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Chapter 21: Problem 83

A radioactive isotope has a half-life of 8 days. If today \(125 \mathrm{mg}\) is left over, what was its original weight 32 days earlier? (a) \(2 \mathrm{~g}\) (b) \(4 \mathrm{~g}\) (c) \(5 \mathrm{~g}\) (d) \(6 \mathrm{~g}\)

Short Answer

Expert verified
The original weight was 2 g.

Step by step solution

01

Identify the relationship between time and half-life

The given isotope has a half-life of 8 days. Half-life is the time required for a quantity to reduce to half its initial amount. We need to determine the number of half-lives that have passed in 32 days. This can be calculated by dividing 32 days by the half-life of 8 days.
02

Calculate the number of half-lives

Divide 32 days by the half-life duration of 8 days:\[ \text{Number of half-lives} = \frac{32 \text{ days}}{8 \text{ days/half-life}} = 4 \text{ half-lives} \]
03

Calculate the original weight using the number of half-lives

If the current weight is 125 mg and 4 half-lives have passed, it would double in weight with each half-life going back to its original weight. Thus, we compute backwards: - After 1 half-life, the weight was 250 mg. - After 2 half-lives, the weight was 500 mg. - After 3 half-lives, the weight was 1000 mg (or 1 g). - After 4 half-lives, the weight was 2000 mg (or 2 g).
04

Select the correct answer

From the calculations, we find that the original weight of the isotope was 2000 mg, which equals 2 g. Therefore, the correct answer is: (a) 2 g.

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Key Concepts

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

Half-life
Half-life is a fundamental concept in nuclear chemistry. It represents the time it takes for half of a sample of a radioactive substance to decay or transform into another material. This constant rate of decay is unique to each radioactive isotope.
The importance of half-life can be seen in several ways:
  • It allows scientists to predict how long it will take for a radioactive material to reach a certain level of activity.
  • Half-life determines the safety and effectiveness of radioactive medications and tracers used in medical imaging.
  • It's a crucial factor in radiometric dating methods, helping us understand the age of fossils and rocks.
In the exercise, the half-life is 8 days. This means that every 8 days, only half the quantity of the radioactive material remains. Knowing this helps to calculate how much of the substance was present before a certain number of half-lives passed. This concept is integral to comprehending how radioactive substances diminish over time.
Exponential Decay
Exponential decay is a process that describes how quantities diminish quickly at first, then more slowly as time progresses. It's a pattern seen in radioactive decay, among other phenomena. The decay of radioactive materials is a classic example of exponential decay.
Understanding exponential decay involves a few key points:
  • The amount of substance decreases by the same factor over equal time periods.
  • In a mathematical sense, the remaining amount after a set number of time periods can be calculated using the formula \( N = N_0 \times (1/2)^n \), where \( N_0 \) is the original quantity, \( N \) is the remaining quantity, and \( n \) is the number of half-lives.
  • This behavior provides a quick method to predict how a radioactive sample changes over time.
In our exercise, understanding that 32 days encompass four half-lives allows us to track the exponential decay of the isotope, determining the original quantity from the remaining 125 mg.
Nuclear Chemistry
Nuclear chemistry is a field that explores the changes in the nucleus of atoms, particularly those involving radioactive transformations. It's a subsection of chemistry closely linked with physics. It includes understanding concepts like radioactivity, nuclear reactions, and the behavior of radioactive elements.
Key aspects of nuclear chemistry include:
  • Understanding nuclear reactions, including fission and fusion, which power stars and nuclear reactors.
  • Investigating the mechanisms of radioactive decay and the various types of emissions like alpha, beta, and gamma rays.
  • Applying knowledge to practical uses such as medical treatments, nuclear power generation, and environmental monitoring.
In the context of the exercise, nuclear chemistry principles allow us to calculate how much of a radioactive isotope was originally present based on its half-life and current amount. This application shows a real-world utility of nuclear chemistry in solving problems related to radioactive decay.

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Most popular questions from this chapter

Select the correct statements: (a) In the reaction \({ }_{11} \mathrm{Na}^{23}+\mathrm{Q} \rightarrow{ }_{12} \mathrm{Mg}^{23}+{ }_{0} \mathrm{n}^{1}\), the bombarding particle \(\mathrm{q}\) is deutron (b) In the reaction \({ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow 56 \mathrm{Ba}^{140}+2\) \({ }_{0} \mathrm{n}^{1}+\mathrm{p}\), produced \(\mathrm{p}\) is \({ }_{36} \mathrm{Kr}^{94}\) (c) In a fission reaction, a loss in mass occurs releasing a huge amount of energy (d) A huge amount of energy is produced during nuclear fission and nuclear fussion reaction

Lead is the final product formed by a series of changes in which the rate determining stage is the radioactive decay of uranium-238. This radioactive decay is a first order reaction with a half-life of \(4.5 \times 10^{9}\) years. What would be the age of a rock sample originally lead free, in which the molar proportion of uranium to lead is now \(1: 3\) ? (a) \(1.5 \times 10^{9}\) years (b) \(2.25 \times 10^{9}\) years (c) \(4.5 \times 10^{9}\) years (d) \(9.0 \times 10^{9}\) years

Hydrogen bomb is based on the principle of (a) artificial radioactivity (b) nuclear fission (c) nuclear fusion (d) natural radioactivity

Which of the following notations shows the product incorrectly? (a) \({ }_{5} \mathrm{~B}^{10}(\alpha, \mathrm{n}){ }_{7} \mathrm{~N}^{13}\) (b) \({ }_{96} \mathrm{Cm}^{242}(\alpha, 2 \mathrm{n}){ }_{97} \mathrm{BK}^{243}\) (c) \({ }_{7} \mathrm{~N}^{14}(\mathrm{n}, \mathrm{p}){ }_{6} \mathrm{C}^{14}\) (d) none of these

In the transformation of \({ }_{92} \mathrm{U}^{238}\) to \({ }_{92} \mathrm{U}^{234}\), if one emission is an \(\alpha\) particle, what should be the other emission(s)? (a) two \(\beta^{-}\) (b) two \(\beta^{-}\)and one \(\beta^{+}\) (c) one \(\beta^{-}\)and one \(\gamma\) (d) one \(\beta^{-}\)and one \(\beta^{-}\)
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