Question

The radio nuclide ${}_{90}^{234}Th$ undergoes two successive $\beta$-decays followed by one $\alpha$-decay. The atomic number and mass number of the resulting radio nuclide are:

• A. $92,234$
• B. $94,230$
• C. $90,230$
• D. $92,230$

Answer 1

Answer: (C)

$90,230$

$\alpha$-decay

${}_Z^{A}X\underrightarrow{-\alpha }{}_{Z-2}^{A-4}Y$

$\beta$-decay

${}_Z^{A}X\underrightarrow{-\beta }{}_{Z+1}^{A}Y$

${}_{90}^{234}Th\underrightarrow{-2\beta }{}_{92}^{234}U\underrightarrow{-\alpha }{}_{90}^{230}Th$

Atomic number of resulting radio nuclide : 90

Mass number of resulting radio nuclide : 230