount of the isotope to decay. It is a crucial concept used in various scientific fields, including geology and physics.

To calculate the half-life, we can use the decay constant, which is derived from the exponential decay formula. Once we know the decay constant, the half-life can be found using the relation:
  • \( T_{1/2} = \frac{\ln(2)}{\lambda} \)

Here, \( \lambda \) is the decay constant. The natural logarithm of 2, \( \ln(2) \), is approximately 0.693. By using this relationship, we can easily calculate the half-life of a radioactive substance if we have determined its decay constant from the observed decay rate.
Exponential Decay Formula
Radioactive decay is best described by the exponential decay formula, which shows how much of an isotope remains after a certain period. This formula is given as:
  • \( N = N_0 e^{-\frac{t}{\tau}} \)

In this formula, \( N \) is the remaining quantity after time \( t \), \( N_0 \) is the initial quantity, and \( \tau \) is the decay constant.

Exponential decay occurs because the rate at which a radioactive isotope decays is proportional to the amount of the isotope that remains.
As time goes on, less and less of the isotope is present, causing the rate of decay to slow down. This predictable decay pattern allows scientists to use isotopes for dating, as it provides a clear timeline for how much of a substance remains over any given period.
Meteorite Dating
Dating meteorites involves determining their age by measuring the concentrations of certain radioactive isotopes within them. Radioactive isotopes are like natural clocks, ticking away as they decay, helping us understand the time that has passed since the meteorite solidified.

To date a meteorite, scientists measure the remaining percentage of a known radioactive isotope, as done in the given exercise, which showed 93.8% of a particular isotope remaining.
  • This information allows scientists to apply the exponential decay formula to calculate the meteorite's age.
  • Knowing the age of meteorites helps us learn about the early solar system's history and the formation of planets and other celestial bodies.
It is an essential tool for geologists and astronomers in their study of the universe.
Decay Constant
The decay constant, symbolized as \( \tau \) or sometimes \( \lambda \), is an integral part of understanding radioactive decay. It measures the probability of decay of a radioactive atom per unit time.

This constant directly influences how quickly or slowly a radioactive isotope decays, and it can be found by rearranging the exponential decay formula:
  • Take the natural logarithm of both sides of \( N = N_0 e^{-\frac{t}{\tau}} \).
  • You get \( \ln(N/N_0) = -\frac{t}{\tau} \).
  • Solving for \( \tau \) yields \( \tau = -\frac{t}{\ln(N/N_0)} \).
Once determined, the decay constant allows us to calculate the half-life of the isotope, offering valuable insights into the rate of decay. Understanding this constant is crucial for accurately timing the age of substances via radioactive decay.

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Most popular questions from this chapter

When any radioactive dating method is used, experimental error in the measurement of the sample’s activity leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity of 0.100 Bq per gram of carbon is measured to within an accuracy of \(\pm 10.0 \%\) Find the age of the fossils and the maximum error (in years) in the value obtained. Assume that there is no error in the 5730-year half-life of \(^{14}_{6} \mathrm{C}\) nor in the value of 0.23 Bq per gram of carbon in a living organism.

The number of radioactive nuclei present at the start of an experiment is \(4.60 \times 10^{15}\) . The number present twenty days later is \(8.14 \times 10^{14} .\) What is the half-life (in days) of the nuclei?

A sample of ore containing radioactive strontium \(\frac{90}{38} \mathrm{Sr}\) has an activity of \(6.0 \times 10^{5} \mathrm{Bq}\) . The atomic mass of strontium is \(89.908 \mathrm{u},\) and its half-life is 29.1 yr. How many grams of strontium are in the sample?

One day, a cell phone company sends a text message to each of its 5800 subscribers, announcing that they have been automatically enrolled as contestants in a promotional lottery modeled on nuclear decay. On the first day, 10\(\%\) of the 5800 contestants are notified by text message that they have been randomly eliminated from the lettery. The other 90\(\%\) of the contestants automatically advance to the next round. On each of the following days, 10\(\%\) of the remaining contestants are randomly eliminated, until fewer than 10 contestants remain. Determine (a) the activity (number of contestants eliminated per day on the second day of the lottery, \((\mathbf{b})\) the decay constant (in \(\mathrm{d}^{-1} )\) of the lottery, and \((\mathbf{c})\) the half-life (in d) of the lottery.

Two radioactive waste products from nuclear reactors are strontium \(_{38}^{90} \operatorname{Sr}\left(T_{1 / 2}=29.1 \mathrm{yr}\right)\) and cesium \(\frac{134}{55} \mathrm{Cs}\left(T_{1 / 2}=2.06 \mathrm{yr}\right) .\) These two species are present initially in a ratio of \(N_{0, S_{\mathrm{r}}} / N_{0, \mathrm{Cs}}=7.80 \times 10^{-3}\) What is the ratio \(N_{\mathrm{Sr}} / N_{\mathrm{Cs}}\) fifteen years later?

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