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Chapter 7: Problem 9

Californium, clement number 98, was first synthesized by bombarding a target with alpha particles. The products were califormium- 245 and a neutron. What was the target isotope used in this nuclear synthesis?

Short Answer

Expert verified
The target isotope used was Curium-242 (\(^{242}_{96}\text{Cm}\)).

Step by step solution

01

Understanding the Components of the Reaction

In the given reaction, Californium is produced after bombarding a target with alpha particles. An alpha particle is composed of 2 protons and 2 neutrons, and can be designated as \(^4_2\text{He}\). The reaction results in Californium-245 \(^{245}_{98}\text{Cf}\) and a neutron.
02

Determine the Mass and Atomic Numbers in the Reaction

Before the reaction, assume the target isotope is \(^A_ZX\). After the reaction: 1. The resultant Californium (^{245}_{98}\text{Cf}) has a mass number of 245 and atomic number of 98.2. An additional neutron \(^{1}_{0}\text{n}\) is produced. 3. The sum of the mass numbers and atomic numbers must be conserved.
03

Set Up Equations for Mass and Atomic Numbers

The initial isotopes (target + alpha) must equal the final isotopes (californium + neutron). The equations are:- Mass number equation: \(A + 4 = 245 + 1\)- Atomic number equation: \(Z + 2 = 98\).
04

Solve for the Target Isotope

Solve the two equations:- Mass number equation: \(A + 4 = 246\) implies \(A = 242\).- Atomic number equation: \(Z + 2 = 98\) implies \(Z = 96\).The target isotope should be \(^{242}_{96}\text{Cm}\), where Cm stands for Curium.

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Key Concepts

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

Alpha Particles
Alpha particles are crucial components in nuclear reactions. They contain two protons and two neutrons, making their composition identical to that of a helium nucleus. Represented by the symbol 8_2 ext{He}8, alpha particles have a charge of +2, due to the two protons.

  • Being relatively heavy and positively charged, they have low penetration power. This means they can be stopped by something as thin as a piece of paper.
  • In nuclear synthesis, alpha particles are often used to transform or synthesize other elements, like in the creation of Californium.
They are important in scientific research, medical applications, and even in some types of smoke detectors!
Californium
Californium is an element with the atomic number 98. It was first synthesized by bombarding curium with alpha particles, leading to the production of Californium-245 and a free neutron. This synthetic element is part of the actinide series and plays significant roles in various fields.
  • Its primary use is in neutron sources, serving as an efficient means of neutron production due to its high rate of neutron emission.
  • Californium has applications in cancer treatment, through a process called neutron therapy.
  • It is also used in detecting metal fatigue and neutron radiography for checking welds.
Although highly useful, Californium must be handled with care given its radioactivity.
Curium
Curium, represented as 8^{242}_{96} ext{Cm}8, is an element in the actinide series with important applications. It has the atomic number 96 and was used as the target to synthesize Californium. When bombarded with alpha particles, it transforms, contributing to the formation of new elements.
  • Curium's notable property is its ability to emit alpha particles, which can be harnessed for generating heat in space missions.
  • It serves as a starting material for producing other transuranic elements like Californium.
The understanding of Curium is vital for the study of nuclear reactions and the development of new materials.
Neutron Production
Neutrons are electrically neutral particles found in the nucleus of an atom. In nuclear synthesis, neutron production is a significant outcome, as seen with the formation of a neutron when Californium is produced from Curium.

  • Free neutrons can initiate further nuclear reactions, leading to the synthesis of new elements.
  • They are extensively used in nuclear reactors as they promote fission, the process crucial for energy generation.
  • Neutron production also plays a critical role in scientific experiments, such as neutron scattering, to explore material structures.
In contexts like Californium synthesis, understanding neutron production helps link nuclear chemistry and practical applications.

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Most popular questions from this chapter

Einstein's equation, \(\Delta E=\Delta m c^{2}\) applies to chemical reactions as well as nuclear ones. An important chemical change studied in Chapter 4 was the combustion of methane, which releases \(50.1 \mathrm{~kJ}\) of energy for each gram of methane burned. a. What mass loss corresponds to the release of \(50.1 \mathrm{~kJ}\) of energy? b. To produce the same amount of energy, what is the ratio of the mass of methane burned in a chemical reaction to the mass loss converted into energy acconding to the equation \(\Delta E=\Delta m c^{2}\) ? c. Use your results in parts a and \(\mathbf{b}\) to coemment on why Einstein's eq?uation, although correct for both chemical and noclear changes, usually is only applied to nuclear changes.

A stockp?le of appraximately 50 metric tons of 20 plutonium exists in the United States as a result of disassembling warheads from the nuclear arms race. What is the likely fate of this plutonium? Hint: Search for plutomium disposal. Try also including Vinited Srares and Department of Energy in yous search string. 2\. Some propose that the plutonium be sent to local nuclear power plants to "burn \({ }^{-}\)as fissionable fuel. What are the advantages and disadvantages of sech a course of action? b. Ouhers propase that it be stored permanently in a repository. Again, list the advantages and clisadvantages.

\(E=m c^{2}\) is one of the most famous equations of the 20th century. Explain the meaning of each symbol in it.

Radioactive decay is accocupanied by a change in the mass number, a change in the atomic number, a change in both, or a change in neither. For the following types of radioactive decay, which change(s) do you expect? a. alpha emission b. beta emission c. gamma emission

The representations \({ }^{14} \mathrm{~N}^{15} \mathrm{~N}\) give more information than simply the chemical symbol N. Explain.
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