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Chapter 19: Problem 21

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn \(\mathrm{T}\). Seaborg, discoverer of the transuranium elements. a. \({ }^{263} \mathrm{Sg}\) was produced by the bombardment of \({ }^{249} \mathrm{Cf}\) with a beam of \({ }^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \({ }^{263} \mathrm{Sg}\) decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \({ }^{263} \mathrm{Sg}\) ?

Short Answer

Expert verified
a. The balanced nuclear equation for the reaction is \({ }^{249} \mathrm{Cf}\) + \({ }^{18} \mathrm{O}\) 鈫� \({ }^{263} \mathrm{Sg}\) + \({ }^{4}\mathrm{n}\). b. The other product resulting from the 伪 decay of \({ }^{263} \mathrm{Sg}\) is \({ }^{259} \mathrm{Rf}\) (Rutherfordium).

Step by step solution

01

Part a: Writing the balanced nuclear equation

To write the balanced nuclear equation, we will follow the law of conservation of mass and particles (protons and neutrons). We are given the reaction \({ }^{249} \mathrm{Cf}\) + \({ }^{18} \mathrm{O}\) 鈫� \({ }^{263} \mathrm{Sg}\). We need to find the unknown product which we will denote by X. Let's follow the law of conservation of mass and particles step-by-step: 1. Conservation of protons: The number of protons in the reactants must be equal to the number of protons in the products. The reactants are \({ }^{249} \mathrm{Cf}\) (which has 98 protons) and \({ }^{18} \mathrm{O}\) (which has 8 protons). The total number of protons in the reactants is \(98 + 8 = 106\). The product \({ }^{263} \mathrm{Sg}\) has 106 protons. Since the number of protons is conserved, the other product X should have 0 protons. 2. Conservation of nucleons (protons and neutrons): A similar way, the total number of nucleons (protons and neutrons) should be conserved throughout the reaction. The reactants have a combined total of \(249 + 18 = 267\) nucleons. The product \({ }^{263} \mathrm{Sg}\) has 263 nucleons, so the other product, X, must have \(267 - 263 = 4\) nucleons. Since the product X has 0 protons, it must have 4 neutrons. Therefore, X is \({ }^{4}\mathrm{n}\). Now we can write the complete balanced nuclear equation: \({ }^{249} \mathrm{Cf}\) + \({ }^{18} \mathrm{O}\) 鈫� \({ }^{263} \mathrm{Sg}\) + \({ }^{4}\mathrm{n}\)
02

Part b: Finding the other product after the 伪 decay of \({ }^{263} \mathrm{Sg}\)

An 伪 decay is the emission of an alpha particle, which consists of two protons and two neutrons (\({ }^{4}\mathrm{He}\)). To find the other product, we can write the balanced equation for the decay process. \({ }^{263} \mathrm{Sg}\) 鈫� 伪 + Z We need to find the unknown product Z. 1. Conservation of protons: The product \({ }^{263} \mathrm{Sg}\) has 106 protons. Since 2 protons are emitted in the form of an alpha particle, the unknown product Z must have \(106 - 2 = 104\) protons. 2. Conservation of nucleons: In the same way, the product \({ }^{263} \mathrm{Sg}\) has 263 nucleons. Since 4 nucleons are emitted in the form of an alpha particle, the unknown product Z must have \(263 - 4 = 259\) nucleons. Finally, the unknown product Z is \({ }^{259} \mathrm{Rf}\) (Rutherfordium). This is the other product resulting from the 伪 decay of \({ }^{263} \mathrm{Sg}\).

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Most popular questions from this chapter

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \(^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

A recently reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is \(15.0\) seconds. If 199 atoms of bohrium- 267 could be synthesized, how much time would elapse before only 11 atoms of bohrium- 267 remain? What is the expected electron configuration of elemental bohrium?

Which do you think would be the greater health hazard: the release of a radioactive nuclide of Sr or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chemical properties of \(\mathrm{Sr}\) and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

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

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 \(\left(t_{1 / 2}=3.11\right.\) 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 \({ }^{222} \mathrm{Rn}\) levels not exceed \(4 \mathrm{pCi}\) 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 \({ }^{222} \mathrm{Rn}\) per liter of air.
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