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.

Answer 1

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:

\[
^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He
\]