Nuclear Physics And Radioactivity Question 371
Question: When $ _{90}Th^{228} $ transforms to $ _{83}Bi^{212} $ , then the number of the emitted a- and b-particles is, respectively [MP PET 2002]
Options:
A) $ 8,\alpha ,,7\beta $
B) $ 4,\alpha ,,7\beta $
C) $ 4,\alpha ,,4\beta $
D) $ 4,\alpha ,,1\beta $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {n_{\alpha }}=\frac{228-212}{4}=4 $ and $ {n_{\beta }}=2\times 4-90+83=1 $
BETA
