Q. $\ _{90}^{232}Th$ an isotope of thorium decays in ten stages emitting six $\alpha$-particles and four $\beta$-particles in all. The end product of the decay is
A) $\ _{82}^{206}Pb$
B) $\ _{82}^{209}Pb$
C) $\ _{82}^{208}Pb$
D) $\ _{83}^{209}Br$
Explanation
(c) New mass number $A^{'} = A - 4n_{\alpha} = 232 - 4 \times 6 = 208$
Atomic number $Z^{'} = Z + n_{\beta} - 2n_{\alpha} = 90 + 4 - 2 \times 6 = 82$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.