Question

Helium decays to form lithium. Which equation correctly describes this decay?

Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 3 Subscript 3 Baseline Upper L i + Superscript 0 Subscript negative 1 Baseline e

Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper H + Superscript 0 Subscript negative 1 Baseline e

Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper L i + Superscript 0 Subscript negative 1 Baseline e

Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper L i + Superscript 0 Subscript 1 Baseline e

Answer 1

The correct equation is:

Superscript 6 Subscript 2 He → Superscript 6 Subscript 3 Li + Superscript 0 Subscript −1 e

Explanation: This is β− (electron) decay. The mass number stays 6 while the atomic number increases by 1 (He Z=2 → Li Z=3); the emitted beta particle has mass number 0 and charge −1, conserving both mass number and charge.