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43.27. Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?

Short Answer

Expert verified
The half-life of the isotope is approximately 6.72 days.

Step by step solution

01

Understand the Decay Formula

The decay of isotopes follows an exponential decay model given by the equation:\[ N(t) = N_0 \cdot e^{-\lambda t} \]where:- \( N(t) \) is the decay rate at time \( t \),- \( N_0 \) is the initial decay rate,- \( \lambda \) is the decay constant,- \( t \) is the time elapsed.
02

Apply the Decay Formula

We have the initial decay rate \( N_0 = 8318 \) decays/min and the decay rate after 4 days is \( N(t) = 3091 \) decays/min. We're given that the time \( t = 4 \) days, which needs to be converted to minutes as decay rates are in minutes. So, \( t = 4 \times 24 \times 60 = 5760 \) minutes. Substitute into the decay equation:\[ 3091 = 8318 \cdot e^{-\lambda \times 5760} \]
03

Solve for Decay Constant \( \lambda \)

Rearrange the equation to find \( \lambda \):\[ \frac{3091}{8318} = e^{-\lambda \times 5760} \]Taking the natural logarithm on both sides yields:\[ \ln\left(\frac{3091}{8318}\right) = -\lambda \times 5760 \]So, \( \lambda = -\frac{1}{5760} \cdot \ln\left(\frac{3091}{8318}\right) \).
04

Calculate \( \lambda \)

Plug the values into the formula to calculate \( \lambda \):\[ \lambda = -\frac{1}{5760} \cdot \ln\left(0.3715\right) \approx 7.16 \times 10^{-5} \text{ min}^{-1} \].
05

Relate \( \lambda \) to the Half-life

The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by the formula:\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
06

Calculate the Half-life

Substitute the calculated value of \( \lambda \) into the half-life formula:\[ t_{1/2} = \frac{0.693}{7.16 \times 10^{-5}} \approx 9678.7 \text{ min} \].
07

Convert Half-life to Required Units

Convert the half-life back into days, since typically half-lives are expressed in days. Divide the minutes by the number of minutes in a day (1440):\[ t_{1/2} \approx \frac{9678.7}{1440} \approx 6.72 \text{ days} \].

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Key Concepts

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

Exponential Decay
Exponential decay is a process where the quantity being measured decreases at a rate proportional to its current value. This type of decay occurs in various natural processes, including the decay of radioactive materials. In mathematics, this is represented by the formula \[ N(t) = N_0 \cdot e^{-\lambda t} \] where
  • \( N(t) \) is the remaining amount at time \( t \).
  • \( N_0 \) is the initial amount.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.718.
  • \( \lambda \) is the decay constant, which determines how quickly the decay happens.
  • \( t \) is the time elapsed.

The exponential nature of the formula means that the rate of change is not linear; instead, it decreases more quickly over time. This makes it suitable for describing processes involving a constant proportional decay, like radioactive decay.
Decay Constant
The decay constant, denoted by \( \lambda \), is crucial in understanding the rate at which a substance undergoes exponential decay. It relates to the likelihood of decay happening.
  • A larger decay constant signifies a faster decay, meaning the substance will diminish more rapidly.
  • A smaller decay constant indicates slower decay, suggesting the substance is more stable.

The decay constant is typically found by rearranging the exponential decay formula to solve for \( \lambda \).
In our example, we calculate \( \lambda \) with the equation:\[ \lambda = -\frac{1}{t} \cdot \ln\left(\frac{N(t)}{N_0}\right) \]
Once calculated, \( \lambda \) provides valuable insight into how swiftly or slowly the material will decrease over time.
Radioactive Decay
Radioactive decay is a stochastic (random) process by which an unstable atomic nucleus loses energy by radiation. It's a natural process that results in the emission of particles or electromagnetic waves, leading to a more stable nucleus.
Key characteristics of radioactive decay include:
  • It follows a predictable exponential decay model that's described using math equations.
  • Each isotope has a specific decay constant, determining its decay rate.
  • The process is spontaneous and influenced by the properties of the radioactive substance.

This concept is essential in fields like nuclear physics, medicine (for treatment like radiotherapy), and environmental science (e.g., dating materials through radiometric methods). Understanding radioactive decay is crucial for both practical applications and theoretical research.
Half-life Formula
The half-life of a substance is the time it takes for half of the original amount of a radioactive material to decay. It's a key concept in understanding how long a radioactive substance remains significant.
The half-life formula is given by:\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
where:
  • \( t_{1/2} \) is the half-life.
  • \( \ln(2) \approx 0.693 \) is a constant value related to the exponential nature of decay.
  • \( \lambda \) is the decay constant.

This relationship shows how the decay constant inversely affects the half-life; a larger \( \lambda \) results in a shorter half-life and vice versa.
In practical terms, understanding this formula is vital for applications like carbon dating, nuclear power management, and medical imaging. The half-life provides crucial information about how quickly a radioactive substance loses its activity, aiding in safety and efficacy assessments.

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Most popular questions from this chapter

43.68. A \(70.0-\mathrm{kg}\) person experiences a whole-body exposure to \(\alpha\) radiation with energy 4.77 \(\mathrm{MeV}\) . A total of \(6.25 \times 10^{12} \alpha\) particles are absorbed. (a) What is the absorbed dose in rad? (b) What is the equivalent dose in rem?(c) If the source is 0.0320 \(\mathrm{g}\) of 26 \(\mathrm{Ra}\) (half- life 1600 \(\mathrm{y}\) ) somewhere in the body, what is the activity of this source? (d) If all the alpha particles produced are absorbed, what time is required for this dose to be delivered?

43.36. To Sean or Not to Scan? It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans using \(x\) rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 \(\mathrm{mSv}\) , applied to the whole body. By contrast, a chest \(\mathbf{x}\) ray typically administers 0.20 \(\mathrm{mSv}\) to only 5.0 \(\mathrm{kg}\) of tissue. How many chest \(\mathrm{x}\) rays would deliver the same total amount of energy to the body of \(75-\mathrm{kg}\) person as one whole-body scan?

43.70. The nucleus \(\frac{15}{8} \mathrm{O}\) has a half-life of \(122.2 \mathrm{s} ; \mathrm{g} \mathrm{O}\) has a half-life of 26.9 s. If at some time a sample contains eqnal amounts of \(\frac{15}{8} \mathrm{O}\) and \(_{8}^{19} \mathrm{O}\) what is the ratio of \(_{8}^{15} 0\) to \(_{8}^{19} \mathrm{O}\) (b) after 15.0 minutes?

43.24. Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. 131 \(\mathrm{I}\) , a \(B^{-}\) emitter with a half-life of 8.0 \(\mathrm{d}\) is one such tracer. Suppose a scientist introduces a sample with an activity of 375 \(\mathrm{Bq}\) and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 \(\mathrm{d}\) (about 2\(\frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 \(\mathrm{Bq}\) , what percentage of the tracer went to that gland? (c) What isotope remains after the \(\mathrm{I}-131\) decays?

43.65. We Are Stardust. In 1952 spectral lines of the element technetium- 99 ( \(^{99} \mathrm{Tc} )\) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of \(^{99} \mathrm{Te}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99}\) Te been in the red-giant star if its age is 10 billion years? (b) What fraction of the original "Te would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^{99} \mathrm{Tchadbeen}\) part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust"

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