In a radioactive decay chain, the initial nucleus is $$^{232}_{90}Th$$. At the end, there are 6 $$\alpha$$-particles and 4$$\beta$$-particles which are emitted. If the end nucleus is $$^A_Z X$$, A and Z are given by:

We have an initial thorium nucleus written as $$^{232}_{90}\mathrm{Th}$$. In the course of its decay it emits six $$\alpha$$-particles and four $$\beta$$-particles. We will now track, one by one, how the mass number $$A$$ and the atomic number $$Z$$ change.

First let us recall the effect of each type of emission:

For a single $$\alpha$$-decay we use the relation

$$A\;\longrightarrow\;A-4,\qquad Z\;\longrightarrow\;Z-2$$

because an $$\alpha$$-particle is $$^{4}_{2}\mathrm{He}$$, carrying away 4 units of mass number and 2 units of atomic number.

For a single $$\beta^{-}$$-decay (electron emission) the rule is

$$A\;\longrightarrow\;A,\qquad Z\;\longrightarrow\;Z+1$$

since a neutron changes into a proton, leaving the mass number unchanged while increasing the atomic number by 1.

Now we apply these changes step by step to the original nucleus.

The starting values are

$$A_{\text{initial}} = 232,\qquad Z_{\text{initial}} = 90.$$

Six $$\alpha$$-decays occur. For the mass number we subtract 4 six times:

$$A_{\text{after }6\alpha} = 232 - 6\times4.$$ Substituting the numbers, $$A_{\text{after }6\alpha} = 232 - 24 = 208.$$

The atomic number simultaneously falls by 2 for each of the six emissions:

$$Z_{\text{after }6\alpha} = 90 - 6\times2.$$ Hence $$Z_{\text{after }6\alpha} = 90 - 12 = 78.$$

Next we deal with the four $$\beta^{-}$$-decays. The mass number remains the same through all $$\beta$$ emissions, so

$$A_{\text{final}} = 208.$$

The atomic number, however, rises by 1 for each emitted $$\beta$$ particle:

$$Z_{\text{final}} = 78 + 4\times1.$$ Therefore $$Z_{\text{final}} = 78 + 4 = 82.$$

Putting these results together, the end product nucleus is

$$^{208}_{\;82}X.$$

Looking at the options, we see that $$A = 208$$ and $$Z = 82$$ correspond to Option A.

Hence, the correct answer is Option A.