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Strontium-90 has a half-life of 28 years. How long will it take for all of the strontium- 90 presently on earth to be reduced to \(1 / 32\) of its present amount?

Short Answer

Expert verified
It will take 140 years for the strontium-90 presently on earth to be reduced to \(1 / 32\) of its present amount.

Step by step solution

01

Understand Half-life

The half-life of a substance is the time it takes for half of the original amount of that substance to decay or reduce to half its amount. Since we are given the half-life of Strontium-90 as 28 years, we can use this to calculate the time it takes to reduce to any fraction of the original amount.
02

Determine Fraction Remaining After n Half-lives

To determine the number of half-lives needed to reduce the present amount to \(1/32\) of its original amount, we can use the formula \( \left( \frac{1}{2} \right)^n = \frac{1}{32} \), where n is the number of half-lives.
03

Solve for the Number of Half-lives

Simplify the fraction \( \frac{1}{32} \) to a power of 2, which is \(2^{-5}\). This is because \(32 = 2^5\). Now we can equate the powers of 2: \(2^{-n} = 2^{-5}\). Therefore, n equals to 5, and it would take 5 half-lives to reduce the amount to \(1/32\) of the original amount.
04

Calculate Total Time

Now multiply the number of half-lives by the length of one half-life to find the total time: time = number of half-lives \(\times\) half-life = 5 \(\times\) 28 years = 140 years.

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Key Concepts

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

Half-Life Calculation
Half-life calculation is a fundamental concept in the study of radioactive decay and nuclear chemistry. Simply put, the half-life of a radioactive substance is the time required for half of the original amount of that substance to undergo decay. This is a constant value for each radioactive isotope. Understanding this concept is crucial for calculating how long it will be before a certain fraction of the substance remains.

When applying half-life calculations, it's often useful to think in terms of powers of two since each half-life reduces the remaining substance by half. To determine the number of half-lives that it takes for a starting amount to decay to a specific fraction of that amount, one can employ the formula: \( \left( \frac{1}{2} \right)^n = \text{Remaining Fraction} \), where \( n \) is the number of half-lives. By rearranging this formula or using logarithms, one can solve for \( n \) and then multiply by the half-life to find the total time needed for the substance to decay to the desired amount.
Strontium-90
Strontium-90 is a byproduct of nuclear fission and is considered a high yield nuclear fallout product from nuclear explosions. It has a notable half-life of about 28 years, which means after 28 years, only half of the original amount of this isotope remains due to radioactive decay. Strontium-90 poses health risks because it's a beta emitter and behaves similar to calcium, incorporating into bones and potentially causing bone cancer or leukemia.

The study of Strontium-90 is important in understanding the long-term environmental and health impacts of nuclear pollution. To model how quickly levels of Strontium-90 will decrease over time, such as how long it takes to reach \( \frac{1}{32} \) of its original amount, requires an understanding of its half-life along with the principles of exponential decay.
Exponential Decay
Exponential decay is a process in which a quantity decreases at a rate proportional to its current value. In the context of nuclear chemistry, this refers to the way in which radioactive substances lose their radioactivity over time. The decay of radioactive isotopes is a perfect example of exponential decay because the amount of substance that decays in each time period is proportional to the amount present at the start of the period.

Graphically, exponential decay can be represented by a curve that starts at an initial value and decreases rapidly at first then more slowly over time, never quite reaching zero. This pattern continues as long as some of the substance remains, making it increasingly difficult to complete the decay process. The concept of exponential decay is critical not only in nuclear chemistry but also in fields such as pharmacology and finance.
Nuclear Chemistry Education
Nuclear chemistry education involves the study of the chemical and physical principles of nuclear transformation, radioactivity, and radiation. This includes understanding the types of radioactive decay, such as alpha decay, beta decay, and gamma radiation, along with their effects and applications.

Education in nuclear chemistry is essential to develop the knowledge required for safely handling radioactive materials, managing nuclear waste, and using radioactive isotopes in various applications, such as medical diagnoses and treatment. It also provides insight into the role of radioisotopes in scientific research, power generation, and the historical and social impacts of nuclear technology. By mastering concepts like half-life calculations and exponential decay, students gain a comprehensive understanding of the processes that govern the behavior of radioactive substances, preparing them for a range of careers in the field.

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Most popular questions from this chapter

What is a homogeneous catalyst? How does it function, in general terms?

The following question is based on Chemistry Outside the Classroom 13.1. The reaction of hydrogen and bromine appears to follow the mechanism shown, $$\begin{aligned}\mathrm{Br}_{2} & \longrightarrow 2 \mathrm{Br}^{*} \\\\\mathrm{Br} \cdot+\mathrm{H}_{2} & \longrightarrow \mathrm{HBr}+\mathrm{H} \\\\\mathrm{H} \cdot+\mathrm{Br}_{2} & \longrightarrow \mathrm{HBr}+\mathrm{Br} \\\2 \mathrm{Br} \cdot &\longrightarrow\mathrm{Br}_{2}\end{aligned}$$ (a) Identify the initiation step in the mechanism. (b) Identify any propagation steps. (c) Identify the termination step. (d) The mechanism also contains the reaction $$\mathrm{H} \cdot+\mathrm{HBr} \longrightarrow \mathrm{H}_{2}+\mathrm{Br}$$ How does this reaction affect the rate of formation of \(\mathrm{HBr}\) ?

The half-life of a certain first-order reaction is \(15 \mathrm{~min}-\) utes. What fraction of the original reactant concentration will remain after 2.0 hours?

In terms of the kinetic theory, why does an increase in temperature increase the reaction rate?

Estimate the rate of the reaction, $$ \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O} $$ given the rate law for the reaction is$$\text { rate }=\left(1.3 \times 10^{11} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\right)\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]$$ for neutral water, where \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\) and \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\)

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