One of the hazards of nuclear explosions is the generation of \({ }^{90} \mathrm{Sr}\) and its subsequent incorporation in place of calcium in bones. This nuclide emits \(\beta\) rays of energy \(0.55 \mathrm{MeV}\), and has a half- life of \(28.1\) y. Suppose \(1.00 \mu \mathrm{g}\) was absorbed by a newly born child. How much will remain after (a) \(18 \mathrm{y}\), (b) \(70 \mathrm{y}\) if none is lost metabolically?

Short Answer

Expert verified
After 18 years, 0.56\(\mu\)g of \( {}^{90} \mathrm{Sr} \) will remain; after 70 years, 0.12\(\mu\)g will remain.

Step by step solution

01

Determine the number of half-lives for each time period

First, calculate the number of half-lives that have passed for each time period. The number of half-lives (n) can be found using the time elapsed (t) divided by the half-life (T): n = t / T. Do this for both (a) 18 years and (b) 70 years. For example, for (a), the number of half-lives is n = 18 years / 28.1 years.
02

Calculate the remaining amount of \( {}^{90} \mathrm{Sr} \) after each time period

To find the remaining amount, use the formula for exponential decay: remaining amount = initial amount * (1/2)^n. Calculate this for both (a) 18 years and (b) 70 years using the respective number of half-lives determined in step 1.
03

Express the final answers with the appropriate units

After calculation, make sure the final amounts are expressed in micrograms (\(\mu\)g) to match the unit of the initial quantity of \( {}^{90} \mathrm{Sr} \) absorbed by the child.

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Key Concepts

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

Half-Life
The half-life of a radioactive substance is a critical concept in understanding how radioactive elements decay over time. It refers to the time required for half of the radioactive isotope in a given sample to undergo radioactive decay. The half-life is a fixed value for each isotope, which means that regardless of the amount of the isotope present, the proportion of the isotope that decays within each half-life remains constant—exactly 50%. In our example, strontium-90 (({ }^{90}Sr)) has a half-life of 28.1 years. If a child absorbs 1.00 microgram of strontium-90, after the passage of one half-life, only 0.50 micrograms would remain if no other processes affected its quantity. Understanding this concept enables us to predict how much of a radioactive substance will stay in a system as time progresses.
Beta Radiation
Beta radiation is one type of radioactive decay where a beta particle, which can be either an electron or a positron, is emitted from an unstable atomic nucleus. The emission of a beta particle results in the transformation of a neutron into a proton, or vice versa, inside the nucleus. This process results in the production of an atom of a new element with a different atomic number, but the same mass number. For instance, in the case of strontium-90, the emission of beta radiation will transform it into yttrium-90, a different element. Beta particles have moderate penetration power, can pose a significant health risk if ingested or inhaled, and are capable of causing lasting damage to living tissue and DNA.
Exponential Decay
Radioactive decay follows a pattern called exponential decay. This means that the quantity of a radioactive isotope decreases by a consistent fraction over equal time periods, described by a mathematical function. In practical terms, over one half-life, the quantity of the isotope halves, and over the next half-life, half of the remaining substance will decay, and so on. Mathematically, the remaining amount after any number of half-lives can be calculated using the formula: remaining amount = initial amount * (1/2)^n, where n is the number of half-lives elapsed. Exponential decay allows for the precise calculation of how much of a radioactive substance will be left at any given time, which is essential for assessing exposure risks, for instance, from strontium-90 absorbed by a child.
Physical Chemistry
Physical chemistry delves into the principles and theories that explain the behaviors of chemical substances, including radioactive materials. It combines chemistry with physics to understand the microscopic behaviors within atoms and molecules. When it comes to radioactive decay like that of strontium-90, physical chemistry explains not just the observable macroscopic phenomena—such as the half-life and the decay process—but also the underlying quantum mechanical events that govern particle emission. This discipline is instrumental in determining the energies involved in radioactive decay and predicting the behavior and interactions of radioactive substances within different environments, be they biological, chemical, or geological.
Nuclear Chemistry
Nuclear chemistry is a subfield of chemistry focused specifically on the reactions and properties of atomic nuclei. This includes studying various types of radioactive decay, nuclear synthesis, and fusion and fission reactions. Understanding nuclear chemistry is essential when considering the impact of nuclear reactions, such as those occurring in nuclear explosions or in medical treatments using radioactive elements. It's also key in managing radioactive waste and in applications of radioactive isotopes, for example, in radiometric dating. In the context of our exercise, nuclear chemistry helps us understand how strontium-90, formed in nuclear reactions, undergoes beta decay and interacts with biological matter, such as being incorporated in place of calcium in a child's bones.

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Most popular questions from this chapter

The half-life for the (first-order) radioactive decay of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\) (it emits \(\beta\) rays with an energy of \(0.16 \mathrm{MeV}\) ). An archaeological sample contained wood that had only 72 per cent of the \({ }^{14} \mathrm{C}\) found in living trees. What is its age?

Here we explore further the idea introduced in Impact \(19.2\) that quantum mechanical effects need to be invoked in the description of the electronic properties of metallic nanocrystals, here modelled as three-dimensional boxes. (a) Set up the Schrodinger equation for a particle of mass \(m\) in a threedimensional rectangular box with sides \(L_{1}, L_{2}\), and \(L_{3}\). Show that the Schrodinger equation is separable. (b) Show that the wavefunction and the energy are defined by three quantum numbers. (c) Specialize the result from part (b) to an electron moving in a cubic box of side \(L=5 \mathrm{~nm}\) and draw an energy diagram resembling Fig. \(9.2\) and showing the first 15 energy levels. Note that each energy level may consist of degenerate energy states. (d) Compare the energy level diagram from part (c) with the energy level diagram for an electron in a one-dimensional box of length \(L=5 \mathrm{~nm}\). Are the energy levels become more or less sparsely distributed in the cubic box than in the one-dimensional box?

Calculate the energies of the first four rotational levels of \({ }^{1} \mathrm{H}^{127} \mathrm{I}\) free to rotate in three dimensions, using for its moment of inertia \(I=\mu R^{2}\), with \(\mu=\) \(m_{\mathrm{H}} m_{1} /\left(m_{\mathrm{H}}+m_{1}\right)\) and \(R=160 \mathrm{pm}\).

Two products are formed in reactions in which there is kinetic control of the ratio of products. The activation energy for the reaction leading to Product 1 is greater than that leading to Product \(2 .\) Will the ratio of product concentrations \(\left[\mathrm{P}_{1} /\left[\mathrm{P}_{2}\right]\right.\) increase or decrease if the temperature is raised? Theoretical problems

Use the virial theorem to obtain an expression for the relation between the mean kinetic and potential energies of an electron in a hydrogen atom.

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