Question

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

Short answer

The solutions to both parts are as follows: a. The balanced equation for the reaction is: ${}^{249}\mathrm{Cf}{+}^{18}\mathrm{O}{⟶}^{263}\mathrm{Sg}{+}^4\alpha$ b. The other product resulting from the alpha decay of ${}^{263}\mathrm{Sg}$ is: ${}^{259}\mathrm{Rf}$.

Step-by-step solution

Identify the reactants and products

In this reaction, we're given that Californium-249 (${}^{249}\mathrm{Cf}$) reacts with Oxygen-18 (${}^{18}\mathrm{O}$) to produce Seaborgium-263 (${}^{263}\mathrm{Sg}$). We need to find the balanced equation for this reaction.

Balance the equation

In a balanced nuclear reaction, the total number of protons and neutrons before the reaction (on the left side) should equal the total number of protons and neutrons after the reaction (on the right side). For the given reaction: ${}^{249}\mathrm{Cf}{+}^{18}\mathrm{O}{⟶}^{263}\mathrm{Sg}+X$ We know that Californium has 98 protons, Oxygen has 8 protons, and Seaborgium has 106 protons. Now we can calculate the number of neutrons in the missing product X, which can be a nucleus that results from the reaction. The total number of protons and neutrons before the reaction: $249+18$ = 267 The total number of protons and neutrons after the reaction (assuming one Seaborgium nucleus and one X nucleus): $263+A_X$ To balance the reaction, the sum of A values should be equal on both sides: $267=263+A_X$ Solving for $A_X$: $A_X=267-263$ $A_X=4$ Since X has 4 nucleons, and an alpha particle consists of 2 protons and 2 neutrons, the missing product X in the reaction is an alpha particle ($\alpha$). Now we can write the balanced nuclear reaction equation: ${}^{249}\mathrm{Cf}{+}^{18}\mathrm{O}{⟶}^{263}\mathrm{Sg}{+}^4\alpha$ For Part b:

Identify the decay process of Seaborgium-263

We are given that Seaborgium-263 (${}^{263}\mathrm{Sg}$) decays by alpha emission.

Find the product resulting from the alpha decay

During alpha decay, the decaying nucleus loses 2 protons and 2 neutrons as an alpha particle (${}^4\alpha$). Let's denote the mass number of the other product as $A_Y$ and its atomic number (number of protons) as $Z_Y$. Since Seaborgium-263 has a mass number of 263 and an atomic number of 106, after losing an alpha particle: $A_Y=A_{Sg}-A_{\alpha }=263-4=259$ $Z_Y=Z_{Sg}-Z_{\alpha }=106-2=104$ The other product resulting from the alpha decay of ${}^{263}\mathrm{Sg}$ is a nucleus with a mass number of 259 and an atomic number of 104, which is Rutherfordium, denoted as ${}^{259}\mathrm{Rf}$.

Key concepts

Alpha Decay

Alpha decay is a fundamental nuclear process where an unstable atom releases an alpha particle. The alpha particle is essentially a helium nucleus, consisting of 2 protons and 2 neutrons. This emission decreases the mass number of the original atom by 4 and the atomic number by 2.
In nuclear reactions, this process causes the original element to transform into a different element. For example, if seaborgium-263 ${(}^{263}\mathrm{Sg})$ undergoes alpha decay, it emits an alpha particle ${(}^4\alpha )$ and forms a new element with a mass number of 259 and an atomic number 104, which is rutherfordium ${(}^{259}\mathrm{Rf})$. Thus, alpha decay is crucial in transforming nuclei into more stable forms. Understanding it helps us predict the products of decay and balance nuclear equations effectively.

Seaborgium

Seaborgium, represented by the chemical symbol Sg and the atomic number 106, honors the chemist Glenn T. Seaborg for his work in discovering transuranium elements. This synthetic element doesn't occur naturally and is created in laboratories through complex nuclear reactions.
Creating seaborgium involves nuclear bombardment techniques, such as colliding Californium-249 ${(}^{249}\mathrm{Cf})$ with Oxygen-18 ${(}^{18}\mathrm{O})$ nuclei. This nuclear fusion yields seaborgium-263 ${(}^{263}\mathrm{Sg})$ with the subsequent emission of an alpha particle. Seaborgium is highly radioactive and serves mainly for scientific research purposes, given its instability and short half-life. Its creation marked a significant achievement in chemistry, expanding our understanding of the periodic table and nuclear chemistry.

Balanced Nuclear Equation

A balanced nuclear equation must satisfy the conservation of mass and atomic numbers. This means that the total sum of nucleons (protons and neutrons) and the total number of protons must be identical on both sides of the equation.
For example, when californium-249 ${(}^{249}\mathrm{Cf})$ is bombarded with oxygen-18 ${(}^{18}\mathrm{O})$, the resulting equation must show that the subsequential reaction products add up to the original reactants' nucleon count:$${}^{249}\mathrm{Cf}{+}^{18}\mathrm{O}{⟶}^{263}\mathrm{Sg}{+}^4\alpha$$Here, the sum of the masses before and after the reaction is 267 = 263 + 4, which balances the equation. This balance confirms that the conducted reaction properly observes both mass and charge conservation principles, ensuring scientific accuracy and validity. Utilizing these principles effectively allows chemists and physicists to explore and understand complex nuclear transformations.