Nuclear Physics And Radioactivity Question 294
Question: What is the respective number of $ \alpha $ and $ \beta $ particles emitted in the following radioactive decay $ _{90}X^{200}\to _{80}Y^{168} $ [CBSE PMT 1995; Pb. PMT 2004]
Options:
A) 6 and 8
B) 8 and 8
C) 6 and 6
D) 8 and 6
Show Answer
Answer:
Correct Answer: D
Solution:
$ {n_{\alpha }}=\frac{A-A’}{4}=\frac{200-168}{4}=8 $ $ {n_{\beta }}=2n_{a}-Z+Z’=2\times 8-90+80=6 $
BETA
