When \( {}^{232}_{92}\mathrm{Th} \) changes into \( {}^{208}_{82}\mathrm{Pb} \), the number of \( \alpha \) and \( \beta^{-} \) particles emitted are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \alpha = 6;\ \beta = 2 \)
Only alpha decay changes the mass number (by 4 each time); a beta-minus decay leaves the mass number unchanged and raises the atomic number by 1. So the alpha particles are counted from the mass numbers first, and the beta particles from the atomic numbers.
\( \displaystyle n_{\alpha} = \frac{232 - 208}{4} = \frac{24}{4} = 6 \)
Six alpha decays lower the atomic number by \( 6 \times 2 = 12 \):
\( Z = 92 - 12 = 80 \)
The final atomic number is 82, so each beta-minus decay must add 1 to reach it:
\( n_{\beta} = 82 - 80 = 2 \)