How many beta particle(s) are emitted in the radioactive decay of $^{198}_{79}Au \leftarrow ^{198}_{80}Hg$?

Analyze the question

The notation ¹⁹⁸₇₉Au ← ¹⁹⁸₈₀Hg is read with the arrow pointing from the parent to the daughter. So the parent is mercury-198 (Z = 80) and the daughter is gold-198 (Z = 79). We need to find how many beta particles are emitted in this change.

Concept

• Beta-minus (β⁻) decay: a neutron changes to a proton, so the atomic number increases by 1 and the mass number stays the same.
• Beta-plus (β⁺) decay (positron emission) or electron capture: a proton changes to a neutron, so the atomic number decreases by 1 and the mass number stays the same.
• Alpha decay changes the mass number by 4, which is not the case here.

Solution

The mass number stays at 198, so there is no alpha emission.

The atomic number goes from 80 to 79, a decrease of 1. This is a change of one unit in Z, so exactly one beta-type particle is involved. Since Z decreases, it is a positron (β⁺) emission:

¹⁹⁸₈₀Hg → ¹⁹⁸₇₉Au + ⁰₊₁e (+ neutrino)

Check: mass number 198 = 198 + 0, and charge 80 = 79 + 1. ✓

Only one beta particle is emitted, whatever its sign. Had the reaction been Au → Hg, it would have been ordinary β⁻ decay, and the count would still be 1.

Answer: A. 1

Tip: For beta decay, count the change in atomic number. A change of 1 unit with no change in mass number means 1 beta particle. Each β⁻ raises Z by 1, and each β⁺ lowers Z by 1.