Question

${}_{90}^{232}Th$ an isotope of thorium decays in ten stages emitting six α-particles and four β-particles in all. The end product of the decay is.

• A. ${}_{82}^{206}Pb$
• B. ${}_{82}^{209}Pb$
• C. ${}_{82}^{208}Pb$
• D. ${}_{83}^{209}Br$

Answer 1

Answer: (C)

${}_{82}^{208}Pb$

Answer 2

New mass number $A^{'} = A - 4n_{\alpha} = 232 - 4 \times 6 = 208$

atomic number$Z^{'} = Z + n_{\beta} - 2n_{\alpha} = 90 + 4 - 2 \times 6 = 82$.