Question
Suppose a uranium-238 nuclide decays into a thorium-234 nuclide by emitting an alpha particle.
Complete the nuclear chemical equation below so that it describes this nuclear reaction.
Complete the nuclear chemical equation below so that it describes this nuclear reaction.
Answers
GPT-4o mini
In a nuclear decay process, uranium-238 (U-238) emits an alpha particle (which is essentially a helium-4 nucleus, \( ^4_2He \)), resulting in the formation of thorium-234 (Th-234).
The nuclear decay reaction can be represented as:
\[
^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He
\]
To summarize:
- Uranium-238 (\(^{238}_{92}U\)) decays into thorium-234 (\(^{234}_{90}Th\)) and emits an alpha particle (\(^{4}_{2}He\)).
- The atomic number of uranium decreases by 2 (from 92 to 90), and the mass number decreases by 4 (from 238 to 234).
So the completed equation is:
\[
^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He
\]
The nuclear decay reaction can be represented as:
\[
^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He
\]
To summarize:
- Uranium-238 (\(^{238}_{92}U\)) decays into thorium-234 (\(^{234}_{90}Th\)) and emits an alpha particle (\(^{4}_{2}He\)).
- The atomic number of uranium decreases by 2 (from 92 to 90), and the mass number decreases by 4 (from 238 to 234).
So the completed equation is:
\[
^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He
\]