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Chapter 18: Problem 5

What are transuranium elements and how are they synthesized?

Short Answer

Expert verified
Transuranium elements are artificial elements with atomic numbers greater than 92 (uranium). They are synthesized through nuclear reactions involving the bombardment of a target nucleus with particles like neutrons, protons, or heavier nuclei to create larger nuclei with higher atomic numbers. The process involves choosing a suitable target nucleus and bombarding particles, accelerating the particles using a particle accelerator, causing the desired nuclear reaction, and finally detecting and identifying the newly created transuranium element through various detectors and spectroscopic techniques.

Step by step solution

01

Introduction to Transuranium Elements

Transuranium elements are chemical elements that have atomic numbers higher than the element uranium (atomic number 92). These elements do not occur naturally on Earth and are created artificially in laboratories through nuclear reactions.
02

Different Synthesis Methods

There are several methods to synthesize transuranium elements, including: 1. Neutron bombardment: Bombarding a target nucleus with neutrons to form a larger nucleus. 2. Proton bombardment: Bombarding a target nucleus with protons to increase its atomic number. 3. Heavy ion bombardment: Bombarding a target nucleus with heavy ions such as alpha particles or other nuclei. In all these methods, the basic idea is to combine the target nucleus with the bombarding particles in such a way that the resulting nucleus has a higher atomic number than the original target nucleus.
03

Selection of Target Nucleus and Bombarding Particles

First, we need to choose a target nucleus and appropriate bombarding particles to create a specific transuranium element. The choice depends on the desired synthetic route and the specific element that we want to synthesize.
04

Accelerating Bombarding Particles

The bombarding particles are accelerated to extremely high speeds to overcome the electrostatic repulsion between the positively charged particles and target nucleus. This is done using a particle accelerator, such as a cyclotron or a linear accelerator.
05

Nuclear Reaction

The accelerated particles collide with the target nucleus, and if the conditions are right, they will overcome the repulsion and combine with the target nucleus. The result is a new nucleus with a higher atomic number, creating a transuranium element. However, several different reactions might occur, so the desired reaction might not always be the most probable one.
06

Detection and Identification

After the nuclear reaction, the newly created nucleus may undergo radioactive decay, emitting various forms of radiation. These radiations can be detected and measured to determine the atomic number and other properties of the newly synthesized transuranium element. This is typically done using detectors and spectroscopic methods, such as gamma-ray spectroscopy or mass spectrometry. In conclusion, transuranium elements are artificially synthesized elements with atomic numbers greater than 92. They are created through various nuclear reactions, such as neutron, proton, or heavy-ion bombardment, followed by the detection and identification of the newly created nuclei.

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Most popular questions from this chapter

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

The only stable isotope of fluorine is fluorine-19. Predict possible modes of decay for fluorine-21, fluorine-18, and fluorine-17.

Photosynthesis in plants can be represented by the following overall equation: $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Light }}{\longrightarrow} C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Algae grown in water containing some \(^{18} \mathrm{O}\) (in \(\mathrm{H}_{2}^{18} \mathrm{O}\) ) evolve oxygen gas with the same isotopic composition as the oxygen in the water. When algae growing in water containing only \(^{16} \mathrm{O}\) were furnished carbon dioxide containing \(^{18} \mathrm{O},\) no \(^{18} \mathrm{O}\) was found to be evolved from the oxygen gas produced. What conclusions about photosynthesis can be drawn from these experiments?

Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \(^{1}_{1}{H}, 1.00782 \mathrm{u} ; \frac{2}{1} \mathrm{H}, 2.01410 \mathrm{u} ;\) and an electron, \(5.4858 \times\) \(10^{-4}\) u. (Hint: Think carefully about how to account for the electron mass.)$$\mathrm{i} \mathrm{H}+\mathrm{i} \mathrm{H} \longrightarrow_{\mathrm{i}}^{2} \mathrm{H}+_{+\mathrm{i}}^{0}$$

Breeder reactors are used to convert the nonfissionable nuclide \(\frac{238}{92} \mathrm{U}\) to a fissionable product. Neutron capture of the \(\frac{238}{92} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?
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