Question

$\underset{92}{238}U\left(IIIB\right)$ changes to $\underset{90}{284}Th$ by emission of $\alpha$-particle. Daughter element will be in:

• A. $IIIB$
• B. $IB$
• C. $VB$
• D. $IIA$

Answer 1

The correct option is A $IIIB$

Reaction:
Element $(58-71)$ and $(90-103)$ are placed in$IIIB$.
$\underset{IIIB}{\underset{92}{238}}U\to \underset{IIIB}{\underset{90}{234}}Th+\alpha$