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Chapter 25: Problem 14

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

Expert verified
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

01

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.
02

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.
03

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.
04

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.

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