In a radioactive decay chain, the initial nucleus is \[{}_{90}^{232}Th\]. At the end there are 6α− particles and 4β− particles which are emitted. If the end nucleus, If ${}_Z^AX$, A and Z are given by:
A. A=208;Z=80
B. A=202;Z=80
C. A=200;Z=81
D. A=208;Z=82

Hint: In this question, we need to determine the atomic number and the mass number of the resulting nucleus when six alpha and four beta particles have been emitted from the nucleus \[{}_{90}^{232}Th\]. For this, we will use the property of the alpha and the beta particles along with the mass number and the atomic number of the same.

Complete step by step answer:
When an alpha particle has been emitted from the nucleus of an atom then, it takes four units of mass number and two units of the atomic number with it. Mathematically, \[{}_b^aX \to {}_{b - 2}^{a - 4}Y\] where, a and b are the atomic number and the mass number respectively of the nucleus X and when an alpha particle has been emitted from X then, (a-4) and (b-2) are the mass number and the atomic number and respectively of the nucleus Y.
According to the question, six alpha particle have been emitted from \[{}_{90}^{232}Th\] which will result in a new particle with (m-4(6)) as the mass number and (n-2(6)) as the atomic number.
$
  {}_n^mX - 6\alpha \to {}_{n - 2(6)}^{m - 4(6)}Z \\
  {}_{90}^{232}Th - 6\alpha \to {}_{90 - 12}^{232 - 24}Z \\
   \implies {}_{78}^{208}Z - - - - (i) \\
 $
When a beta particle has been emitted from the nucleus of an atom then, it gives one unit of the atomic number to the associated nucleus while keeping the mass number constant. Mathematically, ${}_b^aX - \beta \to {}_{b + 1}^aY$ where a and b are the atomic number and the mass number respectively of the nucleus X and when a beta particle has been emitted from X then, a and (b+1) are the mass number and the atomic number and respectively of the nucleus Y.
According to the question, four beta particles are emitted from the nucleus \[{}_{90}^{232}Th\], which will result in a new particle with (m) as the mass number and (n+2(4)) as the atomic number.
$
 {}_n^mX - 4\beta \to {}_{n + 1(4)}^mY \\
 {}_{90}^{232}Th - 4\beta \to {}_{90 + 1(4)}^{232}Y \\
 \implies {}_{94}^{232}Y - - - - (ii) \\
 $

Now, from equations (i) and (ii), we can say that
${}_{90}^{232}Th - 6\alpha - 4\beta \to {}_{82}^{208}X$
Hence, the resultant nucleus is ${}_{82}^{208}X$ when one alpha particle and two beta particles are emitted from \[{}_{90}^{232}Th\].

So, the correct answer is “Option D”.

Note:
We have added two units of the atomic number in the second case as two beta particles were emitted (so, one atomic number for each beta particle has been added). Moreover, students should be careful while applying the addition and the subtraction in the atomic and the mass numbers.