How many beta particle(s) are emitted in the radioactive decay of \(^{198}_{79}Au \leftarrow ^{198}_{80}Hg\)?
Question 1 Report
How many beta particle(s) are emitted in the radioactive decay of \(^{198}_{79}Au \leftarrow ^{198}_{80}Hg\)?
Answer Details
In the given radioactive decay equation, beta decay occurs when a neutron in the nucleus of Hg-198 is converted into a proton, causing the atomic number to increase by one and the atomic mass to remain the same. In beta decay, a high-energy electron is emitted from the nucleus.
The equation for beta decay is:
n → p + e- + anti-νe
where n represents a neutron, p represents a proton, e- represents an electron, and anti-νe represents an antineutrino.
Therefore, in the given decay equation, one beta particle (electron) will be emitted as a neutron is converted to a proton.
Hence, the answer is 1.
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