Nuclear Physics And Radioactivity Question 391
Question: $ _{90}^{232}Th $ an isotope of thorium decays in ten stages emitting six a-particles and four b-particles in all. The end product of the decay is [CPMT 2005]
Options:
A) $ _{82}^{206}Pb $
B) $ _{82}^{209}Pb $
C) $ _{82}^{208}Pb $
D) $ _{83}^{209}Br $
Show Answer
Answer:
Correct Answer: C
Solution:
New mass number $ {A}’=A-4{n_{\alpha }} $ $ =232-4\times 6=208 $ atomic number $ {Z}’=Z+{n_{\beta }}-2{n_{\alpha }} $ $ =90+4-2\times 6=82 $
BETA
