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Lead-210 is used to prepare eyes for corneal transplants. Its decay product is bismuth-210. Identify the emission from lead-210.

Short Answer

Expert verified
Answer: Beta-minus particle (an electron).

Step by step solution

01

Understanding the Radioactive Decay Process

Radioactive decay is a random process where unstable atomic nuclei lose energy by emitting radiation. There are several types of radioactive decay, including alpha decay, beta decay (beta-minus and beta-plus), and gamma decay. In each decay process, a nucleus will release particles or energy in order to become more stable.
02

Writing the Decay Reaction

To determine the type of emission from lead-210 (Pb-210) as it decays into bismuth-210 (Bi-210), we can write the decay reaction as follows: \[ ^{210}_{82}Pb → ^{210}_{83}Bi + X \] In this reaction, the lead-210 nucleus is converted to a bismuth-210 nucleus and some emission X. To identify X, we need to compare the atomic numbers and mass numbers of the initial nucleus (Pb-210) and the final nucleus (Bi-210).
03

Identifying the Emission Type

Let's examine the atomic numbers and mass numbers of the initial and final nuclei: - Initial nucleus: Lead-210 (Pb-210): atomic number (Z) = 82, mass number (A) = 210 - Final nucleus: Bismuth-210 (Bi-210): atomic number (Z) = 83, mass number (A) = 210 From the reaction, we can observe that the mass number remains constant (A = 210), whereas the atomic number increases by one (83 - 82 = 1). This increase in atomic number corresponds to the emission of a beta-minus particle (an electron), which can be represented by the symbol \(β^-\). Therefore, the decay reaction for lead-210 turning into bismuth-210 is: \[ ^{210}_{82}Pb → ^{210}_{83}Bi + β^- \]
04

Conclusion

The emission from lead-210 as it decays into bismuth-210 is a beta-minus particle (an electron).

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Key Concepts

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

Beta Decay
When it comes to understanding radioactive decay, beta decay is an essential piece of the puzzle. Beta decay is a process that allows unstable atomic nuclei to transform into more stable configurations. This occurs through the emission of beta particles. There are two types of beta decay: beta-minus and beta-plus.

In beta-minus decay, a neutron in the nucleus turns into a proton. This process emits an electron, which we refer to as a beta particle, symbolized as \( β^- \). This emission increases the atomic number by one, turning the element into a new element located right next to it on the periodic table. Meanwhile, the mass number remains unchanged.
  • Beta-minus example: As seen in the decay of lead-210 to bismuth-210, \(^{210}_{82}Pb \rightarrow ^{210}_{83}Bi + β^- \), an electron is emitted, increasing the atomic number from 82 to 83.
Beta decay is a cornerstone concept in nuclear physics, essential for students to grasp when studying nuclear reactions.
Nuclear Reactions
Nuclear reactions are processes in which the nucleus of an atom changes its composition or energy state. Unlike chemical reactions, which involve electron interactions, nuclear reactions alter the nucleus itself.

These reactions can occur naturally, as with radioactive decay, or can be induced artificially in laboratory settings.
  • Types of nuclear reactions: There are several types, such as fusion, fission, and decay processes (including beta decay, alpha decay, etc.).
  • Conservation laws: Important principles in nuclear reactions include the conservation of mass number and atomic number. Mass number refers to the total count of protons and neutrons, while the atomic number refers solely to the number of protons.
The conservation of these numbers is crucial for balancing nuclear equations, as shown in the transition of lead-210 to bismuth-210, where the mass number remains 210, and the atomic number increases from 82 to 83.
Lead-210 Decay
Lead-210 decay is a fascinating example of radioactive decay, showcasing various principles in nuclear chemistry. Lead-210 is an isotope of lead, often used in medical applications such as preparing eyes for corneal transplants.

When lead-210 decays, it transforms into bismuth-210 through beta-minus decay. This specific decay process maintains the mass number at 210 and increases the atomic number by one, from 82 to 83. This increase in atomic number indicates the formation of a new element, bismuth-210.
  • Decay equation: The decay can be represented as \(^{210}_{82}Pb \rightarrow ^{210}_{83}Bi + β^- \).
  • Real-world implications: Understanding such decay processes is crucial for applications in nuclear medicine and radiometric dating.
Studying lead-210 decay helps illustrate key radioactive decay concepts and demonstrates the conservation laws that govern nuclear reactions.

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Most popular questions from this chapter

Radium-226 decays by alpha emission to radon-222. Suppose that \(25.0 \%\) of the energy given off by one gram of radium is converted to electrical energy. What is the minimum mass of lithium that would be needed for the voltaic cell \(\mathrm{Li}\left|\mathrm{Li}^{+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), at standard conditions, to produce the same amount of electrical work \(\left(\Delta G^{\circ}\right) ?\)

Balance the following equations by filling in the blanks. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0} n \longrightarrow{ }_{54}^{137}=2{ }_{0}^{1} n+\) (b) \({ }_{90}^{232} \mathrm{Th}+{ }_{6}^{12}\) \(\longrightarrow\) \(1 . n+{ }_{96}^{240} \mathrm{Cm}\) (c) \({ }_{2}^{4} \mathrm{He}+{ }_{42}^{96} \mathrm{Mo} \longrightarrow{ }_{43}^{100}\) (d) \(+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{84}^{210}+{ }_{0}^{1} n\)

Write balanced nuclear equations for (a) the alpha emission resulting in the formation of \(\mathrm{Pa}-233\). (b) the loss of a positron by \(\mathrm{Y}-85\). (c) the fusion of two C-12 nuclei to give sodium-23 and another particle. (d) the fission of Pu-239 to give tin-130, another nucleus, and an excess of two neutrons.

Consider the fission reaction in which U-235 is bombarded by neutrons. The products of the bombardment are rubidium-89, cerium-144, beta particles, and more neutrons. (a) Write a balanced nuclear equation for the bombardment. (b) Calculate \(\Delta E\) when one gram of U-235 undergoes fission. (c) The detonation of TNT, an explosive, evolves \(2.76 \mathrm{~kJ} / \mathrm{g}\). How many kilograms of TNT are required to produce the same amount of energy as one milligram of \(\mathrm{U}-235\) ?

The amount of oxygen dissolved in a sample of water can be determined by using thallium metal containing a small amount of the isotope Tl- 204\. When excess thallium is added to oxygen-containing water, the following reaction occurs. $$2 \mathrm{Tl}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Tl}^{+}(a q)+2 \mathrm{OH}^{-}(a q)$$ After reaction, the activity of a 25.0-mL water sample is 745 counts per minute (cpm), caused by the presence of \(\mathrm{Tl}^{+}-204\) ions. The activity of Tl-204 is \(5.53 \times 10^{5} \mathrm{cpm}\) per gram of thallium metal. Assuming that \(\mathrm{O}_{2}\) is the limiting reactant in the above equation, calculate its concentration in moles per liter.

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