Question

If $_{92}U(IIIB)$ changes to $_{90}Th$ by emission of $\alpha $ - particle. Daughter element will be in?
A. $IB$
B. $IIA$
C. $IIIB$
D. $VB$

Answer 1

Hint: We should always follow the nomenclature while writing IUPAC names of hydrocarbons. Nomenclature is the set of rules or systems of naming organic compounds. For naming organic compounds there are some sets of rules by which we can name the above compound properly as $4,4$- Diethyl-heptane.This question gives the knowledge about radioactive decay. Radioactive decay is the process of impulsive breakdown of atomic nucleus causing the discharge of matter and energy from the nucleus.

Complete step by step answer:
During alpha decay, alpha particles are produced from the nucleus by reducing its atomic number by two. This process leads to the reduction of atomic number by two and the reduction of atomic mass number by four of a newly formed atom. The produced alpha particle is also called a helium nucleus. Alpha particles generally acquire positive charge and their charge is also high.
The general equation for the process of alpha decay is as follows:
$_Z^AX \to _{Z - 2}^{A - 4}Y + _2^4He$
Where $_2^4He$ is the emitted alpha particle, $_{Z - 2}^{A - 4}Y$ is the daughter nucleus, $_Z^AX$ is the parent nucleus, $Z$ is the total number of protons and $A$ is the total number of nucleons.
According to question, the alpha decay equation for Uranium is as follows:
$_{92}^{238}U \to _{90}^{234}Th + _2^4He$
Here the alpha decay process leads to the reduction of atomic number by two and the reduction of atomic mass number by four of a newly formed Thorium atom.
After producing alpha particles, the atomic number of daughter nucleus will belong to $IIA$.

So, the correct answer is Option B.

Note: Always remember that the radioactive isotope consists of unstable nuclei which does not contain sufficient binding energy to embrace the nucleus together. The produced alpha particle during the process of alpha decay is also called a helium nucleus.