The equation you've provided appears to represent a type of nuclear decay, specifically alpha decay. However, it looks like there might be a minor discrepancy in the particle notation.
The decay of radon-223 ( \( ^{223}{88}\text{Rn} \) ) typically releases an alpha particle (which is helium-4, \( ^{4}{2}\text{He} \) ) and results in the formation of another element. In this case, the notation seems to be indicating the formation of radon-219 ( \( ^{219}_{86}\text{Rn} \) ), which is actually incorrect based on standard decay processes.
The proper decay equation for the alpha decay of radon-223 would look like this:
\[ ^{223}{88}\text{Rn} \rightarrow ^{219}{86}\text{Rn} + ^{4}_{2}\text{He} \]
Here’s how to balance it:
- The mass numbers on the left add up to 223.
- The mass numbers on the right add up to \( 219 + 4 = 223 \).
- The atomic numbers also balance: \( 88 = 86 + 2 \).
Therefore, the equation you've written is not fully correct, but with the minor adjustment to include the alpha particle, it correctly represents the process.