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Chapter 21: Problem 36

A long-cherished dream of alchemists was to produce gold from cheaper and more abundant elements. This dream was finally realized when \({ }_{80}^{198} \mathrm{Hg}\) was converted into gold by neutron bombardment. Write a balanced equation for this reaction.

Short Answer

Expert verified
The balanced nuclear equation describing the conversion of mercury into gold by neutron bombardment is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$.

Step by step solution

01

Identify the Target and Product Nuclei

The target nucleus given is ${ }_{80}^{198}\mathrm{Hg}$ (mercury) and it's being converted into gold. However, the exact isotope of gold is not provided, so it's not immediately clear what changes are occurring in the atomic and mass numbers.
02

Understand the Process of Neutron Bombardment

Neutron bombardment refers to the process of firing neutrons at a target nucleus. The neutron is represented as ${ }_{0}^{1}n$. When a neutron is absorbed by a nucleus, it increases only the mass number of the target nucleus (since neutron carries no charge and does not change the atomic number).
03

Determine the Product Nucleus

Once the mercury nucleus absorbs a neutron, it becomes an isotope of mercury with mass number 199 (${ }_{80}^{199}\mathrm{Hg}$). This isotope is unstable and undergoes beta decay, where a neutron in the nucleus is converted into a proton, increasing the atomic number by one and producing an electron (or beta particle, ${ }_{-1}^{0}β$) in the process. This results in the formation of the nucleus with atomic number 81 that is an isotope of gold ${ }_{81}^{199}\mathrm{Au}$.
04

Write Initial and Final Reactions

The initial reaction is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{80}^{199}\mathrm{Hg}$. Then the decay process can be written as ${ }_{80}^{199}\mathrm{Hg}$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$. These two steps can be combined to write the overall balanced nuclear equation.
05

Write the Overall Balanced Nuclear Reaction

The final balanced equation that describes the entire process is ${ }_{80}^{198}\mathrm{Hg}$ + ${ }_{0}^{1}n$ -> ${ }_{81}^{199}\mathrm{Au}$ + ${ }_{-1}^{0}β$.

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