Radioactive Decay Chain of Thorium to Lead Nucleus

To determine the total number of $\alpha$ particles ($N_\alpha$) and $\beta^-$ particles ($N_\beta$) emitted during the radioactive decay chain, we can write the general nuclear equation for the transformation of a Thorium nucleus ($^{232}_{90}\text{Th}$) into a Lead nucleus ($^{212}_{82}\text{Pb}$):

$^{232}_{90}\text{Th} \longrightarrow {^{212}_{82}\text{Pb}} + N_\alpha \left(^{4}_{2}\text{He}\right) + N_\beta \left(^{0}_{-1}\text{e}\right) + \text{neutrinos}$

By applying the principles of conservation of mass number ($A$) and atomic number ($Z$):

1. Conservation of Mass Number ($A$):

The total mass number must be conserved before and after the decay process: $232 = 212 + 4 N_\alpha + 0 \cdot N_\beta$

Solving for $N_\alpha$: $232 - 212 = 4 N_\alpha$ $20 = 4 N_\alpha \implies N_\alpha = 5$

2. Conservation of Atomic Number ($Z$):

The total atomic number (charge) must also be conserved: $90 = 82 + 2 N_\alpha - 1 \cdot N_\beta$

Substituting $N_\alpha = 5$ into the equation: $90 = 82 + 2(5) - N_\beta$ $90 = 82 + 10 - N_\beta$ $90 = 92 - N_\beta \implies N_\beta = 2$

Thus, the total number of $\alpha$ particles emitted is $N_\alpha = 5$ and the total number of $\beta^-$ particles emitted is $N_\beta = 2$.

Correct Options:

• (A) $N_\alpha = 5$
• (C) $N_\beta = 2$