Question

If a radioactive element ${}_{79}Au$ emits $\beta$-particle, what is the new group? ${}_{79}Au$ belongs to group I B (copper family)?

• A. I A
• B. I B
• C. II A
• D. II B

Answer 1

The correct option is D II B

${}_{79}Au{⟶}_{80}Hg+{\beta }^{-}$. ${}_{80}Hg$ belongs to II B(zinc family). There is no change in the number of moles since atomic mass remains unchanged.