Question

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn 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. \(^{263}\) Sg decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg} ?\)

Short answer

The balanced nuclear reaction equation for the production of Sg-263 is: \(_{98}^{249}\mathrm{Cf} + _{8}^{18}\mathrm{O} \rightarrow _{106}^{263}\mathrm{Sg} + _{2}^4\mathrm{He}\). The other product resulting from the alpha decay of \(^{263}\mathrm{Sg}\) is \(^{259}\mathrm{Rf}\) (rutherfordium).

Step-by-step solution

Write the nuclear reaction formula for the production of Sg-263

To find the reaction equation that resulted in the formation of Sg-263, first write down the initial reactants and their respective atomic numbers and mass numbers. We have californium-249 (\(^{249} \mathrm{Cf}\)) with an atomic number of 98, and oxygen-18 (\(^{18} \mathrm{O}\)) with an atomic number of 8: \(_{98}^{249}\mathrm{Cf} + _{8}^{18}\mathrm{O} \rightarrow\) Now, we need to find the atomic number and mass number of the produced Sg-263 isotope.

Calculate the atomic number and mass number of Sg-263

For the atomic number, simply add the atomic numbers of the initial reactants: \(98 + 8 = 106\) The atomic number of seaborgium is 106. The mass number is obtained by adding the mass numbers of the initial reactants: \(249 + 18 = 267\) The mass number of Sg-263 should be 263. So, we need to account for the mass not equal to 263 by adding a particle as a product. The difference is: \(267 - 263 = 4\) This excess mass belongs to a particle with a mass number of 4.

Complete the nuclear reaction formula

Identify the particle with the mass number of 4, which is an alpha particle (\(^4_2\mathrm{He}\)). Complete the nuclear reaction formula by adding the particle as a product: \(_{98}^{249}\mathrm{Cf} + _{8}^{18}\mathrm{O} \rightarrow _{106}^{263}\mathrm{Sg} + _{2}^4\mathrm{He}\) Now, we have the balanced nuclear reaction equation. For part b, we need to find the other product after the alpha decay of \(^{263}\mathrm{Sg}\).

Alpha decay of \(^{263}\mathrm{Sg}\)

Sg-263 undergoes alpha decay, which means that it emits an alpha particle (\(^4_2\mathrm{He}\)). Write the alpha decay equation: \(_{106}^{263}\mathrm{Sg} \rightarrow _{Z}^{A}\mathrm{X} + _{2}^4\mathrm{He}\) Where \(\mathrm{X}\) is the resulting nucleus after the alpha decay, and Z and A are the unknown atomic and mass numbers.

Calculate the atomic number and mass number of the resulting nucleus

To find the atomic and mass numbers of the resulting nucleus, simply subtract the atomic and mass numbers of the alpha particle from the atomic and mass numbers of Sg-263: \(Z = 106 - 2 = 104\) \(A = 263 - 4 = 259\)

Identify the resulting nucleus after the alpha decay of \(^{263}\mathrm{Sg}\)

Now, using the calculated atomic number (104) and mass number (259), we can identify the resulting nucleus: \(_{106}^{263}\mathrm{Sg} \rightarrow _{104}^{259}\mathrm{Rf} + _{2}^4\mathrm{He}\) The other product resulting from the alpha decay of \(^{263}\mathrm{Sg}\) is \(^{259}\mathrm{Rf}\) (rutherfordium).

Key concepts

Transuranium Elements

Transuranium elements are those elements in the periodic table with atomic numbers greater than that of uranium, which has an atomic number of 92. These elements are not found naturally on Earth and are synthesized through nuclear reactions, typically by bombarding lighter elements with neutrons or charged particles. Element 106, named seaborgium (Sg), is a part of this group and stands out as it was named after Glenn T. Seaborg, himself an instrumental figure in the discovery of many transuranium elements.

One effective method to produce transuranium elements, including seaborgium, involves using a particle accelerator to strike a target, like californium, with high-speed particles, such as oxygen nuclei. The delicate balance in creating these heavy, and often highly unstable, elements requires precise conditions and the right combination of reactants, highlighting the intricate nature of nuclear synthesis.

• They extend the periodic table beyond uranium (atomic number 92).
• Produced in nuclear reactors or particle accelerators.
• They have no stable isotopes and decay radioactively.
• Often named in honor of influential scientists or places associated with their synthesis.

Alpha Decay

Alpha decay is a type of radioactive decay in which an unstable nucleus emits an alpha particle, consisting of two protons and two neutrons, effectively reducing the element's atomic number by 2 and its mass number by 4. In our exercise, cf{^{263}_{106}Sg} undergoes alpha decay, leading to the emission of an alpha particle and transforming into a new element with lower atomic mass. This process is described by the equation:

cf{_{106}^{263}Sg -> _{104}^{259}Rf + _2^4He}

The alpha particle is identical to the nucleus of a helium-4 atom, and through this emission, the original element, seaborgium in this case, becomes a different element—in this instance, rutherfordium (Rf).

• One of the most common modes of radioactive decay.
• Produces a new element with an atomic number reduced by 2 and a mass number reduced by 4.
• Alpha particles are helium nuclei.
• Emphasized in safety procedures as alpha particles can be harmful if ingested or inhaled.

Isotope Production

Isotope production is central to nuclear chemistry and has practical applications in medicine, industry, and research. An isotope is an atom with the same number of protons (same atomic number) but a different number of neutrons, resulting in a different mass number. In the given problem, the isotope cf{^{263}Sg} was produced by a nuclear reaction between cf{^{249}Cf} and cf{^{18}O}, where the sum of the reactants' mass numbers (249 and 18) initially exceeded the desired mass number (263). This discrepancy is accounted for by the ejection of an alpha particle, allowing the reaction to balance and produce the intended cf{Sg} isotope. It's a delicate process that requires precise engineering to calculate and control the various conditions that lead to the successful production of a new isotope.

Key Points in Isotope Production

• Used in a wide variety of applications, including medical imaging and cancer treatment.
• It involves altering the number of neutrons in the nucleus of an atom.
• Requires precise control of nuclear reactions.
• Can involve various methods, such as neutron bombardment or particle accelerator collisions.

Understanding and mastering the production of isotopes not only advances scientific knowledge but also opens up a range of possibilities for the practical application of these isotopes, reflecting the fundamental importance of nuclear reactions in science and technology.