The nuclides of ^{197}_{79} text{Au} and ^{19 | vedclass

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(B) To determine the relationship,we calculate the number of neutrons $(N)$ for each nuclide using the formula $N = A - Z$,where $A$ is the mass numbe

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isotones

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(B) To determine the relationship,we calculate the number of neutrons $(N)$ for each nuclide using the formula $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.For $^{197}_{79} 	ext{Au}$: $A = 197$,$Z = 79$. Thus,$N = 197 - 79 = 118$.For $^{198}_{80} 	ext{Hg}$: $A = 198$,$Z = 80$. Thus,$N = 198 - 80 = 118$.Since both nuclides have the same number of neutrons $(N = 118)$,they are classified as isotones.

The nuclides of $^{197}_{79} \text{Au}$ and $^{198}_{80} \text{Hg}$ are called . . . . . . of each other.

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