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.