Question
The tungsten-180 nuclide radioactively decays by alpha emission. Write a balanced nuclear chemical equation that describes this process.
Answers
GPT-4o mini
When tungsten-180 (\(^{180}_{74}\text{W}\)) undergoes alpha decay, it emits an alpha particle (\(^{4}_{2}\text{He}\)) and transforms into a new nuclide. The atomic number decreases by 2 and the mass number decreases by 4.
The balanced nuclear chemical equation for the alpha decay of tungsten-180 can be written as:
\[
^{180}_{74}\text{W} \rightarrow ^{176}_{72}\text{Hf} + ^{4}_{2}\text{He}
\]
In this equation:
- \(^{180}_{74}\text{W}\) is the original tungsten-180 nuclide.
- \(^{176}_{72}\text{Hf}\) is the resulting hafnium-176 nuclide after the decay.
- \(^{4}_{2}\text{He}\) represents the emitted alpha particle (helium nucleus).
The balanced nuclear chemical equation for the alpha decay of tungsten-180 can be written as:
\[
^{180}_{74}\text{W} \rightarrow ^{176}_{72}\text{Hf} + ^{4}_{2}\text{He}
\]
In this equation:
- \(^{180}_{74}\text{W}\) is the original tungsten-180 nuclide.
- \(^{176}_{72}\text{Hf}\) is the resulting hafnium-176 nuclide after the decay.
- \(^{4}_{2}\text{He}\) represents the emitted alpha particle (helium nucleus).