Question

When ${}_{88}^{226}Ra$ emits an $\alpha$-particle, the new element formed belongs to:

• A. third group
• B. zero group
• C. fourth group
• D. second group

Answer 1

The correct option is B zero group

When $\alpha$ particle is emitted, the new element has atomic number $2$ less than original element. Thus atomic number of product is $88-2=86.86$ is atomic number of Radon (Rn) which is inert gas and hence belongs to $0^{th}\left({18}^{th}\right)$ group.