A nucleus \[{}_Z^AX\] emits an alpha particle. The resultant nucleus emits a \[\beta + \] particle. The respective atomic and mass no. of the final nucleus will be
A. \[Z - 3;A - 4\]
B. \[Z - 1;A - 4\]
C. \[Z - 2;A - 4\]
D. \[Z;A - 2\]

Hint: In the given question, we need to find the respective atomic and mass no. of the final nucleus. For this, we have to define the given situation and the process during \[\beta + \] emission mathematically to get the desired result.

Complete step by step solution:
We know that a nucleus \[{}_Z^AX\] emits an alpha particle and also the resultant nucleus emits a \[\beta + \] particle. So, after emission of an alpha particle, atomic mass number decreases by four units as well as atomic number decreases by two units. Hence, we can say that,
\[{}_Z^AX \to {}_2^4He + {}_{Z - 2}^{A - 2}Y\]
Hence, this gives;
\[{}_{Z - 2}^{A - 4}Y \to {e^ + } + {}_{Z - 3}^{A - 4}Y'\]
During emission of a \[\beta + \] particle, we get \[{}_1{P^1} \to {}_0{n^1} + {\beta ^ + }\]
Thus, we can say that a proton can become a neutron. As a result, the charge number drops by one, but the mass number stays the same.

Therefore, the correct option is A.

Additional Information : A pair of protons and two neutrons combine to form an alpha particle. Both protons and neutrons are tightly bonded to one another and combine to form a single particle. The resultant particle and the helium-\[4\] nucleus are identical. Typically, alpha decay results in the production of alpha particles.

Note: To solve this question, it is necessary to give an explanation using mathematical expression. That is we need to define emission of alpha particles and beta plus emission using mathematical expression to get the clear idea of a solution.