Question

Write nuclear equations to represent the formation of a hypothetical isotope of element 118 with a mass number of 293 by the bombardment of lead- 208 by krypton-86 nuclei, followed by a chain of \(\alpha\) -particle emissions to the element seaborgium.

Short answer

The nuclear reaction sequence starts with the formation of the element \(^{293}_{118}X\) through bombardment of lead with krypton. This element then undergoes a series of alpha particle emissions, each reducing its atomic number by 2 and mass number by 4, until eventually transforming into the element seaborgium.

Step-by-step solution

Formation of element 118

To form element 118 (with a mass number of 293), lead-208 is bombarded by krypton-86. This results in the following nuclear equation: \[ ^{208}_{82}Pb + ^{86}_{36}Kr \rightarrow ^{293}_{118}X + 1n \] Here \(^{208}_{82}Pb\) is lead-208, \(^{86}_{36}Kr\) is krypton-86, \(^{293}_{118}X\) represents the new element 118 formed (hypothetical), and \(1n\) is a neutron.

First alpha particle emission

Now the newly formed element undergoes a sequence of alpha particle emissions to eventually turn into the element seaborgium. In an alpha decay, the nucleus emits an alpha particle, which is a helium-4 nucleus \((^{4}_{2}He)\). The atomic number decreases by 2 and the mass number decreases by 4. Thus the first emission can be represented as: \[^{293}_{118}X \rightarrow ^{289}_{116}Y + ^{4}_{2}He\], where \(^{289}_{116}Y\) represents the new element formed.

Continued alpha particle emissions

The process continues with more alpha particle emissions until Seaborgium is formed. The final equation depends on the exact decay chain of the element, but the process remains the same in each step: the atomic number decreases by 2 and the mass number decreases by 4. This will continue until the atomic number is 106, which represents Seaborgium.

Formulate the final sequence

The entire sequence of reactions from lead-208 to seaborgium can now be assembled together as follows: \[ ^{208}_{82}Pb + ^{86}_{36}Kr \rightarrow ^{293}_{118}X + 1n \rightarrow ^{289}_{116}Y + ^{4}_{2}He \rightarrow ....\rightarrow ^{A}_{106}Sg + n( ^{4}_{2}He)\] where A is the mass number of the final Seaborgium isotope and n is the number of alpha particles emitted.

Key concepts

Isotope Formation

Isotopes are versions of elements that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers.

When forming new isotopes using nuclear reactions, scientists often bombard target nuclei with other smaller nuclei. For instance, in our exercise, lead-208 is struck with krypton-86 nuclei to form a new, hypothetical isotope of element 118 with a mass number of 293. This process changes the number of neutrons and protons, resulting in a new isotope, distinct from naturally occurring elements. Such experiments help in expanding our understanding of chemical elements at the nuclear level.

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus emits an alpha particle. An alpha particle is essentially a helium-4 nucleus, consisting of 2 protons and 2 neutrons.

When a nucleus undergoes alpha decay, its atomic number decreases by 2, and its mass number drops by 4. This transformation leads to the creation of a new element. The initial nucleus loses mass and turns into a lighter, often more stable, nucleus.

• For example, an isotope with an atomic number of 118 becomes an element with an atomic number of 116 after emitting an alpha particle.

This process continues in chain reactions, often until a stable or significantly different element is formed.

Nuclear Equation

Nuclear equations represent the changes in the makeup of an atomic nucleus during nuclear reactions. They show the initial and final isotopes as well as the particles emitted or absorbed. These equations adhere to the conservation laws: the total number of protons and neutrons must remain the same before and after the reaction.

In our example, we can write a nuclear equation to illustrate the bombardment of lead-208 by krypton-86 to form element 118: \[ ^{208}_{82}Pb + ^{86}_{36}Kr \rightarrow ^{293}_{118}X + 1n \] Where lead and krypton react to produce a new element 118 and a neutron. Writing such equations is fundamental to understanding how elements transform at the nuclear level.

Seaborgium

Seaborgium is an element with the symbol Sg and an atomic number of 106. It is named after Glenn T. Seaborg, a noted chemist who made significant contributions to the discovery of transuranium elements.

Though seaborgium doesn’t naturally occur on Earth, it can be produced in a laboratory setting through nuclear reactions. This is done by bombarding lighter elements in particle accelerators. In our problem, seaborgium is eventually formed after a series of alpha decays from a heavier radioisotope.

• These transformations reveal fascinating insights into the stability and properties of synthesized elements, pushing the boundaries of modern chemistry.

While seaborgium itself may not have widespread practical applications, its study enhances our understanding of chemical behavior at the atomic level.