Question

The isotope of thorium ${}_{90}^{234}Th$ undergoes alpha decay according to the given equation
${}_{90}^{234}Th{\to }_z^{A}X{+}_2^{4}He.$
Find out the element X from the above nuclear fission reaction?

• A. ${}_{92}^{238}U$
• B. ${}_{88}^{230}Ra$
• C. ${}_{94}^{236}Pu$
• D. ${}_{88}^{238}Ra$
• E. ${}_{92}^{232}U$

Answer 1

The correct option is B ${}_{88}^{230}Ra$

In a nuclear reaction, the atomic number as well as the mass number is conserved.

$\therefore$ $234=A+4$ $⟹A=230$

Also, $90=Z+2$ $⟹Z=88$

Thus the element $X$ is ${}_{88}^{230}Ra$.